1
Part One
Physical Fundamenatls of Mechanics
1.1
Kinematics
1.1.
Method : 1 (Relative approach)
V
Motor Boat
V = Relative speed of
Fixed point
raft
raft
6 km
motor boat w.r.t.
u
river which is constant
distance = 2 × u
V
Observer on raft see that speed of motor boat is constant because duty of motor boat is constant.
Hence if motor boat take 1 hrs in down stream journey then to reach again at raft motor boat
will take 1 hrs in upstream journey. Hence Total time in complete journey = 2 hrs.
Motion of raft : u = speed of river. Then 2 u = 6  u = 3 km/hr. Ans.
1 hr.
Method : 2 (With frame of ground)
Motion of raft : u (1 + t1) = 6 ––– (i)
t1
6 km
Motion of motor boat : (v + u) × 1 – (v – u) t1 = 6  v + u – vt1 + ut1 = 6

v – vt1 + u (1 + t1) = 6; from (i)  v – vt1 + 6 = 6  t1 = 1 hrs.
put t1 = 1 hr in (i) : u (1 + 1) = 6  u = 3km/hr Ans.
Q.1.2: Total distance travel by point is S. Then Time taken in first journey :
S
t1 =
S
S
t
t
2
 V1 2  V2 2  t2 =
Time taken in second journey :
V1  V2
V0
2
2
2
Mean velocity =
Q.1.3
S
t1  t 2
=
S
S/2
S

V0 V1  V2
 Mean velocity =
Method : 1 (Graphical Approach) Given tan  = 
1
  – t 
(t  ) 
 tan 
Displacement
Area of trapzium
2
2 



Time

Total time
2V0 (V1  V2 )
Ans.
V1  V2  2V0
V
t

t

t   1 –
4

Ans.
 – t
2
 – t
2
2
Compact ISC Physics (XII)
Method : 2 (Analytical)
2
  – t 
1   – t 
1   – t 
Total displacement (S) = 2   2     2   t  2   2 






2
2
2
1   – t 
1   – t 
  – t 

  
 t   

4
S 2  2 
2  2 
 2 

Time taken =  =
t =  1 – 


Ans.
S
2m
Q.1.4 (a)
average velocity =
20 s
(b)
2
= 0.1 m/s = 10 cm/s Ans.
20
S
t
1.4
Velocity will be maximum when slope of S (t) curve will be maximum.
V=
1.4 – 0.4
m/s = 25 cm/s
14 – 10
0.4
Ans.
10
14
t
S
(c)
Q.1.5
Instanteneous velocity may be equal to mean velocity when
slope of line joining final and initial point will be same to
slope at point on curve. Ans. : to = 16 s



Velocity of B with respect to A : VB – A  V2 – V1
16 s
approx
t

 
 
Position of B with respect to A : rB – A  rB – rA  r2 – r1
v1
A
v2
A
r1
VB – A
B
r2
B
Particle B will be collide with A if velocity of B with respect A is directed toward observer A.


 
V2 – V1
r2 – r1





 
Then VB – A || rB – A  V̂B – A  r̂B – A Then
|
r2 – r1 | Ans.
| V2 – V1 |
N
Q.1.6
VW – E = 15 km/hr
60º
W
Y
Y
VS – E = 30 km/hr
E
X

S
X
3


VS – E = 30 i; VW – E  15 cos 60 î – 15 sin 60 ĵ 



VW – S  (15 cos 60 – 30) î – 15 sin 60 ĵ  – VS – E  VW – E
– 15 sin 60

| VW – S |  (15 cos 60 – 30) 2  (15 sin 60) 2  40 km/hr  tan  
15 cos 60 – 30
   19º Ans
Method 2
30 km/hr
60º
15 km/hr




we know VW – S  VW – E – VS – E
VW – S

2
2
| VW – S |  VW
– E  VS – E – 2 VW – E VS – E cos 60  40 km/hr Ans.
15 sin 60
     = 19º
30 – 15 cos 60
from figure : tan  =
Person : 1
2 km/hr
Q.1.7
Time to cross the river : t =
d

d
––––– (i)
2.5 cos 
2.5 km/hr
from figure : sin  =
2
4
3
d
  cos  =
 Put in (i) : t =
––––– (ii)
2.5 5
5
1.5
Person : B
x
2 km/hr
d
2.5 km/hr 
Using trigonometry :
x = d tan 1
Time to reach at destination point : t = t1 + t2 Here t1 = Time to cross the river =
t2 = Time to walk on bank =
from (ii) and (iii) :
d
2.5
x d tan 
2
4
d
d 4

  t=
 
And tan =
–––– (iii)
u
u
2.5 5
2.5 u 5
d
d
4d


 u = 3 km/hr
1.5 2.5 5u
Ans.
4
Compact ISC Physics (XII)
d
Q.1.8
tA =
Boat A :
d
time to reach again at same point.
d
d

VW V–W
–––– (i) where V = Speed of boat w.r.t. river
W = Speed of water w.r.t. earth.
2d
W
Boat B :
Time to reach again at same point. tB =
d
V
tA
Also given V =  ––––– (iii) now t B
d
d

    – 


2d
2
2
  –
2
––––– (ii)
V2 – W2

tA
 2 – 1  t 
B

2 – 1
= 1.8s
x
d
1.9: Method : 1
Vm – r
d

Time to cross the river: t = | V
mr | cos 

A
Vr – E
d
Then Drift (x) = (Vm – r sin  – Vr – E) V cos  Since Vm – r < Vr – E . Hence
mr
this is not possible that drift will be zero. Hence we should have to minimize
dx
drift (x). Hence
=0
d
d sec2 –
30º
90º
(Vr – E ) d
Vm – r 1

sec tan  = 0  sin =
  = 30º
Vmr
Vr – E 2
Angle made by boat with flow velocity of water = 30º + 90º = 120º Ans.
Method : 2 (Vector addition method)
Vm – r




Vm – E  Vm – r  Vr – E Hence  will taken any value between
Vr – E

(0 – 180º) hence we can draw a semi circle of radius of length | Vm – r | . Then.
5
C
C2
C3
C1
C4
Vmr


A
Vr – E
B
Vr – E
And resultant is given by C1, C2, C3 and C4, ....... Cn. But for minimum drift resultant must be
Vmr
1

  = 60º
Vr – E 2
tangent at semicircle. Then cos  =
Then   = 180 – 60º = 120º
1.10: Relative acceleration of particle (1) w.r.t. (2) = g – g = 0; Relative velocity
of particle (1) w.r.t. (2) = V 1
= V0
– 2
=
V02  V02 – 2V0 V0 cos (90 – )
2 (1 – sin ) Where 90 –  is angle b/w two velocity..
g
g
V0
2
1
V0
 = 60º
Since there is no relative acceleration of particle (1) hence its relative velocity
does not change w.r.t time then. Distance b/w two particle at time t is : Distance = V0 t
2 (1 – sin )
= 22 m Ans.
3 m/s
1
4 m/s
x (+)
2
Q.1.11 : Method : 1 (Vector)
+ y
Initial velocity in y direction
of both particle zero. Hence vertical velocity of particles (1) at time t : Vy = u + at  Vy = g t; Velocity


of particle (1) at time t = : V1  – 3 î  g t ĵ ; Velocity of particle (2) at time t : V2  4 î  g t ĵ 


 
Since V1  V2  V1 . V2  0 Then – 12 + g2 t2 = 0  t = 0.12 s. Hence distance between two
particle will be : Distance = Vrelative × t = (4 – (–3)) ×
0.12 = 7 ×
0.12  2.5 m
Ans.


Method : 2 (Graphical) : Since V1  V2   +  = 90
  = 90 –   tan  = tan (90 – )  tan  = cot 
tan  × tan  = 1
––––– (i)
gt
gt
And tan  =
 tan  =
3
4

3 m/s



V1

gt
gt
4 m/s
6
Compact ISC Physics (XII)
 gt   gt 
put in (i):      1
 3  4 

t =
0.12
Distance = Vrel × time = 7 0.12  2.5 m
Ans.
1.12 : Method : 1 (Velocity of approach) Since particle A heading to particle B and B to C and C
to A. Then position of all particle at t = dt is as figure 2.
B
60º
Vdt
V
Vdt
a
a
Vdt
V
60º
60º
A
Vdt
C
V
a
Vdt
Vdt
Vdt
Vdt
Vdt
Again position of all particle at
t = 2 dt is as figure 3.
B

A
30º
P
30º
C
A
Again position at t = 3 dt and so on ........ Since at any time
all particle travel same distance then at each moment of time,
all particle will be at equilateral tringle. Then by symmetry you
can say that all particle will be met at centriod of tringle then
path of each particle will as :
Suppose at any instant of time, distance b/w particle A and
centriod P is r. Then, line joing particle A and P make 30º angle
dr
with side of equilateral tringe then
will always constant and A
dt
dr
equal to v cos 30º then –
= V cos 30º, (–)ive sign because
dt
r is dicreasing function. Finally r = 0 while initial r  a / 3
0
a
3
–
dr

v cos 30
a/
30º
a/
2
t
 dt
0

t =
2a
a
 A P =
sec 30º =
3V
2
a
3
Ans.
3
P
7
Method : 2 (Relative approach)
let distance b/w A and B at time t is r then. At any instant of time, rate of decreasement of
distance b/w two particle A and B will be constant as shown in figure.
0º
s6
co 6
0º
V
co
s6
0º
B
60º V
1
V
B
11
B
V cos 60º
V
60º
V
V
V
11
11
V
C
A
V
1
1
A
C
C
A
dr
3V
Then
= – (V + V cos 60º) = –

dt
2
0
2 dr

– 3V

a
dt
 dt
3a
V
t =

Ans.
0
1.13: Suppose at time t distance b/w A and B is r. Then rate of decreasement of r is :
o
– v + u cos 

– dr 
t
t
0 (–v  u cos θ) dt  –l = – vt + u 0 cos  dt.... (i)
t
Now since
0
0
B u
t
t
0 dx  0 (–V cos   u) dt
0 cos  dt =
x
y
cos  dt is not known then to find this integration,
dx
we use rate of decreasement of component: Then
= – V cos 
dt
+ u
– dr
=
dt
0 = ut – V
ut
now put in (1) : –l = –vt + u
V
 ut

V

t
0 cos 
V


x
r
A
dt 

V

  t= 2

V – u2
W
1.14: With frame of train : With frame of train, train appear in rest

B
then distance b/ there two event is equal to l.
A
With frame of earth : When event (1) will happen. Velocity of train is V = u + at = wt. Since
event (2) will be happen after time  then. Distance travelled




u1 t1 + a t21 = wt() + ww  t   Then distance




b/ two events is =
 – w   t   2 


= 0.24 km.
Ans.
event 1

by headlight (A) = Distance travelled by head light (B) =
A
B
event 2
B
A

w (t +  /2)
8
Compact ISC Physics (XII)

It we want that both event will be happen at same point then velocity of reference frame
0.24 km
= 4 m/s Ans.
60
will be Vreference frame =
shaft
2
1.15: (a)
a=1.2 m/s
For observer inside lift at t = 2s. velocity of bolt = 0
Accn of bolt = 10 + 1.2 = 11.2 m/s2. Then assume t time is taken by
bolt to reach at floor. s =
(b)


at2  2.7 =
× 11.2 t2  t = 0.7s


2.7m
Velocity of bolt with respect to ground at t = 2 sec.
V =1.2×2 = 2.4 m/s Displacement then S= ut +
Distance : H =
 1.3 m
 2

at  S = 2.4 (0.7) – ×10 (.7)2  – 0.7 m


.  2.4
= 0.288 Distance travelled = 2 × 0.288 + 0.7
2g
H
2.4m/s
Ans.
0.7m
1.16: Method : 1 (Relative velocity) from figure
V1
V2
tan  = V  cos =
1
V12
 V22
y V2
V2
1 V2
also tan  =   V y =  1 V  BC =  2 – V
1
1
1
1

 V 
 – 1 2 
Shortest distance = CM = BC cos =  2
V1 
1
V12  V22
V1

0
V1
V2
V1
1
2
V2
2
0
y
B
 M
C
Shortest distance
=
|  2 V1 – 1 V2 |
t 
V12
 V22
now t =
V1 1  V2  2
V12  V22
Ans.
AM
V12

V22

AB  BM
V12

V22

 1 sec   CM tan 
V12

V22

V 1  V2  2
V12  V22
9
Method : 2 (Velocity of approach)
At shortest distance velocity of approach = 0. Then V1 cos  = V2 sin   tan  =
V2 t –  
V
 1
In tringle A B O : tan  =
  – V1 t
V2
V22 t – l2 V2 = l1 V1 – V21 t  t =
V2
1
B
  V1    V2
V12
Ans.
 V22
1
A
A
V1t
2
(   – V1 t)  (V2 t –   )
And Shortest distance is :
Shortest distance =
90–
2
O
B
V12  V22
V2
V2t –  2
 
(   – V1 t) 2  (V2 t –   ) 2
... (1). for shortest
1 – V1t
distance
t=
dl
= 0 
dt
  V1    V2
V12  V22
 – 2 ( – V1 t) V1  2 (V2 t –   ) V2
= 0

( – V t) 2  (V t –  ) 2
Ans.
1.17:

1
2

put value of t in (1) :
 
|   V1 –   V2 |
Ans.
V12  V22
Method : 1
Time to reach at point D : T =
2
 1 – V1t
Method :3
V1t
V2t

V1
  V1    V2
V1
V1
V2
m –  tan   sec 

. For
V
V/n
dT

n
minimum time :
= 0 Then –
sec2  +
sec
dθ
V
V
1
× tan  = 0  sec  = n tan   sin  =
. Then


distance BC = l tan   BC =
Ans.
 –
m
m– tan
A
tan
C
B



D
10
Compact ISC Physics (XII)
Method: 2 (Help of light wave)
medium (1)
We know that light travel via that path in which time will be
i
 sin 



length BC = l tan =
C
B
sin i
speed in medium (1)
sin 90  V


less. Then.
=
sin θ
speed in medium (2)
sin 
V

medium (2)

η2 – 1

Ans.

WX
1.18:
2
1m/s
3
1
2
–1m/s
t
6
x
t
Distance
t
1.19: (a)
Mean velocity in irodov is misprint and it is mean speed then mean speed =
(b) Mean velocity =
And 
2R
τ
Ans. (c) We know             ....(i)


2
t2      2   
ii from


τ2
2
2
(i) and (ii) :    2       
 now V = RW then
τ
τ 
2 R
–0
2
Vf – Vc
τ


V= R
. Average accleration :
=

τ
 τ 

πR
Ans.
τ
Avg. Accn =
2 R
τ2
Ans.
A
B
R
Time taken = 
11
1.20: This is one dimension motion be cause direction of position vector r is same as constant
 



vector a . Then r  a t (1 –  t)  a t –  t 2 a
(a)
(b)

 d r 



dv
 
V
 a – 2t a   a cc 
 – 2  a  V
 a (1 – 2 t)
dt
dt



At initial position t = 0  r = 0 Now at final position  r = 0 then. a t (1 – t) = 0
t = 0 ro
t = 
Ans.
B At point B velocity will be equal to zero so
Initial A

that particle will be turn back. Then 0 = a (1 – 2t)
  
r = a   


a
a


4
4
1.21: (a)
t =


Then position B is :

  a

1 – 

  4 . The Distance travelled in up and down journcy is:

a
Ans.
2
V = V0 (1 – t/) 
x
0
dx 
t 

x then : x = V0 t  t – 2τ 


(b)
Ans.

t2 
t

V0 1 –  dt x = V0  t –  
2τ 
0
τ



t
Position at time t is
t=
A
Ans.
B
C
Henc we see that velocity will change dirn at t =  because When t >   v = 
t 

And t <     v = – ive. Then Case I : t <  x V0 t  l – 2τ 
Ans.


Case II : t > 
Total distance travelled = AB + BC
  
 
t 
V0  V0 
t 




= V0   l – 2τ    V0   l – 2τ  – V0 t  l – 2τ   = 2  2 – V0 t  l – 2  

 






t 

Distance = V0  – V0 t  l – 2τ 


1.22 : (a) v =  x   x
a =
1

when t > 
Ans.
differentiate w. r. t. time :
dv 1
– 1 dx
 αx 2
dt 2
dt
 a
1
α – 12
x
αx 2
2
1 2
1 2
α Since acceleration is constant; velocity will be : V = u + at  V =
α t
2
2
12
Compact ISC Physics (XII)
(b)
S = ut +
1 2
1
at  S =
2
2
<V> =

 s

 α2

 2


 2
t


s s
2 s
Mean velocity <V> = t 
2 s
α
t =
Ans.
1.23: Calculation of time : w = – a v ( – i ve sign is used to show deacceleration)
0
dv
= – a v 
dt

v0
dv
v
to
 –

1
a dt

0
t0 =
 V0 2
Ans.
a
0
dv
 – a v 
Calculation of distance :   – a v  v
dx

v
1
x0
2
dv  – a
v0
3
 dx  x
0
0
=
 V0 2
3a
2
1.24 : (a)
(b)
(c)
(d)
1.25: (a)
(b)

r  at î – bt 2 ĵ  x = at  y = – bt2  y = – b

 dr
v
 a î – 2bt ĵ  v  a î – 2bt ĵ
dt

| a |  2b
x
  a2y = – bx2
a 
Ans.

 dv

2
2 2  a 
 – 2b ˆj  a  – 2b ĵ
| v| a  4b t
dt
Since direction of acceleration is toward y dirn then angle made by velocity vector with y axis
y
a
is known as angle b/w two vectors then. tan  =
Ans.
2bt




x
s at î – bt 2 ˆj
a
Vmean  
 Vmean  a î – bt ĵ  | Vmean |  a 2  b 2 t 2

t
t

x
x = at  y = at (1 – t)  y = x 1 –  a 


2bt
Ans.
dv y
dx


= a  ay =
= a – 2 a  t  v  a î  (a – 2a  t) ĵ  | v |  a 2  (a – 2a t) 2
dt
dt

 dv


 – 2a  ĵ  | a |  2a 
 | v |  a 1  (1 – 2 t) 2  a 
Ans.
dt
vx =
13
Method : 1
a (–2a  t)
(c)
acceleration = 2a
Velocity =
a
 
1
– 2a  (a – 2a t)
a .v

1
2
cos  = | a | | v |  2
a  (a – 2a t) 2 (2a ) 2 t = 
Method : 2
– tan 
4

a
 – a = a – 2 a t  t =  
a – 2a t
Ans.
y 
x
x = a sin wt y = a (1 – cos wt)   a – 1 = – cos wt ––– (i) 
= sin wt ––– (ii)
a


1.26 :
2
2
 dx
dy
x y 
î 
ĵ
    – 1  1  x2 + (y – a)2 = a2 Equation of circle of radius a. v 
a  a

dt
dt
  



v  a w cos wt î  aw sin wt ĵ  | v |  a w = const.  Uniform circular motion.
(a)
Distance travelled in time  is = (aw)  Ans.
a
(b)
Since motion is uniform circular motion. Hence only radial acceleration is
present then Angle b/w velocity vector and accn will be  
Ans.
(0, a)
1.27 : y = ax – bx2 differentiate w.r.t. time :
dy
dx
dx
a
– b 2x
Vy = a Vx – 2bx Vx
dt
dt
dt
At x = 0 Vy = a Vx ––– (ii) differentiate equation (i) w.r.t. time :
dVy
dt
a
–––– (i)
dVx
dVx
– 2bx
– 2bVx2
dt
dt
ay = a ax – 2bx ax – 2b Vx2 At x = 0 Given ax = 0 and ay = w ay = | – 2b Vx2 | w = 2b Vx2 –– (iii)
Speed at origin will be : V =
Vx2  Vy2 =
w a2 w
w

V =
(1 + a2)
2b
2b
2b
Ans.
g
1.28 :
(a)
V
0
 
 
1
1
s  u t  a t 2  s  V0 t  g t 2
2
2
Ans.

14
Compact ISC Physics (XII)


s

v


(b)

t


1
 v   v0  g t
2
V
–g
Ans.
 –
 

 
2 u sin 
– v0 . g


g  – v 0 . g  2
we know T =

u
sin

=


v


v

0
g
g
2  g  g
 
  v0 . g


  v   v 0 – g
g2
y
1.29 : (a)
2 V0 sin 
1
S = 0 in y direction  0 = (V0 sin ) T – g T2 T =
g
2
V0 sin 
V0

(b)
At maximum height final velocity in y direction = 0 Vy 2 = uy 2 + 2 aH
02 = (V0 sin )2 + 2 (–g) H H =
R=
(c)
x
 2V0 sin  
V02 sin 2 


Range R = V0 cos  
 
g
2g


V02 sin 2  2V02 sin  cos 
V02 sin 2 

when H = R 
tan  = 4  = tan– 1 (4)
g
2g
g
1
x = (v0 cos ) t  y = (v0 sin ) t –
g t2  y = v0 sin 
2
gx 2
y = x tan  –
Ans.
2u 2 cos2 

 1 

x
x

– g

 v0 cos   2  v0 cos  




V0
V2
 R0 =
g cos 
normal acceleration
Radius of currature =


A
V0
g
V0 cos 
g
RA =

V02
2
cos 
g
Ans.
1.30 : w = g sin  andw = g cos Here  is first dicreasing and then
increasing and projection of total accn on velocity vector

will be (–)ive then. wv = w1 .v̂ = – g sin 
wn = g cos 
w


wn
g
=w
w

in
gs

V
w
2
V0
2
(d)
V0 cos 
wV
V
t

15
y
h
2g

sin

h
2g
M
2gh

V0 = 2gh
s
co
O
sin

1.31 :
h
2g

C

Just before collision
Just after collision
x
Time to collide with incline T =
2 uy
g cos 

2 2gh cos 
g cos 

2 2g
g


Length MO = (velocity in MO) × T =  2gh cos  sin   2gh cos  sin  


 2 2gh cos  sin  ) 2 2gh 
MO


Range OC = (MO) sin  =
=
R = 8 h sin 
g cos 
cos 
1.32: We know R =
Ans.
u 2 sin 2θ
()  sin 2 
5.1 × 103 =
 1 = 32.5º. Also we know that range
g
g
 usin 
g
will be same for 2 = 90 – 31.5 = 59.5º. Then time of flight will be: T1 =
  240 sin 31.5
T1 =
= 24.3s = 0.41min
10
m/s
240
Ans.

  240 sin 59.5
T2 =
= 42.3s = 0.69 min Ans.
10
1.33:
3
5.1×10 m
Method: 1
  º     º. For both particles x and y co-ordinate must be same then.
Particle (1): y = x tan 1 –
Particle (ii): y = x tan 2 –
From (i) and (ii) :
g x2
2 V02
2
/s
250 m
.... (i)
cos 1
g x2
2 V02 cos 2  2
x tan 1 – g x 2
60º
.... (ii)
2V02 cos 2 1
= x tan 2 – gt
= v0
v0
m/s =
250
45º= 
2V02 cos 2  2
16
Compact ISC Physics (XII)
x=
 V02 sin (1 –  2 ) cos 1 cos  2
x
.... (iii)  Time for particle (1) : t1 = V cos θ
2
2
g
cos 1 – cos  2
0
1
x
x  1
1 
TIme for particle (ii) : t2 = V cos θ  Then t = V  cos  – cos   .... (iv)
0 
1
2
0
2
Put value of x on (iv) : t =
2 V0  sin (θ1 – θ 2 ) 


g  cos θ1  cos θ 2  = 11s
Ans.
Method: 2
Particle (1) :
x = V0 cos (t + t) .... (i)  y = V0 sin (t + t) –
Particle (2) : x = V0 cos (t).... (iii)  y = V0 sin (t) –
(iii) and (iv): t =
2V0  sin (θ1 – θ 2 ) 


g  cos θ1  cos θ 2  = 11s

g (t + t)2 .... (ii)

 2
gt .... (iv)  From (i), (ii),

Ans.
y
v0
V0
1.34 : (a) tan = ay  Time to reach at hight y : t = y/V0

vx = ay
y
dx
Also
= ay 
dt
x
t
dx
dx  aV0 t dt
= a [V0 t] 
dt
0
0


0
aV0 t 2
a V0  y 

x=
 
2
2  V0 
(b)
ay = 0
ax =

 a
 x = 
 2 V0
 2
y


Ans.
dVx
a dy

= aVy = aV0  anet = aV0
dt
dt
= aV0 cos  a =
a V0 (ay)
V02  (ay) 2

x
x
a2 y
1  (ay/V0 ) 2
Tangential acceleration = a
Ans.
Radial accn or normal accn : (an) = a V0 sin  an = a V0
   a y 
V0 


Ans.
17
a
 x = at .... (i) Vy = (a t) b
bx
1.35: (a) tan  =
y
0
dy  ab
Then y =
(b)
t
0
t dt
ab  x 
 
2 a
ax = 0

ab t 2
y
2

anet

ay =
Vy
bx=
v
y
 b 
y =   x2
 2a 

dVy
dt
 b
an
Ans.
x
dx
= ab  anet = ab  Then normal accn : an = anet cos
dt
anet = ab Then an = ab

a   xb  


1



R= b
  a  
a = vx
a
a 2  (bx) 2
 R =
V2
(a 2  b 2 x 2 ) 3/2


an
a2 b
/ 
Ans.
1.36:
Method: 1 (Work-Energy)

 
Tangential accn is given by : | w  |  a .  = a cos 
Suppose at time t particl e at poseti on A and in smal l time dt particl e displacement
by ds.

Work done : dw = Ft ds = (ma cos) ds = ma (ds cos)  dw =
(ds) cos  ma
 ds cos
Using work energy theorem : W = max 

mV2 = max 


V =
ds


a
x
0
ma (ds cos) = ma
a x
Ans.
18
Compact ISC Physics (XII)
Method : 2 (Kinematics)
 
Tangential accn : at = a .  = a cos. Also since we know
d| v |
that tangential accn is rate of change of speed then.
dt
v dv
= a cos 
= a cos  v dv = a (ds) cos = a dx
ds
v
0
v dv 
x
0
a dx 
v2
= ax  v =

ds V

dx
Ans.
a x
v
1.37: Angle travel by particle:  = 2n. Speed of particle : v = Rw = at w =
at
dθ


R
dt
t2 =
2n
t
at
 ds   R dt
0
0
at 2
2R
t2 =

  nR

a
R
  nR
2
a 2 4 n R
a2 t2
Now : Radial accn : ar = V R 
=
a
R
a
R
= 4na Tangential accn : at =
Total accn = a
–
dv
V2

dt
R

dt
R

V0
–
0
R
V
 tR
V0
 V
1
dv
v2
= –v
a 2t  a 2r = at  at =
ds
R
dv

v
S

0
ar
(–) sign because
V0
V0 t

R
–v dv =
Where ds = distance travel by particle in time dt. –v dv =
V
a 2  (4 na) 2
Ans.
t
– V –2 dV 
V0
V –1
a 2r  a 2t =
V2
Since at is rate of change of speed then.
R
V

V is decreasing
anet =
dv
= a  Total accn =
dt
1  (4n) 2 = 0.8 m/s2
1.38: (a) At any time t : at = ar =
(b)
2n =
a
ds
R – n
V S

V0 R
Then V = V0 e–S/R  at =
v2
dt
R
v2
ds
R
V 2 e – 2S/R
v2
 0
R
R
at
Given
at = ar
at t = 0
u = V0
19
anet =
V02 2 e – 2S/R
V2

R
R
2
1.39:
 anet =
V02 2
R e 2S/R

V2
R
2
Ans.
Method : 1

2
v = a s Squaring both side : v = as
at
ar
a
Compare with : v2 = u2 + 2 accS  u = 0 acc =
2
Hence motion is constant magnitude tangential accn. Then.
At
as
V2
=
R
R
any time t : at = a/2  ar =
at

Since we know that velocity vector and tangential accn is parallel then.
a
as
2S
r
tan   a  R a/2  R
t
Ans.
ar
Method : 2
dv
ds
dv
 a
 a = a
= av 
2
t
dt
dt
dt
v2 = as  differiate w.r.t time : 2v =
2
ar = V
R
 as
R

1.40 : (a)  = a sin wt now
tan  =
ar
at

2S
R
 = 0    t = 0
d
 a w cos wt
 = ± a   t =  2 , 3 2 , V =
dt
Ans.
y
wt
a sin
=

x
dv
 a w 2 sin wt Then at = – aw2 sin wt
a =
dt
ar =
V 2 a 2 w 2 cos 2 wt

 anet =
R
R
anet = aw2
at t = 0 
sin 2 wt 
a2
R2
cos 4 wt
a 2t  a 2r
–––– (i)
a2w 2
 anet = aw2 Ans.
 = 0  at t =  2 or 3 2  anet =
R
20
Compact ISC Physics (XII)
(b)
da net
 0  cos wt = R
2 a put in equation(i)
dt
For minimum value of acceleration 
2
amin = aw
R 
1–  
 2a 

 R 

 = a 1 – 
 2a


  = a 1 –
Speed of particle when distance is S. V2 = 2aS –––– (i)
1.41:
S=
R2
2 a2
Ans.
(Because a = const.)
2S
1 2
at  t2 =
a
2
a =
a
4
t
wn = b
2
4bS2
 2S 
Then wn = b    2
a 
a
Radius of curvature : R =
w=
Net accn :
a 2T
V 2 2aSa 2
a3


R
=
wn
2bS
4bS2
 w 2n
1.42: (a) y = ax2 diff. w.r.t. time :
 4bS2
a 
 a2

2
 w =




Ans.
2
Ans.
dy
dx
 2ax
 Vy = 2ax Vx
dt
dt
y
Again diff. w.r.t. time : ay = 2ax ax + 2aVx2
x
At x = 0  Vy = 0  Then Vx = v  ay = 2aV2
Since speed is const. its tangential acceleration will be zero. Then
2
dV
V2
1
V
2

ax =
= 0  anet = 2 a V  R =
 R =
a
2
n 2av
dt
2a
(b)
x2
a
2

y2
b
2
2x Vx
= 1 diff. w.r.t. time :
Again diff. w.r.t. time :
At x = 0 and y = b :
2x a x
a2
2 Vx2
a2
a


2
2 Vx2
a2
2 ay
b



2y Vy
b2
2y a y
b2
2 Vy2
b2
Ans.
0

2Vy2
b2
0
V0
0
21
2 V02
As shown in figure : Vy = 0 and Vx = V0 
a
2

2 ay
b
 0  ay =
b
Since speed is Const. tangential accn = ax = 0 at; t = 0  anet =
V02
Radius of curvature : R =
 R =
an
V02
b
a
1.43 :
Given
2
V02
 R =
a2
a2
b
b
a
2
V02
V02
Ans.
d
d(2)
d
2
= w = const. Then
= 2w = const.
dt
dt
dt
A

Hence angular velocity of point A w.r.t. point P is

constnat then. WP = 2W and velocity will be perpendicular
to position of A w.r.t. P then V = R WP = 2RW
Since WP = const.
dWP
0
dt
O

P
Ans.
Then  Tangential accn = 0
Radial accn = R WP2 = 4 RW2 = anet  Direction is toward the centre.
1.44 :
 = at2 
Ans.
d
dw
= w = 2at 
=  = 2a
dt
dt
Tangential acceleration : at = R = 2aR  Radial acceleration
2
2
V
2 2
: ar = RW = R (2at) = 4 a t R
at
ar

And. v = RW = R 2at  2at = v R –––– (i)
Now anet =
a 2t  a 2r = 2aR
from (i) : anet =
1.45 :
1 (2at 2 ) 2
v
1  (2at 2 ) 2 = 0.7 m/s Ans.
t
l
1
Since acceleration is constant. l =
at2 and
a
V
2
Vt
V = at Then l =
–––––– (i) and also. angular acceleration is const. then
2
1
1
 =
 t2 and w = at then  =
wt and since particle taken n turn in its journey..
2
2
22
Compact ISC Physics (XII)
= 2  n
(I)
1.46 :

V
2v

(II) : 2 W  W =

 = at – bt3  w =
Wavg
Ans.
d
= a – 3bt2 when body is in rest then w = 0  a – 3bt2 = 0
dt
a
 in = 0 and final = t (a – bt2) =
3b
t =
a
final –  in
=
=
t
Win = a and Wfinal
And.  =
1.47 :
1
wt –––– (ii)
2
Now 2 =
3b
a
2a 3 

2a
3
a
 2a 
3b  3 
Ans.
3b
 a 
Wfinal – Win

= a – 3b  3b  = 0   avg =
 
t
dw
= – 6bt = – 6b
dt
a
  = 2
3b
3ab
a
a
w
t
 0dw   0 dt
 w =

3ab Ans.
3b
Ans.
y
Given  =  = at To find tangential accn : (a t) : at = R = Rat
To find radial accn : (ar) :  =

at
dw
 ar = Rw2
dt
ar
x
at 2
R a2 t4
at dt 
 ar =
0
2
4
t
R
at

Then
ar
R a2 t4
tan  = a  tan  =
 t = 3
4R a t
t
4
tan   7 s Ans.
a
ar
1.48 :
Given
 –W
 
W    = angular accn.   = k
dW
k W
d
W  k = const.
0
 (–)ive because W is decreasing. –

W0
1
W

2
dw 
 k d
0
23
3
2 W0 2
3k
  ––– (i)
 = Angular displacement.  Also  = k
0
– dW = k
W dt 
w
–1
W
2
dw 
0

W
–1
2
t
k dt  2W0
0
k
 t ––– (ii)
3
Avg. angular velocity : Wavg
1.49: (a) Given W = W0 – a –––– (i). At
d
Also
= W0 – a 
dt
 n
(b)

0
2 W0 2

= <W> =
=
t
t = 0
d

W0 – a

1


2 W0 2 
3 k 
k 


  = 0

W0
3

t
–1
dt 
n (W0 – a)
0
a
W0 – a
W0
 – at   =
(1 – e – at )
W0
a
Ans.
W = W0

 t
0
Ans.
 W0
– at 
Put value of  in equation (i) : W = W0 – a  a (1 – e )  W = W0 e–at Ans.


y
dW
1.50: Given  = 0 cos    = W
= 0 cos  
d
w
0
w dw  0

0
W = ± 2  0 sin 
cos  d  
Ans.
w  2  0 sin 
0
W2
= 0 sin  
2

– 0
   0 cos 
– 2 0
w  – 2  0 sin 
1.51: (a) Instanteneous axis of rotation is passing through that point
which velocity is zero always. If we observe carefully its
y
A
point must be at line joining AB. Then. at time t :

VP – 0 = RW = Rt = yt
B
V
x
24
Compact ISC Physics (XII)
y
x = vt ––– (i) Velocity of point P :
VP = 0
y=
(b)
 v 
v
 v = ty x = v  y  t = y
 
v2
x
P (x, y)
yt
put in (i)
V
O
x
Ans.
Velocity of point O = 0
y
VO – P = Rw  VP = u + at = wt
O
yw = RW
wy
yw = wt  t =
w
Then
(x, y)
x
P
 wy 
1 2
1

x=
at =
w 
2
2
 w 
2
1 w2y2
2 w
 x =
Ans.
1.52 : Since there is no slipping at ground. VC = 0  V = RW where VC = velocity of contact point.
To find distance, we have to find speed of a particle of rim then.  = wt  Time to one complets
v 2  v 2  2 v v cos ( – ) = 2 v sin /2 = 2V sin (wt)/2  Again
journey = 2/w VP =
s
ds
= 2 v sin wt/2  ds  2 v
dt
0

2 w
0
sin wt/2 dt  S =
8v
= 8 R
w
Ans.
y
V
x
w
v

P
C
1.53: (a) Acceleration of point C : aC = w  Velocity of point

v
A
C at time t : VC = u + at = wt
Angular accn of ball about its centre:
w w

R R R  Angular velocity of ball at
w
time t : w = w0  t =
t  Velocity of point A
R
 =
aC

R
C

B
O
w.r.t. centre C : VA – C = RW = w t î  VC – E = w t î  VA – E = 2 w t
Ans.
W
25
A
wt
y
wt
x
A
y
C
Point B :
wt
B

x
wt
VB–C = Rw = wt (– j)  VC–E = wt i  VB–E = wt (i – j)  VB–E = wt Ans.
Point O : VO–C = – wt i  VC–E = wt i  VO–E = O Ans.
(b)
Acceleration Calculations :
A
wt
C
wt
at
0
y
ar
C
Point A:
x
aC–E = wi .... (i)  aA–C = at i – ar j  Where at = tangential acceleration. ar = radial acceleration.
at = R = w  ar =
v2
R
(i) + (ii) : aA–E = 2 wi –
 (wt) 2
R

w2 t2
w2 t2
 aA–C = wi –
j .... (ii)
R
R
w2 t2
j  aA–E = 2w
R
 wt 2 

 
 2R 



Ans.
26
Compact ISC Physics (XII)
y


Point B : a C–E = wi .... (i)  a B–C
 v2

= (–R) ˆj +  R



 î =

C R

B
w2 t2
– wˆj –
î .... (ii)
R

a
(i) + (ii) :
B
x

w2 t2

= w – R



ˆ
 î + w j  aB = w



P oint O: a C–E = w î .... (i)  a
–w î –
O–C
 wt 2 

 
 2R 


 v2
= (–R) î + 
 R


Ans.
y
ˆ
 j =


w2 t2 ˆ
j .... (ii)
R
C

w2 t2
w2 t2
a O–E = –
j  aO–E =
Ans.
R
R
(i) + (ii) :
x
0
A
2V
n
1.54 : Velocity of point A = 2v  Velocity of point B = v   acc
of point A = v
2
R
A
Again
R
V
B v
R
v
2V
Radius of curvature =
2
v
2
 accn of point B = v
(speed) 2
(2V)2
 RA =
= 4R
V 2 /R
normal acceleration
2
2
v
Again an =
cos 45º =
R
v2
2R
 RB =
(V 2 )
v R

v2
 2 2R
V
2R
1.55 :
Method : 1 (axis is rotating):
w1–2 =
w 12

w 22

w 1–0 = w1 î 
B
45º
V
V 2
x
an


w 1–0 = w2 ˆj  w 1–2 = w1 î – w2 ˆj

dw 1– 2
dj
d î
d î

  1–2 =
= w1
– w2
direction of
is toward y axis and
dt
dt
dt
dt
this is rate of change of dirn of x axis.
27
y
y
w2
w2
x
0
w1
x
0
Method : 2 (axis is not rotating)



w 1–C = w1 cos  î + w1 sin  k̂   w C–0 = w2 j  w 1–0 = w1

cos  î + w1 sin  k̂ + w2 ˆj  | w 1– 0 | 
w12
 w 22
w2


w2
z
1
C
w1
d

= w2   = – w1 w2 sin  î + w1 w2 cos  k̂
dt

|  | = w1 w2. Here x axis is directly attached with observer. Then
d î
| d î |
dθ

= – w2 k̂  | d î | 1  dθ 
= w2
dt
dt
dt
But
x
0

 d 
 d 
 d w 1– 0

= – w1 sin   dt  î + w1 cos   dt  k̂
dt




And
y
y
y axis
z
1 unit
d
1 unit
passing
j
di
i
d ĵ

= 0 Because with frame of this observed dirn of axis does not rotating then.  1–2 = –
dt
w1 w2 k̂ 
 1–2 = w1 w2
 dw

1.56 : w = at î + b t2 ˆj   
= a î + 2b t ˆj (a)
dt

(b) |  |  at
cos  
 2ab t 
1 

 a 

b 
1  t 
a 
 
w .
 
 w   = w cos   cos  
w
a 2 t  2b2 t 3
a2 t 2  b2 t 4

| w |  at
a 2  4b2 t 2
   = 17º
Ans


28
Compact ISC Physics (XII)
y
1.57: Method : 1 (axis are moving): It we see carefully
then x axis is moving while y direction is not rotating.
Angular velocity of disc with respect to centre M

V
x
is : WM – O  î .... (i) Angular velocity of centre
R


0

V ˆ
j  OM = R cot 
M w.r.t. origen. WM – O 
OM
1  tan 2  
p
y axis

V
[ î  tan  ĵ ]
quantity it follow vector addition law WD – O 
R

R
v

V
 WM –O  tan  ĵ . Since angular velocity is vector
R

V
| WD – O | 
R
M
V
Ans.
R cos 
y axis
j
di
d
i
z



V  d î
d ĵ 
dw
β

Again Angular acceleration () is :
 β  R  dt  tan  dt 
dt


And since y dirn is constant
 V
β   
R
d ĵ
d î
= 0  But
 | dî | = d
dt
dt
d ˆi
d

dt
dt
= wM O 
dî
V
 –
tan  k̂
dt
R
2
2
V


  tan k̂  β  – V tan k̂  | β |  V tan 
R 
2
2
 
R
R
Method : 2 (axis are not rotating)
angular velocity of disc w.r.t. M : WD–M =
V
cos  î +
R
y
V
sin  ˆj angular velocity of M w.r.f. O
R

V ˆ
V
j
tan  ĵ angular velocity of disc
: WM – O 
R cot 
R


wrt O: W D–O = W
D–M


+ W M –0  W
D–O
V
=
cos  î
R


x
0


D
M
V
z
w0 =


V
V
V

V

V

sin  ˆj  
tan  ˆj  | WM – O |   cos     sin     tan  
R
R
R

R

R



V
R
V
Ans.
R cos 
29


dw D – O
–V
V
 dθ 
 dθ 
dθ
αD– O 

sin θ   î  cos θ   ĵ  0 and
is angular velocity of M w.r.t.
dt
R
R
 dt 
 dt 
dt
0:
d/dt =

– V2
V2
V2
V
ˆj | α
tan   α D – O 
sin

tan

+
cos

tan

|
=
î
D
–
O
R
R2
R2
R2
tan  Ans.
1.58 :
Method : 1 : (Direction of Co-ordinate axis is fixed)
At time t line AB rotate by  angle then
y

WP – A =
0
w0 cos  î + w0 sin  k̂    t ˆj

| WP – A | =
w 20 cos 2   w 20 sin 2    20 t 2 

| WP – A | = w0
 t 
   0 
 w0 
x
A

w0

Now
z

 d 
 d WP – A
d
 k̂   0 ĵ

 – W 0 sin  ( î )
 W 0 cos  

dt
dt
 dt 

= –W0 0t sin  î + W0 0 t cos  k̂ + 0 ˆj  |  | 

|  |   0 1  w 20 t 2
w0
w 20  02 t 2   20 
Ans.
Method : 2 (x axis is moving) : Here we take line AB along

 t 2

x dirn. WP – A  w 0 î  0 t ˆj  | WP – A |  w 0 1 
w2
y
0

w0
P
0


dw P – A
d î
d ˆj
 
 w0
 0 t
 0 ĵ 
dt
dt
dt
But
Ans.
d ˆj
0 z
Here
dt
di
ds
d î

k̂   0 t k̂   = w0 t k̂ +  ˆj
=  t k̂  Then
dt
dt
dt
x
A

1

|  |   0 1  w 02 t 2 2


30
Compact ISC Physics (XII)
F
1.2 The F undamentals Equation of Dynamcis
1.59: Where F is force due to air which will be constant then. mg – F
= mw .... (1). When m mass is taken then F – (m – m) g + (m
– m) w .... (ii) From (i) and (ii) : m =
m – m
2 mw
Ans.
w
F
w
m
mg
(m – m) g
a
1.60 : Net pulling force = (  m) a  m0g – k m1 g – k m2
m2
m1
 m 0 – k (m1  m 2 ) 
g = (m1 + m2 + m0) a  a =  m  m  m  g Ans.
1
2
0 

km2g
km1g
m0
F. B. D of m2 :
a
T
m2
m0 g
k m2 g
T – k m2 g = m2 a. Put value of a then
(1  k) m 0
T = m  m  m m2 g
0
1
2
Ans.
1.61 : (a) Pulling force F = m1 g sin  + m2 g sin  – k1 m1 g cos  – k2 m2 g cos . Then acceleration
a is : a =
g sin  (m1  m2) – g cos  (k1 m1  k 2 m 2 )
.... (i)
m1  m 2
F. B. D of m1 : N = Normal force between two blocks. m2 g sin  + N + k1 m1 g cos  = m1
a .... (ii) put value of (a) in (ii) : N =
2
m
g
a
m2
g
m1
m
1
si
n
Ans.

os
gc
os
N 1g c
m
k1

m2
s k2
co
in
gs


m1
m1

(b)
g
m1
k1
si n
(k1 – k 2 ) m1 m 2 cos α
m1  m 2

For minimum value of acceleration will be zero and friction force will at maxm value then
0 = [g sin  (m1 + m2) – g cos  (k1 m1 + k2 m2)] / m1 +m2  tan  =
k1 m1  k 2 m 2
m1  m 2
31
1 2
at where V = final velocity in this situation Vf = 0
2
1.62 : Upward Journey : We know S = Vt –
Then  =
1
(g sin  + µg cos ) t2
2
–––– (i)
=
0
f
v

t = Time taken in upward journey.

v
a

a=
Downward journey : We know S = ut +

+
µ
)
1 2
at u = initial
2
u=
(i)
and u = 0  now :
a=
(ii)
 2 – 1 
 g sin   ug cos   1




 =  g sin  – ug cos   2   µ =  2
 tan 


1




0
v
1
(g sin  – µg cos ) (t)2 –– (ii)
2

velocity. Then  =
(g
s in

os
gc
(g
)
a
cos
g
µ
–
si n 
Ans.
m2
1.63 : (a) Starts coming down. m2g > m1 g sin  + fmax m2g > m1 g sin  + k m1 g cos   m
1
m2
> sin  + k cos  (b) m1 g sin  > m2 g + k m1 g cos   g sin  > m + k cos  
1
m2
m2
m1 < g sin  – k cos  (c) At rest : Friction will be static : g sin  – k cos  < m1 <
1
1
m2g
si n
a
fm
ax
g
m
ax
fm


m
g
sin
m2g
g
1
m
sin


1
m
fm
ax
a
m2
1
=

k
m
gc
os

sin  + k cos 
m2g
Compact ISC Physics (XII)
s
32
k
=

k
1
m
fm
m2
m2g
gs

1
m
m2 g > m1 g sin Block m 2 has tendency to
m2
Given m   .
1
ax
2
m g  0.66 m1g Then
3 1
in
 Here m2 g = m1  g =
m
and m1 g sin  = m1 g sin 30º = m1 g/2  0.5 m1g
1
g
co
1.64: To find tendency of sliding check value of m2g
move down ward.
Then. Pulling force = m2g – m1 g sin  – k m1 g cos  acceleration a =
a =
g [ – sin  – k cos ]
 1
m 2 g – m1 g sin  – k m1 g cos 
m1  m 2
Ans.
fr = k m 2 g
1.65 : (a) Before no sliding b/w m1 and m2 :
m2
m1
F = at
F
at
F.B.D. of System : Acceleration of both block will be same. w1 = w2 = w = m  m  m  m
1
2
1
2
fr
w
But friction b/w m1 and m2 will be static then. m1
m2
F = at
m1
m1 at
k m 2 g (m1  m 2 )
at
fr = m1 [w1] = m  m < k m2 g  t <
 w1 = w2 = m  m ––– (i)
a
m
1
2
1
1
2
If
t > k m2 g
m2
km2 g
F = at
w2
F.B.D. of m1 :
Assume to =
(m1  m 2 )
. Sleeping b/w two block will be start then F.B.D. of m2 :
a m1
m1
  w2 =
k m2 g
at – km 2 g
m2
  w1 =
k m2 g
m1
–––(ii) Ans.
––––– (iii) Ans.
k m 2 g (m1  m 2 )
 t > to : Slope of w2 >
a m1
w
w2
 a 
Slope of w1 = w2 because of Slope of w2 =  m  from
w =w
 2


a
t
(ii) [After sliding]  Slope of w1 =  m  m  from (i) [Before sliding]
2
 1
1
0
Slope w1 = 0 from (iii) [After sliding]
w1
2
t
33
1.66 :
F.B.D. of block : ma = mg sin  – k mg cos   a = g sin  – kg cos 
Time to reach at bottom : s = ut +
1 2
1
at  sec  =
(g sin  – kg cos ) t2
2
2
A
km
gc
R
k = coefficient of friction
os

mg




t2 =
s in
2  sec 
2

t =
g (sin  – k cos )
g [sin  cos  – k cos2 ]
2
g [sin  cos  – k cos 2  ]
–– (i)
To minimise t  (sin  cos  – k cos2 ) will be minimum also. Assume. x = sin  cos
 – k cos2   Now
dx
= 0 – sin  sin  + cos2  + 2 k cos  sin  = 0
d
cos 2 = – k sin 2  tan 2 = – 1 k
Ans.
Put value of k :  = 49º  Put value of  in equation (i) :
2  2.10
tmin =
1.67 :
10 [sin 49º cos 49º – 0.14 cos 2 49]
 1.0 s.
Ans.
When block pull up Direction of friction will be downward.
F.B.D. of block : N = mg cos  – T sin   fr = k (mg cos  – T sin ). At just sliding:
m k

N
T

sin
g
m

sin



T
mg sin   k mg cos 
 Tmin  
cos   k sin 
T
T cos  = mg sin  + k (mg cos  – T sin ) T =
T
s
co

m µN
mg cos 
mg
(cos  + k sin )  min  minimum value of cos  + k sin  = 1 k 2  Tmin =
mg (sin   k cos )
1  k2
34
Compact ISC Physics (XII)
1.68 : (a) At time of breaking off the plane vertical component of

F must be equal to weight mg. Then F sin  = mg = at sin
F
V

0
m dV

a cos 
V
(b)

0
t1

m
µ=0
at cos  dv

m
dt
mV
1 m2 g 2
m g 2 cos 
mV
t2

 1 

V
=
a cos α
2 a 2 sin 2 
a cos α
2
2 a sin 2 
t dt 
0
m dV

a cos 
at

 t = mg a sin  . Motion equation of block : a1 = Accelration
of block. F cos  = m a1  a1 =
=
t
 t dt

0
Ans.
mV
t2

a cos α
2
1.69 : Motion equation :  = aS 
mg
cos  = m a1 a1 =
3
F
mg
g
cos  =
cos
3m
3
acceleration of block of mass m. a1 =
v
s


=
3
g/
m

m
g
g
v dv 
cos as ds 
  a1 =
cos as 
3
3
0
0
v2
2
v
s

0
g sin as
2g
 v2 =
sin as  v =
3 a
3a
0
s
t


2g
a cos  2
sin as  v =
t 
3a
2m
3
2 3
a cos 
ds
a
ds 
t 2 dt  s = a cos  t  s = m g cos 

cos  t 2 
2m
dt 2 m
2 m x3
6 a 2 sin 3 
0
0
1.70 :
Ans.
Method : 1
Motion of mass 2 m : T – k2mg = 2 mw  T = 2
2m
k2mg
mw + 2kmg
w
T
k
T
m
a
k mg
k
Motion of motor : T – kmg = ma  2 mw + 2 k mg – k mg = ma  a = 2 w – kg
Acceleration of 2 m w.r.t. m : a2m – m = 2w – kg + w = 3w – kg  Suppose in time t both
will be met then.
 =
1
(3w – kg) t2
2

t =
2
3w – kg
Ans.
35
Method : 2
On system : Tension will be internal force hence k2 mg – k mg = – 2 mw + ma 
a = 2w – kg  a2m – m = 3 w – kg then  =
1.71: For observer inside elevator : a1–2 =
But in form of vector :
a1–car
2
3w – kg
1
(3w – kg) t2  t =
2
m 2 g – m1 g  m 2 w 0 – m1 w 0
(m 2 – m1) (g  w 0 )
=
m1  m 2
m1  m 2
 
(m 2 – m1) (g – w 0 )
=
Ans.
m1  m 2
shaft
2T
(m 2 – m1) (g – w 0 )
Acceleration of m1 w. r. t. shaft : a1–shaft = a1–car + acar =
m1  m 2
+ w0    a1–shaft =
Ans.
T
(m 2 – m1) m1 g  m1 m 2 w 0
Ans.
m1  m 2
m1
m2
m1 g
m2g
2 m1 m 2 (g  w 0 )
Tension in string : T – m1 g = m1 a1–shft   T =
m1  m 2
Force applied by pulley on ceiling = 2 T =
4 m1 m 2  
2T = m  m (g – w 0 )
1
2
w0
T
T
4 m1 m 2 (g  w 0 )
 as vector form:
m1  m 2
Ans.
a/2
T
1.72: Motion of body (2) :  mg – T/2 =  ma  2 mg – T
m

= 2 ma .... (i)
Motion of body (1): T – mg sin  = m a/2 .... (ii) from (1)
2g [2 – sin  ]
and (ii) :  
Ans.
[  1]
 a1  a 2 
1.73 : Equationof motion : T = m0  2  .... (i) m1g


s
mg

in
T/2
a
 m
mg
a1 + a2
2
m0
T
= T/2 = m1 a1 .... (ii) m2 g – T/2 = m2 a2 .... (iii)
T/2
4 m1 m 2  m 0 (m1 – m 2 ). g
from (i), (ii) and (iii) : a1 =  m m  m (m  m )
1 2
0
1
2
m1 mg
Ans.
a1
T/2
m2
m2g
a2
36
Compact ISC Physics (XII)
1.74: Motion equation on system: Mg – mg = M a1 + m a2 .... (i) Motion equation
of m : fr – mg = ma2 .... (ii) Since length of rod is  then.  
t2 .... (iii). From (i), (ii) and (iii)
: fr =
m m 
(M – m) t
2
1
(a + a2)
2 1
a1
Ans.
M
m
mg
Mg
1.75:  = 100 cm – T + mg = m a1 .... (i) + 2T mg = m
(i) and (ii) : a1 =
 =
 (–  2) g
 4
a1
.... (ii) from
2
3 a1
. Suppose t time is taken then
2
arel =
T
T
 (  4)
1
arel t2  t =  (2 – ) g
2
1.76 : Motion equations T – mg = ma
a2
Ans.
2T
nm
a1
2
.... (i)   mg – 2T =  m a/2 ....
a1

m
nmg mg
2g ( – 2)
(ii) from (i) and (ii) : a =
. When body
 4
(1) travel h distance then in same time body (2) travel 2h distance
in upward dirn using constraint relation. Nowvelocity of body (2) Just
a
2
before string slack.
2T
V2 = 2a (2h). It body (2) travel x distance again then V2 = 2a (2h) = 2gx
a
T
h
mg
2
x =
mg
2 ah
2 ah
6h
 h 
g . Total hight from ground : 4 = g
 4
Ans.
1.77: F. B. D. of N sin  = m a2 .... (i) F. B. D. of rod : mg – N cos  = m a1 .... (ii) Constraints:
a2 sin  = a1 cos  .... (iii) From (i), (ii) and (iii) :
a1 =
g
1   cot 2 
a2 =
g
tan    cot 
a1
N
a1

a2

m

a2
N

a2
a1

mg

37
1.78: F. B. D. of m : mg – KN – T = m a2 .... (i) N = m a1 .... (ii) F. B. D. of wedge : a1 = a2....
(iv) From (i), (ii), (iii) and (iv) : a1 =
mg
g

M
2m  M  km
2k 
m
KN
a1
M
T
T
KN
m
a1
N
a2
Ans.
m
a2
mg
mg
1.79: F.B.D of bodies on frame of wedge: Since system is stationary on frame of wedge hence :
mg = kma + ma + kmg  a =
g (1 – k)
1 k
Ans.
ma
ma
1 m
a
kmg
ma
m
2m
kma
N
m
mg
mg
1.80: F.B.D. of block (2) with frame of wedge: At maximum accn w : fr1 will be maxm then fr1 = k [mg
cos  + mw sin ]. Since block is under rest with frame of wedge, then equilibrium equation
of block along incline. k (mg cos  + mw sin ) + mg sin  = mw cos   w =
g (1  k cot )
cot  – k
m
2
fr

mw
1

fr1
w
mw


mg
mg
1.81 :
Method : 1
F.B.D. of System a2 = accn of bar w.r.t. incline. Since no force on system in horizontal direction
then O = ma1 + m [a1 – a2 cos ] –––– (1) F.B.D. of bar w.r.t. wedge : m2 a1 cos  + m2 g
sin  = m2 a2 –––– (ii) from (1) and (2) : a1 =
m 2 g sin  cos 
m1  m 2 sin 2 
Ans.
38
Compact ISC Physics (XII)
m2
m2
m1
2
m1
a2
1

a1
m2
m2a1

m2g
Method : 2
F.B.D. of bar : with frame of wedge, bar has zero accn in perpendicular to incline then. N + m2
a1 sin  = m2g cos  –––– (i) F.B.D. of wedge : N sin  = m1 a1 –––– (ii) from (i) and
(ii) : a1 =
m 2 g cos  sin 
N
m1
2
m1  m 2 sin 
m2a1


a1
90 – 
m 2g
N

N
Method : 3
F.B.D. of bar with frame of ground observes : Motion
a1
equation in perpendicular dirn of incline then : m2g cos  –
a2

N = m2 a1 sin  ––– (i)

m2g
v
1.84: v = 360 km/hr =
360  1000
 100 m
s
3600
C
v
B
R
A
At point A : N – mg =
g+
mv 2
70 [100]2
mv 2
 N = mg +
= 70 g +
= 70
R
500
R
70  100  100
= 70 g + 140 g = 210 g = 2.1 kN
500
At point B : N =
v
Ans.
mv 2
––– (i) put value of m, V and R N = 1.5 kN
R
N
mg
Ans.
N
mg
2
2
At point C : N + mg = mv /R N = mv /R – mg ––– (ii) put value of V, m, R
in (ii) N = 0.7 kN Ans.
N
39
1.85: Tangential acceleration (at) : mg sin  = m at  at = g sin 
m
Radial acceleration (ar) : Energy conservation : mg  cos 
1
=
mv2  v =
2
anet =
a
t
2
a
r
2
2 g  cos   ar = v  = 2 g cos  

mg
v
 g sin 2   4 cos 2  anet = g 1  3 cos 2 
Now : T – mg cos  =
(b)
T

2
mv 2
= 2 mg cos   T = 3 mg cos 

Ans.
Component of velocity in y dirn: vy = v sin 
vy =
y dirn.

2 g  cos  sin   v2y = 2 g  cos  sin2 
For vy maximum v2y will be maximum. Then x = cos  sin2 
will
be maximum. dx d = 0 = 2 cos2  sin  – sin3  = 0  2 cos2 
= sin2   tan  =
= 2 g 
(c)
1
3

2  sin  =
1
2
cos  =
3
3
 v2y
 1 
2
4g
 = mg
v max

y
 T = 3 mg cos  = 3 mg 
3 
3 3
 3
If no acceliration in y direction then. ar cos  = at sin   2 g cos2
3

1
 = g sin2   tan  =
2  cos  =

v
3
Ans.
ar

Ans.
1.86 : Extreme position : Only tangential acceleration is prerent at extreme
at

position. at = g sin 
Lowest Position : Energy conservation : mg [ –  cos ] =
 v2 = 2 g  (1 – cos )  ar =
1
m v2
2
v2
= 2 g (1 – cos )


A/C condition : ar = at  g sin  = 2 g (1 – cos )  sin  + 2
cos  = 2  cos  = 3 5
  = 53º

Ans.
v
mg
40
Compact ISC Physics (XII)
A
1.87: Particle will break off sphere when normal reaction will be zero.
mv 2
 v2 = Rg cos . Energy conservation : mgR
R
1
1
(1 – cos ) =
mv2  mgR (1 – cos ) =
mRg cos  
2
2
mg cos  =
cos 

2
1 – cos  =
 2
2 Rg
Rg  3   v =
 
3
cos  =
A
R

v
R 
mg
2
3 cos 

= 1  v2 =
3
2
Ans.
1.88: Top view : Spring force = F =    N = normal reaction on sleeve. Since no acceleration
in tangential direction. N sin  =    cos  ––– (i). Equation in radial direction: N cos 
x   cos 
(cos ) +    sin 
sin 
m
= m ( + ) cosec  (w2)  x   = m  + m   w2   =
Ans.
x – mw 2
+    sin  = m r w2 = m ( + ) cosec  w2 from (i):

w
l
w
r
N
 


v

r

1.89: k = k0 = 1 – R  friction coefficient. Suppose cyclist at radius of r. then friction provide centripital


r
mv 2

force to motion on circular path. Then where kmg = friction force. k0 1 –  mg 
 v2
r
 R
 r2 
dx
r2
= k0 g  r –  . To maximum value of v :
x = r–
will be maxm. Then
= 0 
dr
R
R 

R R  k0 g R
2r
1
2
1 –  0  r = R  Vmax  k 0 g  –  
k 0 gR
 Vmax =
Ans.
2
2
4
4


R
2
r
O
R
kmg
mv2 /r
41
1.90: Tangential and radial both acceleration is only provided by friction because
friction is acting as external force. Then maximum value of friction = k mg.
v2/R
Velocity of car after d distance travel. v2 = 2 w d. Then ar = Radial
R
2
2 w d
v

acceleration =
  at = Tangential acceleration = w  
R
R
2
anet =
 2w  d 
4d 2

  w 2  w  1  2  Fnet = m w
 R 
R
 4d 2 
2 
R

w
1
2
2

Squaring :
2  = k g  d =

2
R


 kg

w
 
4d2
1
R2
w
= k mg
2

 –1


 
1.91 : y = a sin x α  k = friction coefficient. Centripital force
m v2
=
and centripital force will be provided by friction.
R
m v2
 k mg  v2  k Rg.
At limiting condition :
R
Ans.
y
x
For v maximum  R will be minimum. And we know : speed
y
is constant also we are secing that radius of curvature will be
minimum at maximum point of curve then. ay =
d2 y
dt 2

v
v2 a
ay
2
 
sin x α . At maximum value of curve : sin x α = 1 
ay =
v2
v 2 α2
R = a  2
v a
y
v2 a
α2
2
R = 
a
x
. Then v2  k
α2
g   v   kg
a
a
Ans.
1.92: F.B.D. of differential element of length dl. Equation of motion : 2T sin  – N cos  = (dm) Rw2
––– (i)  N sin  = (dm) g put value of N in (i) : 2 T  – (dm) g cot  = (dm) Rw2 2 T
w

mg
N

T
T
 = very small angle.
sin  = 
N
T


T
(dm) g
42
Compact ISC Physics (XII)
 –
m
m
m
m g cot  m R w 2

(R 2 ) g cot  = (R 2 )
Rw2  T –
T =
[Rw2
2R
2R
2
2
2
mg
+ g cot ]  T =
2
1.93 :

Rw 2 
cot  

g 

Ans.
m2
m1 = 0   T1 – m1 g = m1 a ––– (i)  m2 g – T2 = m2 a ––– (ii)
Relation between T1 and T2 : (T + dT) sin d 2 + T sin d 2 = dN
a
Td = dN  dfr = µ dN = µ Td ––––– (i)
T2
Since  is very small. sin d  d  cos d  1
m2
In horizontal direction : (T + dT) cos d 2 – T cos d 2 = dfr
m2g
T2
dT = dfr ––– (i)  dT = µ T d 

T1
 µ d
0
m2
m1
y
dfr
T + dT
d
T
(b)
x
g –a
g – a 

   

ga
g  a 
0   = 0  eµ = 0  µ =
m1g
   n (T2 – T1 ) = µ 
T2
µ
from (iii) : T  e 
1
Then eµ =
m1

dN
T2 = T1 eµ ––– (iii)
T2 m 2 (g – a)
from (i) and (ii) : T  m (g  a)
1
1
dT

T
T1
––– (iv). Since pulley is fixed : Before skiding a =
1
n 0

g – a 
0 g – a
[n – 0 ]
from equaiton (iv) : en0 =   g  a     g  a  a = g
  n0


Ans.
43
y
Hence velocity along y axis is not responsible for circular
1.94:
motion only velocity along Z-axi s i s responsibl e.
d2 x
1.95: x = a sin wt  a x =
dt 2
2
y = b cos wt  ay =
d y
V0
v0
m VZ2
m V02 cos 2 
Then N =
VZ = V0 cos  N =
R
R
O

R
x

Z
2
= – aw sin wt 

= – bw2 cos wt  a net  a x î  ay ˆj
dt 2
w
y

2
2
= – a w sin wt î – b w cos wt ĵ  F  – mw 2 [a sin wt î  b cos wt ĵ] 



r  x î  y ĵ  a sin wt î  b cos wt ĵ  F  – m r w 2
x

where r = position vector of particle. F = mw2 (a sin wt) 2  (b cos wt) 2
 F = mw2 
x 2  y2
1.96 :
(a)
Ans.
Method : 1 (Impluse equation)

We know  P = Impluse in time t =
t
 F dt
0
t
=
 – mg dt

 P = – mg t
Ans.
0
V0
g
90+
V0
m
(b)
g



2 V0 sin 
2 V0 sin 
 
 
 P  mg

=
V
g
cos
(90
+
)

 – V0 . g = V0
V
.
g
0
0
g
g
 
 

– V0 . g
V0 . g
g sin   = 2 m V0 sin   V0 sin  =
  P   2m
Ans.
g
g
T=
Method : 2 (Kinematic)


y
V0  V0 cos  î  V0 sin  ˆj  Vf  V0 cos  î  (V0 sin  – gt) ĵ




where Vf = final vel oci ty vecto r t hen  P  m Vf – m V0 

 P  – m g t ĵ Ans.
(a)
V0

x
44
Compact ISC Physics (XII)
(b)
T=

2 V0 sin 
2 V0 sin  ˆ

P –mg
j 

 P = – 2 m V0 sin  ˆj
g
g
From method : 1 :
 

 2 m V0 . g
– V0 . g
ˆj
V0 sin  =
  P 
g
g
Ans.
m
1.97 :
 
F  a t (T – t)
(a)
(b)



V

d
V

a
 

2

 w  [ t – t ]  dV  a
F  m w  w = linear acceleration.
dt
m
m
0

 3 

a   t 2 3 

V 
–   m V  a   Pf Ans.

m  2
3 
6


V

a
w 
[ t – t 2 ] 
m
S

dS 
0

0
t
t  

  t dt –
t 2 dt 


 0
0



t

 a  t
 a  t 2
t3 
  t dt – t 2 dt 
dV 
V


–


 
m 
m  2
3  
0
0



 3
  2

 a   3 .   4 
4
4
a  t
t
 dt –
dt   S 
–   S  a   S = a  Ans.

m  2
3 
m  6
12 
m 12
12 m
0
0



 
F0
dV
sin wt 
1.98: F  F0 sin wt  a =

m
dt
S
t


V

0
F
dV  0
m
t

0
t
sin wt dt  V = – F0 cos wt
mw
0
F
F
1
F0
dS  0
[1 – cos wt] dt  S = 0  t – sin wt  
[1 – cos wt] 
V =
mw
mw  w
mw

0
0
S =
F0
mw 2
[tw – sin wt]
distance
distance will be increasing function w.r.t. time then
dS
= (+)ive or > 0  w – w cos wt > 0 
dt
cos wt < 1  0 < wt <   2 < wt < 3 
4 < wt < 5




wt
45
0
dv
F
 a  0 cos wt 
1.99: F = F0 cos wt 
dt
m
π
sin wt = 0  wt =   t =
w
0
Now :

dV 
0
S=
– F0
mw

F0
mw
0
F0
m
t
 cos wt dt 0 =
0
cos wt  V =
0
s
t


F
F0
ds  0
sin wt dt 
si n wt .... (i ) 
m
w
mw
0
0
s =
2F0
velocity will be maximum when sin wt = 1
m w2
Ans.
V
t


dV – r
dV
–r

dt

V 
1.100 : (a) F = – rV 
 n V
V
m
dt
m
V0
0
V = V0 e–r/m t  V will be zero when t  
(b)
F0
sin wt 
mw
Ans.
t
(1 – cos wt) at t = π
w
2
 Vmax =
F0
m

dV 
V
s


V
V0
 –
r
t
m
Ans.
r
dv
r
r
r
v dV  – v ds
a= –
V = v
 
 V – V0 = –
S  V = V0 –
S
m
ds
m
m
m
V0
0
Total distance travel by particle is S1 then final velocity = 0  0 = V0 – r m S1 
S1 =
(c)
Ans.
V0
m V0
r
= V0 –
S2  S2 =
r

m
t2 =
1.101 :
m V0
r
  –1


  
–––– (i)  Also
S
V ( – 1)
m
n  –––– (i)  <V> = 2  0
t2
 n 
r
– kv 2
F  v2  F = – kv2  a =

m
V
V

V0
dv
v2
V0

= V0 e –
Ans.

–k
m
t
 dt

h
0
–1
– kt
 V0 – V  m
1
1
kt



–




 V V  k  t ––– (i)
V V
m
V0
V
m
0 

0
V0
V
r
mt2
46
Compact ISC Physics (XII)
– kv 2
dv
– kv 2 dx
 v
Also a =
  v dv =

m
dx
m
k=
V
 V0 – V 
h
m

n V put in (i) : t = 
V
V
V
h
0
0  n 0

V

V0
m dv
 –k
V
h
V
V0 = – kh
 dx  m n
0
Ans.
V
1.102: Suppose at time t distance travel is x then F = mg sin  – ax
mg cos   mw = mg sin  – ax mg cos  where w =
acceleration of block. w = g sin  – ax g cos  .... (i) v = g
m
in
gs
K

c
mg

os
k = ax

sin  – ax g cos  
xm
0

v dv
= g sin  – a x g cos  
dx
v dv 
0
 (g sin α – axg cos α) dx  0 = (g sin ) x –
0
2 g sin 
2
a x 2 g cos 
x = a g cos  x =
a
2
tan . Velocity will be maximum when dv dt = 0 = acceleration  g sin  = ax g cos 
1
x =
a
v max

t an   no w
ya tan 
v dv 
0
1
 2 g cos   1

 tan   –
 tan  
a
2
a




 (g sin α – a x g cos α) dx

Vmax


0

 Vmax =
g sin 2 α
g

sin  tan 
a cos α
a
= g sin 
Ans.
m
K
1.103: K = friction coefficient  Block start sliding when.

F = at
F = at1 = k m g  t1 = k m g a . Assuming over time start from block start sliding then.
Suppose acceleration of block is w then. mw = at – kmg = a (t1 + t) – kmg = a t  w
v
a
=
 t 
m

0
dV 
a
m
t

(t d (t)  V 
0
a
a
S=
(t)3 =
(t – t1)3
6m
6m
a
t 2 
2m
s

ds 
0
a
2m
t
 t
2
d (t) 
0
t = t – t1
Ans.
t=0
t = t1
t=t
47
v=0
2
V0
h
1.104 :
Upward journey :
V0
mg
Fnet = mg + kv2  a = g +
0
v dv
mg  kv 2
kv 2

 –
ds
m
m
h
mv dv
 mg  kv 2
F = Kv
 –
v0
 ds

h=
0
 mv dv = (mg – kv2) ds
KV02  mg
m
n
.... (i)
2k
mg
2
Downward Journey : F net = mg – kv2
v

h
mv dv
 mg – kv 2

0

ds  h 
0
v0
 v =
1
KV02
v
kv
mg
 a =
mg – kv 2
m
m
KV02  mg
m
KV02  mg
m
mg
n
n

n

2k
mg
2k
mg
2k
mg – kv 2
Ans.
mg

1.105 :(a) Position vector of particle : r = r cos  î + r sin  ˆj

 At time t :  = wt  F  F r̂  F (cos  î  sin  ĵ)
F
m
w

F
[cos  î  sin  ĵ] 
 a 
m

V

0

F
dv 
m
t

(cos wt) dt î 
0

speed = | V | 
(b)
= v dv/ ds
F
mw
F
m
r

0
t
 (sin wt) dt
0

 V 
F
[sin wt î  (1 – cos wt) ĵ] 
mw
(sin wt) 2  (1 – cos wt) 2  Speed =
2F
sin wt 2
mw
Ans.
s
t


2F
ds
2F
sin wt 2 dt
 V
sin wt  ds 
Distance is calculated by speed. Then
2
mw
dt
mw
0
0
S = Distance S = –
2F
mw
2
 2 cos
wt
2
t
S =
0
4F 
wt 
1 – cos

2 
2  .... (ii) velocity of particle
mw
48
Compact ISC Physics (XII)
will be zero : 0 =
S=
2F
sin wt
2
mw
sin wt

 0  put value of t = 2π
w in (ii)
2
8F
Distance
8F/mw 2

(1
–
cos
)

S
=

Average
speed
=
Time
2/w
mw 2
mw 2
4F
Average speed = 4F  mw
Ans.
1.106: K = tan  = friction coefficient  When = W = net
tangent ial
accelerat ion/
Wx
=
Accel erati on
w
mg sin  cos  – k mg cos 
m
in x direction. Now W =

s
g co
Km
 
mg sin
x
= g (sin  cos  – K cos ) = g [sin  cos  – sin ] 
Acceleration in x dirn: Wx =

v
W = g sin  (cos  – 1)... (i)
mg sin  – k mg cos  cos 
 Wx = g sin  [1 – cos ] .... (ii)
m
From (i) and (ii) : W = –Wx 

d Vτ
 –
dt

d Vx
 V = –Vx + cos  at t = 0
dt
V
= V0 and Vx = 0  V = – Vx + V0 .... (iii) Then const = V0  Also Vx = V cos  ....
(iv) From (iii) and (iv) : V =
V0
1  cos 
Ans.
1.107: Tangential force on system is F then  F 
 (dm) g sin  =  ( R d) g sin  = Rg
t
 sin  d 
 = R f  
in = 0 = –  Rg
0

cos 
f


 f =  R  f =
 F =
R
R g
gR
 R   a =  mR g 1 – cos  R  = x  1 – cos  R   a =  1 – cos  R 
Rg 1 – cos 


y

dm=  R d
 d

R
(dm)g
f
F
x
dm

d
49
1.108: At time of break off. normal reaction will be zero. Where mw0
= Psuedo force because observer at sphere.
mV02
= mg cos 
R
R
w0= acceleration
– m w0 sin   V20 = Rg cos  – R w0 sin .... (i)
Energy equation: Wall forces = Kf – Ki  Wpsuedo + Wmg = Kf –
Ki m w0 R sin  + mg [R – R cos ] =
1
m V20 – 0  V20 = 2w0
2
 – V02 

R sin  + 2 g R [1 – cos ].... (ii) From (i) and (ii) : V = 2R 
 R 
2
0
+ 2 g R  V20 = 2 g R 3 
1.109 :
V0 =
R g
 cos  
3
mw0
0 R
0
R
V0
mg
  w0
5  g w 0 
3
 g 
  w  
3 1   0  
  g  



Ans.
1
K
Given F  n   F = n Where K = Constannt.
r
r
A particle is said to steady if we displace particle away from origin, particle want to regain
its position and also it we displaced particle toward origin, particle want to regain its original
position. Then. At steady state (mean position) : F =
m V02
K
m V02
Then n 
.... (i)
r
r
r
m V2
> 0  K r–n – mV2 r–1 > 0  Differentiate both side
r
with respect to r for small increase of r: –n k r–n–1 dr + mV2 r–2 dr > 0 and
It we increases r then. F –
– n K  1   mV02   1 
  
  > 0 .... (ii)  From (i) and (ii) n < 1 Ans.
r n  r   r   r 
v
v0
r
0
F
O
r
50
Compact ISC Physics (XII)
1.110: At steady state: mg sin  = m (R sin ) w2 cos   cos 
=
g
. Case (i) : If Rw2 > g then cos
g
is defined and
Rw
R w2
only one equilibrium position will be exist and will be steady.
g
Case (ii) : If Rw2 < g then only
= 0º will be equilibrium
R w2
position because tangential fore along arch of ring due to mg
2
Y
R
N


will be greater than that of pseudo force and object will come
at lower position of ring.
90–

Y
mg
2
m (R sin  ) w
106
Compact ISC Physics (XII)
Q: 1.247: Method 1: Torque equation on whole system about centre

2 
 m R 2  m R 2  mR  
m
g

km
g
R

2
1
1
of pulley

 2
2 




a1
m1
T2
T2
disc

T1
km2g
T1
g (m 2  km1 )

R (m 2  m1  m
)  Acceleration of mass m1 :
2
a1
m2g
g (m 2  km1 )
1
1  R 
S
a 1t 2
m  displacement in time t :
2
m 2  m1 
2


 km1g 2 (m 2  km1 ) t 2 
 km g 2 (m  km )  1


2
1
2
1
t 
 Workdone of friction = km1gs  
 W   m  2 (m  m ) 
1
2
 m m  m
 2




2
1
2


mR 2
Method 2: m2g - T1 = m2a1 -----(i) T1-km1g = m1a2 -----(ii)  T2  T1 R 
 ----(i) a2 =
2

R -----(iv) from (i) and (ii) and (iv) : 1 
W  ( km1g ) s 
km1g 2 ( m 2  km1 ) t 2
m  2( m1  m 2 )
g ( m 2  km1 )
m
m 2  m1 
2

 S
1
a 1t 2 
2
Ans.
Q: 1.248: F.B.D. of cylinder : Force equation : KN1 + N2 = mg ------(i) KN2 = N1 -----(ii)
Torque equation about centre of cylinder : KN1R  KN 2 R 
mg
-------(iii) from (i) and (ii) : N 2 
2
2
2
1 k
2 and N1 
: W  W0  2   0  W0  2   
mR 2
mR 2
  KR [ N1  N 2 ] 

2
2
2kg (1  K )

put
in
(iii)
:
. We know
R (1  K 2 )
1 k 2
kmg
w 0 2 R (1  K 2 )
w 2 R (1  K 2 )

no. of turn 
 0
4kg(1  K )
2
8kg(1  K )
N2
Cylinder
KN1
W0
W
R
KN2
N1
107
Q: 1.249 : Calculation of torque : Where df = friction force on differential region = This is scalar
 m

2K mgx dx

2 xdx  g 
sum because are going to calculate torque. df = k [dN]  k 
2


R
R2



  I 
R
2 kmgx 2
The differential torques : d  xdf 
R
dx      d  
2
2 kmgx 2 dx
R
0
df
x
4 kg
3WR
tt
3 R
4kg
R
Q: 1.250:    w
Ans
  cw1/ 2    kw1/ 2 . Time calculation :
0
Angle calculation : wdw  kw1/ 2d   w
w
0
1/ 2
dx
W
0
1 / 2 


t1
2
kmgR
3
mR 2
2
4 kg
  kmgR   
2
3
3 R
We know W = W0 + t  0  W 
=

2

w0
dw
w1 / 2
0
3
dw    kd    2 W0
0
3C
2
t1
1/ 2
  kdt  t  2 w 0
1
0
C
. Man angular velocity
3
2W0 2
w0

1
 2W 2 
3
0 
3C
 C 


Ans
m
2 M
2
Q:1.251: Moment of inertial of chord : I   (l  x ) R  xR 
l
 l

l-x
M
M
R
0
R
 MR 2

mgx

R

 mR 2  
 I = mR . Toque about 0 :  = I  l
 2



x
2
2 mgx
   R l ( M  2 m )
x
mgx
l
Ans.
Q:1.252: (a) Torque equation about ICOR : (mg Sin ) R 
7
mR 2 
5

5g sin 
7R
fr 
2
5gSin
10
mR
mg sin  ----- (i). At slipping : fr = K mg
 fr 
5
7R
35
Torque
about
com:
Rfr 
2
mR 2
5

m,R

fr
OR
IC mg
Sin
mg


acm
108
Compact ISC Physics (XII)
Cos ------- (ii) from (i) and (ii): kmgCos  
(b)
KE 
10
10
2
mg sin   K 
tan   K  tan 
35
35
7
1 7
1
7
25g 2Sin 2 5mg 2Sin 2 2
2
2
MR 2 t 2

t
I ICOR W 2  KE   MR  t  KE 
2 5
10
14
2
7 x 7R 2

 
Q:1.253: (a) Torque equation of ICOR which is attached with point of
cylinder
T
3
2g
2
string. mgR  mR    
 a cm  R  2 g 3 . Force equation
2
3R
on cylinder : mg-2T = m2g/3  T = mg/6
R

(b)
 2g 
2
mg 2 t
P  F.v  mg[a cm t ]  mg  3 t   P 
3


T
Ans.
mg
Q:1.254: F.B.D. of cylinder in frame of lift : Torque about ICOR = m (g  w 0 ) R 
3
mR 2 
2
2
2  
2(g  w 0 )
1
1
 W  R  ( g  W0 ) . In terms of vector : w  (g  w 0 ) . Torque about
3
3
3R
 
m (g  w 0 )
m (g  w 0 )
mR 2  2 


F

F

com : FR  I 
. In vector form :
Ans.
R g  w0 
3
3
2  3 
  


F
F
w0
cylinder
R

m, R
ICOR
1
w
mg
mg + mw0
Q:1.255: Torque about ICOR point P : (mgsin) r = (I+mr2)
   
T
mgr sin 
P
I  mr 2

acm = R = r
mgsin
mgCos
2
a cm 
mgr sin 
I  mr 2
Ans.

acm
109
Q:1.256: At time of slipping friction will act at maximum value KN then.
Force equation in y dirn : KN = macm  K(mg+F) = macm -------(i). Torque
equation about I COR :
a cm  R 

2F
3m
 3

FR   mR 2  
 2


put in (i ) k ( mg  F)  ma cm 

acm
x
KN
mg
F

  a  2 kg
cm

2  3k

m

r
mR1
mR 2
1  2T1 
- mR
2
1
2
mR 2
mR 2
----(i). Torque equation on sphere (B) : 2T1R 
 2  2T1 
-2
2
---(ii). From (i) and (ii): 1 = 2 . Force equation on (B): mg-2T1 = maB ------(ii)
3mgR  MgR
9 mR 2  I  MR 2
aA = 3R  a A 

P
2
9mR 2  I  MR 2
9 mR  I  MR
2
2 T1
2 T1
m1R
Q
2
(B)
mg
M
P
R
. Hence acceleration of particle :
aA 

(A)

gR ( M  3M )
3R 2 g ( M  3m )
acm
T2
1
Constraint : Acceleration of point P = R1 = aB-R2 . Then aB= R1 + Ra2 =
2Ra1 . Put in (iii): mg-2T1 = m2Ra1 = 2mR1 -----(ii) from (i) : mR1 = 4T1 put in
(iv) mg - 2T1 = 8T1  T1 = mg/10
Q:1.259: Troque about point P for system :
 

2
I = mR
F2 t 2 (Cos  r / R ) 2
1
a cm t 2  W 
2
2m(1  v)
Q:1.258: Torque equation on sphere (A): 2T1R 
MgR  mg ( 2 R  R )   m ( 3R ) 2  ( I  MR 2 ) 


Ans.
fr
F( RCos   r )
mR (   1)
W = (F Cos) displacement  ( FCos )


F
a
 Fr  Rfr  Ia  mR 2 cm   Fr  Rfr  mR a cm
R
------(ii) from (i) and (ii) : a cm 
y
2F
3
kmg
3kmg
F

2  3kmg
2

a cm  
2
kmg  F  k  
2

3
k

k
3  2  3k
3

3
Q:1.257: (a) Force equation along x axis : F Cos
– fr = m a cm -----(i) Torque equation about com:
(b)
2F
3mR
cylinder
m
3g ( M  3m )
I
M  9m 
R2
Ans.
mg
m
mg
110
Compact ISC Physics (XII)
F
Q:1.260: (a) Cylinder  Force equation on system : F = (m1 + m2) acm  a cm  m  m . Torque
1
2
equation about frame attached with centre of sphere. FR 
2F
m1R 2
    m R . Acculeration of
1
2
F
2F
F(3m1  2m 2 )
point K : a k  a cm  R  m  m  m  a k  m ( m  m )
1
2
1
1 1
2
(b)
1 2 (3m1  2m 2 ) t 2
 1

2
K
.
E
.

F


WF = Work done by F = Grain in K.E. K.E.  F

 2 akt 
2
m1 ( m1  m 2 )


A
m1
m1
acm
k
m2
B

2macm
F
F
m2
Q:1.261: Acceleration of plank = acm - R. Then force equation on system : (Frm ground frame).
F = m2 acm + m1 (acm - R) ----(i) Torque equation with frame attached with centre of sphere :


2
F
m1a cm R  FR   m1R 2  m1R 2   ------(ii) from (i) and (ii) a  2
Ans.
cm
5
7 



2
 m1  m 2 


7


And acceleration of blank : w1 = acm - R -----(iii) put acm in (i) and calculate  and put in ----(iii) :
w1 
F
2
m1  m 2
7
Ans.
sphere
y

m2
acm
R
x
m1

m2
m2acm
R
F
m1acm
m1
F
Q:1.262: (a) Torque about bottom point is zero. Hence angular momenton will be consermed about
bottom
point
then.
 
 
L  L cm  mR  Vcm

1
L final  mR 2  MR ( RW ) 
2

3
mR 2 w -------(ii)
2
1
3
mR 2 w 0  mR 2 w  w  w 0
from (i) and (ii) :
3
2
2

 1

L initial   mR 2  w 0  O
 2

m
-------(i)
N
cylinder
R
m1
mg

Vcm
R
w
w0
k
Slipping
kmg
111
 mR 2

Torque equation about com : kmgR  
 2

w0
2g
1 w 0R
 w0 
t  t
3
kR
3 kg
Wfr  
1
mw 0 2 R 2
6
(b)

2g

 
. We know : w = w0 - t

kR

Wfr  k f  ki 
1
2
 3

1
 mR 2  w 2  m ( RW0 ) 2
 2

2



Ans.
Q:1.263: Suppose at angle ball have the sphere then energy equation : Loss of P.E. = Gain of K.E.
1
1
mV 2cm  I cm w 2 
2
2
mg( R  r ) (1  cos ) 
V 2 cm 
mg Cos   0 
Vcm 
1
1
mV 2cm  I cm w 2
2
2
10
g (R  r ) (1  cos ) ------(i) force equation in direction of N
7
mg Cos  N 
:
mg( R  r ) (1  cos ) 
mV 2cm
(R  r)

m1 r
N

for
leave contact N =
m2
0 then.

mg
Vcm
R
V 2cm
mV 2 cm
 Cos 
. Put cos in (i) :
g (R  r)
Rr
10
g (R  r )
7
w
N
Q:1.264: Suppose at time t solid cylinder make  angle with vertical
energy
eqaation
:
R


mg
1
2
2
 3
  V 


 mR 2   0   1  3 mR 2  w 2  mg( R  RCos )  
 2
  R 


2 2

 


3
3
mV0 2  mR 2 w 2  mgR  mgR Cos  -----(i) Force equation in dir n of normal reaction : mg Cos
4
4
3
7
mV0 2  mR 2 w 2  mgR 4
4
-----(i). For stopping Bouncing N: will be minimum when W will be maximum and this condition will
be come when  = . Hence we have to avoid bouncing at  =  then there will be no bouncing
3
4
2 2
2
before = . Now for minimum V0 put N = 0 and then R W  Vo  gR . Put = and R2w2
7
7
- N = mRw2  mg Cos = N + mRw2 put mgCos value in equation (i) : N 
2
value in (i) : V0  
4
7
gR  gR Cos   V0 
7
3
gR
(7 Cos   4 )
3
Ans.
112
Compact ISC Physics (XII)
Q:1.265: Since angular velocity(w) is
y
w
w
continuously decreasing then there is chance
mg
cm
V
of minimum value of normal reaction at ground m
mg
R
x
when particle is at maximum height or top of
N
m
hoop. Then force equation in y direction on
system from reference attached with com of
hoop. 2 mg - N = mRW2 [Here w=final angualr velcoity]. Since N = 0  2mg = mRW2  w2 = 2g/
R ----(i). Now energy equation : (from ICOR):
0
2
2
1
1
2   V0 
 mg( 2R )   2 mR 2  m ( 2R ) 2  w 2  2 mV0  2mgR  1 (6 mR 2 ) 2g
 2 mR  

2
2 
 R 

2
2
R
 V0  8gR
m/2
Q:1.266: Velocity of top point : VP = 2v. Hence kinetic energy of
v
1  m 
2
crawler : k  2  2  ( 2 v)  k = mv2 Ans.


y
Q:1.267: We know kinetic energy from fixed axis rotation is
1
1
1
2
2
2
given by: K  I xx Wx  I yy Wy  I zz Wz . If
2
2
2
reference frame is not traslatory. And if reference frame is
translatory then we know to use kinetic energy of COM.
m
v
R
0
z
v
v
1
1
and Wy  and Wz  0 . Then K  I zz Wx 2  I yy Wy 2  0
r
R
2
2

1
2
2
2
x

Wx 
 2
 

 
 mr 2   v   1  2 mr 2  MR 2   v 
 5
 r 
2  5

 
  R 
m/2

7
2 r2
2
K

mv
1



10
7 R2



 Ans.

Q:1.268: (a) From this reference frame centrifugal force on particle of mass
(dm) is : dFcf  ( dm) ri w 2  Fcf   (dm ) ri w 2 
w
2
 ( dm ) ri 
Position of CoM 
Fcf  mw 2 R c
 
 2dm [ v  w ]
Corolis


Fcor  2 (dm) v  w  2(dm)v   w

Vc  Velocity of CoM
force:
=
dFcor
dm
P
y
 i
r
 (dm ) ri 
 where R   (dm ) ri
Fcf  w 2 m 
c

m
m


(b)
V

x
0
x
z


 (dm)  v  

2m
  w  wher e Vc  ( dm )  V 
m
m


113
Q:1.269: Method : 1 dc = (dF) x Cos = w2x2sincos dm =
w
l
2
w sincosx
m
l
2
2
2 m


w
sin

cos

x 2 dx 

c
dx 
l
l


2
w (dm) xsin
1
l mw 2 sin 2
mw 2 l 2 sin 2 
c 
 c 
24
24

 ml 2
 dL ml 2
d
w sin    

w sin 
Method : 2 L 
12
dt
12
dt
2

C
x
2
w
in
ws

y
2
 ml
ml

( w sin ) w cos    
w 2 sin 2
12
24
wcos
x
Ans.
Q:1.270: Torque of centrifuglal force
w
l
2
about 0:  d   (dm) x sinw x cos  
0
0

l 2 mw 2

sin  cos  ------(i) Torque of
3
mg about 0 :  mg  mg

l/2
x
mg
l
sin  -----(ii)
2
from (i) and (ii) : for equilibniom:   mg 
(dm) xsin w2
(centrifugal force)
dx
l 2 mw 2
sin  cos   mg
3
l
2 Sin 
cos  
3g
2 w 2l
Q:1.271: where k = coefficient of friction. x0 = Perpendicular distance b/w line of mg and N. Here
N = mg. Since body is in rotational equilibrium then Thorque about COM will be zero. Kmg a/2 =
NX0 = mgx  x 0 
ka
. Suppose at time t cube is displaced by x distance then Torque about origin
2
ka
 mg a 2  mg a 2
is:  0  N ( x  x 0  a / 2)  mg( x  a / 2)  mg( x  x 0  a 2 )  mgx  mg a 2  mg
2
N
y
mgka
 0 
. Because of this torque angular
2
velocity of centre will be decreasing
continuously.
a
x
0
x
z
a
x0
a/2
kmg
x
mg
114
Compact ISC Physics (XII)
w0
Q:1.272: Method : 1 Suppose velocity of sleeve is Vr when its position from
axis is x and angular velocity of rod is w then Centrifugal force: mxw2 = max
Vr
x
l
v r dVr
2
2
V2 l
 mx w  m
  Vr dVr   w xdx  r   w 2 xdx -----(i).
dx
2
0
0
0
Vr 2

2
Vr  lW0 1 
3m
M
momentum conservation :
l
l

o
M 2 l 4 Wo 2 x dx
( 3mx 2  M l 2 ) 2

1  ml 2 
1
Ml 2  2 1
2
W0 2  ml 2 
 w  mVr -----(i). Using angular
2  3 
2 
3 
2

Ml 2
Ml 2  ----(ii) put w in equation (i) :
W0   ml 2 
w

3
3 

 Ml 2

w0

 2 Ml  
1  ml
2 1
3

w0 
ml 
 2
2  3
3   2 Ml 2


 ml 
3

2
Q :1.273:
M
B
Ans.
Method : 2 Energy conservation :
2
 Vr  lw 0 1 
Vr
A
Calculation of w : Using angular momentum conservation :

Ml 2
Ml 2 
Ml 2 W0
W0   mx 2 
w  W
put in (i) :

3
3 
3mx 2  Ml 2

m


  1 mv 2
r
 2



ml 2 w 0 2
M 2l 4 w 0 2

3

Ml 2
9 ml 2 

3





 mVr 2
3m
M
Linear
impluse
J  MVcm  Vcm  J m .
equation
Angular
:
m, l
impluse
J
ml 2
6J
ww
equation about com : J l 2 
.
12
ml
F.B.D. of half part of rod : Fx  Centripital force 
Vcm
l/2
CM
Just after impact
m  l  36 j2
9J 2
  2 2  Fx 
2 4 m l
2ml 2

l/4
m
w
Just before impact
m l  2
 w 
2 4
l/2
 m

Fx 
, l 2 
 2

115
Q:1.274: (a) Top view : Angular momentum conservation about hinge : mv
---- (i) Equation of e : v 
l
l ml 2
 mv ' 
w
2
2
3
12mv
 3m  4M 
l
v'  
 V Ans
w  v' ------ (ii) from (i) and (ii) : w 
l
(
3
m

4
m
)
2
 3m  4m 
l
M
l
l
2
m
m
V
V
m
V'
M
M
Just before impact
(b)
F.B.D. of rod :
Just after impact
Fy
8MV2
 Ml  2
Fy  
W


2
 2 
 4M 
l 1 

 3m 
w
Q:1.275: (a) Angular momentum conservation :
3mv 0
Ml 2
mv 0 l 
w  w 
-----(i). Using energy
Ml
3
conservation
after
collision
:
M

w2 
6g
(1  cos  )
l

w
6g
(1  cos )
l
M
6g
(1  cos )  V0 
l
m


m
m, V0
2
1  Ml 2 2 
W   Mg (l  l cos  )  lw  g (1  cos  )

2  3

6

m<<M
M, l
from (i) :
3mv 0

Ml
6g
(1  cos  )
l

2
gl sin 2  2 Ans.
3
V0 
Ml

3m
(b)
l
P  (m  M)Vcm  mV0  (m  M) w  mV0 because = Hinge force put w and V 0 then
2
P  M
1
gl Sin 2
6

2
Ans.
116
Compact ISC Physics (XII)
(c) For momentom of system constant, hinge force will be zero. And for
this hinge velocity just after collision will be zero. Angular momentum conservation W
x
Ml 2
w ------(i). Using linear momentum conservation M
about hinge. mV0 x 
3
: mV0  M
V0
m
2l
l
w -----(ii) Divide both equation : x 
Ans.
2
3
 MR 2

MR 2

2

mR
w

w 
Q:1.276: (a) Using angular momentum conservation :  2
 0
2


M

2m
w  1 
M

WF 

1
2

 w 0 (b) Work done by force F = Change in K.E.


 MR 2

 2

w0
m
R
R



2
 w 2  1  MR  MR 2  W 2

 0
2  2



F
2
2


mR 2 w 0 2
1  MR 2   2m 
2 1 MR
 1 
 mR 2  w 0 2  WF 
 w0  

2
2  2  
M 
2  2


1  2m

M





Ans.
 
Q:1.277: (a) Since initially L  0 and no torque on system about centre of disc then
d

L will be constant always.  I1W1  cons tan t  0   I1 dt  0    I1 d  0 or
I11 + I22 = 0 (from ground frame) where 1 and 2 is angular displacement. Then
0
R
'
m1
m1'
2 m1 '


m 2R 2
2
m2
m 2  2 m1 Ans.
  m1R (' ) 
 m1
2
2
(b)
Using angular momentum conservation : 0 
m 2R 2
w  m1R v' (t )  RW  
2
 m R2



2 m1V ' ( t )
m1RV ' ( t )  w  2
 m1R 2   W 
 2

( m 2  2 m1 ) R


I 
m 2R 2
2

2 m1

 R ( 2 m1  m 2 )

2 m1
dV ' ( t )
   ( m  2m ) R
  =
dt
2
1
 dV ' ( t )
 m m R
1 2




 2 m1  m 2
dt


 dV ' ( t )


dt

Ans.
117
Q :1.278: (a)
Angular momentum conservation about centre of disc:



I1w1  I 2 w 2



I1W1  I 2 W2  I1  I 2 w  w 
I1  I 2

I1

w1
I2
w2
(b)
Workdone by friction : Work 
Work 
1
1
1
I1W12  I 2 W2 2  (I1  I 2 ) w 2 
2
2
2
 
1
1
1
I1W12  I 2 W2 2  ( I1  I 2 ) w.w
2
2
2
1
1
1

I1W12  I 2 W2 2  ( I1  I 2 )
2
2
2
Work  
 
I12 W12  I 2 2 W2 2  2 I1I 2 w1.w 2
( I1  I 2 ) 2
  2
1
I1I 2
( w1 .w 2 )
2 ( I1  I 2 )
Ans.
Q:1.279: Linear momentom conservation : mV =
nm Vcm + mv'  nVcm + v' = v ----(i). Equation
nm, l
A e=1
l
w  v' -----(ii).
2
Angular momentum conservation about inertial
V
of restitution : V  Vcm 
A e=1
Vcm
w
A e=1
V
v'
m
m
Before collision
m
lw
ml 2
ml


point (A): 0  0  0  Lcm  m r  V cm  0  0 
w
Vcm  Vcm 
-----(iii) from (i), (ii)
6
12
2
v(4  n )
and (iii) v '  4  n ;
12 v
( 4  n )l for v' = 0  n = 4 and for reverse direction : v' = (–) ive 
w
n>4. 1800 rotation : Angularmomentum conservation about 00’ axis. 0 + I0w0 = IW - I0w0  w 
2I0 w 0
.
I
 1
1
1
1  2I w
2
2 1
2
Iw 2  I  0 0
Work done by motor : Work   2 Iw  2 I 0 w 0   2 I 0 w 0  Work 
2
2 
I


 Work 
(b)
2I 0 2 w 0 2
I
2
Ans.


dL
 
 I0 w 0
dt
I w
  I 0w 0 0 0
I  I0




y
dk̂
dt
 I0 w 0 w
I0 2w 0 2
 
I  I0
Ans.


L  I 0 w 0 k̂
x
z
118
Compact ISC Physics (XII)
0' 

Q:1.280: (a) 900 rotation: Moment of inertia of
motor + platform = I. Angular velocity of shaft =
w0. Moment of inertia of sphere = I0 . Angular
momentum about 00’ axis will be conserved.
Because torque about 00’ is zero. Then
B'
M
I w
0  I 0 w 0  I  I0 w  w  0 0 . Work =
I  I0


NA
B
P
I
0
final
initial
 1
1
1
2  1
2
2
2
change in energy.   2 I 0 w 0    2 Iw  2 I 0 w  2 I 0 w 0  

 



I 2W 2
1
1 I 0 2 W0 2
(I  I0 ) 0 0  
2
2 (I  I0 )
(I  I 0 ) 2
Q:1.281: F.B.D. of rod : Here Ny = mg
w0
N

1
(I  I0 ) w 2
2
1 I 0 2 W0 2
 Work done by motor =
2 I  I0
Nx 
m l0 2
l
w    mg 0 . F.B.D. of hinge axis : Here
2
2
l0 2
F 2 = Ny = mg  F1  N x  m w . Torque about point A :
2
m
l0 2 l
mg l 0
2g
2
w
w

 w 
l
2
2
2
A
l0
2

F1
w

m,l
B
l
mg l0
 
2
2
F2
Ny
2
F1
2g
Ans.
l
w
l
Ans.

Nx
mg
A
mg
Nx
Ny
0'

ml 2
w sin  Ans.
Q:1.282: (a) Resolved angular velocity : L 
12

Lf

(c)
 
m12 w 2 sin 2 
Ans.
24
m,l
y

x
C
w

in
ws


in
ws

os
wc



(b)

ml 2
m2
Li  
w sin 2  ˆj  1 w sin  cos  î 
12
12
2
m12
2 ˆ m1

w sin  j 
w sin  cos  î
12
12
 ml 2
m2
L 
w sin 2  î  L  1 w sin 2
12
12


dL
dL̂
m2
 
 L
 LW cos   1 w 2 sin  cos 
dt
dt
12
w

initial
final

119
  
Q:1.283: (a) Here w’ = Angular precession velocity. Then gyroscopic torque   w ' x L
  = w’ IW Sin ---(i). Torque of gravity about hinge : = mg l sin ----(ii). Since
w'

(b)
F.B.D. of rod : FH = centrifugal force. FH  ml sin w ' 2
m
mg
FH
F.B.D. of rod :
m (l sin ) w'


Q:1.284: Moment of gyroscopic forces :   w1 x L   = 2n Iw
m(g+w) l -----(i) w 
wl
nR 2
w
I
2
w' = 2
w
R
0
mR 2
 2n
w  nMR 2 w . Torque of psuedo force about origin:  =
2
w
l
mgl
we know that gyroscopic torque is provided by gravity then mglsin = w’Iwsin w ' 
Iw
Ans.
m(g+w)
from (i) and (ii) : m(g+w) l = nmR2w 
( g  w )l
nR 2
Ans.
w
Q:1.285: tan   g  Fnet  m 2 g 2  m 2 w 2  m g 2  w 2 . Now gyroscope motion : Torque
vertical axis
equilibrium

  
about fixed point C    w ' L ----(i) and   w ' Iw sin 1
w'
2
 w'
ml g  w
IW
2

l
-----(ii) from (i) and (ii) : w' Iw sin 1  m g 2  w 2 l sin 1

mw

mg
Ans.
w, I
Fnet
m g2  w 2
equilibrium axis
2

2

Q:1.286: w1  w1ˆj  w
 wk̂ . Moment of inertia about own axis : I  5 mR . Torque of gyro

2
2

2
2
scope :   w ' L    w ' mR w . Since bearing distance is l then Fl  mR ww '
5
5
 F
2mR 2 ww 1
5l
y
w1
Ans.
l
w m1R
x
z
120
Compact ISC Physics (XII)
Q:1.287: Here w’ is angular presession velocity and hence w ' 

w'
d
2
2t
 m
Cos
.
dt
T
T
Torque
of
2
 2  m cos 2t  mr

w . And this is equal to

T  2
 T
bearing torque. Fl 
2 m
mr 2 w cos 2t
T
 Fmax  cos 2 t  1 then Fmax 
2
2 t
    m sin w ' t    m sin
T
T
gyroscope
  
  w ' L
=
   w ' Iw
l
Radius of bearing = l
w'
m, r, w

I
w
mr 2
2
  m mr 2 w
Tl
Ans.
I, w
v
Q:1.288: Hence w = 2  and velocity of angular presession is w’
 v 
  
= v/R . Then Torque of gyroscope is :   w ' L     R  Iw


v
2nIv
I 2 n   
 
R
R
R
Ans.
w'
Q:1.289 : Here w = 2n and angular precession
velocity is w ' 
v
. Here dirn of w is in
R
negative y dirn while direction of w’ is in z
direction. We know Torque of gyro scope =
  
  w' x L 
v
v

(Iw ) 
I 2 n . Let gyroscopic force
R
R
is F then

Fl 
2 nIv
2 nIv
 F
R
Rl
y
l
v
w
I
x
R
z
w'
121
1.6
A
Elastic deformation of solid body
A
F
F
Q:1.290: If temperture is increased by T then increase in length
l
of rod : l = l × T ----(i) . Since no increament of length hence for F should decrease its length
F/ A
E
stress
 l  F  E l
E

in same amount. Using hook’s law:
where E = young’s modules 
A
strain
l
l
from (i) : F A  E [T ] 
F
Pressure =
A
 E T
Ans
T
Q:1.291: (a) Top view = is very-2 small. Suppose
r
/2
internal pressure is P then. F = P(area) = P (R) l
And this force is balanced by tensile force T. Then
l
2T Sin /2 = F  T = RPl  T = PRl ---- (i).
If breaking strength is  m then T = (lr)  m
from (i) :  m
r
 PRl
R
F

l
T
R
T
Here R = r
T
(b) Take a cone of apex angle . F  P r sin  / 22 
 2
Pr 
. And this force
4
2
 2
T  P r   2
 Sin   Pr 
Pr  
T





is balance by T :
 T
---4
2
4
 2

4

r

F
P

--(i). Now T   m 2rSin  2 r   m r r ----(ii) From (i) and (ii) :
T
2
 m r r 
Pr 
2 m r
P
4
r
Ans
Q:1.292: rod ruptures where pressure is maximum and this point is P. To find force at point P :
F.B.D. of half of rod : Contrifugal force =
Stress 
ml  2
 l  l 2
Al 2 w 2
  w    A    w . Then F 

2  4
 2  4
8
2 2 m / 
2 2 m / 
F l 2 w 2

 m  w 
 2 
  
l
l
A
8
2 m / 
l
Ans. where m
= Braking strength of rod   = Volume density of rod
w
P
A
A
l
F
 l  l 2
 A    w
 2  4
122
Compact ISC Physics (XII)
Q:1.293: F.B.D. of arch making  angle at centre. Here  is very-2 small. m =
mass of differential element =  volume = (A) (r) where A = area of cross
section of wire 2 TSin
T
 w 
Ar

2
 w
2
 mrw
2
 2 T

2
2
2
w

2
 ( Ar  ) w r  T    Ar  w
1 T
 
r 2  A 
T
2
w
1
r
2

1
   2 

r
where   Breaking strength of wire 2n 

1

2
2 r
r


r
2
2
2
T
Ans
Q:1.294: R  d 2 where R = radius of wire. Here 2T Cos 
P
A
A
l
= mg -----------(i) Again we know Stress/strain = young
moldules = E   T 
mg
Ed 2 2
 (cos ec   1) =
4
2
l
mg
Ed 2
(cos ec   1) ----(ii) from (i) : T 
---(iii) from
2
cos

4
(ii) and (iii) 
h
m
h
T
T
mg
mg
Ed 2  1


1 -----(iv). Here

2 cos 
4  sin  
l
cot  ----(v) from (iv) and (v) which calculation is difficult: h  l 3
2
Q:1.295: Area of cross section = s young modules = E a 


F0
m.
mg
2 d 2 E
Ans
S
l
m
F0
Since plank is accelerating and at each cross-section force will
be different. So stress will be different then. Now force
F
Stress
s
E
F0 x
m  F0 
d
(

x)
F

Strain

Now
F  mxa  x  
l
l m
dx
 d(x) 
Fx
F
dx  0 dx 
ES
lEs
F0
l
 d(x)  Els  xdx 
0
x 
F0 l
2 ES
F
x
dx
Ans
123
w
Q:1.296: Area of cross section = s. F.B.D. of dr part : - df (dm)
F
r
m 
  dF    dr  rw 2

0
l l
rw 2 
mw 2  r 2 
l 1  2 
l 

2
F
 r2 
1 w 2l 3
1   dr

r


2
 l 
3 E


S
E
l
l
D
Also
D
l
l
µ
r2
r

l
l
r
 D 

where
dr
m
sl
Ans
F = 1000N
 Vold = Sl 
Fl
l
F
Fl


 l new  l  l  l 
SE
l
SE
SE
l 
m 2
w
2l
F
F
Q:1.297: We know
F
S E
 Now Stress Strain  E 

d ( r )
dr
l
2
F
l
dr   mw
2
SE
SE
0
 d(r) 

m, l
l
 Fl 
 l l 

 SE 
l
Cross Section = S
D
D
µDl µDF
D µR



 Now
l
SE
D
SE
2
 D 2 
(D  D) 2
D 2 l   D    l 
l
Vold  
V

(
l

l
)

1



 

new
 4 
4
4 
D   l 



D 2l 
D   l 
1  2
 1   
4 
D  
l 
 2 D l 
Vold 1 
  
D
l 

F
Sl F
 l 2 D 
V  Vold  Vnew  Vold  

(1  2µ)  V  Fl (1  2µ)
  Vold (1  2µ)
D 
SE SE
 l
E
Ans
Q:1.298: (a) area of cross section = s. F.B.D. of part x : Now
T
T
Stress
Strain  E 
S
d (x )
dx
E
where d (x) is elongation
mxgdx
mgl 2  m  gl 2

E
d
(

x
)


x

 

in part dx. 
ls
2SEl  SL  2E
0
l
l
T
dx
x
x
weight of part x
mx
g
l
T
mx
g
l
124
Compact ISC Physics (XII)
where
m
1  gl 2
  then x 
Sl
2 E
Vnew 
(b)
This
part
is
same
as
Q:
Vold 
1.297.
D 2
;
4
V
 Vold
 2  2D   l  V
l
(D  D) 2
 new
D l 1 
 1  
(1  2µ)
(l  l ) 
=
. Then
Vold
4
D 
l  Vold
l

4
take value
D
of using poisson’s ratio .
D
P
l1
l2
y
Q:1.299: (a) Here we have to find decreasement of
volume due to x dirn, y dirn and z dirn pressure and add
P
l3
all three which will be answer. where S is surface area
0
z
S = l 1l 2
x
P
l
P
 P
 E  3   l3 new  l3 1  
l3
l3
E
 E  Also
 l
3
P

l1

l3
l1
µ
l3
P
l1
l2
l1 µP
 µP 
 µP 
new

l new  l1 1 
 l2 1  
  l2
l1
E . Similary : 1
E
E 



 µP   µP 
 1 

 Vnew  l1l2l3 1 
E 
E 

2
l3
P   2µP 

 l1l 2 l3 1   1 
 
E
E 

 
Vnew
V
V
 Vold
2µP P
P
P
 1
  new  1 
(2µ  1)  new

(2µ  1) 
Vold
E
E
Vold
E
Vold
E
Vold  Vnew
V P
V P


(1  2µ) 

(1  2µ)  This is only due to force from x.
Vold
V
E
V
E
Similarly :
V P
V P

(1  2µ)  This is only due to force from y 

(1  2µ)  This is only due
V
E
V
E
 V 
to force from z  Total decreasement of forces are applied by all direction :  V   sum of all

 net

(b)
3P
(1  2µ) Ans
E
Compresibility
:
 dV  1


 V  dP


1 dV
V dP

dP
dV
3P
3P
 Volume strain 
(1  2µ)  dP  P   
(1  2µ) . Then
dP
E
PE
=

Applied
pressure
3
(1  2µ) Ans
E
125
(c)
Compressibility can not zero if we applied forces from all direction. Then  > 0 
3
(1  2µ)  0
E
 µ < 1/2
Q:1.300: This is a cantilever and bending moment of cantiliver is given by : M  E I R (At point (x,y) where
radius is R). F.B.D. of cantiliver in dotted part b= width of cantiliver. Torque about point P :
  mg
(1  x ) 1
 bhg (l  x ) 2 and I = Glometrical moment of inertia w.r.t. neurtal line See F.B.D. of cross
2
2
h
section of rod :
I
2
y 2 bdy 

h
2
bh 3
Ebh 3
M

.
Then
. This moment must be equal to torque of mg for
12
12R
Eh2
Ebh3 1
2
R

 bhg (l  x )  at x  0 
rotational equal
6l 2g
12R 2
y
y
b
dy
h/2
x
0
y
h
M
l-x
h
P
h/2
b
(x, y)
mg   (l  x ) bhg
Q:1.301: (a) N( x)  EI
d2y
d 2h
d2 y
N
0
. F.B.D. of length of (l-x) : Now N(x) = No  EI 2  N0  2 
EI
dx 2
dx
dx

N x
dy
No
dy
dy

x  C1 at x  0 
 0 as shown in figure. Then C = 0   Now
 o 
1
dx
EI
dx
dx
EI
y
N x2
N l2
No x 2
At x  l  y  0
 C 2 . At x = 0  y = 0  C2 = 0 then y  0
from Q: 1.300:
2EI
2EI
2EI
3
a4
I  bh 12 Here b  h  a  I 
12
x
N(x)
l-x
(x,y)
No
No
y
126
Compact ISC Physics (XII)
F1
P
(b)
l, x
l, x
(x,y)
F.B.D. of length (l-x): Here F1 = F
F
N(x) = F
N(x)
F
(l-x)  EI
d 2y
 F(l  x) 
dx 2
at x  0 
d2 y
dx 2

2
F
F
 dy 
(l  x ) dx  dy  Flx  Fx  C1
(l  x)  d   

EI
EI
 dx 
dx
EI 2 EI
dy
dy Flx Fx 2
Flx 2 Fx3
 0  C1  0 



 C1 at
 y
dx
dx
EI 2EI
2EI 6EI
Flx 2 Fx3

. Put
2EI 6EI
C2 = 0  y 
x = l
y
x=0  y=0 
Flx3 Fl 3 Flx3
Flx3


 y
2EI 6EI 2EI
3EI
Ans
Q:1.302: By symmetry F1 = F2 = F/2. F.B.D. of half of length : Then F3 must be equal to F1  F3 = F1 = F/2. Now
Fl 3
Fl 3
Fl 3
our question is as : Similar as Q: 1.301 : y  0 0 Here l0  l 2  F0  F 2  y 
 y 
3  2EI  8
48EI
3EI
F1
F
x
y
F2
F1
x
F/2
F3
Q:1.303: (a) F.B.D. of (l-x): N(x ) 
dy
bhg (l  x)3

 C1x  C2 
dx
2EI 12
y
d 2 y bhg
d 2 y bhg
bhg (l  x )2
(l  x)2  2 
(l  x)2 
 EI 2 
2
2EI
2
dx
dx
x 0
 C1x  C 2 at x = 0  y = 0  C2 = 
dy
bhgl3
 0  C1 
. Again
dx
6EI
y 
bhg (l  x ) 4
2EI
12
bhgl4 bhgl4
bhgl4
bhgl 4


at x = l  y = C1 l + C2 
24EI
6EI
24EI
8EI
bh3
3 bhgl 4
y
12
2 8EI
( l x )
( l x )
2
l
l-x
(x,y)
x
N(x)
l
and I 
y
l
l
 (l  x ) bhg
2
127
(b)
Here F1 = mg/2  F.B.D. of half of portions  Similar as part (a)   
F1
F1
F1
5 gl 4
Ans
2 Eh 2
l/2

Because F + F1 = 0
F = mg/2
mg
d 2h
1
Q:1.304: Torque about origin : N( x )  lh  x 2dx  lh (l 3  x 3 ) ----(i). Now we know N( x )  EI 2
3
dx
x
3 2
3
3 2
lh d y 1
lh
Elh d h
then N( x ) 
 lh (l 3  x3 )
and I 
------(ii) from (i) and (ii): E
12
12 dx 2
12 dx 2 3
l
4
 3
4  l 3x 2 x 5 
 l x  x   C1 at x  0 dy  0  C  0
y


 C2 at x = 0
. Now
1

4 
20 
dx
Eh 2  2

4  l 3 x 2 x 5 
dx

x
put x = l  0
 y = 0  C2 = 0  y 
2 2

20
Eh 

  maes volume
dy
4
 dx 
Eh 2
y
9  l 5
Ans
5 Eh 2
dF = (dm) a  ( lh dx) (x)
Q:1.305: (a) Here 1 = shear strain. Take a differential length (ds) on circumference. If we rotate this element
by a angle then will be shear strain like above figure. Then
dF  1 (r ) ds .
Then
F
Stress
 1 
   Stress  1 
(

r
) ds
1
d  rdF   1 r rds  Net
torque

=
N
then

2 r
N  1r r  ds  21 r 2 r
also
we
know
0
l
r
r  l1    1  1 
r
l
 r 
2
 N  2r r  
 l 
1


l






N

 2 r 3r    = Torsion angle


(b)
2 x
Take differential ring at radius x. then  dN  2 x  dx
l
 =  = Torsion Angle
 = Shear coefficient

r
ds
dF
from part(a) 
x
r
 2l 
2
3
2  r 4

 N
 dN  2 x l  x dx  N 
 
4

r
4
l


0
dx
128
Compact ISC Physics (XII)
d2
from Q.No. 1.305:
Q :1.306:
N
2
 dN  l

2
3
 x dx
d1
G 
2 
d 24  d14  Here  = G then N 
d 24  d14
32l
4x8l
from
Q :1.307:
(Power ) max  NW 
Q.No.
N
1.305:
 r 4 
l
d 4  d 4 
1
 2

 16

2  
4l
2



N




G  d 2 4  d14
Ans
32l
N 


 N. w 
Power  torque . w
 r 4w
G r 4w
 Here   G  (Power ) max 
2l
2l
Q :1.308: dF on differential element of length dr : dF  (dm) a t  (dm) r 
m(2rdr) r
r
2
 d   rdF 

r
2m
3
 r dr   
2
 r12


m r 4  r2 4 

r 4  r12    
. Note in integration
4r2 2  r12 
2 r2 2  r12 
2m
r2 2  r12 r2
range take from r2 to r1 because  elastic torque will be equal to torque of Pseudo force in region (r2-r)
m

r

r2
Q:1.309: we know 
elastic energy
1
r1
dF
elastic energy
volume  2 E (strain)
elastic energy 
2
1
volume  2 stress x strain . Also we know stress = E strain 

1
m
E 2
Ans
2
p
1
A
l
T
r 2
 (l  x)g we know
Stress
Stress

E
1  2 (l  x ) 2 g 2
1 (stress) 2
 r 2dx
x r 2 dx  U 
2
E
2
E
1 2g 2r 2 l
 2 r 2 g 2
(l  x)2 dx 

 2
2E
E
0
m
  .
p
m
 E  Strain 
volume  2 stress x strain  Strain
dU  elastic energy 
1

  E 2 
2


1
1 2
2
E  elastic energy   E   Al
2


2
Q:1.310:(a) T  r 2 (l  x)g  Stress 
elastic energy
r
dr
l
 (l  x ) 3 
1 2 2 23


 3   U  6E  r g l Ans

0
r
l
x
T
dx
129
0
(b)
l
0
 (l  x ) dx
g (l  x ) 2
 (l  x)g
d(dx)
Stress

l

d
(
dx
)

E 
 Strain 

 

E
2
E
Strain
E
dx
0
l
l
2
 l 
2
r 2l  l 
2r 2
r 2l
 l 
gl 2
l
gl 2
El   Ans.

gl 2 
 2E   U 

from (a) : U 
6E 
l 
3
2E
l
2E
6E
 l 
Q:1.311: (a) Make F.B.D. of bending: If radius is R then 2r  l  R 
+ x  length of neutral surface = R2   strain 

l
increase in length of AA’ = x
2
2 x x

Volume  (R 2) (dx ) 
R 2 R
2
2
Eh 3
x
1
dU  E (strain) 2  (R 2) (dx )  U   E  R   R 2 dx  12R . Now : for complete circle : 2 =
 
2
 2
dx
1 2 3
2   = =  R = l/2 put then U   h E / l
6
l
l
m
x
A'
Neutral Surface
Here is very-very small
R
S
h
h

Q:1.312: dE 
1
Gr
(stress x strain) r 2l and Stress  G 
and this
2
l
stress will be constant in next d rotation then. dE 
rd  r
r
d  d
l
d
l
1 Gr
 r 2 d
2 l
4 2
Gr 2 
1
r 

Gr r 2  d  E 
  d  E  Gr  Ans
2
l
2
l 

4l
Q:1.313: from Q: 1.312: E 

r
d

Gr 4 2
Gr 42
dE 4Gr 42

dr 

 dE 
4l
dr
4l
l
r
2 2
dE 1  G r
dE
Energy Gr 42 dr


 dV  2 
2
(2rdr)l volume
l (2rdr)
 l


 Ans

R
dr
130
Compact ISC Physics (XII)
Q:1.314: We know   
pressure then. then
1 dV
. Here dP is pressure at depth x due to weight of water or hydrostatic
V dP
1
dV 1
dV
 (stress )  strain when B= Bulk modulus =
 [ xlg] . Also we know :

V
2
V
r
1 1
dE
1
1
 ( xg) 2 

  ( xg ) 2 . If x = h

B (strain) 2 
2 
dV
2
2
l
Cross Section = S
dx
dE
1

 (hg) 2 Ans
then
dV
2
1.7
x
Hydrodynamics
Q:1.315: Make F.B.D. of AA'BB'CC' :
F1
A
A'
F2
0
R
Centre .
Since system is in circular
motion, (F1-F2) will provide required centripital
0
force. And hence F1  F2  P1  P2 because F1
Centre
B
A
A'
B'
and F2 will only arised due to pressure difference.
'
A
Now, if we study motion of all particle in this differC
A
ential region then.
C'
With whatever mechanism, if particle at A go toward A’ its kinetic energy will be increased because of work
done F1 is greater than that of F2 Hence VA '  VA  V2  V1 . Since velocity is increased, density of
stream line will be increased.
(3)
Streamline density variation analysics : Here P3 = P4 = P5 = P6  V3 = V4 =
(1)
V5 = V6 . And hence density of stream line (Distance b/w two streamline) at
line joining (3-4) and (5-6), at each point will be same. Now P1 > P2  V2 > V1.
And density of streamline will be increased if we are going from (1) to (2).
(2)
(4)
(6)
(5)
Using continutity equation at (1) and (3) V1 > V3 and continuty equation at (2)
and (4): V2 > V4. Now if we take force concept alongle 2-4 then P2 > P4 and P1
> P3
Q:1.316: Using bernaullis equation at S1 and S2 :
h
1
1
P1  V12  P2  V2 2 -------(i) Using continuity
2
2
x1
V1
S1
P1
S2 V2
equation : S1V1 = S2V2  V1  S . Put in (i)
1
 2

1
1 S 2V 2
1
2 S
P1   2 22  P2  V2 2   V2  22  1  P2  P1 
2
2
2
S1
 S1

x1
V2 2 
2P2  P1 
S 2 
  2  1 
S 2

 1

V2
S2
P2
131
2gh
2gh

S 2 
S 2 
 2  1
  22  1
S

S 2

 1

 1

V2 
rate 
d
 S2 V2  S1S1
dt
. P1  P0  gx1  P2  P0  g ( x1  h )  P2  P1  gh . Volume flow
d
d
Q
 S2S1
Q
where

volume
/
s

dt
dt
S2 2  S12
2gh
2gh
S2 2  S12
Q:1.317: At point (1) point (2) pressure difference is only arised due
to dynamic pressure. Now bernaullis equation at (1) and (2) :
gas
S
x1 2
1
1
2
V1 2  P2  0  P2  P1  V1 ----(i). At line M-M pres2
2
sure
will
be
same.
Then
x2
P1 
P1  g(x1  x 2 )  0gh  P2  g(x1  x 2 )  P2  P1   0gh .
Now put in (i) :  0 gh 
Volume flow rate 
1
V12
2
d
 SV1  S
dt
 V1 
h
M
M
liquid
2 0 g  h
.

 
2g  0   h Ans
  
Q:1.318: P0 = atmospheric pressure  Bernaullis equation between (1) and
P
1
1
2
2
P


V


gh

P


V

0
(2):
. Here V 1 << V then
1
1 1
1 1
0
1

Kerosine
2
2
h
1
P1  1gh1  P0  1V 2 . Now static pressure equation between (1) and atmo- water (1)
h
2

1
2
(2)
sphere : P1  P0  h 2 2 g  Put in (i) : P0  h 22 g  1gh1  P0  1V 
2
2g (h 2 2  h11 )
V2 
. Caution : even if we use bernaullis equation in two different liquid, give same result
1
but solution is wrong. Because bernaullis equation use in same liquid (P=const). In above equation a wrong
1
2
method give correct answer because of V1 is very-2 small which provide 1V  0 if V1  0 then answer
2
will be wrong.
0
2
2
1
1
Q:1.319: Velocity of efflux : Ve  2g ( h  x ) . Time to reach at bottom :
x
1 2
gt  t 
2
2x
g . Now R = Ve t

2g ( h  x )
2x
g . Now
h
Ve
x
R
132
Compact ISC Physics (XII)
R 2  4(hx  x 2 ) for
R max  R 2  max : Then
d(R 2 )
 0 . Then
dt
h  2x  0  x  h 2

 h h2 
R  max  4 h 
  h 2  R
max  h h= 50 cm
2
4


(2)
Q:1.320: Bernaullis equation at point (1) and (2) along stream line
P1 
h
1
V 2  P0  O  g(h  h 0  x) -----(i) Here P1  P0  gx
2
h0
free water surface
V2
1 2
 h0
in (i) P0  gx  v  P0  g(h  ho)  gx  h 
2g
2
x
V
(1)
Q :1.321: Bernaullis equation between (1) and (2) : P1 
1
1
(0) 2  gh  P  V3 2
2
2

1
P  P0  gh  V32 ---(i) Now continutity equation between (3) and (2)  A 2V2 = A 3V 3 
2
2R1V2  2rV3

 R 2
P  P0  gh 1  12

r



 Ans

V3 
R1V2 R1 2gh

r
r
put
in (i) :
1 R 2 2gh
P  P0  gh   1 2

2
r
(3)
(1)
h

r
R2
(3)
r


R1

(2)
 = Thickness of orffice
R1
Q:1.322: Method 1: Here we know Wall forces = K.E.  WF + Watmosphere = K.E. Here work done by atmosphere = 0 Because Power due to atomosphere : Power 
dw
 P0SV1  P0 sVe from continuity equation :
dt
S
F
V1
(1)
s
(2)
Ve
P0S
A
V1
P0s
Ve
133
SV1 = sVe  Power of atmosphere = 0 then  WF = K.E. Bernaullis equation between (1) and (2)
along
Ve 
P1 
streamline.
2P1  P0 


1
1
(V1 ) 2  P0  Ve2
2
2
P1  P0  F / s 
and
2
(F / s)  Since F = constant Ve = Constant 

in differential term : dWF 
2P1  P0 
 Ve 2


2( F / s )
-----(i) . Now work done

1
1
1
2
2
2
(dm ) Ve 2  WF   (dm)Ve  MVe  WF  V Ve -------(ii)
2
2
2
Also, using volume conservation : V  (Ve t) (s)  Ve 
3
V
. Put in (i) : WF  1 V
st
2 s2t 2
Method 2 : Work done by F : WF = F (displacement) = F (V1t) 
Ve 2
 V
  
in method (2) F S 
2
2  st 
2
 WF 
Ans
F
FV
(SV1t ) 
. Also from (i)
S
S
 V2
1 V 2
V

2 s2t 2
2 s2t 2
Ans
Q:1.323: Suppose at time t, x length of water is inside tube then. Ve  2gx .
Now

using
continuity

Again
0
1
  2 x 2

h
s
S
equation:

dx
s

2g  dt 
dt
S
0
2g   2 h 
SV1  s Ve
 V1 
s
2gx
S


1
s
  (dx ) x 2  S
0
V1
h

x
2g  dt
s
S
0
s << S
Ve
s
s
2g    
2g Ans
S
S
Q:1.324: Method : 1. Bernaullis equation w.r.t. rotatory axis :
0
W
1
1
P  Vr 2  (rw ) 2  gz  const . equation b/w point (1) and (2):
2
2
1
1
1
1
P0  V12  (l  h ) 2 w 2  0  P0  V2 2  l 2 w 2  0
2
2
2
2
1
V12  0
2
1
1
1
V2 2  l 2 w 2  (l  h) 2 w 2
2
2
2
then
(1)
A
(2)
V1
V2
B
h
here

l
0'

2 2  2l
2l
 1 Ans
V2 2  w 2 l 2  (l  h ) 2  w 2 (l 2  l 2  h 2  2lh ) = w h  h 1  V2  wh
h




134
Compact ISC Physics (XII)
1
2
1
2
2
2
Method: 2 Bernaullis equation between (2) and (3): P2  (0)  0  P0  V3  0 
1
P2  P0  V32 -------(i)  Calculation P 2-P 0 : dF = (dm) xw2  (dP) S  (S dx ) xw 2 
2
P2

2

2
 dP  w  xdx  P2  P0  w l 2  (l  h ) 2
P0
2
lh
w2h2  2l  1
2l
2
 1  V3  V3  wh
1
2 n  2
h


w 2
(h 2  2lh )
2
put in (1) 
Ans
w
w
(1)
(2)
B
A
(3)
h
x
l
dx
Q:1.325: At differential element : 
dV
dP
 gCos 
  vdv    g (cos ) ds   dP
dt
ds
(2)
  g  (ds ) cos    dP 
ds
dh
V2
z2

P2
 vdv   g  dh   dP 
V1
z1
dh
P1
s
V1
V 2 V 2
 2  1  gz 2  gz1  P2  P1
2
2
Then
F   AV 2
F1  SV12  S2gh  2sgh
dv
 ƒ  P
dt

dv
dP
 ƒ
dt
ds
g
Z1

h

V1  2gh
F2  S(2g(h  h ))  2sg(h  h )  Fnet = F1F2  Fnet  2sgh Ans


(1)
1
1
2
2
 P1  V1  gz1  P2  V2  gz 2 Ans
2
2
Q :1.326: We know force
V2
Z2
ds
F2
V2  2g ( h  h )
F1
h
135
Q:1.327: Here V1  2gx Now force is : dV   AV 2  (b dx ) 2gx 
l
x
h
 dF  2gh  xdx  F  gbl (2h  l ) Ans
h l
dx
l
V1
b
Q
Q:1.328: Using continuity equation : Q  AV1  V1  . F.B.D. of tube : A = r2 = Area of crossA
section F1  AV 2
also F2  AV 2 . Then F1  F2  
Q2l
Q 2 Q 2
 2  Torque about O :  0 
A
r
r 2
F2
l
0
F1
0
V1
r
V1
Q:1.329:
F.B.D. of water inside tube : where F = force due to change in momentum Fnet on tube
water : Fnet  P0  gh  S  P0s  P0 (S  s)  F  Fnet  ghs  F -----(i) Calculation of F :
F
dP dm

 V1  V2  Q  V1  V2  ----(ii) where Q = volume flow in one second. Here
dt
dt
V2  2gh And V1 
Q  sV2  s 2gh 
Fnet 
 S  s  gh(S  s)
2
dP
Ss 
 s 2 2 gh 
  F  Then Fnet  ghS  s 2gh sS  
S


dt
 sS 
gh (S  s)2
S
h
dP
Q Q
Ss 
 Q 
   Q 2 
  And
dt
s
 s
 sS 
Q
Q
; V2 
 Put in (ii) :
S
s
2

P0 (S  s )
( P0   gh ) s
S
V1
s
V2
P0 s
F
V1
dm
dm
V2
136
Compact ISC Physics (XII)
Q:1.330: Method : 1 (a) Dotted line express isobaric surface. Then F.B.D. of one of the dotted line.
Bernaullis equation between point (1) and (2) from rotatory reference frame.
 w2  2
x2w 2
1
1


P0  0  0  P0  0  gh  (xw ) 2 . Then gh  x 2 w 2 . Here x = r then y 
 y   2g  r
2g
2
2


(b)
1
1
2
2
Bernallis between (1) and (3): P0  0  0  P0  0  ( xw )  P  P0  0  ( xw ) Ans
2
2
y
y
w
(2)
0
x
y
(1)
x
x
Method : 2 Pressure equation from rotatory
1
1
2
2
frame : P0  ( xw )  P1  P0  ( xw )  P2
2
2
w
P0
(2)
(1)
y
 2
x


Ans (b)
1
P3  P1  ( xw ) 2  
2
(1)
x1
1
2
(a) P2  P1  ( xw )  gy . Here P1 = P2 = P0 
2
 w2
y 
 2g

(2)
x
x2
(3)
1
P3  P0  ( xw ) 2
2
Q:1.331: Top view Viscous force at (dx) element at lower surface : dF  A
dV
xw
 (2x dx )
dz
h
Net viscous force at (dx) element from both lower and upper surface: dFnet  2dF 
4w 2
x dx . Torque
h
due to this force : d  xdFnet 
(Power ) net 
4wx 3
4wx 3
dx  Power due to this torque dP  wd 
dx 
h
h
4wx 2 R 3
wx 2 R 4
x dx  (Power )


h
net
h
0
w
h

dF
w
x
h
R
x
dx
dx
137
dx
Q:1.332: Cross-sectional view viscous force on this differential
dr
R2
V0
r
dv
element: dF  2rdx 
where 2rdr = curve cross-sectional area.
dr
Now
dF
dv
dF
 2r
 cons tan t  C then
in laminor flow
dx
dr
dx
2r
dV
dr
 C  2 dV  c   2V  C ln r  C1 ----(i). Using
dr
r
R1
dr
x
dx
boundary condition : r  R1  V  V0 then 2V0  C ln R1  C1
------(ii) and r  R 2  V  0 then C1  C ln R 2 from (ii) and (iii):
C
2V0
ln R1 R
and C1   C ln R 2 put in (i)
V
2
V0
R
ln 1
R2
ln
r
R2
 dw 
 2rdx . Torque due to this force :
Q:1.333: (a) dF  r
 dr 
dr
r
dx
d
 dw 
 dw 
d  rdF  r 2 
 2 r 3 
 2rdx 
 . For laminor
dx
 dr 
 dr 
R1
w2
R2
dr
r
d
dw
dr
 C now C  2r 3
 c
 2 dw
flow this torque is constant then
dx
dr
r3
 2w 
C
r2
 C1 ----(i). To find C, C use boundary condition: (i) r  R  w  w
1
2
1
C
C
 C1 ---(iii) from (ii) and (iii) find
(ii) r  R 2  w  0 then 2 w 2  
 C1 -----(ii) 2  0  
2
R 22
R1
 R 2R 2 
 1 2 
C1 and C2 and put in (i): w  w 2  2 2 
R R 
 1 1 


1 
 1

 2
 Ans
r 2 
 R1

r 2 


r 2 
Q:1.334: (a) v  v 0 1  2   dQ  (2rdr) v  (2rdr) V0 1 

 R2 
 R 


R 
R2 R4 
r3 
V R 2
dQ

2

V
r

dr
Q

2

V



  Q 0

0 

Ans
0 
2
2
R 
0 
2
 2 4R 
138
Compact ISC Physics (XII)
l
, 
d
 dw 
 2 r 3 

dx
 dr 
(b)

dw
dr

 w R 2R 2 
 2 1 2 

r R1
 R 2 R 2 
1 
 2
v
r
w  w2
R
R12R 22
 1
1 


R 2 2  R12  R12 r 2 

dr
r

dw w2 R12R22

dr
R 22  R12

 2w R 2
d
3 
2 2
. Then dx  2R1 
2
2
R
R
R 12 R 2 2  R 12
 1 2  R1
2w 2 R 2 2
 2 

 
R 3
 1 



 2
 3 
r 



 

2
2
d 4w 2 R 1 R 2

dx
R 2 2  R 12
(c) from option (a) : Q 
R 4 P1  P2 
1
V0 R 2 ---- (i) And from poiseuille’s law : Q 
. Comparing
2
8l
4lV0
R 4 P1  P2 
1
V0 R 2 
 P1  P2 
Ans
2
8l
R2
(i) and (ii) :
Q:1.335: If we see carefully, we get that heights are
different in pipe (1) and (2). While velocity of flow at
bottom of both pipes are same because cross-section
area of horizontal pipe is same at each point, then using
continuity equation, we can say that velocity of flow
is same at each cross-section of horizontal pipe. This
is because of friction loss and bent loss in pipes still
we can say that dotted line show isobaric surface. Height
h
3l
l
3h
pipe (1)
h3
B
above isobaric surface provide, velocity at efflux. Then
h
pipe (2)
c
Ve  2g (AC ) 
h2
l
h1
Vc
l
2g (h 3  h ) . Calculation
3
2
of h : Using similar tringle properties. h  2l  h  2  2 x 20  30 cm  h3-h = 35 cm - 30 cm
2
2
1
2
= 5 cm  V  2  g  5  10 2  1 m / s  dk  (dm ) v
e
2
K 
l V02
2

r2 
1
2
2
 r1  2  dr  K  l R V0
6
R 
0 
R
Ans
 
1
r 2 

(2rdr ) l  V0 1  2  
2
  R 
139
Q:1.336: We know Reynold’s no. for circular cross-section is
defined as Re 
vl
where l = length of charactersitic and for

circular tube with full of water. D = 2R  l = D  Re 

vD

R
Re1 V1D1 
vR

 1 1 ------(i) Also we know :
Re 2 V1D 2  v 2 R 2
Q  V1R 12  V2 R 2 2
R 2 


 2 
Re 2  R 2 
 1 
Re1
put
in
(i)
:
R 22
V1

V2
R 12
R1
R2
v2
v1

R  R
r e x
Re1
 1  2  0
 e x Ans
( 0)  Re
R 
R
r0e
2
1
 2
x0
g
Q:1.337: Maximum on value of Reynold’s no. for glycerin for laminor flow : R 
Rg 
VmaxD


1v12r1
w 2v2 2r1
. Reynold no. for water : R 
1
2 We know : Reynold’s no. for turbulent flow
> that of laminor 
vr
 r  
2 v2 2r1 1v1 2r1
v2  1 1 2 2  v2  v1 1 2 2  Ans


2
1
2r21
 2r21 
Q:1.338: F.B.D. of sphere :  = density of glycerin at maximum velocity, terminal velocity will be
4
3
4
3


3
3
attained then. Fnet  0  6rV  R g   R   g  Vmax 
R
2
l   gr2 Rynold no. :
9
VmaxD
1 x 2r 2
 l  P gr2  92  8r    gr 2 ----(i). Put density of
 0.5  
l
2

9

l  1.26 103 10 6 m 3  1.26  103 kg / m 3  lead    11.3 103 kg / m 3 put in (i) : get r 
D = 5 mm
4
R 3 g
3
6rVmax
D

r

mg
5
mm
2
140
Compact ISC Physics (XII)
4 3
R 0g  0 Because R  3 103 mm which is
3
2
Q:1.339: Where 0 = density of olive oil. Since
very-2
small

Fnet  mg  6rv  we
Then
Vmax 1 100
mdV  (mg  6rv) dt 


0
Vmax
m ln
(mg  6rV) 100  t1
 6 r
0

1.8

4 3
R 0 g
3
6rV
mg
mg
mg  6rVmax  0 Vmax 
6  r put in (i) 
 d2
ln n Ans
18
Relativistic Mechanics
Q :1.340: We kn ow

t1
mdV
  dt
mg  6rV 0
dv
 mg  6rv 
dt
mg  6rVmax 100
m
ln
 t1 .
 6r
mg
Calculation of Vmax : At time of Fnet  0
t
Fnet  m
kn ow
1
v 2  l 
 1  
c 2  l0 
2
1 
1

 v2  2
  v2  2 

l


 1  1  
l  l0 1  v 2  l  l0  l  l0 1  1  2   
 c2 
l0
c
  c  




v2
2l
2 l c 2
1 2  1
Here l  l0
V2 

,
----(i )

l0
c
l0
2
l
l 0.5
 0.5  2
100  0.5 

V2  2 
 c  V = 0.1C
,
put
in
(i)
:
l0
l 100
 100 
Q:1.341: (a) Here l1  a sin 60 
 lapp 
2
a 3
length lƒ appeared from reference frame (x-y): lapp  l1 1 v 2
c
2
2
a 2 3a 2

1 v2 c2
4
4

a
2
1  v 2 c 2   Now l0     l app 
2
a 3
2
V
l0
A


a
v2
43
2
c2
A
y
a
l0
a
l1
l0
lapp
v
0
c
60
a
B
x
B
c
a/2
141
Perimeter
(P)
=
l0  l0  a  a  a 4  3 v
2
c

2 
P  a 1  4  3 v 2 
c 


2
Ans
for v  c P  a (1  2)  3a
Length AC : lAC  a 1  v2 / c2 . There will no change of length of BD while there is change
(b)
in length of AD: lAD 
a
2
1  v 2 . Now appeared length AB : l AD 
c
2
l 2 AD  a 2
3
a
2

4v 2
c
4
2


2
2
v2
v2
Now perimeler (P) : 2lAB  AC  a 4  v 2  a 4  v 2  P  a  4  c2  1  c2  Ans
c
c


v << c
P = 3a Ans
B
B
B
y
lAB
a
a
a
v
a sm 60 
a
0
A
x
0
60
60
a
A
C
60
0
a cos 60  a 2
a 
a 3
2
A
D
D
lAD
0
Q:1.342: Here only length in x dir n is changed. Actual length of AC = l0 then l Cos 45  l0 1  v
l0 
l
2
1
v2
2
2  l 
 =
. Then original length AB is : lAB  l0  
c2
 2
l2
21  v

2
2
2
 2 v2 


1 v 2 sin2 450
1 v 2
2
l2 
1 
1 
c 
c
2c
l l
 l
1 

2
 lAB  l
2
2
2  1  v2   AB
2
v
v
1 v 2
1 2
1 2
c2 

c
c
c
y
B
l
B
l
0
l Sin 45
2
A
l Cos 450
C
A
x
v
l0
3
2

c2 

2
c2
l2
2 =
142
Compact ISC Physics (XII)
r0
Q:1.343: (a) BD = radius of cone = r 0  length of AD = height of cone = l0  tan   l . Then
0
lateral surface area S0  r0l 0 sec   from this reference frame length along x axis doest not appear
c
to change. Hence from this reference frame : BD = r 0 And length l0 will be appear as l : l  l0 1 v 2
c
r0
 tan'  
l
r0
2
put value of  and v in above function : 1 = 590
2
l0 1 v c
A
A'
y

  450
l0
B
(b)
r0
'
v

S0  4 m 2
tan 1 
l
4
v C
5
D
x
0
B'
C
r0
tan 
1  v2 c2
C'
D'

A1B1 = l sec 1  Lateral surface area :   r0l sec1  r0l0 1  v2 c2 1  tan2 1
 r0l0 1  v
2
c
2

tan 2  
1



2 2
 1  v c 
1
2

2
2 2 
S0
2 2  Sec   v c 

1 v c
1

Sec
2 2 2 
 1 v c


 Sec 2  v 2 c 2 
S0
1  v2 c2 

   
2 2
Sec 

 1  v c
1
2
1

1
2
2
1
 v 2 cos 2   2
 . Put value of v and  and
 S  S0 1 

c2


S0  S = 3.3 m2
Q:1.344: Time measured by moving dock : t; Actual time of moving clock : t- t 
we
kn ow
v2 c2 

t 2  ( t  t ) 2
t
2
t  t
t
1 v2 c2

( 2 t  t ) t
t
2
2

t  t  t 1  v
c
2

 t  t 
2 2

  1 v c
t


t  t

 v  c 2  
 v = 0.6 × 108 m/s
t  t

l0
Q:1.345:
2
v
v1 = 0
t 
l
l
------(i) v 
Now . Proper length = l0 =
v
t
length of rod appeared by that frame from which rod will be appear to stationary. Then

143
t
t1 
l2
1 2 2
c t
l
1
l
0
 now t  v 
Then
v 1  v2 c 2
1  v2 c 2
l
l  l0 1  v 2 c2  l0 
l2
 t 
 1 2 2   
 t ' 
c t
2
y
  t 2 
2
2
2 2 

 t 
l

c

t
1



l

c

t
1




  t '   
 t ' 


y
l0
v
v1 = 0
v
x
x
Q:1.346: Assume velocity of particle w.r.t laboratory frame is v then
 1  v
2
t 
c  0 
 t 


2
2
2
v2
Q:1.347: (a) We know

(b)
1  v 2 c2
2
t 
2
Ans
t 0
1 v2 c2
  t 0  t
2 2  t 0
 1  v c  
 t
1  v 2 c 2  t 0 




2
y
v = 0.99c

l
1  v 2 c2
V
l
k
t 
x
l
l

0.99c v
Actual distance travel is : distance = Actual velocity x proper time = vt 0  l 1  v2 c 2
Q :1.348:

t 0
  t0 
t 
  v  ct 1   0  . Distance travel in laboratory
 2  1  

 t 
c
 t 


 t 
frame is : Dis tan ce  Ct 1   0 
 t 


t 0 2  t 2 1  v 2 c 2
t 
l 
3 / 4C  t
1 v2 c2

3
C t
4
-----(i)
vt
1 v2 c2

l0 
We
know
l  l0 1 v2 c2 
vt
1  v 2 c2
3/4 C
Ans
l0 
Ans
l
1 v 2 c 2
T
A
R
G
E
T
3/4 C
l

1

Q:1.349: We know proper length of rod does not change w.r.t. reference frame then x  l 1 v2 c2 2
1 0
144
Compact ISC Physics (XII)

2
-------(i) x  l / 1 v
2 0
 x
v  c 1  1
 x
 2




 x
 x1
1
 1  v 2 c 2  v 2 c 2  1   1
2 2 -------(ii) dividing both :
 x
c
x 2
 2

put in (i) :
l0  x1x 2
Ans
y
0
1
2
3


 

x2
y'
v
4
v
 x1
x
v
x'
 Vr 2 
l0  l0 1  vr 2 c2
2
2


V

t

l


l
0
0 1  2  
Q:1.350: Time to move from A to 0: t 
 r
c 
Vr

l0
2l
Vr  0
t
l0
y
l 2
1 0
c2t 2
Vr
Ans
0
x
A
l0
l
Ct A  Vt A  l 2  t A 
2(c  v)
Q :1.351:
v 2
1 r
c2
l0

l
Ct B  Vt B  l 2  t B 
 Here t A < t B means later count occur first. t  t B  t A
2(c  v)

l
l

2(c  v) 2(c  v)
t 

l vc
l c  v  c  v
lV

t



. Assume c / v  
2  c2  v2 
C(1  v2 c2 )
c 2  v2
v
l
C(1  2 )
Ans
B
A
l/2
Q:1.352: (a) Since rod is moving with speed V with respect
to this frame it length will be appear less than proper length
Then
xA
-
l
=
v
rest
0
Vt A
xB  VtB
xA  xB   l  V(tA  t B)
l  (xA  xB)  V(t A  t B) now proper length :
l0 
l/2
y
V

l
1 v
2
A
B
(0,0,0)
c
2

(l, 0, 0)
x
l0 
x A  x B   Vt A  t B 
1  v 2 c2
Ans
145
(b)
Here
tA  tB 
xA  xB  l0  xA  x B  l0  xA  x B   l0 Then
l0 
l0  Vt A  t B 
1  v3 c 2

 l  Vt A  t B 
l0 
l0 
2 2
l  0
2 2
1  1  v c  or 0
 t A  t B  1  1  v c  Ans
3 2
1 v c
V
V


Q:1.353: (a) Reading of clock at B: t ' ( B) 
l
l
 0 1  v2 c2
V V
l0
A'
l0
B'
V
A
l0
B
l0
1 v2 c2 . When B’ will be
(b) Appeanent length of AA’ from A: lAA '  l0 1 v 2 c 2  t(A) 
V
at B then distance traul by point A’ in frame of rod A’B’ will do Then
Q:1.354: Lorentz transformation of time
If take t =0 for k frame then
t' 
t' 
t(A') 
l0
Ans
V
t  xV c 2
1 v2 c2
.
 xV c 2
1 v2 c2
. If x  0  t '  0  If x  0  t '  0
Q:1.355: Since both show zero reading at origin. If clock (k) reads time t and clock (k’) reads time
t’ then. according to lorentz transformation.
t' 
t  xV c 2
1  v2 c2
 Vx 
t' 
t  Vx
t
differentiate w.r.f. time
t  xV c 2
1 v2 c2
according to queation : t’ = t. Then
V
c2 
c2  1
2 2
1  1  v c  we know   v c
 Vx 
2 2
V

1 v c
K'
c 
2
1  1    Ans


Q:1.356: Suppose a shot is made and it hit the target after time
t then from k’ frame (mouing with V velocity)

t ' 
V
K
t ' 
K frame (stationary)
target
t  x v c 2
1 v2 c2
x
x=0
t=0
x=d
t = t
t  d V c 2
1  v2 c2
. If means target is hitted before shot is fired then. t '  0  t '  dV c 2  0
146
Compact ISC Physics (XII)
 d 
2
 t '  dV c 2  dV  c 2 t    V  c

t
 
if
d
 Vx . Then V V  c 2 . It is possible only of
x
t
one of the Vx > C or V > c which is not possible. It prove that target will be after shot made.
12
'
'
Q:1.357: (a) We know in variant formula : c 2 t12
 x12
at
same
point
then
'
x12
0
then
2
2
2
2 in frame k’ both events occure
 c 2 t12
 t12

ct 2  ct1   x 2  x1 
(b)
2

16
2
2

 

2
c 2 t '12  c 2 t12
 x 2  x1 2  c 2 t 2  t 1 2  x 2  x1 2
= 6  1  5  2   16 
'
t12

t1 
8 2  12
310 
4
3 10
8
=
 1.3  10 8 s  13ns
2
1
'
2
2
in frame k’ if both occure simultaneously then t112  0   t12
 c 2 t12
 x12
 16  t12  4m
Q:1.358: We know
frame : Vnet 
Vx' 
Vx  V
V V  Vy ' 
1 x
c2
1  Vx
V
 Then net velocity appear from k’
c2
Vx  V 2  Vy 2 1  V 2 c 2 
2
V ' x  V ' y 2  Vnet 
y
Vy 1  v 2 c 2
1  Vx V c 2
Vy
Ans
y'
Vx
K
K'
x
x'
V
v2 = 0.75C
Q:1.359:
v 1 = 0.5C
Velocity of approach is taken from laboratory frame hence
v2
Velocity of approach = V1 + V2 = 0.5C + 0.75C = 1.25C Ans
K'
Relative velocity : (Vr) :
Q:1.360:
Vrel 
V1
V  (V )
1  (V )

( V)
c2
V2
2V
1 V
2
c
2


V' x 


V1   V2
V1  V2
V1  V2  V' x 
1
1  V1 V2 c 2
2
c
2V
1   2 now
l0
l0
V
V
147
apparent length is : l  l0 1
1   
1   
V 2 rel
c2
  l  l0 1 
2V 2
 l  l0
1    c
2 2 2
4 2
1

1   
2 2
2
l  l0
2
Ans
VAx  V1
Q:1.361: V  0
Ay
VBx  0
y
V2
A
VBy  V2 From reference frame A : velocity
B
V1
x
0
components are as shown :
'
VBx

0  V1
 V1
1  o V1 c 2
'
 VBy 
Vnet  V' 2Bx  V' 2By  V  V 2  V 2 1  V 2 c 2
net
1
2
1


Q:1.362: To find velocity of particle in k frame assume
components of velocity in k frame is Vx and Vy
then
using
lorenz
transformation
:
y
V2 1  V12 c 2
1  0 V1 c 2
V V 
V12  V22   1 2 
 c 


 Vnet 
V'
Vx ' 
1  Vx V c 2
from k’ :
0
; Vy ' 
1  Vx V c 2
Vy '  V'
K
x
V
1  Vx V c 2
Th en
 Vx  V
 V' 
Vnet  V 2  V' 2 (1  V 2 c 2 ) 
Ans
y
Vx '  0
Vy 1  V 2 c 2
Vx  V
2
V'
K'
x
Vx  V
 V2 1  V12 c 2 Then
V 2  V' 2 
Vy 1  V 2 c 2
1 V V c
2
 Vy  V' 1  V 2 c 2 . Now
V 2 V' 2
. Hence we say that Vnet is velocity from k frame
c2
then. For proper time t0, we have to choose another frame k” which is attached with particle. Then
Reference frame k will be appear to move with Vnet speed then time internal will be appear to increase
t 0

t

frame this frame (k) then.
. Distance travel in k frame :
1  V 2 net / c 2
y
K
K
x
Vnet 
2
V 2 - V1 
12
2
V V
C2
x
Vnet
148
Compact ISC Physics (XII)
Dist  Vnet t  Vnet
t 0
2
 2

12
2 V1

2
 v  V  V
c



1  V 2 net / c 2
t 0
2
1
2
V 2  v1  V 2 V1 / c 2
c2
2
t 0 V 2  V1 (1  v 2 / c 2 )
Distance =
(1  v 2 / c 2 ) (1  v 2 / c 2 )
Vy 1  v 2 / c 2
V cos  V
V sin 


1  v2 / c2
Q:1.363: V' x 
----(i) Vy ' 
Vx V
( v cos ) V
Vx V
V
cos

V
1
1
1
c2
c2
c2
Vx  V
tan ' 
Vy'
V sin 

1  v2 / c2
Vx ' V cos  V
y
Vy'
y'
Ans

K'
v
V'
V

'
x'
x
Q:1.364: Here in k frame length of rod will be contracted because
Vx'
y'
l
it is moving with V velocity then l  l0 1 v2 / c2  Velocity of
(l,0,0,0)
(0,0,0,0)
K'
rod : V x = V  Vy 
Vy1 1  (V / c) 2
1  Vx1 V / c 2

A
v ' 1  ( v / c) 2
1  ov / c 2

V
y
t
t ' x v / c
1 v / c
t 
2 
V=Vx
l0v / c2
2
1 v / c
x'
O'
Vy  v' 1  v 2 / c 2 . Time difference of measuring for A and B:
1
B
v'
2
 y  Vy  Vyt  l0V v' / c2
A
K
B
Vy
l
x
2
tan 
y
l V v' / c
V v'
 0
tan  
Ans
l l 1 v2 / c2 
2
c 1 v2 / c2
0
B
y
A

D
l
149
Vx ' 
Q:1.365: (a) We know
Vx  V
y'
y
 now differential
1  Vx V / c 2
k'
1  Vx V / c 2  dVx   Vx  V   (dVx ) V / c 2 



dVx '  
equation :
1  Vx V / c 2 


V
K
x'
w
v
x
 dVx 1  Vx V / c 2  Vx V / c 2  V 2 / c 2  1  Vx V / c 2 

 



dVx ' ( dVx )

(1  V 2 / c 2 ) (1  Vx V / c 2 ) 2 ------(i)
dt '
dt '
dt ' 
dt  (dx) V / c 2

3 / 2
1  Vx V 2 / c2 3

w 1 V2 / c2
 w ' 
we
t' 
know
t  xV / c 2
1 V2 / c2


 dVx 
2 2 3/ 2

 1 V / c
dVx '
(dVx) 1  V / c (1  V / c )
dt 

 
2
2 2
2
dt '
(dt V / c ) (1  Vx V / c )
 VxV   VxV 
1 
 1 

c2  
c2 

2
1 V2 / c2
Also
Vx  v
2
2
2
Ans
y'
Q:1.366: Co-moving frame(k) is that frame in which particle is appear to
instanteously rest. But its accn of particle in this frame may or may not
zero. now we know : w ' 
w' 

w 1 V2 c2
1  V 2 c 
2 3/ 2

dv  w ' 1  V 2 c 2
x
c
w'
Vx' = 0

2 3/ 2
from Q:1.365 Here Vx = V then
1  V 2 c 
2 3/ 2

 w 1 V2 c2
y
3 / 2 


V
earth K
x
3 / 2 dt 
w ' t dt

3 / 2
2 =0
w  w ' 1  V 2 c Vx'
x


0 (1  V
2
2 3/ 2
c )

 w ' dt  
0

w' t
V
t
dv


2
c2 

 w' t 
x
1 
  1


1
/
2
w' 
 c 

0 1  ( w ' t / c) 2


t
 dx  
0
3 / 2

w 1 V
2
 w' t 
1 

 c 
cv 1 c 
 


c
2  w' t 
2
2
An s

150
Compact ISC Physics (XII)
Q: 1.365 (b) we know Vx' 
Vy 1  V 2 c 2
 Vy' 
dt ' 
1  Vx V c
dt  (dx) V c 2
1 V 2 c2
2
Vx  V
1  Vx V c
2  Vx' 
 Vy 1  V 2 c 2

0V
 V
1 0
dVy ' dVy

1 V 2 c2
dt '
dt '
from
(dVy / dt )
option


k'
V
x'
K


(w) 2


0

  w ' t  2
1   c  c

 
dt
(w) 2

 
1
0
c2  (w ' t)2
dt

 w' t 
1 

 c 
2
0
put in (i)
c2

dt  

2
c
w'c
 w'   

ln 
 1 

  0  
  0 
2


w
'
C
c


 w' t 
0


1 

 c 
m0
m
m
Q:1.368: We know
Here   v c  
2
v
1
2

x
w' t
V
0
1
w
(1  V 2 c 2 )
w
1 V2 c2
2
 dx 
 w '  w 1  V 2 c 2 . Here Vx = 0
1    V c 2 1  Vx V c
 dt 
 0   1  (v / c) 2 dt -----(i) from Q: 1.366
0  
(a) :
y
dVy'
dVy
1 V 2 c2
 dt ' 
2
dt  (dx) V c

Q:1.367:
1  Vx V c 2
y'
v

Vy 1  V 2 c 2
a x'  0  Vy' 
c 2  w ' t 2
dt

2
c
w' 
 w'   
ln 
 1 

 c
w'
 c  


m0
m0

1  2
(1  ) (1  )
Ans

c
0.01
C
C
v
100



c
C
1
m0
m
1 +  = 2  m 
2 (1  )
0
.01
 m 
100
Q:1.369: Density is defined as  
1
2(1  )
 70 Ans
m0
m0
m0
rest maes


-----(i)    l ' l ' l '
l1l2l3
volume
volume
1 2 3
2
Here l 2  l 2 '  l3  l3 '  l1 '  l1 1  v 2
c
 
m0
l1l2l3 1  v
---- (ii) from (i)/(ii):
2
c2
151
y'
y'
l2
l3
l2
x'
V=0
0
2
 1 v 2   Here
c

V
C
(1  )
V
0
1

 1 
  (1  ) 0 
 Then
1
2
 1 v 2
c
1 
 (2  ) Ans
Q:1.370: Assume mass of proton is m0 then P = mv where v = velocity of proton P 
v
Squaring
both
side
P
m 20 
:
P2
PC

m 02 C 2  P 2
C2
 m 2C 2 
CV
V
1
CV




 1  1
 1  1  0 
2
C
C
C


P
m2C2


1 0
2
P
Q:1.371: We know Newtonium momentum : P
2P 
m0
1 v

V ------(ii) (i)
(ii ) :
2
1

2
1 v
2
=
1
m 20 C 2
P
1 2
1 v
C

m0v

2
c2
Now
2
Ans
m 0V -----(i) Reletivistic momentum:
v2
3
1
v2

C  v 3C

1



2
2
2
2
c
4
c
2
c
c2
m0
1
2 2 1
2 2
(.64  .36) c 2
Q:1.372: Classical mechanics : w  Kf  Ki = m 0 (0.8) c  m 0 (0.6) c =
2
2
2
m0
x 0.28 c 2  
w  0.14 m 0 c 2 . Relativistic mechanics :
w  m f c f 2  m i c i 2
2
m0
m0

0.8 2 c 2 
0.6 2 c 2 =
m0 × 0.42 c2  w = 0.42c2
2
1.082
1  0.6
Q:1.373: m0c2 = Rest mass energy  mc2 = Total energy  Kinetic energy = mc2 - m0c2 according
m0
 2m 0
2
1
2
3
to question : m0c2 = mc2 - m0c2  m = 2m0  1  v 2
 1  v 2 
 v 2 
c
4
c
4
2
c
3
c
 v 
Ans
2
152
Compact ISC Physics (XII)
Q:1.374: Using classical mechanics : T 
2T
T
2
2
m 0 . Relativistic mechanics : T = mc - m 0 c 
Vc 
T
2
 1  v 2 
2 
c 
m 0c
1
2
1  1 
v2
2c 2
c2
 2
 3 T
1 
 4 m c2
0

T
m 0c 2
T
V  Vc
Vc
 




1

(1 / 2) (1 / 2  1) v 4
1  
2
c4
v2
v 2  3 v2 
1 
2  approxim ately
2c 2  4 c 
v2
1
m v 2 where Vc = velocity calculated by chassical mechanics
2 0 c
2c 2

m 0c 2
T
T
m0c2
 3 T
1 

m 0 c 2  2 m 0 c 2
2T


T 
2T
1  3
m0 
4
 m

m 0c 2 
0


2T
m0
T
then
m 0c 2

m 0c 2
1 v
 1
 m 0c 2
2
v2
2c 2

3 v2
1
8 c4


v2  3 T 
1



2c 2  2 m 0 c 2 


T
4
 
2

3
  m c
0

at max Ans
2

=



 3 T 
T

V

2
1

m
  
0 
2 m c 2   Now

0 

Q:1.375: We know E 2  m 20 c 4  P 2c 2 -----(i). Let us kinetic energy is T then. E  m 0 c 2  T


c2

put
2
m c
in (i) m c 2  T  m c 2  P 2 c 2  0 4  T 2  2m 0c 2 T  m 02c 4  P 2c 2  P 2c 2  T T  2m 0c 2
0
0
c
 P 
1
c

T T  2m 0 c 2

Ans
Q:1.376: From question no. 1.375 : P 
n then ne = I   

1
c

T T  2m 0 c 2
 of no. of particle collide per second is
I
dP
I
 p 
momentum transfer per second is : F 
e
dt
Ce


T T  2m 0 c 2 . Power
= Energy radiate or absorb persecond  Power = n (kinetic energy of one particle) =  T  Power 
I
T
e
153
Q:1.377: Sphere Reference frame Velocity of gas particle in frame of sphere
Vx ' 
Vx = 0
in
 Vx’ =
m ( v)
dP 

1 v2 c2
-v
(mv)
v.
 V =
Momentum transfer
Vx  V
1  Vx v 2
c
Here
one collision
:
2mv

1 v2 c2
1  v2 c2
. Since volume will be decreased by a factor of 1 v 2 c 2
But actually no. of particles of gas does not change but its occupied volume is decreased and hence
dA
Before collision
v
dA
After collision
V=v
v
v
apparent no. of particles per unit volume is increased
per second :





2
N app  
 (dA ) 
2
2
 1 v c 


dF
2mv2

dA 1  v 2 c2
Pr essure 

Gas Refrence frame : We know Vx ' 
v 
collision)

V
2


1
v 2 v' 2
c4
v  V'
V'
1 v
c2
app 
dF 

1 v 2 c 2
. no. of particles collides

dP 
2 mv

dt  1  v 2 c 2












 (dA ) V .
2
2
 1 v c 


2mv2
1  v2 c2
Vx  V'
Vx '   v (velocity of sphere w.r.t. gas frame after
1  Vx V' c 2
Vx  v (velocity of sphere after collision which does not change)
2v 3
3
 v4

V' 
2 vv ' 
2
v
2
2

 V  v' 2 vv '  V' 
 1 
 2 v 
v2
2
 c4

1
c 2 
c


c2
v
V=v
V=v
m
Before collision
mV'=
After collision
m
V'
154
Compact ISC Physics (XII)
m 2V
dF  2 mV 
V
momentum transfer in one collision = 1  v 2
 dA  
1  v 2 c 2 
2

c
P
2mv 2
1 v2 c2
Ans
Since in gas frame mass of gas particle and its density does not change
Q:1.378: At time t, letus velocity of sphere is V then F = ma 
t

t
F
2 2
 dt   1  v c dv 
m0 0
0
s
ds
v

dt
Fct
m 20 c 2  F 2 t 2
t
ds  FC 
 
t dt
 S
dx
c2 t

Q:1.379: Given x  a 2  c 2 t 2  v 
dt a 2  c 2 t 2
dV
a 2  c2 t 2

 Momentum :
dt
a 2  c 2t 2
F
m 02c 2  F 2 t 2
2 2 22
0 m0 c  F t
0
P = mv 
V
Fct
v

m


m0

 dV


2
2
 1  v c  dt



m02c2 F c2 t 2

Q:1.380: (a) and (b) : we know
c2

dv
dv
and
 a t  a. v  Then
dt
dt
Ans


1 / 2 
 
  m d  V
v2 

1



0 dt  


c2 





1 / 2


2
dV   1    v 2  3 / 2   2 V  dv 
  F  m 1  v 2 

    We know
   V 1 

0 
c 
dt  2  
c2 

 c 2  dt 
a
F
v 
1  v 2  m  dV   m v d 1  v 2 





0 dt 
dP
c 2  0  dt 
c2 

m0c2
 F
put
value
of
dv/dt
:
F

2
dt
1  v

a


c2 





 dP d  m 0 V
F

dt dt 
2
 1 v 2
c

m 0c 2
Ans
dV
c 2 t ( 2c 2 t )
 (a 2  c 2 t 2 ) c 2 

dt
a 2  c2t 2
m0
2
1 v
Al so


a
V



F  m0


v2
c2
1


2
c


  


2
 1  v 2  Va.v

c 




155







a
V

 

 

F  m0 

a.v
 If a || v  a.v.  av then
2 2
2
2 2 32
1

v
c
c
1

v
c






m 0a
m 0a v v
m 0a
m0 v2 c2 a

F



3
3
1 v2 c2 
1 v2 c2 1 v2 c2 2
2 2 2
1

v
c









m0 a

v2 c2

1 
2 2 
1 v c
1 v2 c2


F




m 0a


F

3
 
1  v2 c2 2

Ans


Q:1.381: Assume velocity of particle in K frame is Vx then we know
m0
Px 
1
and
Vx 2
c
E
m0c2
1
2
Vx 2
c
Px 

2
m 0 c Vx
and
E
c 2  Vx 2
m 0c 2
c 2  Vx 2
.
Also we know from invarient theorem : (ds)2 = c2(dt)2 - (dx)2 = constant = c2(dt)2 - (Vxdt)2 


(ds ) 2  c 2  Vx 2 (dt ) 2 
Then
Px ' 
Px ' 
Px 
c 2  Vx 2 
m 0 c Vx
dt    (i)
(ds )
ds
 constant in any inertial frame of references 
dt
 dt 
E  m 0 c 3      (ii )  Vxdt = dx  Also
 ds 

m 0 c Vx ' dt ' m 0 c

dx '  m 0c  dx  vdt
ds
ds
ds
2 2
 1  v c


 

m 0 c  dx  m 0 cVdt
Px
V


 
1   2  ds 
1   2 ds
1  2
1  2
v 2
c
 m c3

0
P
'


 x
2
2 
2
1 
1    c  Vx 2
Px
E'  m 0 c
3 dt '
ds





V
Px
1  2


m 0 c 3  dt  (dx ) V c 2 

ds 

1  2



m0c

 2
2
 c  Vx


c2




E
1  2
E' 
Px  E
 Px ' 
V
1
c2
2
proved 
m 0 c3
3
2
 dt  m 0c dx V c
 

ds
1   2  ds 
1  2

156

E' 
Compact ISC Physics (XII)
E
1  2

E
1
m 0 CV  dx 
  
1   2  ds 
PxV

2
1
2 
E
1  2
E' 

m 0 CV  Vxdt 

 
1   2  ds 
E
1  2


m 0 CVxdt  V

ds
 1  2



 


E  PxV
1  2
Ans
Q:1.382: We know E 2  P 2 C 2  m 0 2 c 4  For photon. rest mass m0 = 0 then E 2  P 2 C 2  E =
C
EV 

 Px  2 
c 
'  
PC  From K frame :   Px C  Px   / c form k’ frame = '  Px ' C 

2  
 1 


V

2
1 

1
V 3
c  (1   )
' 
' 
 
  '  
Ans  For
   
Ans
2
2
1


2
1


c 5
1 
1 
Q:1.383: From Q 1.381 We know Px  m 0 c
2
Then Px  m 0 c
dx
ds
 dt 
dy
dz
E  m 0 c 3    Py  m 0 c
 Pz  m 0 c
ds
ds
ds
 
dz
2 dy
2 dz
-----(i)  Py C  m 0 c
-----(ii)  Pz C  m 0 c
-----(iii). Squaring and add
ds
ds
ds
2 4
((i)+(ii)+(iii))
:
Px 2  Py 2  Pz 2  c2  m(ds0c)2 dx 2  dy2  dz2 
2
(dl) 2
 dt 
E 2  P 2 C 2  m 02 c 6    m 20 c 4
 ds 
(ds) 2
 P 2 c 2 
m 02c 4
(dl) 2

2
 2 2
2
2 2
2 4 c dt  dl
  m 20 c 4  cour
 E  P C  m 0 c 

ds 2


(ds)
2
E 2  P 2 C 2  m 02 c 4 Ans
Q:1.384: We know P = mV 
V PC

 For
C
E
system :
C T (T  2 m 0 c 2 )
Vcm



C
T  m 0c 2
mc2
V Also we know P 
c2
Vcm PsystemC

c
E system 
T
 
T  wm 0 c 2
EV
c2
PC  T (T  2m 0 c 2 )  Now

2
2 
 T (T  2 m 0 c )  0 (0  2 m 0 c )  C
Vcm



2
2
c
(T  m 0 c )  m 0 c
Vcm
T
 C
C
T  2m 0c 2
Ans
157
(a) Since maes of both particles are same hence magnitude of momentum of both particles from com
frame will be same. Velocity of each particle from com frame = 0- Vcm = –Vcm. Energy in com frame:
E
2m 0 c 2
1
~
T 
 2 m 0 c 2 (T  2 m 0 c 2 )
2
Vc m
c2
~ ~
. Using equation : E  T  2m 0 c 2 


~
T
2 m 0 c 2 T  2m 0 c 2  2m 0 c 2  T  2m 0 c 2  2m 0 c 2
1  


2m 0 c 2





2
~
 2m 0c
P  
 1  v2 c2



 Vcm 


m0
Q:1.385:
assume
1  Vcm 2 c 2
m0
energy
m 0c 2  C
m0

m0
does
2 m 0 2m 0 c 2  T
Vcm
not
V
losses.
  m 0  C1
We know  E 2  P 2 C 2  In var iant M 20 C 2 . We
m c  T  m c   T T  2m c   m C 
2 m  2 m c  T  Ans. Also we know from Q: 1.384
2
0
0

0
2
2
EV
C T T  2m 0c 2
P  2  v  PC  (PC)  v 
c
E
E
T  2m 0 c 2
0
2 2
0
2
2 2
0
2

 v C
Q:1.386: Use invariant equation : E 2  P 2 C 2  Invariant
T
Ans
T  2m 0c 2
T1
T
T
2m 0c 2  T1 2  T1 T1  2m 0c 2   2 2m 0c 2  T 2  0 . Because
m0
m0
momentum of system from com = 0.
Inertial K frame
=
m0
m0
COM frame
2
2
2
T12  4T1m 0 c 2  4 m 20 c 4  T12  2T1m 0 c 2  4T 2  8Tm 0 c 2  4 m 20 c 4  4T  8Tm 0 c  2T1m 0 c 
T1 
4T ( T  2m 0 c 2 )
2m 0 c
Q :1.387:
2
1
 T 

2T T  2m 0c 2
E1  E 2  E 3  m 0 c 2
2m 0c
2
 Ans
 P1  P2  P3  0 .
E 2  E 3 2   P1  P2 c 2  m 0c 2  E1 
2
Take
 P12 c 2  Invariant
system
of
(m 2 +
m 3)

m0 c 2  E1 2  P12c 2  m 2  m3 2 c 4 .
158
Compact ISC Physics (XII)
Because this invariant is same in all frame of reference hence from com frame of (m2 + m3). In
variant
m 2  m 3 2 c 4 
=

Now
m 20 c 4  E12  2m 0 c 2 E1  P12 c 2  (m 2  m 3 ) 2 c 4
m 20 c 4  m12 c 4  2m 0 c 2 E1  (m 2  m 3 ) 2 c 4

E2
m0


m1
m2
 2m 0 c 2 E1  m 02  m12 c 4  m 2  m 3 2 c 4


=0
v=0
m3
E1
 m 2  m 2  m  m   2
1
2
3
E1 max   0
 c
2m 0


E3
Ans
Q:1.388: Suppose velocity of ejected mass is Vx from a frame having
u
Vx  V
Vx ' 
v
zero
velocity then
we
know
1  Vx
2
dm
m
v
c
Vx  V
Vu
u 
 Vx 
v
uv . Change in momentum of (dm) mass : dP = -(dm) V + dm Vx
1  Vx
1
c2
c2



vu 
   dm
 dm  v 
 1  uv 

c 2 


u  u v 2 
uv
dP
(dm) u
c

  Here 2  u then 
 
 Force on
uv
c
dt
dt 1  uv
 1


c 2 
c2
m dm
v
 uv 
u
dv
 dm 

u

1  2  dv
F



m




gas particle then force on rocket :

dt
c 
 dt  1  uv
m0 m
0
2
c
1  m m 0 2 u / c
v

2u
c
 m  C


1  
 m0 
159
Part Two
Thermodynamics and Molecular Physics
2.1
Equation of gas state process
Q: 2.1: Suppose gas is at pressure P, volume V and tamp. T. Then we know PV=
m
nRT. Also PV  RT where m = mass of total gas M = molar mass. Now V
M
= const. and T = const dP  V 
PV 
P.V.T
(dm )RT
M (dP )
V ----(i). Calculation of M/RT : We Know
. Then dm 
M
RT
M
P
m

RT  PM  RT 
. Given at NTP (normal tempreature and pressure) : T0 =
RT

M
00C = 273 K P = 1 atm = P0 = 1×105 N/m2. Then
P
P (dP ) V
PV
M
 0 put in (i) dm  0
 m  0 P
RT



V
Q: 2.2: Total no. of mole of gas is n and due to heating n mole
of gas goes in other chamber.
PV
P1V  nRT1  n  1
RT1
P1
t1
V
Value
P=0
n
Initial
(A)

(B)
V
P V
P2 V  (n  n ) R T2  n  n  2 ---(i) (P  P) V  nRT 
2
2
RT2
n 
P2 
t2
P2
t2
n- n
n
P2- P
Final
1  1
P1 P
(P2  P)V
P1V (P0  P)V P2 V
P2 



 
put
value
of
n
and
n
in
(i)
:

T1 T2 
RT2
RT1
RT2
RT2
 T2 

PT
P1T2 P
P 1  P1T2
P2  P  1 2 
 
 P 

.
Increase
of
pressure
of
Vessel
B
:
2T1
2
2  T1
2T1 T2

Q: 2.3:
H 2 ; He
P; t , m; V
Let mass of H2 gas is m1 and that of He is m2. no. of moles of H2 : n2 = m/
 m1 m 2 

 Rt
2. no. of moles of He : n2 = m2/4. Also m1 + m2 = m ----(i). PV  n1  n 2  Rt  PV  
4 
 2

PV 

m1 m 2 PV
m 2 2 PV
m2
2PV




 m
 m1 
----(ii) Now (i) - (ii) :
 m 2  2  m  2
Rt 
2
4
Rt
2
Rt
2
Rt

put in (i) :
PV
PV m PV  m
4
m

m1
Rt m 2
Rt
PV

 Rt 4 
m1  m  m 2  4
m  m
m PV Ans
PV  m PV

2


Rt
2 m  2

2
Rt
m
Rt
Rt


1
160
Compact ISC Physics (XII)
N 2 ( m1 )
Q: 2.4:
T; P0
CO 2 ( m 2 )
We know PM = RT----------(i) Where M = Molecular weight of mixture.
 = Density of mixture. Calculation of M : no. of mole of N2 : (n1) = m1/M1  no. of mole of CO2
n M  n 2M 2
m  m2
: (n2) = m2/M2 where M1 and M2 molecular weight of N2 and CO2. Then M  1 1
 1
m1 m 2
n1  n 2

M


1 M2


P
m

m
1
2


0 
m1  m 2 

m
m 
P
  RT 
 Put in (i) :
RT  1  2  Ans
 m1 m 2 
M

M

2
 1


 M1
Q: 2.5:
M2 
(a) V, T1, v1, v2, v3. Suppose molar maes of O2, N2, CO2 are M1, M2 and M3. Then PV =
nRT PV = (v1 + v2 + v3) RT P  v1  v2  v3 
RT
V
Ans
Total mass
Total maes of mixture: M = v1M1 + v2M2 + v3M3 molar mass of mixuture = total no. of mole
v1M1  v2 M 2  v3 M 3
 M 
Ans
v1  v 2  v3
(b)
Q: 2.6: Here ('1) v'  (  1) v ----(i)  F.B.D. of piston : P2-P1 = mg/A
now P1v  RT0  P1 

Since
RT0  1 
RT 0
RT0
1  
 P2 
 P2  P1 
V   
v
V
P2  P1  P2 'P1 '  mg / A
Similar P21-P11 = mg/A
Similar : P2 ' P1 ' 
RT0  1 
RT 
1
1   
1  
V  n
V'  n ' 
then
 n 2  1

n ' 
 T
RT0  n  1 
RT  n '1 
RT
( n '1) ( n '1)
T

0

 





2

V  n 
V'  n '  (  1) V
n'
 n ' 1 


n=1
T0
V
T
n=1
n' V'
T0
n=1
V
T
n=1
V'
Initial
RT  1 
1  
v '   
Ans
P1A
P2 = P1 + mg/A
Final
P2A
mg

161
Q: 2.7: This question is based on operation of an engine. In this engine, first, piston pull right side
and during pulling piston, value is opened and gas is filled in vacent space. then value is closed and
gas between value and piston is removed. And piston moves left way. After that piston is again pull
right way and value is opened and gas comes with piston. This process continues.
First stroke right PV = P1 (V + V)  P1 
PV
------(i)
V  V
First stroke left : P1, V
PV 2
PV 2
Second stroke right P1V  P2 (V  V )  V  V  P2 (V  V)  P2 
-------(ii)
(V  V) 2
In 3rd stroke : P3 
PV 3
(V  V) 3
In n
th
stroke : Pn 
ln 
PV n
( V  V ) n
P
Vn
1


n


( V  V )
ln 
n

V
 V  V 

V 
  ln  
 n ln
 ln  1 

ln 
( V  V)
V 
V 




Ans

P, V
V
Piston
Valve
Q: 2.8: Suppose at time t, pressure is P and in next dt time
dP pressure is released. Then PV = (P+dP) (V-Cdt)  PV =
PV + VdP + CPdt + C(dP) (dt) -VdP = (CP + CdP) dt  CP
Isothermal
V, P0
P
t
P0
0
dP
+ CdP = CP -VdP = CPdt   V  P   C dt  P + dP
< P because P is decreasing function hence dP = (-) ive.
ln P / P0 
c
ct
t  P  P e v Ans
0
v
Value
piston
162
Compact ISC Physics (XII)
Q: 2.9: from Q: 2.8: P  P0 e
ct
v 
ct
P0
v
ct
 P0 e v  ln  
 t    ln 

v
c
Q: 2.10: Let assume pressure inside gas chamber is P F.B.D. of piston. Force
Ans
P0S
m0
equation : P0S+ P(S-S) + m0g + (m-m0) g = PS + P0(S-S)  P  P0  mg S
l1
P
----(i) . After increasing T of tempreature, final temp. is T + T. since from
(i) pressure is constant, process will be isobaric. Then
V
V V  V
 const  
T
T T  T
l2
m-m0
 V 
T  T
V
  V
 1

 T  T 
= Sl1 + (S - S) l2
initial
T  T
V
 V 
P0
P0 S
Change in volume : Vinitial = Sl1 + (S -S) l2. If l2 increase, l2 decrease in
same amount. Then l2 l  l2 -l Then Vinitial = Sl = V. Then T 
T
Sl 
V
S-S
PS
m0g
P (S- S)
mg 

 V  RT  Now
----(i) Also, PV = nRT . Here n =1   P0 
S 

T
1 
mg 
(  S)l

P0 
T 

R
V
R 
S  put in (i) 
Q: 2.11:
(i)

  T  ( mg  P S) l

0
R

(a) P = P0-V2 -------(i) No. of mole of gas = 1. We know PV = nRT  P 
P0 (S- S )
Ans
RT
V
put in
3
2
RT
dT
 P0  V 2  T  P0 V  V ------(iii). For T maximum,
 0  0  P0  3V  0 
V
dV
R
R
R
R
V
(b)

 P0  mg

S

(m-m0) g
P0
P0
put in (ii): Tmax 
R
R
P = P0 e-V 
P0
2P0
 Tmax 
3
3
P0
3
Ans
RT
dT
P
 P0 e  V  T  0 Ve  V -----(i). Now
 0 for maximum Temp. Calculate
V
dV
R
V and put in (i) Tmax 
Q: 2.12:
P0
P 
  0
R
R  3 
P0
eR
Ans
T. If v is increasing, T will increase and hence
T  T0  V 2 ------(i) We know PV = nRT = RT
P will increase. Hence calculation of Pmin: T 
PV
PV
 T0  V 2 ------------(ii)
put in ----(i) 
R
R
163
2


dP
P  R T0 V 1  V   To P
 0  0  R  T0 V     V 


min :
dV

P
R
P
T0
T
 2 T0  P
 T0   0  2T0 
min  2R T0


R 
 = const
N2
dx
Q: 2.13: We know PM = RT  (dP) M = R(dT)  -(gdx) M
= R dT  
dT gM
dT dT  Mg




Ans
dx
R
dx dh
R
Q: 2.14: Here
x
 
 const  C  P   n     P 
 C
 
P
n
1
n
 Also PM =
dx
1
x
1
P n
1 1
RT
RT P n C n 
 Differentiate :
 RT  PM   
M
C
1  1n  P
1
 n 1
(i)   n

n
1
(dP) c
s
n
R

dT -----(i) Also we know dP = - gdh put in
M
 1/ n 1
1 1
R
P
c 2 (gdh ) 
dT  dT   (n  1) P 2 c 2 Mg

M
dh
n

1
 (n  1)  12 12
  n

 1
 P
  
C
Mg    n  1  P 2 Mg   P 
 n 
 n 
  n 


dT
(n  1)

Mg Ans
dh
n
Q: 2.15: We know dP  gdz   
  dP 

PM  RT 

gdz



P0
put in (ii)
g
Ans
P
T0

dP   Mg
P
RT
h  5  10 3 m
g  9.8m / s 2
dP
Also PM = RT 
gdz
In question dz = dh   PM 
h

 dh 
P  P0 e
0
R  8. 3
 P  0.5  10 5 N / m 2  0.5 Atm
RT
dP 
gdh
Mgh
RT . Above earth surface :
T  27 0 C  300K
M  78  10 3 kg
 n 1 
 Mg 
  
 n 
. Put in above equation
T = const
M = const
dh
x
164
Compact ISC Physics (XII)
Below earthsurface :
h  5  10 3 m
g  9.8m / s 2
T  27 0 C  300K
n  78  10 3 kg
R  8. 3
 put P  2 10 5 N / m 2  2 Atm
Q: 2.16: PM  RT  (dP) M  (d) RT ----------(i) Also we know
: dP  gdz ---------(ii) then from (i) and (ii):

 
0
d


 Mg h
 dz
RT 0
0
    e
0
zh
0

Mgh
RT
h
Mgh
RT

RT
 1
 h1  Mg
e e
 Mgh 1  RT
 
(b)
  0

      (1  ) 
0
 0 (1  )  0 e
0
h2 
 0e
e
T = const
g = const
M = const
dh
Mgh
RT
(a)
e
Then
dz  dx
(d) RT
 gdz
M
Mgh 2
RT
Ans

RT
RT 
RT
ln (1  ) 
 h2 
Mg
Mg
Mg
Q: 2.17: Suppose pressure at height x is P then in (dx) volume
PV = nRT in differential volume Psdx = (dN) RT We Know
M, T, g
h
dx
N
 dm 
m
dm
 Ps dx   M  RT -----(i) dN 
. Also we known
M
M


dm dm

dP  gdx  
dV Sdx
dP  g
P  P0 e
dm
SdP
 dm 
s
g
x
Mg

-----(ii)
m
RT from (ii) :
when x = h
P
 dm  
0
P0
  dm 
 g dx
dP  
 Sx 
put
P0
i
(i)

PS dx 

s
P
RT  SdP 
 Mg x
dP

 
dx


P  
M   g 
RT 0
P0
S
SdP
m
P  P0
g  
g
x



Mg
P0S 
RT
m
1  e
 Ans
g 



x


x



 Mg
P0S 
RT
1  e

g 



165
Q: 2.18: We know x cm 
P  P0 e  Mgx RT And
dm 
 xdm
. Also we know from Q: 1.215 
 dm
P(sdx ) 
M
MS
PS dx 
P0
RT
RT
dm
RT 
M
x
P0
e  Mgx RT dx 

MSP0 x  Mgx RT
 RT e
x cm 
MSP0  Mgx RT
 RT e
 xe

 Mgx RT
s
dx
0
e
 Mgx RT
dx
 x cm 
RT
Mg
0
Q: 2.19: (a) T = T0 (1-ah)  Psdx 
dm
RT -----(i)
M
And also
Ps dx 
x
RT
M
P
 Mgdx
dP
 R T0 (1 ax)   P
0
P0
 s 

dP 
 g

x
P = P0 at h = 0
Also T  T0 (1  ax
P0
Similiar as above : P 
(1  ah ) n

s
x
 Mg
 RT (a ) ln (1  ax)
0
 Mg
ln (1  ax)
a RT0
 ln P / P0 
0
and x  h  P = P0 (1-ah)n
P = P0 (1-ax)  Assume : Mg / aRT0  n
(b)
M, g
dx
s
 dm 
(dp) -------(i)
 g dx   g ( dm )  dm 
dP   g dx   

g
sdx
s



M, T, g
dx
 ln P / P0 
Ans
Ans
dP  dm  2
 dm 
2
 rw .
Q: 2.20: Force equation of dr element dF  (dm) rw 2  dP  S   S  rw  dP  
S




 s
dm  
 rw 2
PS(dr) 
 dm 

 RT 
 dP also we know P(Sdr )  
M 

RT
M
r
 S 

 dP   Mw 2  rdr 
 rw 2 
0


P
RT  dP
Mw 2 r 2
MW 2r 2
 RT ln P P  P  P e 2RT Ans
0
0
2
P0
P

M, T
P0
s
r
dx
This end is
open in air
166
Compact ISC Physics (XII)
P  ?
Q: 2.21:  , T
M = (12+32) × 10–3 kg = 44 ×10–3 kg. Idial gas equation :  
We know PM = RT  P 
P
RT
 M


 b
 





2 a

  P  P
M2

M2
 P
RT
2a

M b M2
T
(b)
kg
m3
 M


 b   RT

 P


V
M


Ans
RT
. Using vander wall
V


 P  a  ( v  b)  RT  P2  RT  a . Here

V  b V2
v 2 

P1  P2  (1  )
 500
 2a
Q: 2.22: (a) Using idial gas equation : PV  RT  P1 
equation :
10  3 m 3
RT
500  8.2  300

 280 atm  Vanderwall equation : (for one
M
44 10 3


 P  a  (V  b)  RT    M


V
V2 

mole gas) :
500 3 kg

RT  RT
a 


(1  )
V  V  b V 2 

(1  ) a / v 2
(1  ) a (V  b)V
a (V  b)(1  )

2
T
1  R R

V
R
(
nV

V

b

v
)

VR (nv  b)
Vb V
P1  P2
 
P2
V
n=1
 R (1  )R 
a
  (1  )
T 
V
V

b
V2



Ans
RT
a
Put value of this temp in vanderwall equation. P2  V  b  2 Ans
V
Q: 2.23:
n 1
V
T1, P2


a 
a 
  P1  2  (V  b)  RT1 ----(i)   P2  2  (V  b)  RT2 V 
V 


T2 , P2
2 T P T P 
-----(ii) from (i) and (ii) : a  V 1 2 2 1 T2 T1 
T T 
Ans  b  V  R 2 1 P2  P1  Ans
167
dP
Q: 2.24: We know bulk modulus of a gas is given by : B  dv . While compressibility is given
V
dv
by :

isothermal:

V
dP

T

a 
1  dV 

 in vanderwall equation :  P  2  ( v  b)  RT if process is
V  dP 
v 

=
const.
differentiate
 dP  a 

a 

 (V  b)   P 
 0

 dV V 3 

V 2 




(i):
dP

dV




 dP   a dV  (v  b)   P  a  (dV)  0



3
2
V
V 





 P  a 2 
  PV 3  aV   a (V  b)
a
V 



 
(V  b)
V3
V 3 (V  b)

1
V 3 ( V  b)
V 2 ( V  b)


x




V  (PV 3  av)  a (V  b)  (PV 3  aV)  a (b  V ) ----(ii) put value of P from (i) in (ii)


V 2 ( v  b) 2
[ RTV 3  2a ( v  b) 2 ]
Ans
Q: 2.25: Using ideal gas : PV = RT; V= RT P-1 
2.24: x 
2.2
1 RT
V
V

  RT P  2  x1 
from Q :
2
RT
VP
P
V 2 (V  b) 2
V 2 (V  b) 2
V
a


According
to
equation
:
x
>
x

 T  R
3
2
3
2
1
RT
b
RTV  2a (V  v)
RTV  2a (V  v)
The first law of thermodynamics, Heat capacity
Q: 2.26: We know that room is an open thermodynamic system, in which
no. of molecules may change suppose no. of mole of gas is n. Then PV
nRT PV

= nRT. Also we know Internal energy is given by: U  nCVT 
.
v 1 v 1
Here  
Cp
cv
Windo
w

U
V
P
 constant and room pressure and volume is also constant. The Internal energy also const.
U = Const if P and V are const. And not depend on T because when T increase, no. of moles will
be decreased in room because PV = Const. Hence in room internal energy (U) = Const.
168
Compact ISC Physics (XII)
Q: 2.27: Suppose no. of moles of gas = n. Directional kinetic energy of
M
CP

CV
1
(nM) V 2 . When Vessel sudden stop, then after long time this directional
2
kinetic energy of gas is converted into randm kinetic energy when thermodynamic
gas =
equililnium will be achieved and then
V
(v  1)
1
MV2 (v  1)
MV2
(nM ) V 2  nC V T  T 
 T 
2R
2
2
R
Ans
Q: 2.28: Method 1: When value is opened and thermodynamics equilibrium is attained then. no.
P1V1 P2 V2 PV1  V2 
P1V1 P2 V2 PV1  V2 


of moles will be constant. Then RT  RT 

------(i). Also
RT
T1
T2
T
1
2
we know, in whole system : Q  0  U  W  Q  0 Because vessell is insulated. And also.
 W  0 Because gas does not work on atmosphere. Because vessel closed. Then U System
= 0 then n1c v T1  n 2C v T2  n1T1  n 2 T2 
T1T2 P2 V2  P1V1 
 T  P V T  P V T
1 1 2
2 2 1
P1V1
T  T1   P2 V2 T  T2   0 ------(ii)
T1
T2
Vessel (1)
put in (i) Ans 
(2)
Value
V1 , P1 , T1
V2 , P2 , T2

P V P V
P  1 1 2 2 Ans
V1  V2
V1 , V2
P
T
Method : 2 From method : 1, we see that there
will be no change of internal inergy of system. Hence
Initial Internal Energy = final internal energy.
Also U 

PV
P1V1 P2 V2 P V1  V2


then
v 1
v  1 v 1
v 1
 P1V1  P2 V2 

 V1  V2  
 V V

1
2 


V1 , P1 , T1
P1V1  P2 V2
 P  V  V
.
1
2
Also we know PV = nRT
T
 P1V1 P2 V2 
P V  P V  T T

 RT

T  1 1 2 2 1 2 Ans

 RT

P1V1T2  P2 V2 T1
 1 RT2 
Q: 2.29: Method: 1: We know Suppose initial temp. is T1 and final is (T1-T) then.
PV
PV
V
Ui  1 ; Uf  2  U 
P2  P1  ----(i).
v 1
r 1
r 1
also nR 
P, T ,
V1  V2
V2 , P2 , T2
nRT1
P1V  nRT1  P1 
V
P1V
V T  nRT2 nRT1 
nRT2
nR
U 


 
P2 
T  


T1
r 1  V
V 
V
v 1
Sealed vessel
V
T
169
U 
P1VT
T1(v  1)
P VT
Here P1  0  T = T  U  0
1
0
T0 (v  1) Ans U = Increase in P.E.  Q = U
P0 VT
+ w Since vessel is realed then w = 0  Q  U  T (v  1) Ans
0
Method : 2 U  ncvT  n
R
P V
T ----(i) Here P = P P V = nR T  V = v nR  0
put
0
0
0
T0
(v  1)
P0VT
P0 VT
in (i)  U  T (v  1) Ans Also Q = U + w  w = 0 then  Q  U  T (  1) Ans
0
0
PV
 A . Also W = PU + A
v 1
Q: 2.30: We know : Q = U + W =
 Q 
Q
 v 
A

 A  Q  A

v 1
 v 1 
Q: 2.31: Isobaric process: We know  Q = U + W also we know
Q
PV
PV
 W  PV = nRT = RT  W  Q 
v 1
v 1
U 
P = const
N2
PV RT

v 1 v  1
Q: 2.32: Initial
n=1
T
Ans  Q
Ans
P0 ; T0 ; V  isochoric  P0

2
; T1 ; V  isobaric  P0 ; T0 ; V1  Here,

 2

2RT0
P1 T1 '
PV = nRT  P0V = 2RT0 V  P . Isochoric process : P  T ' 
0
2
2
n  2
T0
V  const
P0
Pf  P0
2
T0
P0
T
V1 ' T1 '
V
 0  T1  T0
2  V  2V


.
Isobaric
process
:

. In whole system : from
P0
T1
1
2
V2 ' T2 '
V
T
1
0
2
initial
to
final
position.
Q  U   W
Here
U
P
P V P 2RT0
W  0  0 V  0  0
 W = RT
T0  Q  RT0
2
2
2 P0
 
=
0
final
temp.
is
zero.
170
Compact ISC Physics (XII)
Q: 2.33: Method : 1 v1 = 2 O2 (r1)  v2 = 3 CO2 (r2). Assume adiabatic exponent of O2 and CO2
are r 1 and r 2 . Internal energy of system : U  v1
r1
r11

v  v 
r2
 1 2
r2  1
r 1

v1 r2  1  v 2 r1  1
v1  v2
r1  1 r1  1  r  1
r  1 r2  1 v1  v1   1 r  v1 r1 r2  1 
r 1

v1 r2  1  v2 r1  1
v1 r2  1 
Method:2 C v (mixture) 
v2 r2 r1  1
v2 r1  1
v1c1v  v 2 c v2
n1c 2  n 2 c 2

n1  n 2
v1  v 2
R
R
RT
T  v2
T  v1  v2 
r1  1
r2  1
r 1

r1  1 r2  1 v1  v2 
 r  1 
v1 r2  1  v2 r1  1

Ans
Cp - Cv = R  C p (mixture) 
v1c1p  v 2c p2
v1  v 2


R 
R 
v1 R 
 v 2 R 


r1  1
r2  1
C p v1c1p  v 2c p2


r

 Cp - R = C v  r 


Ans
R
C v v1c v  v 2c v
v

v
(
r

1
)
1
2
1
1
2
r1  1
Q: 2.34:
n1 
5
R
2
7
C1P  R
2
C1v 
N2 :
7
14
3
R 
2
5
C 2P  R
2
20
n2 
2
10
Ar : C 2v 
C v( mix ) 
3
15 
 R   2 
22 
 2   0.42 J
n C1  n 2C 2P
C P(mix )  1 P
9 m.k.
 C v(mix ) 
1
n1  n 2
2
2
n1C1v  n 2C 2v
n1  n 2
Ans 
5 
17 
 R   2 R 
22 
 2   0.65 J
C p(mix ) 
9 m.k .
1
2
2
Q: 2.35: Suppose at time t, piston moves by x distance and in next dt time, piston moves dx distance.
Since T = Const  P1V1 = P2V2  P0 V = (P0 - P) (V+xs) Here P = loss of pressure
171
P0
P V
 xsP0
P  0  P0 
. Force applied by person on piston :
V  xs
V  xs
F  PS 
x xs 2 P dx
 xs 2 P0
0
work done in ‘dx’ distance : W   dW  
V

xs
v  xs
0
x
= P0 s   
0

air
(xs  v  v)
dx 
v  xs
P0Sx  P0SV
x
n 1
T  const
V
l
s
final volume = nv
V = Sl
x
v
P0s  dx  P0SV  
dx
0
0 v  xs
ln v  xs x
v  xs
W  P0 sx  P0 V ln
------(i) Final volume : nv = (l+x) s =
0 
S
v
nv = v+xs  sx = (n-1) v  initial volume : v = sl. W  P0 V (n  1)  ln n   RT (n  1)  ln n 
Ans
Q: 2.36: Suppose at time t, piston move by x distance, and in next dt time, piston move by dx, distance.
P0 V0
P0 V0
Since T = const. PV = const  P0 V0  P1 V0  xs  P0 V0  P2 V0  xs   P1  V  xs P2  V  xs
0
0
 1
1 
2xs
 P  P2  P1  P0 V0  V  xs  V  xs   P0 V0 2
. Force applied by external agent:
V0  x 2s
0
 0

xƒ
F  ( P) S  2P0 V0 x s 2 V0 2  x 2 s  Work done : W   Fdx  
0
V1  V0  x f s 
2 P0 V0 xs 2 dx
V0 2  x 2 s
(1  n ) V0
2V0
2V0
V  V0
 Sx f
 V0  0
 x f  (1  n) S
1 
1 
1 
 (n  1) 2 

W  P0 V0 ln 
 4n 
V1
s
--------(i) Here
Put in (i) and integrate
 V1
P1
V0 , P0 , T0
V0 , P0 , T0
P2
V1   V1  2V0
V1 
2V0
1 
3RT0
and P0 V0  P1 5V0
Q: 2.37: Q = Total heat transfer Isothermal process : P0 V0  2RT0  V0  P
0

P1  P0
Q  W  nRT ln
5  1
  Q
v2
v1
1
n  3; T0 ; P0 ; V0
A
T0  const
= 3RT
T0 ln5 -----(i) Isochoric process : Q2 = nCvT
n  3; T0 ; P1; 5V0
B
n  3; T; P0 ; 5V0
5V0  const
C
172
Compact ISC Physics (XII)
P0
P1 T1
5  T0  T  5 T

--------(i) Also P

0  T = 5T0-T0 = 4T0 from (i) : Q2 = 3CV4T0 = 12
T2
P0
T
2
C vT 0

v 1 
Q1  Q 2  3R 0 ln 5  12C v T0  Q

Cv 
Q  3RT0 ln 5
R

12T0
v 1

12RT0
12RT0
v
 1 Ans
Q  3RT0 ln 5
Q  3RT0 ln 5
Q: 2.38: (a) Isochoric : V = const P & T Isobaric : All lines intersect at same point P = const.
Because of P, and T are same then  V = same PV = nRT  Isothermal T = const if n = const.
(b)
P
T = const
V
=
t
ns
co
V
T = const
isobaric
V = const
P = const
T
T
Isobaric P = const
V
Q: 2.39: (a) We know PVv = const. --------(i) V 
vRT
put in (i)
P
O2
V1
T0 , P0
Adiabatically
 P0 , v, T1

 vRT 
1 r 4
P
  const  P T  const  P01 r T0 4  P0 1 r T1r 
 P 
1 r
1
T1   
 
 
r
vR
r 1
r 1
T0  T0 n
r Ans (b) A = – vC T (work done by gas) A1 = vC T (work on gas)
v
v
r 1


T0 r  T0  


A' 
RT0
r 1
 r 1 r 
 1


Q: 2.40: Adiabatic process : w a  vC v T 
Here v  1 Ans
vR
r
T  PV  const .
r 1
PV  vRT 
vRT
v
r 1
v
vRT
 const .  Tv r 1  const .   T1  0
Then
v
 




vRT0  n r 1  1

 ------(i)
 w a 
r -1
P0 ; V0 ; T0
 T0 V0 r 1  T1  T0 r 1  T  T0  n r 1  1


N2
N2
P0 ; V0 ; T0
Adiabatically
P1;
V0
; T1

P2;
V0
; T1

Isothermally
173
Isothermally : w T  vRT0 ln
w a
n r 1  1

w T (r - 1) ln 
v2
vRT0  n r 1  1 RT0  n r 1  1

w



 
a

v1  vRT0 ln n -------(i)
w T (r - 1) vRT0 ln  (r - 1) RT0 ln 
Ans
Q: 2.41: Work done by gas on system is zero
here temp. is only increased because of work
done by external agent Wext = Usystem. Calculation
at
conducting piston
v2
T0 , V0 , P0
T0 , V0 , P0
R
T gas
r 1
t
:
of Usystem : U system  (v1  v2 )
equation
s
v1
time

x
Initial
T
V
T;
 V; P0
P2
Final
2V0
nV  V  2V0  V 
1 
P0 V0 P1 V0  xs 
P0 V0 T
P0 V0
V  xs  P  P0 V0 T

 P1 
 P2 0
2

. Force applied by
T0
T
T0 V0  xs 
T0
T
T0 V0  xs 
P0 V0T
ext agent: F  P2  P1  S  T
0
in (dx) displacement : dw 
2P0 V0 TS2 x
dU 

T0 V0 2  x 2 s 2

dx 
 1
P V TS
1 

F  0 0

S
 V  xs V  xs  
T0
0
 0



2xs


 V0 2  x 2 s 2  . Work done


2 P0 V0 TS 2 x
dx  dU  v1  v 2  R dT  On system : dW =
ext
2
2 2

r 1
T0  V0  x s 


v1  v2  R dT
r 1
 r 1 
=

P0V0
RT0

P0 V0
RT0

x
xdx
1 T dT
2
RdT
s




2
r 1 T T
r 1
0 V0  xs
0


 (V  xs ) (V  xs )   2 
4V0 2
2

0
0

(
V

xs
)
(
V

xs
)


V
.
v


v

0
0
 T  T0 
-----(i)
put in (i)

(1  ) 2
V02


r1
2
 (  1)  2
 T  T0 


4 
Ans
174
Compact ISC Physics (XII)
Q : 2.42: Bernauli’s equation for gas particle at same level :
P
1
dU
dU
V 2 
 Const. Where V = directional velocity 
2
dVolume
dVolume
= internal energy per unit volume and for gas
2Cp T
M
nC v T
1
 0  V 2  0
Volume
2
 v2  V 
Q: 2.43:
2CpT
M
RT aR
P

V
V2
(2  r)
r 1
RT
m
1

C v T  V 2 
M
mVolume
2
2 TR
(  1)M
PV  RT  PV 
RT C v T 1

 V 2 
M
M
2
T
a
. We know Q = dU + dW
V
R
T -----(i)  dw  PdV  w   PdV ------ (ii) PV = nRT = RT
T
r 1
v2
put
in
(ii)
:
w  
v1
aR dV   aR 1
V
v2
 Q = U + w

v2
1
1 
 aR  

 v1 v 2 
v1
RT
 RT  RT
r 1
1 r 1 


 r 1 


Ans
Method : 2 We know heat capacity of gas :
C
V
(2)
Ans
a
v = 1 = no. of mole 
T
 a
a 
w   
 R  R[T ]
 V1 V2 
Q  RT

 V
Method: 1 V 
    dU  C v T 

(1)
dU
 0 . Now bernauli’s equation between (1) and (2):
dVolume
liquid :
P1  0 
dU
 0  while for
dVolume
Ra
V
C
R
R

. Here
r  1 x 1
V
a
a
T

T
V
 PV 2 = aR = constant. Compare with PV x = const  x = 2 
R
R
R
R T (2  r)
R 
(2  r )  Q  nCT 
(2  r ) T  Q 
r 1
r 1
r 1
r 1
Ans
175
Q: 2.44: dW dU v = no. of mole of gas. dW = KdU K = constant PdV = KvCvdT P 

vRT
V
R dV
dT
R
vRT
PV
dV  KvC v dT T 
 KC  v   T  KC ln V  ln T  ln K1  K1 = constant
V
vR
v
v

1R
ln V R KC v  ln K1 T 
KCV
K1T  V R KC v 
K1
1 R
R KC v
PV
KC V
V
 PV
 const
vR

 n  const  PVn = const Ans
Q: 2.45: PVn = const. -------(i) We know
n 1 = no. of mole of gas  dQ = dV + dW  n1CdT
= n1CvdT + PdV ------(ii) differentiate equation (i) : nPVn-1 dV + VndP = 0 nPdV + VdP= 0 ------n RdT
---(iii) Also : PV = n 1RT  PdV + VdP = n1RdT from (iii): PdV-nPdV = n 1RdT  PdV  1
1 n
n RdT
n R
R
R
R

put in (ii) : n1CdT  n1C v dT  1
 n1C  n1C v  1  C  C v 
 C 
1 n
1 n
1 n
r 1
n 1
P
Q: 2.46: Argon : V0  V0  P0  0  . If this is Polytropic process then PV x = const
P0
x
V0 x
 P0 V0 

C
R
R

ln

r 1
1
ln 

V0 x 
V0 x

 x
V0 x

  x  
ln 
 x 
ln  .
R (n  r )
R
R
ln 

Assume = n  ln   C 
 C  (n  1) (r  1)
r 1 n 1
Q: 2.47: n = 1.50  C 
T hen
Ans
(n  r) R
R
R

(a) Q  vCT Here v  1 = CT  Q  (n  1) (r  1) T
r 1 n 1
(b) Compare : C = CV + C0. Here C0 is constant which provide work done dW = vC0 dT Here v
= 1  dW = C0 dT  C 0 
R
1 n
Then W 
R
T
1 n
Ans
Q: 2.48: Method : 1 P = V  Pv-1 = . Compare with polytropic gas equation : PVn = const
n = -1
C
Then (c) Molar heat capacity of gas is given by : C 
 1
R
R
1
R 1  r 

 R
   C


r 1  1  1
2  r 1
 r 1 2 
Ans (c)
R
R

-----(i)
r 1 n 1
176
(b)
Compact ISC Physics (XII)
vR
 R 
dT 
dT  w  vR T -----(ii)
Work portion in equation (i) : dw  v 

2
  1  1
2
Calculation
of
T
final : Pf   V0  Tf 
PV
:
=
P V
V02
initial : P0 V0  vRT0  T0  0 0 

vR
vR
 V2 
V02
 V0  0 



T


V


put in (i )

 vR 
vR


vRT
 V0 V0
vR
  2    V 2   2  1 V 2
VR 
 0  
 0
w 
2
vR
2
V2
vR  2
U 
    0 
 vR
r  1 
Method
:
2
(a)
  2  1 V0 2


2
Ans
V = vCVT  U 
(a)
P V
V0 2
T0  0 0 
vR
vR
  2  1 V0 2


r 1
Ans
V 2
T  Tf  Ti   2  1 0

 vR
(b) dw =
dV

Tƒ 
V0 2 2

vR
nV0
 2 V0

W


VdV

V


2
V0
V0
(c) nCdT  nC vdT  PdV ------(i) Also PV = vRT
T  V2 = vRT

vRdT
vRdT
2VdV = vRdT  2   dV  vRdT  nCdT  nC v dT 
 PdV 
 
2
2
C  CV 
vR
T 
r 1
  2  1 V02


r 1
 U  vC v T  vR T 
r -1
 W 

put in (i)
R
R
R
R (1   )


 C
Ans
2  1 2
2(  1)
Q: 2.49: Method : 1 (a) d = - nCVdT  Here molar heat capacity is C then nCdT = -nCVdT
 C  C v 
v-1  x 
 TV
v 1
2
R
 1
Ans
(b) Also we know C 
v 1
Then PVx = const  Put
2
R
R
R
2
1




 2x-2 =
  1 x 1  1
v 1 x 1
P
vRT
vRT x
Then
V  const  TV x-1 = const
V
V
 const Ans (c) We know d = dU + dW  Here d = -dU Then dW = -2(dU)   2
 W  
2vR
T
r 1
vR
dT
r 1
177
r 1

r 1
Calculation of  T: T0 V0  2  Tƒ V0 2
 
 Tƒ  T0  1 
 
v 1 

   2 
2RT0
W 
2vRT0 
1


W

1

 
v 1

 v=1
v  1    




r 1
2
r 1


2


1 

 T  T0  
 1
 

 



v 1 

   2 
1   1  
2RT0
      W 
v 1
   


Method : 2 (a) d = – nCvdT  nCdT = – nCvdT  C 
R
r 1
v1 

1   2 




Ans (b) Also we know d =
nCvdT + PdV  – (nCvdT) = nCvdT + PdV  PdV = –2nCvdT. Also we know P 
nRT
R
dV  2 n
dT
V
r 1
r 1
TV 2
 (r  1) dV
dT
 

2
V
T


 r 1 

 ln V  ln KT .
 2 
r 1
1
  const   TV 2  const
K
Ans (c) dW  2nCv dT 
2 
2RT0 
r 1 


W

1

(

)
calculate T as method : 1 and put.
r 1 


Q: 2.50:
C
 PV 
PV
P = T -------(i)  T 
put -------(i)  P   

vR
 vR 

 P 

W 
R
R
R
R
R
C
 

 R (1  )


v 1
1 v 1
v 1 x 1
1
vR
1 
 1
T  vR (1  ) T
Here v = 1

2nR
 2nR
dT  W 
T
r 1
r 1

 
v R
x
P1  CV   P  C1V 1  PV  1  Const  Compare with PV = Const
(b)
r 1
V 2
Ans


KT 
nRT

V
PV 
x

 1
 R 
Ans
W  R (1  ) T
(a) dW v 
 dT 
 x 1
Ans
178
Compact ISC Physics (XII)
Q: 2.51: U = aV. We know that U 
Compare with PVx = const x = 1-
(a)
Given vCvT = U 
PV vRT

 Then
r 1 r  1
(b) C 
PV
 aV   PV1- = a(r-1) = const.
r 1
R
R
R
R
R
R




 C 
r 1 1    1 r  1 
r  1 x 1
Ans
 R R   r  1
vR
(r  1) U
T  U  T 
   vCT  v  r - 1     vR  T
r -1
vR



 1
 r  1
1
(  r  1) U
  U 1 

     r  1    (r  1) U 





  Ans   



 r  1
W    U  U 1 
  U 
 

W 
U(r  1)

Ans
dT
Q: 2.52: (a) T  T0 e V  dT  T0  V dV  dV 
 We know dQ = dU + dW ------(i)
T0e V
PV = vRT  PdV + VdP = vRdT ------(ii) Here T = T0eV 
PdV 
vRT0 e V
V
 dT 
vRdT

 
V  from (1)
 T0 e V 


PV
 T0 e V
vR
vCdT  vC v dt 
P
vRT0 V
e   Now
V
vRdT
R
 C  Cv 
V
V


Q: 2.53: P  P0  V  dP   2 dV (a) We know dQ = dU + PdV  nCdT = nCvdT + pdV----(i)
V
  dV 

dV  nRdT
 nRdT  PdV 
We know PV = nRT  PdV + VdP = nRdT  PdV  V 
2 
V
 V 
nR
 PdV + (P 0–P) dV = nRdT  P0 dV  nRdT  dV  P dT put in (i) 
0


R 
 R
R
 
  nRdT
C
  P0  
C
 R 1 
nCdT  nC v dT   P0  


 vP 
r 1 
V  P0
r 1
V  P0

0

rR
R
C

Ans (b) P  P0 e V  dP  P0 e V dV  dQ = dU + dW  nCdT = nCvdT
r  1 VP0
V


+ pdV ----(i)  PV = nRT  PdV + vdP = nRdT  PdV  V P0 e dV   nRdT 


nRdT
PdV  V P0 e V dV  dV  nRdT  PdV + VPdV = nRdT  PdV 
put in (i) 


1  V
nCdT  nC v dT 
nRdT
R
R
R

 C  Cv 
 C
1  V
1  V
r  1 1  V
179
(b)


PV = RT   P0  V  V  RT


P0 V  
 T
R
P
R
U  C v T 
T  U  0 V2  V1
r 1
r 1

V
 W  P0 V2  V1    ln 2 V1
P V  V1 r
V
Q  0 2
  ln 2
r 1
V1


C  Cv 
(b)
P0
P
V  0 V2  V1
R
R


P0
V2  V1   P0 V2  V1    ln V2
r 1
V1

Ans
dT
 Also dQ = dU + dW  vCdT = vCvdT + PdV

PdT
P
P RT
RT

 C  Cv 
----(i)  PV = vRT
T 
 C  C v 


v
v
V
V
RT0
RT0
RT
R
rR RT0
T0  V  C 

R  C

 C  C p 
V
r  1 V
r  1 V
V


V2 


W

  P0   dV
Ans  dW = PdV 
V
V1 
Q  U  W 
Q: 2.54: T = T0 + V (a) dT  dV  dV 
 vCdT = vCvdT +
 T 

Ans

RT0 
T = T0 + V  dT = dV  dQ = vCdT  Here v = 1  dQ = cdT   C p  V  dT


V2
V2
RT0

RT0 
dV  Q  C V  V   RT ln V2
 dV  Q  C p  dV  
  C p 
p 2
1
0
V1
V 

V1
V1 V
Ans
Q: 2.55: (a) C = Cv + T  dQ  vCdT  v C v  T  dT  dQ  vCdT  v  TdT ----(i) Also dQ
= dU + PdV -----(ii)  Compare (i) and (ii): PdV = v TdT  P 
vRT
vRT
dV  vTdT
put 
V
V
 
T
V
dV 

 e R  Ve T R  C  const
  dT  ln V  T  ln C1 
1
C1
V
R
R
(b)
C = CV + V  dQ = vCdT  dQ = vCvdT + vVdT  Compare with dQ = vCvdT + PdV
PdV = vVdT 
R
Ans
vRT
dV  dT
vRT
1 
dV  vVdT  
 R
dV  v VdT 
   R ln C1T
2
V
T
V
V
V
R
 C T  e V  Te V  const
1
Ans
180
(c)
Compact ISC Physics (XII)
C = Cv + ap  PdV  vapdT   dV  va  dT  r = vaT + C1  For 1 mole gas : v = 1 
V = aT + C1  V-aT = C1 = const
T0
Q : 2.56: (a) from (i) :
 W   ln  
R
T0 (n  1)
r 1
pdV  C v dT  PdV 


 RT
 C v  dT 
dV
T
V


C v ln T 
T0

dT  C v dT
T0 T
T0
W   PdV  
(b) C 
W   ln   C vT0 (  1)


 dQ  vCdT  v dT . Compare with dQ = vCvdT +
T
T
RT

dT -----(ii)  P 
V
T


 (C v   T) dT   R 
put i n (i)  C v dT 
dV
V


RT

dV  dT 
V
T
C v dT
dT
dV

  R
2
T
V
T



C
R
 R ln KV  ln  T v  KV   
 T C v KV R  e  T  T C v V R e  T  const


T
T
C v R
R
C

 PV  v R R
 ( r 1)
Cv
PVCv
PV


V
e

const

PV
e
 Const 
 
r PV  const

PV
e
R


Ans

a 
Q: 2.57: We know vandervall gas equation :  P 
(v  b)  RT ----- (i) W   dW   PdV from
V 2 

V2 
V2  b  1
1 
RT
a
RT
a 
a
  Ans

dV  W  RT ln
(1) : P  v  b  2  Then W   


V1  b
2
V
 V2 V1 
V1  v  b V 
Q: 2.58: (a) Internal energy of one mode: U  C v T 
temp. T U1  C v T 
a
a
 U 2  C vT 
V1
V2
a
VM
where VM = volume of 1 mole gas at
 1
a
a
1 
 U  U 2  U1  V  V  a  V  V 
1
2
2
 1
Ans
2 


a 
RT
a
 P  n a  (V  nb)  nRT
(b) We know 
 n  1   P  2  ( v  b)  RT  P  v  b  2
2 
V
V 
V



v2
v2
 1
v2  b
RT
a
1
 W  dW   v  b dv   2 dv  RT ln v  b  a  v  v  .
Now
Q  dW  U 
 2
1
1 
v1
v1 v
RT ln
 1
1
v2  b
1
1 
v b
 a     a     Q  RT ln 2
Ans
v1  b
v1  v
 v 2 v1 
 v1 v 2 
181

a 
a
a
Q: 2.59: (a)  P  2  (v  b)  RT -----(i)  U  C v T  ----------------(ii)  dU  C v dT  2 dV .
v
V 
v

 RT 
For adiabatic process : dQ = 0 = dU + pdV  -dU = PdV  C v dT  dV 

V  b
C dT
dV
C
 v

 v ln TK  ln( V  b)  V  b  ( TK ) C v R  V  b  T C v R  K C v R 
RT
Vb
R
V  b  T Cv
R
 K C v R  T ( V  b ) R C v  Const
Ans

dQp = CvdT +PdV  CpdT = CvdT + PdV ------(i)   P 




a
a 
differiate w.r.t. V:  0  3 dV  (v  b)  P  2  dV  RdT 
V
V 



(b)
 adV

V
2

abdV
V
3
 PdV 
v
C p dT  C v dT  RdT 
P
a
ab
V
3
2
dV
dV  RdT

dV
ab
dV R
ab
R

 
3
dT
dT P V 3 P
V
Cp  C v  R 
2abR
3
V P
 R
PdV  RdT 

Cp  C v  R 

ab dV
V 3 dT
Cp  C v  R 
a 
RT
a
 (V  b)  RT  P 

2
V  b v2
v 
 ( v  b)
a
adV  PdV 
dV  RdT
3
V
V2
abdV
V3
put in (i) ---------(ii) 
-------------(iii)
from
(ii)
ab  R
ab 
abR Rab

 R

3P
3 
V 
V P
V 3P V 3P
:

2abR
R
C  Cv 
a   p
1 2a ( V b ) 2

V 

2
V

b
RTV 3
V 

3  RT
Q: 2.60: We know : dQ = dU + dW  Here dW = 0  dQ = 0  Then dU = 0  Ui = Uf  
vC v T1
 1


1
 v2
 v 2a
v 2a
 vC v T2 
 C v T2  T1   va  v  v  v   C v T  va  v ( v  v ) 
v1
v 2  v1
2
1 
2 
 1
 1 1
vav 2
vav 2 (r  1)
 T  C v (v  v )  T  Rv (v  v )
v 1 1
2
1 1
2
Value
v1
v2
Ans
v
vaccum
v 2a
v2a
Q: 2.61: We know : Q = U + W Calculation of U : U1  vC v T 
 U 2  vC v T 
V1
V2
1 
2 1
 U  v a  v  v  . Calculation of W : w   PdV  zero . Because work done against Vaccum
 1
2 
1 
 1
  v 2  v1 
is zero Q = U + W  Q  v a  v  v  Ans  Q  v a  v v 
 1 2 
2 
 1
Ans
v
T  const
v1  v 2
182
Compact ISC Physics (XII)
2.3 Kinetic theory of gases, Boltzmann’s law and maxwell’s distribution
Q: 2.62: P = 4 × 10-15 atm at room temp. We know
: P = nKT. Where n = no. of gas particles/volume 
P
4 10 15  1.01 10 5
n

/ m 3  110 5 / cm 3
8 .3
KT

300
23
P = 4×10–15 atm at room temp.
N2
.
6.02310
Since in 1 cm3 volume no. of molecules are = 105 one molecules occupied volume is = 10-5 cm3. This
volume will be like cube then side of cube is :
1
l   v
1
3
 10
5 3
 1 
 l   10 106  cm  10 3   102 cm  l = 2 × 10-2 cm = 0.2 mm Ans




Q: 2.63: Suppose initial no. of molecules of gas is v1 then PV = v1RT 
  10  5 


1
3
cm ,
3
v1 = PV/RT. no. of (molecules + atoms) in gas finally : (1  )
PV
PV
 2
.
RT
RT
Due to disassociation, no of particle is increase by 2 times. v f  (  1)
N2
PV
,
RT
V, m, T
(  1) PV
RT  p ƒ  (  1) P -----(i). Calculation of initial pressure (P):
RT
mRT
(  1) mRT
RT  P 
Pf 
put
in
(i)

Ans
MV
MV
p ƒ V  Vƒ RT 
m
PV   
M
Q: 2.64: Suppose no. of molecules per unit volume of He is n 1 and that of N2 is n2. Also molar mass
of H1 and N2 are M1 and M2. Then We know density of mixture = Total mass/Total volume
m1  m 2
 1   2    n1M1  n 2 M 2 ------(i) Also : P = nKT P = (n + n ) KT -------(ii) from
1
2
v

PM 2 
P


P
  



KT
KT
M
M
KT
 
2
2
 P

n 
n  
M1
(i) and (ii):   n1M1   KT  n1  M 2  1
 1
Ans
M1
M1  M 2
1
1


M2
M2
Q: 2.65: momentum transfer is one callision = 2mvcos  no.
v, n

of molecules collides per second =  (VA) 

 dP 
   V (2 mv Cos ) A  F = 2nmV2ACos  F/A = P =
 dt  net
2nmv2Cos   Ans
Q: 2.66: We know Vs  Vsound 
to contribute energy.
2
i

wall


P
v s2 P
2



 1
where i = modes of degree of freedom

P
i
v 2
 s 1 
P
i
2
v s2
1
P
put value i = 5 Ans
183
v
Q : 2.67:
v
v rms
1 2i

3

RT
M
(2  i )
3i
vrms 

3RT
M

v
r

vrms
3
Ans (a) Monoatomic : i = 3 
Vibrational mode will be not active. i  5
v

vrms
  1
 Here
v
5

vrms
9
25

3x5
7
15
2
2
1
i
i

Ans (b) Rigid diatomic :
Ans
i
KT . i = no. of translational digree of freedom + no. of rotational
2
DoF +2 (no. of vibration DoF). Linear N-atomic molecules : i = 3+2+2 (3N-5) = 6N - 5 
Q: 2.68: Kinetic energy (K.E.) =
 6N  5 
 KT  3N  5 2 KT . Network (non linear atomic) i = 3 + 3 + 2 (3N - 6) = 6N + 6 KE  
2


1
12 = 6N - 6  K.E.  (6 N  5) KT = (3N-3) KT = 3(N-1) KT Ans
2
R
i
2
 R    1 
Q: 2.69: Molar heat capacity (C) at constant volume : C  C v 
r 1 2
i
2 9
7
r  1 
(a) Diatomic gas : i = 3 + 2 +2 (3 x 2 - 5) = 7  C  R Ans
Ans
7 7
2
2
6N  3
6N  5

R  r  1 
(b) Linear-N-atomic : i = 3 + 2 + 2 (3N - 5) = 6N - 9  C 
6 N  5 6N  5
2
6N  5
R  3( N  1) 
(c) Non linear-N-atomic : i = 3 + 3 + 2 (3N - 6) = 6N - 9  C 
2

  1

2
6 N  4 3N  2 N  2 / 3



6 N  6 6 N  6 3N  3
N 1
Q: 2.70: For isobaric process :
Ratio 
dQ = nC pdT  dW = nRdT 
N
P  const
dW R
6N  5

dQ C p  C p  R  C v from Q 2.69: C v  2 R (linear ) 
Cp  R 
6N  5
R
2

Rx 2
2
1
(6 N  3)R 3


 (2 N  1) R  Ratio 
3
(
2
N

1
)
R
3
(
2
N

1
)
(
3
N

3 / 2)
2
2
R
1
Ans For Cv = 3(N-1)R (For nonlinear) Cp = R + 3 (N - 1) R = (3N - 2) R  Ratio  (3N  2)  3N  2
Q : 2.71: C v = 0.65  C p = 0. 91   
Cp
Cv

0.91
2
 1
0.65
8

2 0.91
0.26

1 
i
.65
0.65

0.65 2 x 65

 5 . Calcalation of molar mass : We know R = 8.3 joule/mole K. C - C = 0.91 p
v
0.26
26
0.65 = 0.26 J/gK  For 1 gm, value of Cp - Cv  is 0.26 For 1 mole, value of Cp - Cv  is 0.26
M where M is molar mass then 0.26 M = 8.3  M = 32 Ans
i  2
184
Compact ISC Physics (XII)
Q: 2.72: (a) Cp = R + Cv Cv = Cp - R =
 Cp 
V 
i


R  i  2  R  1 Ans (b) PT = const P  nR   const
2
 


R
1
R
R
R
 Pv 2  const  Pv x  const  x = 1/2  Also C    1  x  1    1  C  x  1 
1
C
1
C


1 
1
C
1

2
 
R
x
 1  i  2  
  1  1
R x  1  Ans
 1 R x 1

i
3
5
v1 R  v 2  R
2(v1  v2 )
3v1R  5v 2 R
R
2
v 1 

Q: 2.73: C v (mix)  2
 C v (mix) 

3v1  5v2
2 (v1  v2 )
r 1
v1  v 2
2v1  2v2
5v1  7v2
 v  3v  5v  1  v  3v  5v
1
2
1
2
P
Q: 2.74: We know
T 
Ans
RT
R
1
T -----(i) Calculation of T:: C v T  mv 2 
 P 
M
M
2
1 mv2
P  1 mv2 

 put in (i): P 
2 nC v
M  2 nCv 


P
 v2M
P
M v2




P
i RT
P
i RT
Q: 2.75: (a)
one drop
4 d
m     
3  2
Q: 2.76: Adiabatic process V1 
N2
K.E. 
t  27 0 C  300K
3

1
KT (b) Vrms 
2
4 d3
x
3
8

3RT

M
d 3
 Vrms 
6
3RT1
V1

 V2 
M

T1  T2 2 -----(i) Also we know PVv = const 
vRT
V
3RT
M
3KT 6
d 3




v 1
V
 n2   2
 V1

2
1 1
i




V   const 
2
V2
V i
i
  2   n 2  V  n
1
 V1 
Where m = mass of
Ans
2KT
 Vrms  3 d 3
3RT2
divide both equation :  
M
v 1
v 1 from (i) :
T2 2 V1v 1  T2 V2v 1 
TV v 1  const  T1V1  T2 V2
V
 2
 V1

v  100 m / s
Ans
3RT
Ans
M
Vrms 
1 R v2
v2

2 i / 2R
i
T1
T2
185
Q: 2.77:
3RT1
 Vrms 
M
Vrms 
3RT2
M
m
Q    C v T -----(ii)  from (i) :
M
Q 
1
Here T1  T    
T1
2
T2  T2   T1 ------(i)
 T2  T1  T    1  2  T1   2  1 T put in (ii) :




m
m i
C v   2  1 T   Q 
R   2  1  T
M


M 2 

m, N 2 , T
Ans

1
1
2
2
I yy Wyy

I zz Wzz
2
2
= Wzz = Wsq  Iyy = Izz = I.
y
Q: 2.78: Rotational kinetic energy : K.E. 
x
By symmetry it is assumed that Wyy
1
1
2
2
2
K.E. 
IWsq
 I Wsq
 IWsq
------(i). A/c to equipartition law of
2
2
energy : K.E. 
1
KT
2

We know PVv = const 
TV v 1  const

vRT v
V  const
V
V 
T0 V v 1  T  
  
Diatomic
KT
I
Ans
Adiabatic
v 1

V, T0
V/, T
T  T0 v 1 ----(i). Mean kinetic energy of rotational motion : K.E. 
motion  K.E. 
z
(no. of degree of fredom) no. of degree of freedom of rotation motion of diatomic
1
2
KT (2)  KT -----(i) from (1) and (2) : KE  IWsq
 Wsq 
2
gas = 2  K.E. 
Q: 2.79:
T
1
(i) KT i = 2  Rotational
2
2
1
1 1
(2) KT0v 1 Ans 
 K.E.  KT0 2 / i
i
K
.
E
.

KT

2
0
Ans
1
nv
4
= const 
Q: 2.80: We know no. of collision per second per unit area of wall : v 
Where n = no. of molecules per unit volume. We know TVv-1
Ti V v 1  Tf (V )v 1
2
1 1
i
1
 Tƒ  Ti  

1 N
8RT
 v1  4  V 
M
 
v 
1
 Tƒ  Ti  

diatomic gas
2/i
-----(i) 
8RTi
N 8RTi
1 N 

v2  


M
4V M
4  V 
8RTƒ
M
1i
v2

v1
1

1
 
n
2/i

v2
1

v1

1/ i
1
 
 
 
1 i
v2
1
  

1
i
 v1

 i
Ans
v2
1
 v  
1
Tƒ
Ti 
186
Compact ISC Physics (XII)
Q : 2.81: We know
1
3

 2  3x  3
x 1
2
Ti V
2/3
Vƒ 
 Tƒ ( v)
1 N 


4  V 
2/3
C
R
R
R
R
5
1


  R  i 
 i  5   1 
v 1 x 1
2
x 1
2
x 1
PV 5/3
 1 
 Tƒ  Ti  
 
 
8RTƒ
M
 Now
Vƒ
Vi
=
const
vRT 5 / 3
V
 const TV 2 / 3  const
V

2/3


1

Also
Tf
1

Ti

we
 1 
 
 
 

know
2/3

1/ 3
1


 1 
 
 
 
Vi 
 1 
 
 
 
1N
 
4  V 
4/3

8RTi
M
Vƒ

1
  

Vi

4/3
diatomic gas
( i 1)
1
 
 
 
i 2
1
  

51
5 2
4
1 3
   
 
V
V
T0
Tƒ
for diatomic gas
1 N
Q: 2.82: No. of collision per second per unit area : v  4  V 
 
8RT
N
v

M
4
8RT  1 1 / 2 
v T 

M 
1/ 2
1  PV 
 const  P1 / 2 V 11 / 2  const PV-1 = const
Since v = const  V T = const  V 

 vR 
R
R
R
R
Ri R R

  (1  i)
Compare PVx = const  C 
 x = -1  C  i 
 C 
v 1 x  1
2
1 1
2 2 2
-1
Q: 2.83: Vp 
 vp 
1/2
1g
3
RT P
2RT

 We know PM = RT 
put P = 1.1 × 105 N/M    1  1 kg / m
M 
M
2P
v 
 
8RT
 v
M
8P

Ans  Vrms 
2
v
dN
4
e  u u 2 du Here u 
Q: 2.84: d (ƒ( u ))  N (u ) 
And
v

p
3RT
M
3P

 Vrms 
Ans
dN (u )
 sum of fraction of gas
N
 v 
V  Vp
V


 1  
      
molecules having ratio  V  between u to u + du (a)     V
V
p
p
p


Then sum of fraction of molecules ƒ(u ) 
1 n 4

1n 
2
e  u u 2 du  ƒ( u ) 
8

e n
Ans
187
Here    
(b)
V  Vs
V
   1    
 1  n Also Vs 
Vs
Vs
Vs 3
3
  Vs  Vp
Vp 2
2
V
 1    V
p
integration : ƒ(u )  12
3 3 / 2
e

2
Q:
V
 3  2
2KT
m
Vp 
(a )
2.85:
m

K
T  T 
3
V
(1  n ) 

2
Vp
Vp 
3
(1  n )
2
2RT
M

from (a)
Ans
3KT
m
Vrms 
 Vrms  Vp  V 
3KT
2 KT

m
m

mV 2

K 3 2

2
Ans (b) F (u) will be maximum at maximum probable
2 KT
mv 2
 T 
Ans
m
2K
velocity. Hence Vp  v 
Q: 2.86: (a)

2
 1   
3
3RT
and
M
 m 

F(u)  
 2KT 
We know
3/ 2
 mv 2
e
2 KT 4v 2
. Let temp T at which for v1 and v2,
 mv12
3/ 2
3/ 2
 mv 2
1


2 KT 4v 2   m 
2 KT 4 v 2 . Here range will be taken
e
e
1 
2

 2KT 
 mv12 mv 22
m v 22  v12 
2



v
same for both e 2 KT 2 KT  2 2  T 
Ans
v2
4K ln v
v
 m 

F(u) are same. 
 2 KT 
1
(b)
 m 


 2 KT 
mv 22
3/ 2
Q: 2.87: v p 
e
2RT

M

1
 1


2K  mN
m0

v

T
2T0K


m

4 v 2  
 2 KT 
0

2KT
N
 vp 
m




v 2 m N

m N 
2 K 1 

m0 


1
2
Ans
T 
 mv 2
3/ 2
e
2KT
o
 v p 
mN
T 
v 2
2K
mN
m
0 
2 T0K
4 v 2  v 
3KT0  ln 
m
 1
2KT
v pN  vpo  v 
m 0 
m0
mN
 
T
v 2
2K
m
2KT

mN
m Nm0
0 
mN
Ans
2 KT
m0

2

188
Compact ISC Physics (XII)
 mH

 2KT

Q: 2.88:
 m
 H
 mHe





3/ 2

e




m Hv 2
2KT
3/ 2
e

 e
3KT ln  m 2 m 
1

m 2  m1
v

T 1 
3/ 2
3
m

e
m He 2
v
2 KT
 m H  m He
2 KT
2
H
 2 ln m  v 
He

3/ 2


 Here mH = m2  mHe = m1 

m 2
v
2
KT
e
4 v 2 Here v = const but T is variable then for f(u) maximum
3/ 2 
2
 3 / 2  2 KT v 
2
T
e

 4v




m
mv 2 1 
mv 2 1
2
T

T  3

mv

2

 (1)T  e 2K

 T 5 / 2
e 2K

 2K 


mv 2
3
K
T
mv 2
3K
 2 
2
e
 mVx 2
occupied this volume
3
2 kT
 T 7 / 2 

mv2 3 5 / 2
 T

2K
2
dN
 ƒ( Vx ) ƒ( V ) ƒ(V ) dV  dV  2V dV dVx
N
Probability distribution function
1
d ƒ( u )
0
dT

Ans
2.90: Volume of differential region
 m 
f (Vx )  

 2 kT 
4 v 2 

 m 

ƒ(u )  
 2K 

ﾞ
0T
3 / 2
m Hev 2
2KT
 m He 

4 v  
 2KT 


2
Ans
 m 

Q: 2.89: f (u )  
 2KT 
d ƒ( u )
 0
dT
mH 2
v
2 KT
ddV
V
fraction of molecules
Vx
dN
 ƒ(Vx ) f (V ) f (V1) dV 
N
dN  m  2 mv 2 2 kT

2V1dV dVx
 e
N  2kT 
3
dVx


dN  m  2  Vx 2  V  y 2  V 2 z

dv 
 e
N  2kT 
dV
V
dVx
189

VxdN
2.91: We know Mean velocity =  Vx   

 N
 m 

 
 2 kT 
1

Vx N

m 2

V
m 1 / 2 e 2 kT x
2 kT
 
dx
N

m
2 
KT   2 kT Vx

 e
 0 Mean speed = <|V |> 
x
 m 

2
1

m
2
 m  2  2kT Vx
1
2
2  Vx N 
dVx
 e
 m  2 KT  m Vx 2 kT 
2

kT

2
e





0
m
0
| Vx | 
 2kT 
N
| Vx |  2
m
k 2T 2


2 kT
m2
4 kT

2 m
1

 m 
 | Vx | N  2kT 


Note : | Vx |  
N
2

e
m 2
Vx
2kT
dVx
Q : 2.92:
We
know
2
 Vx  
Met hod

N
:
2
 V 2 x    Vy 2    Vz 2  

 m 

 2  V x 
 2 kT 
0
2
 Vx 2  
1
 m 

2  Vx 
 2 kT 
0
=

Ans


2  V2 x dN
2 kT
m
| Vx | 
1
2

e
3RT

M
2
m 2
Vx
2 kT
dVx
3KT
m


e
m 2
Vx
2 kT
dVx
2
  Vx  
 Vx 2  
KT
 V2 
2
 V x 
Ans
m
3
KT
m
3KT
m
Ans
Al so
Vx
dA
Q:2.93: no. of molecules/Volume = n. Fraction of molecules having velocity
dN
Vx to Vx + dVx is :
no. of molecules/volume having velocity Vx to
N
 dN 
  no. of molecules approach per second toward wall with velocity Vx : =
Vx + dVx is = n 
 N 
 dN 
 dN 
n
 (volume travrse in one second with velocity Vx)= n 
 (V×dA). Total no. of molecules
 N 
 N 
approach per second, per unit area of wall =
v
n
4
8kT n
 V
m
4
Ans
 dN 
m 

 nVx  N   v   nVx 
 2kT 
3
2
mVx 2
e 2 kT dVx 
190
Compact ISC Physics (XII)
Vx
Q:2.94: No. of molecules per unit volume = n 
Momentum transfer in one collision due to one molecule = 2mVx  Fraction
of molecules having velocity Vx is
dA
dN
 no. of molecules per unit volume
N
 dN 
  no. of molecules collide per second with velocity Vx
having velocity Vx to Vx + dVx = n 
 N 
 dN 
 dN 
 (volume travel in one second) = n 
 Vx dA  Momentum transfer with Vx velocity
is : n 
N


 N 
 dN 
 Vx (dA ) ( 2m Vx )  Net moment transfer with walll per second
per second n 
 N 

dF 
 dN 
dF 
 m 
  2nmVx2 

 n  N  Vx (dA) (2mVx)  p 
dA


 2kT 



Q:2.95:

1

v
Q :2.96:
v
1
 
v
 V1 (dN)
0

=
N
2m
 4  v 
KT
We know :

 dN 
 V1  N 

0 

=
0
3
1 m 


V 2 kT

1
2
m
e 2 kT
2  m v2
e
2 KT

Vx 2
dVx  P = nKT Ans
4v 2 dv 
Ans
dN  m 


N  2kT 
2
d
d  mvdv vdv 
m
m
3
2
mv 2
e 2 KT 4v 2 dv

 K.E.   
dN  m 


N  2kT 

 32
1
mv 2 differentiate :
2


2 d
e KT 4
m m



3
dN
 2 kT  2 e KT
N
 d . For maximum proble value of K.E. at which no. of molecules will be

maximum f  2  kT  3 2 e  KT
speed :

df
1
 0   p r  KT . K.E. corresponding to most probable
1
  d
2
1  2KT 
m
  KT  Ep1r
2  m 


d
3
3
dN
Q:2.97: from Q. 2.96
 d  En . And   KT
 2  kT  2 e KT  d ----- (1)  n 

2
N
 32 3

3
2
2
dN
 2 
 and KT 
 2   
e
  n

put in
(1)


KT 2
3
N
 3 
dN
 2 
 2  
N
 3 
32
e
32
n 

dN
 3 6 e
N
3
2
n
Ans
191

dN
3
KT
dN
 2  KT  2 e
 d where
= fraction of molecules which kinetic
N
N
energy lies between  ––– +d. Now we want sum of fraction of molecules whose kinetic energy 
Q:2.98: from Q. 2.96

dN

> 0. Then  N   2  KT 
0
3

2
e
KT
 d 
dN
2
 
N KT  3 2
   KT
e
 d
Ans
0
2
Q:2.99: (a) F  Av 3 e  mv / 2 kT . Probable velocity is that velocity at which no. of molecules
will be maximum. And for maximum no. of molecules F will be maximum and hence.
 v

1
3kT
2
(b)   mv  v 
2
m
dN
 2 
 A 
N
m
3
2
e
 
kT
hole
dF
0
dv
2
dN
2
 Av 3e mv / 2 kT dv
----(i). And d = mvdv -----(ii). Then
N
m
d
 2 
 F' ()  A  
2m
m
3
2
 1 


 2me 
1
2  
e kT
 For F'() maximum :
dF' ()
 0   KT Ans
d
Q: 2.100:
no. of molecules making solid angle d on centre. dv'  (dN)
d  sin d(d)  formula 
d
 formula and
4
no. of collision persecond at anlge  on uni t area:
 dN

d  ( v cos )
dv  (dv' ) ( volume)  (dv' ) ( v cos ) 1  dv  
 4

3


1
 m  2  mv2 / KT
  N
4v 2 dv
v cos  d =
 e
  2kT 
 4


3

2
 m  2
 mv 2 / KT 3
N
e
v
cos

(
dv
)
sin

d



 d
 2 kT  v o
0
y
dv  2KT 


 
N  m 
1
2
dA
sin  cos d

V
dV

d
x
192
Compact ISC Physics (XII)

2
Q: 2.101: Smilar like Q:2.100 dv   (dn )
0
(d )
v cos 
4
3


2


2
2
m
1


 mv / 2 KT
4v 2 dv 
v cos  sin   d
 e
 dv   N 
  2KT 
 4
0




3

dv  m 


N  2KT 
3
2
e
 mv 2 / 2 KT 3
/ 2
dv
 m 
v dv  sin  cos d 


N
 2KT 
 0
2
Q: 2.102: Suppose potential energy of level (1) is zero. then U0 = 0 and n0
= n 
According to botlz mann’s formula n  n e
0
 ( v v)
KT
2n  n e
 ln 2 
KT
ln 2
h
T
n
level 1
2n
level 2
h

KT
U
 U = KT ln2. Since filed is uniform hence force will be uniform
KT
and work done by force = - (Uf - U0)
F
(u u0 )
2
e  mv / 2 KT v 3dv
KT
ln 2  Magnitude of force :
Fh   U f   U   F 
h
Ans
3
Q: 2.103: On droplet, force is arised due to mass changed and hence F  mg 
from Q.No. 2.102 F 
 n f  n 0 e
 ln n 
 Uf  Ui
KT
4 d
d 3
   
g
3 2
6
6RT ln n
( d 3p) hg
R
KT
d3pg

ln n 
 K
 N A  3
6T ln n
NA
d  phg
h
6
Ans
Here Uf = mgh  U f  mgh Here n0 = n nf  n f  nn f e  mgh / KT
mgh  d 3 gh 



KT
6


 NA

 RT
6RT ln n

  N A 
d 3gh

Q:2.104: Also n1 '  n1 e M1gh / KT  n 2 '  n 2 e  M 2 gh / KT 
n1' n1 (M 2  M1 ) gh / KT
n

e

 e ( M2  M1 ) gh / KT where
n2 ' n2
no
M1 is mass of H2 molecule M2 is mass of N2 molecule
N2
H2
n1
n1 '
n1
 n0
n2
n2
h
n1'
n2 '
n 21

193
Q:2.105: Also n1 '  n1 e  m1gh / KT  n 2 '  n 2 e  m 2gh / KT  Here
n1 '  n 2 '

n1 e  m1gh / KT  n 2 e  m 2gh / KT
Then
n1

n2
m1, m2
T, g
h
n2 > n 1
n2
n
KT ln n1
ln 1  (m1  m 2 ) gh / KT  h  KT ln
 h
n2
g (m1  m 2 )
g(m 2  m1 )
Q:2.106: We know P = nKT
n1
n2
n = density of molecules
n f  n 0 e  mgh / KT 
Pf
Pi mgh / KT

e

KT KT
Pf  Pi e  mgh / KT 
At temp T = T0  Pf '  Pi e  mgh / KT 0 
At temp T = T0  Pf "  Pi e
Pf "
m2 > m1
 mgh / KnT 0
T
h

1 
 mgh / KTo  1
   . At bottom h = 0 Pf'' = Pf' = not change Ans
Pf '  e
 (dm )gx where m = mass of one molecule  n = no.
Q:2.107:  U 
0
N
of molecules per unit volume. Where N = Total no. of molecules  dm =
m 0 (Adx) n
=
 P 
m0A 
 dx
 KT 
dx
T
x
mgh

m0A

P0e KT Adx  
KT
A

m gh / KT
dx
x e 0
moA

mgh
/
KT
P0 e
gxdx  m g 0
e  mgh / KT

o 
P
KT
N   nAdx   0
Adx Then  U  
m gh / KT
P
dx

mgh
/
KT
0
KT
x e 0
Adx
 KT e
0
 <U> = KT.
Constant and not depends on type of molecule.
nf
mwl / KT
1
Q:2.108: Here effective accleration will be : geff = w  n f  n 0 e  mwl / KT  n  1  e
0
 n  1  e  mwl / KT  ln(n  1)  
mwl
KT(ln(1  n ))
 W
KT
ml
l
(1)
Ar
T
geff
(2)
w
KT ln(1  n)
KTn
Magnitude of acceleration : W  
 Also we know ln (1+x)  x  W 
Ans
ml
ml
194
Compact ISC Physics (XII)
w
Q:2.109: Excess force on particle is arised due to change in
density. Then F = (dm) rw2 where
m
= Volume of particle

w
r2
r1
m
      0  rw 2 m = mass of one molecules. Then po
2
tential energy U 
M (  0 ) r w
2
2
 n  n 0 e m ( 0 ) r
2
dr2
2
w / 2KT

dr1
n ( r2 )
ne
n ( r1 )
m ( 0 ) w 2 2 2
r2  r1
2KT



m(  0 ) w 2  r22  r12 
2KT ln n
N A 2KT ln n

  m
 M  NA m 
ln n 
2KT
(  0 ) w 2  r2 2  r12 
(  0 ) w 2  r2 2  r12 




2RT ln n

(   0 ) w 2  r2 2  r12 


w
Q: 2.110: Centrifugal force on particle of mass m. Where m is mass of one molecule of
T, l
1 2
2
CO 2 . Then Potential energy: F  mrw and P.E.  mr / 2w   Then
2
2
n ƒ  n 0e mr
ln n 
2
/ 2 w 2 / KT

m2w 2
 w
2KT
nf
n0
 n  e ml
2(ln n ) KT
ml
2
2
2
/ 2 w / KT
w
= e ml 2 / 2 w 2 / KT 
2KT ln n
ml
2

2KT ln n
Ml 2
2
Q:2.111: (a) n ƒ  n 0 e ar / KT  where n0 = no. of molecules/volume at
T
0
n0
centre. n ƒ = no. of molecules/volume at distance r from centre.
2
dN  n 0 e ar / KT 4r 2 dr
r2
r1
2
U = ar
Then dN  n ƒ (4r 2dr)
(b) At most probable distance, temp = const. then for maximum dN.
df
2
f  e  ar / KT r 2 will be maximum then dr  0  r 
KT
(c) Fraction of molecules in region r  r+dr..
a
2
n e ar / KT 4r 2 dr
dN
3
 0
dN  a  2
N 
2 ar 2 / KT
2
2 ar / KT

dr Ans (d) Compare Ans (a) and
 4r e


n
4

r
e
dr
0

N  KT 
0
 a 

Ans (c) n 0  N 
 KT 
by n 3 / 2 Ans
3
2

Tin
If Temp decreases n times Tƒ 
 Then concentration increases

195
dU = 2ardr   r 
Q:2.112: (a) U = ar2
 dN  2 n 0 a
3
ƒa
2.4
2
1
3
U 2 e
U1 / 2 e
2
U
KT
U
KT du
Ans
will be maximum then
U
a
dr 
 U  dU
dU
e  U / KT
then  dN  n 0 4   
2ar
 a  2 aU
(b) For most probably value of U at constant temp.
df
1
 0  U  KT Ans
dU
2
The second law of thermodynamics, entropy
T2
T2
Q:2.113: n  1 
  If T1 increased by T then new efficiency. n ƒ  1 
=
T1
T1  T
T2 T2
T2  T
T (1  T) 1
1 2
= 1  T  2 T  If T 2 by T then  n"ƒ  1  T
=
T1
T1
1 T1
1
1
T1
W
T2 T
T2 T T
T2

---(ii) from (i) and (ii) : T  T  T  1 because T2  T1
T1 T1
1
1
1
T2
Then efficiency of second case will be more than that of first nƒ” > nƒ’ Ans
nRT r
V  const  T V r-1 =
Q:2.114: (a) We know PV r = const. 
V
T1
 n r 1  efficiency () 
const  T1V0 r 1  T2 ( nv 0 ) r 1 
T2
r
T
1
RT 
1 2  1
 1  n 1 r (b)
r
P
PV  const    p  = const 
T1
r 1


1 r
P01 r
T1
r
P 
 0
 2 
efficiency ()  1 
T2
r
 T2
P1 r T r  const  
 T1
1 r
T2
1 2 r
T1
2
T1
3
1
4
T2
v0
T1
1 r
r

T
  21 r  2  2 r
T1

   1  2 (1 / r 1)
P
nv0
V
P0

T2
P0 /2
V
Ans
P
Q2
Q1  Q 2
Q
T
T
 1  1   1  1    1  1  
Q:2.115: (a) Engine :   Q
Q
T
T
2
2
2
2
T2
Q1
Refrigerater : n R 
nR 
Q1
W
1
1 n
1
1  n R  n
1 n
Q1
1
nR 

Q
Q 2  Q1
2 1 
Q
Ans
nR 
T1
1
T1
1 
T2
V
P
Q2
T2
T1
Q1
V
196
Compact ISC Physics (XII)
Q:2.116: Let initial pressure, volume and temp is P0, V0, T1 Process 1-2 : Q
= U + W  U = 0 (Isothermal process)  Q1 2
Here
P
T1
1
P0
V
 W  nRT1 ln 2
V1
2
T2
3
4
6
V2
 x  Q
12  nRT1 ln x  V2  xV1
V1
5
T3
V
V0
Process 3-4 : Q12  nRT1 ln x  Calculation of P1V1T :
P V
P
P
P2 V2  P0 V0  V2  0 0  x  0  P2  0  P V r  P V r  TV r 1  const.
x
2 2
3 3
P2
P2

1
 T1  r 1
T
r 1
r 1
T1V2
 T2 V3  V3  V2  
 xV0  1
 T2 
 T2
T 
V5  V4   2 
 T3 
1
r 1
T 
 x 2 V0  1 
 T2 
1
r 1
 T2 


T 
 3
1
r 1



1
r 1
 V4  xV3  x 2 V0  T1 
T 
 2
T 
 x 2 V0  1 
 T3 
V
Process 5-6 : Q 56  nRT3 ln 6 V  nRT3 ln
5
1
r 1

1
r 1
 T1V0 r 1  T3 V6 r 1  V6  V0  T1 
 
V0  T1 T 
 3
1
r 1
 T3 
1
r 1
x 2 V0  T1 T 
 3
1
r 1
 Q 56   URT3 ln x That
given to system : AQgiven  Q12  Q 34 = (nR lnx) (T1 + T2)
Work Done : W= Qnet = nR lnx (T1 + T2 - 2T3)  efficiency (n) 
  1
W
Q given

T1  T2  T3
T1  T2

2T3
Ans
T1  T2
Q:2.117: Suppose initial pressure, volume and temp. are P0, V0, T0 Process 1-2: Q1 2  vC v T2  T0  Here T2 > T0
Process 3-4:
Q 3 4  vC v (T4  T3 )
W  Q1 2  Q 3 4  vC v (T2  T0  T4  T3 )
Heat given : Q1 2  Q given  vC v T2  T9 
P
2
Process 2-3 Q = 0
Here T 3 > T 4 
.
3
P0
T0
1
4

V
T T
w
efficiency ()
 1 4 3
Q given
T2  T0
V0
nV0
197
Process -1: T0 V0 r 1  T4 (V0 ) r 1  T3  T2 1 r
Process 2-3:
T0 V0 r 1  T3 (V0 ) r 1  T3  T2 1r  put in (i) :   1 
T0  T2  1r  1  1r
T2  T0
   1  1 r Ans
P
Q:2.118: Process 1-2 : Q = 0 Here (T3 > T2)
Process 2-3 : Q 2 3  vC p T3  T1  
Process 4-1 : Q 4 1
nP2
vrR
T3  T2  Process 3-4 : Q = 0
r 1
P0
vrR
T0  T4  
 vC p T1  T4  
r 1
w  Q 2 3  Q 4 1 
3
2
4
1 T0
V
V0
vrR
T3  T2  T0  T4   Heat given : Q given  vrR T3  T2  
r 1
r 1
T  T2  T0  T4
T  T4
efficiency ()  3
 1 0
----------(i)
T3  T2
T3  T4
T3 r nP0 1 r  Tr 4 P0 1 r
Calculation of T 3 and T 2 :
r
1r
Again T4 P0 
r
 T4 P01 r  T2  T0 n
1

1 r
1
1
r
T4  T3 n
 T3 n r
1
T  T3 n
  1 0
r  put in (i) :
T3  T3 n
1
1
  1 n r
Ans
1
1
r
1
1
r
n
2
Q:2.119: Process 1-2: Q1 2  vC v (nT0  T0 )  vC v (n  1)T0
Process 2-3: Q 23  vC p (n 2T0  nT0 )  vC v n (n  1)T0
Process 3-4: Q 3 4  vC v (T4  n 2 T0 )
2
n T0
3
T0
P0
1
V0
4
V3
V
2
Q C v (n  1)T0  C v n( n  1) T0  C v (T4  n T0 )  C p (T0  T4 )

--(i)
W
C v ( n  1)T0  C v n ( n  1) T0
P4
T4
Relation b/w T4 and T0 : P  2
----(ii)
n T0
3
(n  r )
1  rn
(1) 
Q 41  vC v (T0  T4 ) Work done (w) Q1 2  Q 23  Q 34  Q 41 Heat given
(Q)  Q12  Q 23 efficiency () 
  1
1
1
r
P
nT0
Process-1:
1
 1

Ans
P4
nT

----(iii) from (ii) and (iii) T4 = n3 T0 put in (i)
P3
T0
198
Compact ISC Physics (XII)
Q :2.120: (a) T2  nT0 (Higher temp) T1  T3  T0 (isothermal process)
Q12  vC v (nT0  T0 )  vC v T0 (n  1)
Process 1-2:
Q13  vRT0 ln

v0
Process
P
2
3-1:
1
v 3  Work done (w) = Q12 + Q13  Heat Given (w) = Q12 
Q
w Q12  Q13

 1  13
Q
Q12
Q12
  1

R ln v 0 v
3
V
V0
----------(i)
C v (n  1)
3
T0 = const
v 
r 1
 T0 V3 r 1  n   3 
Using adiabatic process : T2 r0 r 1  T3 V3 r 1  nT2 r0
 v0 
r 1
v0
1
 v  r 1 put in
n
3
1
1
( r 1)
1
R (r  1)
 1
ln n 1 r
ln n Ans
(i)    1 
   1
ln 1 n r 1
( n 1)
n 1
R (n  1)
v
(b) Process 1-2: Q12  vRT0 ln 2 v1
Q 23  vC p (T3  T2 ) 
Process 2-3 :
vC p (nT0  T0 )
P
=
Q23
nT0
2
3
vC p T0 (n  1) . Work done (w) = Q12 + Q23 Heat given (Q) = Q23
T1  T2  T0
R ln v 2 v
w Q12  Q 23
Q
1
 1  12    1 
 effficienc y ()  Q  Q
Q 23
C p (n  1)
23
R (r  1) ln
= 1
v2
v1
vR (n  1)
v2
(r  1) ln
   1
v1
r (n  1)
T0
isothermal
T3  nT0
V
----- (i)
Calculation of v2/v1 : Take help of adiabtic process : Trr-1 = const  (nT0 ) ( v 3 ) r 1  T0 (V1 ) r 1  
1
1
T3 T2
n
1
 1  r 1
 1  r 1
V3  v1  
  Also V  V (isobanic)  v  v  nV2  V3 Then nV2  V1  

3
2
3
2
n
 
n
r
1
V2 1  1  r 1
(r  1)
ln n
ln n
 1  r 1
      1
ln  
 1
   1
Ans
V1 n  n 
r (n  1)  n 
n 1
n 1
Q:2.121: (a) Process 2-3: Q23  vRnT0 ln
V3
V2
Process 3-1: Q 31  vC v (T1  T3 )  vC v (T0  nT0 ) 
Work
done
(Q)  Q 23  Q 31
(Q)  Q 23  efficiency ()
  1
C v T0 (1  n )
RnTo ln
V3
V2
=
1

Heat
(1  n )
V3
2
isot
herm
al
3
Q 31
w
 1
Q
Q 23
=
n(r  1) ln
given
P

T2  T3  nT0
T1  T0
1
V
V2
----(i)
199
V3
Calculation of
1
r 1
 T1V1r 1  nT2 V2 r 1  T0 V3 r 1 
V2 : using adiabatic process: T2 V2
1
1
V3
 n r 1
V2
V2 n r 1  V3 
put in (i)    1 
(1  n )
1 n
(1  n )
nn r 1    1 
   1 
n ln n
n (r  1) ln
n ln n
P
(b) T2  T3  nT0  T1  T0
2
V
Process 2-3: Q23  vRnT0 ln 3 V2
isothermal
T2  T3  nT0
T1  T0
Q 31  vC v (T1  T3 )  vC v (T0  nT0 ) 
Process 3-1:
Work
done
(w )  Q 23  Q 31

Heat
3
1
V
given
C (1  n )
r (1  n )
Q 31
w
  1 v
 1
V3
(Q)  Q 23  efficiency ()  Q  1  Q 
Rn ln V
(r  1) ln V3 V ---------(i)
23
2
2
T0
V1
1
V3
V
Calculation of 3 V2 : using is isobaric process: V  nT  n  V1  n using adiabatic process:
3
0
T1 V1
r 1
 T2 V2
r 1
 V3 

 T0 
 V2 
r 1
V 
 nT0 V2 r 1   3 
 V2 
 n.n r 1  n r 
r(1  n ) r
(1  n )
(1  n )
   1  (r  1) n ln n r 1  1  n ln n    1 
n ln n
Q :2.122:
Process
2-3:
Q23  vRnT0 ln
V3
r
r 1
V3
 n r 1
V2
put in (i)
Ans
P
2
V2
isothermal
3

R 
Process 3-1:  Q 31  v  C v  x  1  T0  nT0  Work done


adiabatic
V3
V2
T3  T2  nT0
1
V
(w )  Q 23  Q31 Heat given (Q)  Q 23  Q 23
Rn ln
Q 23  Q1
w





Q given
Q 23
T1  T0
Polytropic
R 

R 

  Cv 
 Cv
 (1  n )
 (1  n )
x
1 
x

1



   1 
---(i)
Rn ln V3 V
Rn ln V3 V
 
2
2
1
Process: T1V1r 1  T2 V2 r 1   T V r 1  nT V r 1  V  n r 1 V ----(ii)  TV x 1  const 
0 1
0 2
1
2
1
nT0 V3
x 1
 T0 V1
x 1 
1
(n x 1 ) V3  n x 1 V2 
1
1

V3
n 1
 n x 1 x 1 put in (i) :   1 
n ln n
V2
200
Compact ISC Physics (XII)
P
Q: 2.123: (a) Process 2-3: Q 23  vC p (T3  T2 )
2
nP0
Process 3-1: Q 31  vC v (T1  T3 )  Work done
(w )  Q 23  Q31  Heat given (Q given )  Q 23 
Q 31
Q 31
w
efficiency () 

 1
Q given Q 23
Q 23 
3
adiabatic
P0
1
V
  1
C p ( T3  T2 )
C v (T1  T2 )
adiabatic process:
nP0 T3
process: P  T
0
1
---(i). Calculation of T1 and T3 : Using
P11 r T1r  P21T2 r

P01 r T1r =
1 r
(nP2 )
T2
r
1
 
n
1 r
r
T1  T2 --(ii) Using isochoric
1 r




1 r
r nT1    T1 
n


 n  T3  nT1 --(iii) put in (i) 

  1  r ( n  c)
  1
T1  nT1
n r 1
P
(b) Process 2-3: Q 23  vC p (T3  T2 ) Process 1-2:
3
P0
Q1 2  vC v (T2  T1 ) . Work done (w )  Q 23  Q1 2  Heat
Q1 2
w
 1
given (Q given )  Q 23  efficiency ()  Q
Q 23
given
C p  (T2  T1 ) 
   1  C  (T  T )  ----(i) Calculation of T2 and T3 : Adiabatic
v 3
2 
P0/n
1
2
V
1 r
1 r
P
1




r
1

r
r
r
process: P31 r T3 r  P21r T1r P0
T3  n   0 
T1  T3  T1  
--(ii) isochonic
n
n
 
1 r


r
P3 T3
T3
T

T
n
3
process: P  T  n  T2  n --(iii) put value of T2 and T1 in (i)   1  r  3 / n
2
2
T3  T3 / n

r

n 1
  1
r 1 Ans
r (n  1)n


P



 


3
Q: 2.124: (a) Process 2-3: Q 23  vC p (T3  T2 )
Here T3  T2 (Q 23  ive) Process 1:
Q31  vR T3 ln
V1
V2 Here V1  V2 (Q 31  ive)
T3  T1  isothermal
2
V0
1
nV0
V
201
Process 2: Q1 2  vC p (T2  T1 ) Here T2  T1 (Q1 2  ive)  Work done
(w )  Q 23  Q 31  Q12  Heat given (Q given )  Q 23  Q 31 
efficiency () 
C p (T1  T1 )
Q12
w
  1
 1
C v (T3  T2 )  RT3 ln V1 V
Q given
Q 23  Q 31 
2
  1
C p (T2  T1 )
C v (T1  T2 )  RT2 ln V1 V
 Here T1 = T3
---------(i)
2
Calculation T1, T2:
V
T2 and 1 in (i ) :  
V2
Process 1-2 :
CP
nV0 T1

 T2  T1 n
V0
T2
CV
T1  T1 n  R Cv Tn1 ln n
(b) Process 1-2: w  Q12  vR To ln
V0
n
---(ii)
V1
 n -----(iii) put value of
V2
r (n  1)
   1  n  1  (r  1)n ln n Ans
V0
  vRT0 ln n ---- (iv)
P
Process 2-3:
Q 23  v C p (T3  T2 )
(ive be cause T3  T2 )
2
3
Process 3-1:
Q 31  v C p (T0  T3 ) (ine, T3  T0 ) . Work done
(w )  Q1 2  Q 23  Q 31 
T1  T1  T0
T0
P0
1
V
V0/n
V0
w   vRT0 ln n  vC p (T3  T2 )  vC v (T0  T3 ) 
C (T  T )  RT0 ln n
w
 1 v 0 1
Heat given (Qgiven )  Q23  Q31  efficiency ()  Q

C p (T3  T2 )
given
Cv
  1
R
C p ( T0  T3 )  Cp ln n
T3  T2
Calculation of T0, T3 : isochoric 
  1
T3 P3
T
P

 3  2    (ii) (P2  P3 ) T = n To put in (i) 
3
T0 P1
T0 P1
(n  1)  (r  1) ln n
(n  1)  (r  1) ln n
1
Ans
r (1  n)
r (n  1)
202
Compact ISC Physics (XII)
P
Q: 2.125: Process 1-2:
3
Q1 2  vRT0 ln V0 nv   vRT0 ln n  ive .
0
4 T0
Process 2-3: Q 23  vR (T3  T2 )  ive
2
Process 3-4: Q 34
nV
 vRT0 ln 0  vRT0 ln n  ive .
v0
1
T0
V
Process 4-1: Q 41  vC v (T0  T0 )  vC v T0 (  1)  ive 
V0
nV0
Work done (w )  Q1 2  Q 23  Q 3 4  Q 41 
w  vRT0 ln n  vR ln n  vC v T0 (  1)  vC v (T3  T2 )  Heat given
(   1) ln n
RT0 ln n  C v T0 (  1)

(Q given )  vC v (T3  T2 )  vR ln n   1 

C v (T3  T2 )  R ln n
 ln v  ( 1) r 1
Ans
P
Q: 2.126:
nP0
Process 1-2:
Q12  vC p (T0  T0 )
Process 2-3: Q 2 3  vR T0 ln
V3
3
4
2
 ive
T0
V2
 ive
P0
1
Process 3-4: Q 3 4  vC p (T0   T0 )  ive
Process
 Heat given
  1
Q 41  vR T0 ln
4-1:
V1
V
V4

Work
done
(w )  Q1 2  Q 3 4  Q 41
Q1 2  Q 23
w
 1
(Q given )  Q 34  Q 41    efficiency ()  Q
Q 3 4  Q 41 
given
C p T0 (1  )  RT0 ln V3 V
2
T4 V4
V
---- (i) Calculation 3 V2 and V1 V 4 : T  V    V4  V3
C p (T0   T0 )  RT0  ln V1 V
3
3
4
 Isobaric process :
T1 V1

   V1  V2
T2 V2
P0 V2  nP0 V4  V1 V1  n -----(iii) put in (i)
V
Isothermal process : P0 V2  nP0 V3  3 V 2 
  (  1) ln n  ln v  ( 1)
Q:.2.127 Process 1-2: This is process in which
( 1)
Ans
P
2
x
P  V  PV 1  const   compare with PV = const  x = -1 

R 
 T0 (   1) 
 Q 1 2  vC ( T0  T0 )  v  C v 
x  1 

 R
R
v ( r  1) R
 Q 1 2  v 
  T0 (   1) 
T0 (   1)  ive
r

1
2
2 ( r  1)


1
---(ii)
n
1
T0
T0
3
V
203
Process 2-3: Q 23  vC v T3  T0  (()ive) T3  T0 
Process 3-1: Q 31  vC p T0  T3  (()ive) T0  T3  Work done (w )  Q1 2  Q 23  Q 31
Heat given (w given )  Q12  efficiency () 
1
Q12  Q 23  Q31
Q 2 3  Q 31
 1
=
Q1 2
Q1 2
C v (T3  T0 )  C v (T0  T3 )
V3 V3
(r  1)R

---(i) Calculation of T3 and T0 : isobaric process :
----(ii)
T0 (  1)
V1 T1
2(r  1)
Polytropic process (Pv-1=const) : TV 1  const  T1V1 2  T0 xV3  2 
1
1 



C T0  2  T0   C p  T0  T0  2 
r 
V3
T
T



  1 2
  2  3  3  T3  T0  2 put in (i) :   1 
( r  1) R
(
1

r )(1   )
V1
T1 T0
T0 (  1)
2(r  1)
1
1
2
Q:.2.128 Carnot Cylce:   1   Usinng clausis inequality :
1
dQ
 T  0 (irreversible cylce) 
Q1
Q
Q
T
Q '
 2  2  min -----(1) Irreversible cycle : '  1  2 For irrewrshile cycle :
Tmax Tmin
Q1 Tmax
Q1 '

dA
0
T
Q1 '
Q '
Q '
Q '
Q ' T
Q ' Q
 2  0  1  2  2  min ----(2) from (i) and (ii) : 2  2 therefore
 T
T
T
T
Q
'
T
Q1 ' Q1
max
min
max
min
1
max
n > n’ Ans
Tmax
Tmax
Q1'
Q1
Control = Reversible cycle
I
Q2'
Q2
Tmin
= Irreversible cycle
Tmin
204
Compact ISC Physics (XII)
dA d1 – d 2

2.129: Efficiency () =
d 1
d1
1–
A
 2
   1–

=
 1
 1
P
d1
3
4 T+ dT
2
1
d2
P
T
T
T
 (dT)  V from


.... (i)  A = (  P) (  V) = 
T  T
T  T
T
  T V
V
A
A
 T Also we know :   = U + PdV  
 T =  U + P dV  T   P  =  V =
(i) :  1 = 
3–4
3–4

T

T




  T V
  U
P
P
  U
 +P  V = T –P
 U + pdV  T   T  = 

V 
V

 V   T V
2.130 : v = 1  C O2 (a)
 T0

 ds    C
T0
V
Ans.
Iso choric Process : Tin = T0 ; Tf = T0 We know : ds =
d
 d =  Cv dT
T
dT
 S  C V nn  S = C nn = R
nn Ans. (b) Isobaric Process : d
T
V 
 –1 
=  Cp dT 
 ds    C
2.131 : S =
Δθ
T
P
R
dT
S = CP  nn  S =
nn Ans.
 –1 
T
(isothermal process)  S = T S .... (i). Also we know  S = S + w  TS = 0 +
S
Vf
S
Vf
Vf
 e R Ans.
 R T  n V Here U = 0 , because T = const  R  n V 
Vi
i
i
dθ
2.132 : Process 1 – 2 : ΔS12  T 

dθ

Process 2 – 3 : ΔS2 3 
T

T1
 C
T2
V
T2
 C
T1
V
dT
T
  C V n 2
T1
T
dT
T
  C V n 1
Net entropy change : S = 
T2
T


T
T
T
CV n 2 T –  CP n 2 T =    n 2 T1  (CV – CP) .... (i)
1
1


Calculation of T1 and T2 :
P0
P0
n

(i) S = (  nn) (CP – CV) Ans.
T0
T
T2
1
 1
T2
T2  T1  n put in
P0
T0
=2
1
P2 = P0 n
P2
2
3 T0
V
205
2.133 : Process 1 – 2 : Adiabatic process : d = 0 alwalys S1–2 =
Process 2 – 3 : Isobaric process : S2–3 =
n
T3
T2
Calculation of
T3
T2

d
 S2–3 =
T
: Process 2 – 3:
T3

T2
T
T2
 3
n V0
V0
= S1–2 + S2–3  Snet = – Cp  n n Ans. Snet = –

P
d
=0
T
adiabatic
3
 C P dT
=  CP
T

1
2
V
nV0
V0
T3
1
  S = –  C  n n  S
2–3
P
net
T2
n
m rR
nn
M r –1
Ans.
2.134 : Since entropy is state function hence entropy change can be found by another process
joining (1) and (2) : Now take a random process as :
P
Process 1 – 3:
dT
T
  C v n 3
T1
T

S1 – 3  C v
T1
T2
Process 3 – 2:

 S 3 – 2  C P
T3
Calculation of
Process 1 – 3:
T3
2
P0/
dT
T
  C P n 2
T3
T
V0
V0
V
P
P0
1
T1 :
P0
T
 1
P0
T3

V0
T3
Process 3-2:  V  T
0
2
Snet  – 
1
P0
T3


T3
T1

1

T2
  Then
T3
2
P0/
V0
V
V0
S net  C V n 1  C P n  

R
R
R
R
n  
n  
(  n   n )  Snet 
(  n  – n ) Ans.
 –1
 –1
 –1
 –1
T1
2.135 : V2 = V3   T   ;
2
T3
Process 1-3 :
 S1 – 3  C P

T1
Calculation of
T3
T1 and
T2
V2
  . For entropy change S – S take a other process 1 – 3 – 2
2
1
V1
dT
T
 C P  n 3
 S 3 – 2  C V
T1 Process 3-2 :
T
T3 :
T3
V1 T1
V1 T1 1
Process 1-3 : V  T  V  T    T   ––– (ii)
3
3
2
3
1
T2

T3
dT
T
 C V n 2
T
T3
P
2
1
3
V
206
Compact ISC Physics (XII)
P
T3
T2
1
put in (ii)   T    T   –––(iii)
2
3
Process 3-2 : T1 =  T 2
put in (i)  S net  C P n  – C V n  
2  T0
R
R
n  –
(n   n ) 
 –1
 –1
T0
1
V

n  
 Ans.
 S net   R  n  –
 – 1 

2.136 : We know
 ds  
d
T
 S 

d
 d = C dT S =
T
R 
R (n – r)
R
R
 R
–
n 

 n   S =
 S = 
(r – 1) (n – 1)
r –1 1 – n
 r –1 n –1
C=
 T0

C
T0
dT
S = C n  
T
Ans.
2.137 : P  V  P = C1 V  P V– 1 = C1 Compare with PVx = Const. x = – 1 Vf =  Vi
Tf
S =

Ti
(Given)
 C dT
R
R
1  R (r  1)
T
 1
  C n f –– (i). Hence C =
–
R
 
Ti
T
r –1 –1–1
r
–
1
2  2 (r – 1)

Tf
Calculation of
T V
 f   f
Ti  Vi
–1
Ti : PV = Const.
2

   2

2.138: We know S =
 R T V –1
= Const.  T V– 1 = Const.  Tf (Vf)– 2 = Ti (Vi)– 2
V
 R (V  1) 
 R (V  1)
2
n 
put in (i) S =  2 (V – 1)  n   S =
(V – 1)



d
. For S maximum, dq must be (+)ive and when d  willPbe negative S will be
T
decreasing. Then we know d = du + dw  d = du + P dV  0  du  – P dV 
CV dT  – P dV   CV
Ans.
dT
 –P
dV
(Since dV = (+) ive)   CV
P0
P = P0 –  V
dT
 –P
dV

dT
–P

––––– (i) Also
dV  C V

P
=
P0
–

V
tan  = 
P
RT
dT
dT
dT
2V
 0 –
= P0 –  V  v R T
= P0 V –  V2  v R
= P0 – 2  V 
V
dV
dV
dV  R
R
(i) 
V
put in
P0 2  V
–P
–PR
–R
– R P0 R  V
–


 P0 – 2  V 
=
(P0 – V)  P0 – 2  V 
 P0
R
R
 CV
CV
CV
CV
CV
207
V

R  
R
1 
 

 C   2   C  V 
CV
V 
V 

Vmax =
P0 r
 1  r 
P0 (R  C V )
P0 r
 2  CV  R  

  V   1  r  Maximum Value of Volume
CV


Ans.
2.139 : S = a T + CV n T  ds = a dT +
CV
dT  Also we know d = dU = dW
T
And d = T ds  T
T
V


dV
CV


RT
RT
a dT  R
dT  = C dT +
ds = dU + P dV  T  a dT 
dV
T
a
dT
=
dV

V
V
T
V
V


T0
V0
R
n V
 a (T – T0) = R n V V  T = T0 +
V0
0
a


2.140 :
dS 

0
d
1
T  S = T
V2

V2
P dV  V 

  =
V1
V1
RT n
2.141 : S =
– b)
 d

0
 RT
a

–
 VM – b V 2
M


dθ

T

T1
–––– (i)
a
V
2
dV 
1
  And   =
T

V – b
a
 dV   U = RT n  2

M



 V1 – b  VM

V2
U =
V1
a
Calculation of   U : U = CV T – V  U =
M
Ans.

a 
a
dU  pdV
.... (i)  dU = CV dT + 2 dV Vander wall equation :  P  2  (V
T
V 
V

RT
a
– 2
V–b V
RT dV
a
–
dV
V – b V2
T
T2
(Because isothermal process)  S =
 V2 – b 
V –b


 R n  2
put in (i)   = RT n  V – b   S =
T
 1

 V1 – b 
= RT  P =
C V dT 
CV n

 1
(V2 – b)
1 
   U
 a 
–
(V1 – b)
 V2 V1 
 a
a 
–
–

V
V
1
 2
Ans.
V –b

 R n  2

 V1 – b 

p d V =
RT dV
a
– 2
V–b
V
T2

 S =

T1
Ans.
C V dT

T
put in (i)  S =
V2

V1
R dV
v–b
  S
=
208
Compact ISC Physics (XII)
a T3
a T 3 dT
 S =
3
T
 T2
d
cdT

 m
 b d T   S = m  d dT 
2.143 : d= m C dT = m (a + b T) dT  ds =

T
 T

T T
 1
T


 S = m  a n 2 T  b (T2  T1 ) 
Ans.
1


2.142 : d = C d T 
ds =
d
cdT


T
T

ds 

dT
=
T
C


T
2.144 : T = a s  s =  
a
n
T
C =
na
T

a





1 –1
n
1
n
dθ
 ds =
T
1 T
 
=
n a
1
n
T

a

1
ds =
na
 a Sn

 a

1
 C=
n




1




1 –1
n
n
 C=
1
dT  Then
na
S
2.145 : P V = const
 S–S 
0


T
T
C 

 e
 S = S0 + C n


T0
T0
2.146 : (a)
S=

T
 dS = –
T2
(b)
T2
 d   C dT
T1
T2
U =
 CV dT

–
T10
α
T
2
T = T0
dT = –
Tα
T2






b dT 

T1

T2

1 –1
n
C dT
T
dT 
S
 C < 0 if n < 0
n
R
R
–
and C =
 Then  ds 
r –1
x –1
S0
x
T

a

Ans.
T
dθ
C dT
T
T   T
T0
C>0
 S – S0 


e C 
dT  d = –
C<0
S
α
α
dT = C dT  C = –
T
T

T
T
dT   n 1
T2    n 1 T Ans.
T
2
Ans.
(c) = U + W 
T
= CV (T2 – T1)  W = – U  W =  n 1 T – CV (T2 – T1)
2
Ans.
T1
2.147 : (a) We know d = T ds  net =
 d   T ds = Area under T VS
S
T
n T0
2
1


curve  net = (T2 – T3) (S2 – S1) = (n – 1) (S2 – S1) T0 . Also  = U + w.


One complete cycle : U = 0  net = w  given =

T ds  if ds =  the
through proces and this in 1  2 process  given =
 T ds = n T
 
1
Δ θ net
Δw
n –1

 
  1 – 
 
n
Δθ given
Δθ given
2n
Ans.
0
3
T0
S
(S2 – S1)  Efficiency () =
209


(b) net =
(n T0 – T0) (S3 – S1)  given = Area under process 1 – 2 =
(T

 0
Δ θ net
n –1
 
Δθ given
n 1
– n T0) (S3 – S1) efficieng () =
T
2 n T0
Ans.
1
3
T0
2.148 : Thermally Insulated ressel Initial volume = V0 Final Volume =
S
P
final
P2, 3V0, Tf
3 V0  Since value open suddenly, hence process is not slow hence we can
not use PV =  R T for intermediate process. But we know entropy can find
Initial
P1, V0, T0
from intial and final state hence we take another slow process in which change
in volume occure very slowly. Here final temp (Tf) : Change in internal energy
V
of system = 0  Ui = Uf   CV T0 = CV Tf  Tf = T0 . Take on imaginary
process
final
in
and
which
heat
may
be
exchange
statges are same.
vC v dT
d
dT
p dV
vRT dV
 v Cv




T
T
T
T
VT
T0
initial
dT
nV0
but
Valve
ds
=
v1= v
v2= 0
S
=
V0
2V0
dV
 ds   v Cv T  v R  V  S =  R  nn
T0
V0
Ans.
2.149 : Internal energy calculation : Process 1
2 : Ui = Uf because
= 0 And w = 0  0 = U + 0  Ti = Tf = T2 Process
1
2
adiabatic process V2 = 2V0 ; V1 = V0  U = Uf – Ui  = v CV Tf – v CV Ti = v

–1
1
:
2
=1
–1
CV (Tf – Ti)  Ti = T0 = T2  PV = const  T V = const  T0 (2 V0) = Tf V0
 Tf = 2–1 T0 U = CV (T0 2–1 – T0) =  CV T0 (2–1 – 1)  Unit = 0 +  CV T0
 R T0
R T0
(2–1 – 1)  Unet =
(1–1 – 1)  Put  = 1  Unet =
(2–1 – 1)
 –1
 –1
, T0
V0
Vaccum
V0
Fext
2.150 : (a)
adiabatic chamber
If process is made sudden in first part while slow in
P
second part then. Fext in case (a) will be more than part (b) Hence W(a) > W(b) (on
gas)  Here U(a) > U(b) (of gas)  Tfinal of process (a) > Tfinal of process (b)  Vfinal
 R Tf
= Vfinal (b)  P =

(a)
Vf
V
Pfinal (a) > Pfinal (b)
P
P f, V0, Tf Final
Entropy Calculation : Here we are seing that initial and final volume is same. Then
d
take a imaginary process with constant volume.  ds   T 
Tf

T0
P0, V0, T0 Initial
dT
 Cv
T S = 
V
210
Compact ISC Physics (XII)
R

S
=

n 2–1  S =  R  n2 Ans.  Put  = 1 
T0
γ –1 
S = R  n2 Ans.
CV  n
Tf
 S N 2   ds  
T0
T0
i C V   2 CV
= CV put in (1)  Tf = T0 N2 Gas :
1   2
T0
n V0
n V0  V0
 i C V dT
1 R
pdV
d




S

=


N
2
T
V = 1 R  n (n + 1) U2 Gas :
T
T0
V0 T
T0
 2 C V dT

T
 S02  
nV0
V0
2.151 : Cv = 5/2 R  N2 as well as O2 At final equalibrium Ui = Uf  i CV T0
mix
+ 2 CV T0 = (1 + 2) CVmix Tf .. (i)  C V 
 2, T0
 1, T0
n V0  V0

nV0
pdV
T =0+
(n 1)V0

nT0
 2 R dV
 n 1
  2 R n 
 S
=  S N 2 
system
V
 n 
 n 1
  S
 S 0 2  1 R n (n + 1) + 2 R n 
= 1 R n (n + 1) + 2 R n   1 n
system
 n 


2.152 : m1 = 300 gm  t1 = 97º C  m2 = 100 gm  t2 = 7ºC . Let us final temp is Tr . Then Heat given by (m1)
= Heat taken by (m2)  m1 c1 (t1 – tf) = m2 c2 (tf – t2)  tf =
For m1 :

ds 

d
  Sm =
1
T
For m2 :  Sm 2 = m2 c2  n
tf
tf

t1
m1 c1 t1  m 2 c 2 t 2
.... (i)
m1 c1  m 2 c 2
m1 c1 dT
t
= m1 c1  n f t
1
T
t 2  Ssystem = m1 c1  n
tf
t1 + m2 c2  n
tf
t 2 . Where c1 and c2 are heat
capacit is of coper and water
Case 1 : 2.153 : If value is opend : Let us final temp is Tf  Ui = Uf (For system) 1 = 2 = 1  1 CV
T1 + 2 CV T2 = (1 + 2) CV Tf  Tf =
d
Tf
 ds   T  
T1
C V dT
+
T
2 V0

T1
R dV
V
T1  T2
2
For chanber (A) :
 SA = CV  n
 n2 For Chamber B : Similiar SB = CV  n
tf
A
 1= 1
tf
t1 + R
T1
CV
Same ideal gas B
 2= 1
V0
T2
CV
V0
t 2 + R  n2  System = SA + SB  System = CV

Tf
T 
(T  T2 ) 2
 n f  + 2 R  n2 = C  n 1
n
+ 2R  n2 (T1 + T2)2 > 4 T1 T2 (Using AM  GM) 
V
T1
T2 
 T1 T2

S  O
211
Case 2 : If value is conducting Tf =
T1  T2
For changer A :  S =

Tf

T1
C V dT
+
T
 dV Here V =
 (T1  T2 ) 2 
Tf
T 
  S >
0  SA = CV  n  f  For chanber B : SB = CV  n
 Snet = CV  n 
net

T

T
T
T
2
1 2 
 1

0 from above log function.
2.154 : Cubic vessel (a) Probability to find a molecule in one half part of vessel = 1/2. Probability to find N
1 1 1

molecules on one half part :       N time  P =
N (b) Velocity of gass molecules is : v = 105
2 2 2

1
= 10–5 s.  no of time to cross the vessel = t τ where t = 1010
t
1
year. Then there will be a certainty to occopis this half valume will be 1.  N  1  2N  t τ 
τ
2
cm/s  Time to cross the vessel  
N=
n t/
n2
105
Ans.
N!
n
2.155 : We know statical wright of n molecules among N molecules : C N  n! (N – n)! . Since in half part N/
N!
2 molecules will be occupied then. Statical weight of n = N/2 molecules. C N/2

N 
N N
!
!


2
2
 N!  1
1


Probility (P) = C N 2
Ans.
N P= N
N
N
N ! 2
2
!

2 
 2

 
2.156 : There is N molecules, and all are free from each other. Suppose probality to one molecule to enter
successfully in one part is P  Then probability of unsuccesful attempt is 1 – P. Then probality of n
n
mole celes to enter in one part :
Pnet =
C  P n (1 – P) N – n
Here P =
N!
1
 Pnet =
n ! (N – n)! 2 N
2
 Put N = 5. and n = 0, 1, 2, 3 and get different result
N!
2.157 : Like Q : 2.156  Pnet = n ! (N – n)! pn (1 – P)N–n. Where P = Probability to find one molecule in
certain part.

R3 n0  Where no = Loschmidt’s const = no of

1
 d3 n 0
n0  N =
. We know relative fluctvation is
= n  Then
N
6
2.158 : No of molecules in sphere of diameter d N =
molecules /volume. N =
 d
  
 2
3
212
Compact ISC Physics (XII)
1
1
1
 η2  N 
 Then
N
η2

η2
 d3 n 0
 n =
6

 d 3 n 0 . Also are know. Avrage no of
1
molecules inside sphere : < n > =
η2
Ans.
dθ
T0  ΔT
2.159 :  = 1 monoatomic gas T0, V0  Tf = T0 + T  Vf = V0  We know  ds   T 

T0
 C V dT
T
T0  T
T  T

 S = CV n 0
Also we know S = K n 2 – K n 1 = K n 2  . Then CV n
=
T0
1
T0
iR


   ΔT  2 K
 ΔT 
i
iR
 = n
n 1 

k n  . Here CV = R where i = Degree of free dom
  = 1 


2K
T0 
2



T0 


i
N

 ΔT  2 A

  = 1 
NA = Avogadro no

T0 

2.5
Liquids, Capillary Effects
2.160 : (a)
Mercury drop : P =
2T
Hence T = = Sorface tension R =
R
3
2
1
2α
4α
2α
d/2.  P =
 P =
(b) Soap bubble : P1 – P2 =
P2 – P3 =
d/2
d
R
R
R+
dR
R
2α
2
4
8
4α

 P1 – P3 = P =
=
 P =
R  dR
R
d/2
d
R
2.161 : Force applied by surface tension : F = ( R2) h  g =
h
 2
4
2

 R2  h =

h
=
pdd
 R 
 g d2
h
d = diameter
 
2.162 : Pinitial inside bubble : Pin = P0 +
bubble : Pf =
isothermal
P0
8
+

df
:
PV
R
8
4
= P0 +
.... (i) Pressure final insider
d
R1
And df = d  Pf =
=
1.
1
Const
P0
8
+
.... (ii) Since process is

d
(Inside
soap
bubble)
P
8  4
8  4

  (  R 1 ) 3    P d (1 –  /n)
 R 13   0 
 P0 

0
d
3

nd



 3

Ans.

 – 
P0
R1
R1=d/2
213
P0
4α
2α
2.163 : Pinside = P0 + h  g +
Pinside = P0 + h  g +
d
R
h
4α
2
4α
2.164 : Pboltom = P0 + h  g +
= P0 + h  g +
 Psurface = P0 +
. Hence if
d
R1
d
4
4
 R 13 = P
 (n R 1 ) 3 
process is isothermal then P bottom
surface
3
3

 P0  h  g  4 

d


 × 1=


P0


 P0  4   n3  h = [P (n 3 – 1)  4α (n 2 – 1)/d]
0


nd


g
h
2.165 : Method: 1 (Force Method) F.B.D. of liquid coloum  2 R cos = (h  g)
A  2 R cos = h  g R2  2 cos (– = h  g R  h =
Hence d = 2R  h =
–2  cos 
g R
 2 R
–4  cos 
. First coloumn : h1 =
 gd
138º=
–4  cos 
 g d1

h
R
r
Second column : h2 =
–4  cos 
4

h
=
h
–
h
=
cos
1
2
 g d2
g
(h  g)A
 1
1 
4  | cos  |  d 1 – d 2 




 d – d   h =
g
 d 1 d 2 
1
 2
P
r
Method : 2 (Pressure) Pa = Atmospheric pressure. Pressure at point C : (Using
excess pressure equation) Pc – Pa =


 Pc = Pa +
R
sec 
r
.... (i). Pressure at
point c : (using hydrostatic equation) Pc = Pa – h  g .... (ii) From (i) and (ii) : h  g =
h=
 cos 
4  cos 
4  cos   d 2 – d 1 


h
–
h
=
 d1 d 2 
1
2
g
Rg
dg


R
d

r
2r
2  cos 
4  cos 
(a 2p R) cos q = (h r g) A = h r g pR2  R = h  g  d = h  g
2.166 :
Method : 1 (Force Method) r cos q = R Þ cos q =
 cos  
2
hgd
hgd
d
.... (i)
=
 r hg
4
4

2r
R C


cos  Similar
R
 2 R
 r
R
h
(h  g)A
Ans.
214
Compact ISC Physics (XII)
Method : 2 (Excess pressure equation) P =
2
2
= h  g  r = h  g
r
Ans.
2.167 : Pressure at point (1) : since surface are coincide we assume
A
4
  0  cos  = 1  P1 = P0 +
Using PV = const (T = cosnt)
d
 –x
1
P0

x
4 


 ( – x) A   x =
   P0  A =  P0 
P0 d
d  


4
2.168 : Pressure at point (1), Calculated by Boyles law : P0  A = P1 (  – h)A
d
P0 
 P1 =
.... (i)  Also
–h
1
From (i) and (ii) :
4α
P1 = P0 – hg +
cos  .... (ii)
d
P0 
4α
= P0 – hg +
cos   P0
–h
d
 h
hg
h
 
h g
4α
–1

 

–
h
P
d

0 

4α
cos  P0   – h  P   d cos   

0 

2.169 : F. B. D : At equilibrium : (d1 + d2) =

P
[ g h  P0   – h] d
4 cos 
d1
d2
 g h
(d1 + d2)
4
d2
d1
h

(d2 – d1)  h = g (d – d )
2
1
2
gh
d2
4
2
–
d1
4
2.170 : Fy = (cos ) dx  Fnet = 2 Fy = 2 cos dx. Since
y
every differential elements are free from each other. Then Fnet = weight of liquid
column of differential length dx. 2 cos  dx = g [y x d dx]  y =
(2α cos βo)
 g x d
Fy
Fy
x
d
y
Ans.
x
dx
2.171 : Pressure at point (1) and (2) are : P1 = P0 –
4π
4π
 P2 = P0 +
= P0 +
d
d/n
V1
d
V2
4 π
π d2
π d2
V1 
V
Continuity equation at (1) and (2) :
2 2  V1 = n 2 ...
d
4
4n
(i)
Bernaullis equation between point (1) and (2) :

d/n





 P0 – 4 α    g   1  V12   P0 – 4 α n   0  1  V 22 .... (ii)  V d ....




1
d 
2
d 
2



(iii) from (i) and (ii) calculate V1 and put in (iii). 
π d2
4
2gl – 4α (n – 1) / d
n4 –1
.1
V2
Ans.
.2
215
2.172 : R1  R2  h/2. Pressure difference between point (1) and (2) : P1 = P0 +
2α
P2 = P0 +
 P = 2
R2
= h  g  R =
h3  g
8
h
.2
Ans.
1 1
P =      r1 cos  = h/2  r1
 r1 r2 

r1


1

  cos  1 
  = 
  Henc
P =  
R
 h/2 cos  R 
 h
h
and r2 = R  Then
2 cos 
P =
.1
 1
1 
R –R 
2 R
1
2 


= h  g 
 R – R  = h  g  2  

h h
1
R
R
 2
 1 2 

2 2
2.173 : Method : 1 We know
=
2α

R1
r2
h/2
r1
1
1
 
R
h

 2  cos  
2  cos 
 R2. Also V = R2 h 
. Normal force due to surface tension : N = (P) R2 = 
h


h
 2  cos  
 2  cos  


1


R = V/h  N = 
 V = m1g m1 = mass of plate. N = 
 V = (m1 + m) g  mg
2
12
 h

 h

2
= (2 cos ) V
 1
1

– 2
 12
h
h




2
Also h   h n 2  m =
(2 cos )  R 2 (n 2 – 1)
gh
Method : 2 Surface energy = U = (–sa + se)  R2  U = ( cos ) (2  R2). Also  sa
we know volume will be constant while R will be changed if surface area in-  
creases. Then V =  R2 h   R2 = V/h. U = ( cos ) 2V/h Normal force N
– dU
2  V cos 

Then N = 2  V cos 
dh
h2
=
 n2 –1


R h) cos   h 2  =  mg 


2
m=
 1
1 

 h  2 – h 2  = mg  2  (


2   R 2 cos  2
(n – 1)
h
2.174 : Normal rxn at surface from question no Q.2.73 : N =
is complete cos  = 1  N =
2 V
h
2
Hence V = m/ N =
2 m
 h2
 s
 s –  sa =  cos 
Ans.
2  V cos 
h2
Ans.
. If wetting
h
Ans.
216
Compact ISC Physics (XII)
2.175 : Normal force
 V
h2
N=
2
2  ( R h)

h2
 N=
2  V cos 
h2
2 R
h
R
. For complete wetting cos  = 1  N =
Ans.
d

2.176 : ho = height of liquid 2   =  (d  ) ho g  h0 =  d g . Pressure
change at x height : P = x g Force on dx thickness : dF = (x  g)  dx  F
h0
= g 

0
 g   2 
x dx  F = 2   d g 



dx
h

x

2 
 F=
 g d2
Ans.

2.177 : Suppose at time t, radius of bubble is x. Then pressure at point (1) : P1
4
 4   r
dθ
dθ
( P)  r 4
4

= P0 +
Also we know
 dt   x  8  
dt
8 
x


=
h
R
2

1
2
r
 d
R
  r 4
  r 4
dt. Suppose in dt time dx radius is decreased than d = (4 x2) dx =
  x3 dx =
8  x
8  x
R
t


4
4
2

 r4
3
 r4
x
dx

dt  16 R  α r t  t = 2 R η 
dt  

 x
4 2
η
α r4
0
0
Ans.
F (Fore due to surface tension)
2.178 : Calculation of energy due to rise of coloumn:
r
h
  r2 g 2
E = (r h g) – (2 r h) =
h – 2 r 
2
2
2
h
Also we know
h =
2
gr

E
=
h
mg
E=0
E=?
4  g 2 4  r 2
  r 2g
4 2
2
2  2 4  2
2  2
–
–
2

r

–
 –
=
=
. This show that
2
2 2 2
gr
2
gr
2g
g
g
g
 g r
2  2
energy is decreased and hence heat will be released.  =
g
  d  

2.179 : (a) U = 4  2   = d2
   
Ans.
  d  

(b) U = 4  2   2 = 2 d2
  

d
217


 

4 d
4
d

 
  new  
2.180 : A =   4   d   = 2 d2. Since volume is conserved then     2 




3  2
3
 2 
2



1/3
dnew = 2
 d new 
 = 22/3 d2 U = Anew – A
d  Anew = 4 
 2 
2/3
2
2
2
+
d
=
d
dnew
2/3
= (Anew – A)  U = (2 d – 2 d )  U = d (2 – 2)
U = 2d2 (2–1/3 – 1)
Ans.
2.181 : P = P0 + 2 α R V =
4
 R3 PV = n RT. Work done by
3
external agent : W = Work done against increase pressure from P0 to

V
R
1
Final
Initial

   (  4 R 2 ) 2
α  

2
R 
 0

P0  2 α R  + Work done against increase surface energy W = –  n RT n  P

P0+2/R ;n
P0, n
P0


V
P 
 W = – PV n  0  + 8 R2 . Work done by gas = n R T n 2 V = – (Work done by external agent).
1
P
 
Note : Here W is net work done by (external agent + outside atomosphere)
2.182 : We know inside pressure is : P = P0 +
4
. Also we know that no of molecules inside
r
4 4
4


 r3 =  RT differentiate equations   P0 
 4
bubble is not change then  P0 
r 3
r 


r2 dr +
r
4
 4

8
8


 r 3  –
dr  = v R dT   P0 
 4 r dr =  R dT   P0 
 dV =  R dT dV =
2
2
3
3
r
3r 



 r



R
 R dT


. Also we know d =  C d T =  CV dT + p d V  C = CV + (P0 +   r )  P  8 α   C – Cp
8
0
3r 

P0 
3r
4α
8α 

R
 P0  r – P0 – 3r 
= C – C V – R = (P 0 +   r ) P  8 α
– R = R 
  C – Cp =
8α
0
3r


P0 
3r


4α
3r
8α
P0 
3r
R  R/2
1 
3
8
P0
r
α
 Ans.
218
Compact ISC Physics (XII)
2.183 : We know
A
T – dT
 1–
Where  A = work done

T
d
P
T
in cycle Where a = area of soap bubble    = Given energy
in cycle 
T – dT
a + da
V
dα
 dθ 
(da) (d)
T

    –T

dT
dθ
T
 da 
dα
Proved.
dT
d
Heat given  = (q ) 2
dT
d
Δθ
entropy change : S =
= – 2 ()
. Also we know
dT
T
 dα 
 + 2  U = 2    – T
– W = – 2T  
 dT 

2.184 : We know q = – T
2.6
 – d
a
A
T

.... (i). Also we know  A = work

T
done due to given heat  A = (– d d – da = – (da) (d) 
q= –T

There is two surfaces in bubble = –2T 
d
dT
U + W Here W = –2  6  U =
dα 

dT 
Ans.
Phase Transformations
2.185 : T = 100º C = 373 K = t m = 0.70 gm . We know vapour will be condensed in
liquid at constant temperature and pressure. Then work done on gas = P V = W = A.
Piston
Where V = decrease in volume also PV = n R T  P V = n R T (Here P and T are const)
 W = n R T  W =
m
RT . Put values in S.I. unit W = 1.2 J Ans.
M
2.186 : Suppose volume of vapoar and liquid are VV amd V   V = VV + V  m = mV +
1 and 1  m =
m  . Also suppose specific volume of vapour and liquid are VV
V
V
VV
1
VV

V
1  m =
1
1
1
VV = mV VV
 V1  V m  V  . Then V = mV VV + m V  V = mV VV +
(m – mV)
V1  V = mV
1 +m 1
VV
V
– mV V1  mV
=
V – mV1
1
VV
– V1
1 =
. Put V = 6 It : m = 5kg ; V1 = 1 it/kg ; VV
50 l/kg  mV  20 gm. Volume (VV) = 20 × 10–3 × 50 = 1 lt . Ans.
2.187 : Hene PV = Const. Means if we compressed then pressure increased and according to phase diagram
curve vapour will be condensed to get saturation equalibrium. Since temperature is 100ºC and pressure will
219
be equal to atomospheric pressure p0. Now volume of vapour
freeze = V0 – V and equivalent mass will be equal to mass of liquid.
V =5 V0=5 
t = 100ºC
t = 100ºC
Δm
P (V – V) M
P0 (V0 – V) =
RT  m = 0 0
Put P0 = 1.1 × 105
M
RT
n/m2  T = 373º K  R = 8.3  V0 = 5 × 10–3 m3 V = 1.6 × 10–3 m3
 m = 20 gm Ans.
2.188 : Given
Vf=V = 1.6 lt
Final
Initial
V
Vv1
= N Suppose V = initial volume of vapour then initial Mass (m) = 1 assume mass of
V
Vv
liquid and vapour finally is mx and mv. Then
V
Vv1
= m  + mV 
V
N V1
= m  + mV  V = m  N V1 + mV
V
V
V
V
N V1 .... (i) Also
= V  + VV 
= m  V1 + mV Vv1  = m  V1 + mV N V1 
= m  V1
η
η
η
η

1
V
n –1
m V1
n –1

  Ans.
+ N mV V1 .... (ii) (i) - (ii): V  –  = m  V1 (N –1)    

V/n
N –1
 
V/n
N –1

(1  N)
1

Hine final volume :  
n
=
2/1+N
put
Ans

2
N 1
2.189 : We know UW  At at atomospheric temp and pressusre = mq where q = latent heat
of vapourisation  W = P (V) = n RT  W =
RT 

q –
 and S =
M

dθ
 T . Since T = Const  S =
m
m
RT
T  Then U = mq –
RT
T  U = m
M
M
mq

Ans.
 S =
T
T
2.190 : Heat required to increase temp. by T = 100ºC, up to pointing
P0
point = m C T where C = specific heat of water Remaining heat 
=  – m C T. Also we know UW Here W = P0 (Sh)
where h = height rise by piston U = (m) q (Increase internal energy
P  V M 

RT


of gas) U = q 
 qM 
  h=
= POSH  
RT 

 U = q
P0 Sh M
 Then – mC T
RT
θ – mC ΔT
 qM 
P0S 1 

RT 


m = 20 gm (water)
2
S = 410 cm
220
Compact ISC Physics (XII)
2.191 : Temp of Vapour is T = 373º K  Heat required to increase temp t0 to
T :  m C (T – t0). This heat will be transfer from gas to liquid. = mV q 
t0=22ºC
mC
Then mV = q (T – t0) = mass of Vapour. Work done : P V = n RT  w
=
mC (T – t 0 ) RT
mV
RT = work done on gas. w = q
M
M
2.192 : We know P =
=
m = 1 gm = mw
mV = 1 gm
 RT
V 
M

 V
  

d=




 2

 r





P =
4 M
put
  RT 
 V  4

   d

   pvapour  


dN

mass of condense unit area in unit second µ = m n 

put values




r
d  0.2 µm
1
2.193 : no of molecules condense unit are a in unit second : dt   4 n  v

2.194 : From Q. 2. 193 : µ = P0
 m RT
 M
V
 vapour

dN
1
   

n
dt
4

KT 

2  m  m = mass of one molecule µ = P0
M
Here = 1 be cause other part in vacuum Then P0 = µ
2 RT
8 KT
m
M
n
2  RT
2RT
M
P0 = 0.9 n pa

n 2 a 
RT
a

– 2
2.195 : We know vander wall’s equation :  P  V 2  (V – nb)  nRT Assume n = 1 P =
V
–
b
V


Here a  is arised due to molecules attractions. Then change in pressure due to molecules attractions vanish.
 P =
a
V2
Ans.
a
2.196 : We know internal pressure Pi =
VV

dw 

V
VV
P dV 

V
(For one mole liquid) Work done by internal pressure :
V2
a
a
a
a
dV
–
2

w
=
But
we
know
V
>
>
V


w
=
= qM
V
V

V VV
V
(M = mass of 1 mole of liquid) Pi =
a
V2

qM
 Pi = eq Ans.
V
RT
a
– 2 .... (iii) Also we know in P – V diagram critical point act
V–b V
  P 
 P
RT
2a
 = 0 Then O = –
 =0
as inflection point. Then 
as wall as 

2


2

V

T
(V – b)
V3
  V T
RT
2a
2 RT
6a
RT
3a

–


.... (i) Again differential w. r. d. V :
=0 
.... (ii)
(V – b) 2
V3
(V – b) 3
V4
(V – b) 3
V4
2.197 : We know for one mole gas. P =
221
Solving (1) and (ii) : V = 3b = VM.Cr put V in (ii) : TCr =
8a
put in (iii) :
27 Rb
P Cr =
a
27 b 2
Also
PCr VMcr
 3
8
TCr
Tcr
8a
0.082  304
....
(i)
T
=
....
(ii)
dividing
(i)
(ii)
:
b
=
R
 b =
2
cr
8
P
27
b
R
  8
27 b
cr
2 2
 R Tcr
 R 2 Tcr2
8a
 8 Pcr  a =
= 0.043 lit / mole put value of b in (ii) : Tcr =

a
=

Pcr
 Pcr
27 R 2T
2.198 : We know Pcr =
a
cr
1
2.199 : Specific volume : Vcr 
3 R Tcr
1
in (i)  Vcr  8 M P
cr
Ans.

a 
2.200 : We know  p  2 
V 

 P
a   V
b

 
P  V P  V – V
2 cr   cr
cr
 cr
2

 π  27 b

V2






3
  
v2


3 R Tcr
Vcr
V
 cr .... (i) Also we know Pcr Vcr =  R Tcr Vcr =
put

8 M Pcr
M
M
(V – b) = RT .... (i)     P V = 3 T .... (ii) dividing (i) and (ii)
cr
cr
8 cr
 8  T 
V
T
P
 


; v
; 
Here




T 
Pcr
3
V
T
cr
cr

 cr 
1 8

v –  
3
3 Also b

=
V cr /3

2

 π  27 Vcr

g V2

 
  v – 1   8 
3
 
3




 v – 1  8      3 2  (3v – 1) = 8  put = 12 ; v = ½  = 3
3
3
2
v 


2.201 : We know at critical temp   = V and after that only gas phase exist. Then maxm volume of liquid
phase is : VMcr = 3 b × n where b = Volume of 1 mole of liquid = 0.03 lt. n b =

× 0.03 lt. Here 0.03 lt is



 0.03 V = 5 lit . Also P is maximum vapour pressure in
volume of one mole gas. VMcr = 3 
mcr
cr



a
na
a
saturated water and vapour equilibrium  Pcr =
 Pcr
2 . For mole gas : Pcr =
2  Pcr =
27 b
27 n b
27 b 2
5.47
=
= 225 atomospers
Ans.
27  0.032
2.202 : We know Tcr =
8a
8  3.62



 304 K  Also Pcr V =
R Tcr  Pcr =
cr
27 Rb
27  0.082  0.043


 Pcr M
 Pcr 
M 8
M
44
M  a
27 Rb  8

 RT
 P =


 =
Tcr  Pcr =
 Pcr =
= 0.34
2
cr

 R Tcr
8b 3
3b
3  43
R  27 b
8a  3
M
gm/cc Ans.
222
Compact ISC Physics (XII)
2.203 : We know Pcr Vcr = . R Tcr (for one mole gas). For n mole gas Pcr Vcr =
3 m
  R Tcr  Pcr Vcr =
8 M
V
8 Pcr M
3  Vp 


 R T   
. Where Vcr is maximum volume of vapour at liquid-vapour
cr
Vcr
3 R Tcr P
8 M
equilibrium.
P
4
2.204 : Here 1 – 3 – 5 is a phase transition line at which Pressure and temp are
same.
T
While curve are isothermal.
For reversible process.
I
5
3
2
 ds   du   p dV   ds = 0 because P and T are same on line 1 – 3
–5
V
 du = 0 because curve is isothermal. Hence  p dV . But incycles :  p dV  0 Which tell us this
is irriversible same 3 – 4 – 5 are irrieversible. But cycle 1 – 2 – 3 – 4 – 5
phase change from I to II phase.
 p dV = 0. Which tess at at point 3
Ans.
2.205 : Fraction of water turn into ice : f =
Ct
Where t = temp of liquid C = specific heat of water
q
q = latent heat of fusion f = 1 × 20/80 f = 0.25 for f = 1  1 =
  t
t = – 80º C Ans.
80
q12
dp
2.206 : We know clausius – clapeyron equation dT 

T (V21 – V11 )
=
II
1

dT 
T (V21 – V11 )
q12
 dp  T
T (V21 – V11 )  P
. Put value in S. I. unit. T = – 0.0075 K
q12
T (V21 – V11 )  P
2.207 : From Q 12. 206 : T =
q12
volume of liquid we know V11 << V21  V21 =
2.208 : Here we know
P =
q12
P. V21 = Specific volume of vapour  V11 = Specific
 0.9
q12 ΔT

 10 – 3 = 1.7 m3/kg
 V21 =
373
3.2
T ΔP
q
dp
q12
dp
q
1 >> 1 
 12  dp = 12 dT 


Also
V
V
1
1
2
2
1
dT
dT
T V2
T (V2 – V1 )
T V21
 T .... (i)
T V21
Also
P0 V2 =
m
RT 
M
V21 
V2
RT

m
P0 M
q12 T P0 M
P M q12 T
P0 M q12 T
 0
 P = P0 + P = P0 +
 P = P0
2
TRT
RT
RT 2
put in (i)  P =

M q12 T 
 
 Ans.
RT 2 

223
2.209 :
dP
Mq

P
dT
RT 2
= n m + n T + n
(from Q : 2.208) 
R
M

dP
Mq
m

dT .... (1) Also PV =
RT   n P +  n V
P
M
RT 2
dP dm
dT



P
m
T
dm
 M q  dT
 
– 1
Put value in S. I. units
m
 RT
 T
P = P0 e
Mq
R
 1
1


T – T
 0

 P = P0 e
Mq
R
dm
M q dT dT

–

m
T
RT 2
dm
= 4.85% Ans.
m
P
dP
M qP

2.210 :
(from Q. No. 2.209)
dT
R T2
dm
dp dT

–

m
P
T

P0
 T – T0

 TT
0

dP
Mq

P
R




T
dT
 T 2  n
T0
P
P0
 –
Mq
R
 Mq
 If T – T0 < < T0  P = P0 1  RT

1
1 
 –

T
T0  

T – T0
T0

 This

will be reasonable.
2.211 : Here P  T. We know P =
T  V1
T Then T = T
= lowerq12
T (V21 – V12 )
q12
ing of melting point. Also suppose total mass is m1 and mass ofice is m2. m1 C  T = m2 q12
m2
CΔT
CT Δ V1
CT Δ V1
η



η



2
2
m1
q12
q12
q12
2.212 : (a)
Ans.
P0
b
log P = a – . At triple point pressure and temperuture of solid CO2 and
T
vapour CO2 will be same log P = 9.05 –
1.80  103
1.3  103
.... (i) (Sublimation) log P = 6.78 –
.... (ii)
T
T
(Vaporisation). Solving (i) and (ii) : T = 216 K  P = 5.1 atm (b) Here  n P = a –
 Mq

Compare with Q. 2.210 : P = P0 e  R

8.31 × 2.303 ×
b

b
P =  a – T  .... (iii)
e
T
 1
1  
b
Mq
Mq



 b 
 T – T   .... (iv)  
T
R
R
 0

8.31 2.30 3  1310
1800
j

 570 j
= 783 gm 
gm Ans.
44
44
liqued ges

P

sub
Rb
=
M
224
Compact ISC Physics (XII)

S 

2.213 :
T

2.214 : S =

t2

t1
mcdT
mq
 t2 
mq

T
t 2 S = mc n  t   t Heat Given () =
2
 1
Δθ
m qm

=
T
T
t2

t1
t1
 mcdT + mq
t1
mq V
m qm
mq V
m c dT
t
 mc n 2 

S =
Ans.
t
1
T
T
T
T
2.215 : Suppose final temp is T. Which is greter than 0ºC. Then entropy change of S =
 S = mc n T
t2

d

T
Where C = specific heat of copper Put values S = – 10 J/K Ans.
T

t1
mc dT
T
Calculation of T : C = 0.39 J/g.k  Cw = 4.18 J/g.k. Heat given by copper to make temp. of mixture zero ºC.
= mc T = 90 × 0.39 × 90. Heat taken by ice to convert into 0ºC temp of liquid 2 = 50 × 4.13 × 3 + 50 ×
80 × 4.2 ; 1 = 50 × 4.18 × 3  Here 2 > 1 and 1 > 1. Final temperature will be 0ºC = 27 3º K.
2.216: (a)
t2 = 60ºC. Let us final temp is t then this 0ºC < t < 60ºC. Now
100 × 80 + 100 × 1 × t = 100 × 1 × (60 – t)  (Heat taken by ice) 80 = 60 – 2t
m1= 100 gm (ice)
t < 0. Which tell us, some of ice is not melt. Then Again assume, m mass
t1 = 0º C
m2 = 100 gm (water)
t2

300

= 75 gm finally mass

4
of ice remain (m1) = 25 gm mass of water (m2) = (200 – 25) gm = 175 gm.
of ice is melt m × 30 = 100 × 1 × 60  m =
Now Sice =

d
1
m c (t – t )

  2 2 2 1  S =
water
T
t1
t1

d

T
t1

t2
m 2 c 2 dT
t
= m2 c2 n 1 t 
T
2




System = m2 c2  t 2 – 1  n t1   m 2 c 2  t 2 – 1 – n t 2  Ans.
t
t
t2 
t1 
 1

 1

(b)
t2 = 94ºC Suppose all ice is melt and temp at equailibium is T. Then
0ºC < T < 94º C 100 × 80
dθ
m q
T
1
 
+ 100 (1) T = 100 × 1 (t2 – T)  80 + T = 94 – T  – 6 = – 2T  T = 3º C  Sics  T  t
1
t1
T
m1 c 2 dT
T
m1 c 2 dT
m1 q
dθ
 Sics =
+ m1 c2  n T/t1  Swater = 
= 
 Swater = m2 c2  n T/t2  System =
T
t1
T
t2
m1 q
T
T
m1 q
+ m1 c2  n
+ m2 c2  n
 Ssystem =
+ c2
t1
t1
t2
t1

t 
T
 m1 n
– m 2 n 2 

t1
T

225
dθ
t1
m dT
c
2.217 :  T   T
t2
S ice =

1

 t1

dθ
T
1
t2
 d Here t2 = melting temp. of lead. Slead = m c  n t1/t2 – 1 mq .... (i)
t2
1
t1
Since T is constant because amount of water is very high. =
t2


m c dT 
t1

mq 
1
1
1
mq
t


–
t 1  = t (mc) (t2 – t1) + t  System = mc n 1 t + mq  t
t
1
2
 1
2
1

1
1
 + mc
–
 System = mq 
t 2 
 t1
2.218 : We know
 t2
 – 1 – n t 2
t
t1
 1
 d
= +

t

 + mc  2 – 1

t


 1

m = 5 gm ; t2 = 327ºC




t1 = 0º C
dp
q

where V1 = specific volume of gas. = volume/kg Also
dT
T V1
PV = RT  P dV + V dP = R dT  Again d = dU + Pd V  Cd T = CV dT + (R dT – V d
 q 
dP
  C = C – m q Where m = mass of vapour
P) C = (CV + R) – V
 C = CP – V 
1
P
dT
T
TV 
since amount of vapour taken is 1 mole m = M C = CP –
Mq
T
2.219 : Here at T2 tempretore, all water is in saturated vapour from which is boiling temp of vapour of vapour
if you decrease temp. from T2, vapour will be concurted into liquid. Then  S =
mq
  CP
=
T2
T2

T1
dT
T
dθ
T
Δθ

T2

dθ1
 S
T
T 
mq
 2
 S = T   C P  n  T    Here
2
 1 
 = 1  m = M  S =
Mq
Mq
T

T
 C p n  2 
  C p  n  2 T   S =
1
T2
T2
 T1 

2.7
Transport Phenomena
N
– s/
2.220 : (a) We know fraction of molecules traversing distance S without collisions. N  e
. Now we
0
S
S
want to find sum of fraction of molecules having   S <   1 
<  . Assume x =  1  x < 


N
 N = e–x Now
0


x 1
N

N0


e
x 1
–x


x 1
N

N0


1
e – x dx 


x 1
N
N0

e–x
–1


1
1
 .
e
226
Compact ISC Physics (XII)
2
(b)
Similiar :

x 1
2
N

N0
e
1
–x
1
1 
dx =  –
  0.23
e
e–2 

2.221 : Suppose Intensity of molecules is N0 at S = 0 while N at S =   then 

N
1

 e – Δ/λ   n
N0
η
Δ
Δ
 
Ans.
n η
λ
2.222 : Consider a molecule having velocity v. P(t) = Probability of a molecule experiencing no collision in
time interval (0, t) dt = Probability of a molecule suffers a collision between time t and t + dt. Here is
function of speed (v). 1 – dt = Probability of a molecule suffer no collision between time t and t + dt. Then
dP
 dP 
P (t + dt) = P(t) (1 – dt)  p (t) +   dt = P(t) – P(t) dt 
= – dt 
 P(t)
 dt 
dP
 P(t)

 –  dt
  n P(t) = – t + C  P(t) = – t + C  P(t) = e–t + C at t  0 P(0)  1 because when time interval is
very small, probability of no collision will be approach 1.  1 = eC  C = 0  P(t) = e–t
(b)
Collision time or mean time or Relaxation time () : Probability that a gas molecenle survive for time t
and suffer a collision in next dt time : f(t) = P(t) ( dt)  f(t) = e–t dt


0
Mran time () = 
 0 f (t)
– t
0 e t dt

= 
 =
– t

 0  e dt
f (t) t
2.223 : (a) Mean free path of a gas molecule  
 Then  
 Put value 
Also we know : Mean time ( ) 
 
.  10 – 8
8  1.3  10 – 23  273
1
2  d 2n
. Also we know P = nK T
1.38  10 –23  273
KT
2  d 2P
Ans.
2  (0.37  10 – 9 ) 2  105
= 6.2 × 10–18 m

 Where v = Average speed =
|v|
= 0.13 ns (b) Similar like part (a)   
  28  10 – 3
= 6.2 × 106 m   =
.  106
8  1.3  10 – 23  2  3
  28  10 – 3
= 3.8 hous
Ans.
 KT
m

.  10 –23  273

2  0.37  10 – 9
 10

–9
227
2.224 : We know at STP volume of 1 mole gas is 22.4 lt. Volume occupied by one molecule is :
V1 × 6.023 × 1023 = 22.4 × 10–6 m3  V1 =
22.4  10 –6

.  10 23

22.4  10 –6
R3 =
R3  10–29  R3 = 10 ×

.  10 23
10–30  R = 3 10 × 10–10 distance b/w two molecules = 2R = 2 3 10 × 10–10 m  = 6.2 × 10–8 m

λ
10 –8

 100 time Ans.
2R
10 –10


d
 
2.225 : We know vander wall constant b is four time of actual volume of gas molecule. b = 4 NA
 2
1/3

KT
b
 3b 
KT

3
  and  =
 d = 4 N   d = 
 =
2
2
/ 
 d 
2 d P
A
 4π N A 


 2 N A
 = 
 3b




2/3
3b
2 P

4

 N A 
 KT 


0
 2 P 
0 

Ans.
 RT
Then
M
2.226 : We know acoutic wave speed : v =
 Where =

2
 d n
2.227 : (a) We know =
=
2
 f =
 d n
KT
2
2 πd P
γ RT

M
> P<
=
v
1
 =
f
f
2  d 2 P0 NA
RT
KT
2  d2 
γ RT
1
f =
M

γ RT
M
2γ
γ RT
 f = d2 P0 NA MR T
M
0
Put value P < 0.7 Pa (b) now P = n KT  n
0.7  1.1  10 5
P
m 3  n < 2 × 1014 / cm3 and valume will be assume for one
 if P < 0.7 Pa  n <
– 23
KT
1.38  10
 273
molecules is V1 =   then 2 × 1014 × V1 = 10–6  2 × 1014 ×   = 10–6   0 = 0.2 µ m Ans.
2.228 : (a) We know second, no of collision is 1  1 second no of collision is f =


2
 d n

λ
=
v

  d2
n < v >. Then f =
2
10
 d n < v > put values f = 0.74 × 10 /s
(b) Suppose there are n molecules present then total no of collision persecond (N) =
d2 n 2 < v > 
1
. Also


is taken because in one collision, two molecules are participated.



f (n)  N =


Ans.

228
Compact ISC Physics (XII)
2.229 : We know d = effective diameter of molecule = const 
V = n [  3]  for iso choric process n =
1
2 π d 2n
Total no. of molecules
= const Since V and n const   will
Total Volume
constant hence d = const then const no of collision pur unit time (v)  v =
v  T
v=
(b) Isobaric process : V =


8 RT
1

M
CT
2.230 : Pf = n Pin  
v
1

λ
λ
8 KT
1

m
λ

(a)
  d2 n

2
Ans.
8K PV
 f  P  f increase
m  NR
 KT
2
 decreans
n time
1

time f =
n

Diatomic gas :  d = const  n =
2 d n
 V = Volume of gas. Hence V Ans. Frequency (v) =
For adiabatic process : T V–1 = const = C1  v =
 v  V
  1 
–

 2 
 V
–65
8 RT

M
for isochoric proces d and n will beconst = const  f = v =
2 d P
2.231 : (a) 
v
1



KT
N RT

 T
P = n KT Since d = const  
P
  d2 P
8 RT
1
 v 
M
T
(b) Iso thermal process : 
(a) Isochoric process :
1
CV
Ans.
8 KT
 f increases n time
m
N
where N = Total no. of molecule
V
v
1


CV
8 RT
where C = const.
M
8 RT C1

1
 v 
  1/2
 –1/2
 M V  –1
V.V
V
Ans. v = K V–6/5 .... (i) Where K = const
(b) PV7/5 = K1  V (K1 P)–5/7 put in (i) v = K (K1P)6/7  v P6/7 Ans. V  P–5/7
Ans.
(c) Similiar T –5/2 T3 Ans.
2.232 : Similar like Q : 2.231 (a)  V  v V–(n+1)/2 (b) P–1/n  v P(n+1)/n (c)
 T(n+1)2(n–1)
T1/1–n  v
229
2.233 : (a) No of collision/time = ( f ) =
d2 N
V
2
v
=

2
 d n
8 RT
 f=
M
N
2
 d n V
8 RT
 f=
πM

8 RT
PV
f

= C P1/2 V–1/2 where C = const. No. of frequency per unit volume : f  
= C P1/
πM
NR
V
V–3/2 = const  PV–3 = Const. Compare PVx = const x = –3  C =
R
R
R
R
–


 C=R
r –1
x –1
r –1
4
 4  r – 1
R  3 r 
R  3  1  2/i 
R  4i  2 
R

 . Also we know r = 1 + 2/i  C =

 



 
C=
4  r –1
4  1  2/i – 1 
4  2  
4
 4 (r – 1) 
(2i + 1) Ans.
(b)
f = const PV–1 = Const  x = –1. Similiar as (a) C =
2.234 : no. of particle approach per second toward wall v =
time : dN =
n

 –
n0
n = n0 e
2.235 :
v2 =
also
t

n < v > no. of leaks out from hole in dt


0
 Where 
< v > dat
 v  St
n
1  v S
–
4V
 –
dt  n
t  n = n0 e
n0
4
V
(1)
(2)
T0
n T0
P0, V
P0, V
T0  v12
n T0  v11  v12
P0 V
P0 V  1  n 
1
v11  v12 = v1 + v2 = RT  n  .... (ii) From (i) and (ii) : v 2  n RT

0
0 


< v >    


 KT

m
= n0 e
– t

8 RT
M
4V
 < v> =
S(v)
P0 V
P0 V
v1=
Also n1 < V1 > = n2 < V2 >  v11

RT0
nRT 0
2.236 : D =
Ans.
dN
1


n < v > sdt. Then decrease in concentration : – dx =
V
4

dn
1  vS

n
4
V
– t

R (i + 2)

1
 D= 3
2 d 2 n
1
 1 n

 n
8 KT

πm
n
.... (i)

v1
  2  1  n
v2
1 n

KT
2π d 2 P
 C
T 3/2

P
230
Compact ISC Physics (XII)
 PD 

T= 
 C 
2/3
1
1
< v >  P  n =
3
3
.... (i)  Now n =
 n 
 T =  
 C1 
P 
 2   f 
 Pi 
2
/ 
  2/3 


P
D 
   f  f 
Di 

 Pi
/ 


P
 f Ans.

Pi
C T 3/2
n = C1 T1/2 (a)
P
Iso thermal process : n = const Ans. D =
(PV/VR) 3/2
 D V3/2
P
C T 3/2
V
 RT
 D increase n3/2 times  n = C1
/ 
 n V1/2 n increase n1/2 times Ans.
2.238 : We know from Q: 2.236 D =
= C1  D = C2
3 – 7/5
V 2
 C1 
 r –1 
V 
1
C T 3/2
C T 3/2 V
C


T1/2 V. For adiabatic process : T Vr–1 = const
P
NRT
NR
V = C2
V r –1/2
  –1

1 – 
2 
V 
 C2 V
3– 
2
. For diaatomic gas :   = 7/5  D = C2
 C 2 V 4/5 . If V decrease n times then D will be decreased by 4/5 Also we know  = C3 T1/2 = C3
/ 
v
  = C3 C1 2 V
2.239 : From Q: 2.239 (a)
n = C1 T
x = const
1/2
1– r
2
v
  = C3 C1 2 V–1/5 . If V decrease n times then n will be increase n1/5.
C T 3/2
C  PV

D=
P
P   R
 PV 
= C1   R 






/ 
 P1/2 V3/2 = const  PV3 = const  n = 3
/
 PV = const  n = 1 Ans. (c)
x=

< V > CV  x = n CV  If

 n = const  same as part (b)  n = 1 Ans.
2.240 : We know
2
3
KT
PM

n = C1 T1/2
RT 
2π d P P
2
/ 
 n 
 nt
 PD 
   

 
 C 
 C1 
 ni
.... (ii) From (i) and (ii) :
D n Ans. (b) Isobaric process : D = C
(b)
1
2
2.237 : From Q : 2.236 D =
 PV 
 vR 


8 KT

πm
 
KT
PMRT
2
=
2
m RT π d P
3
1
< V >
3
 
KT M K
 d=
m R π d2
 
2

3

1
3
8 KT
m

1
2
2 d
2
n
=


KT

=
m  d2 n
/ 
KT M K 
m R π 
. Put values d = 0.18 nm
231
1
1
< V >   CV = C xAr  3
3
2.241 : Assume C = 8.7  x =
C
1
3



 

8 KT
 M He
M
M
Ar
He





 Ar
2  d 2Ar

 He
  Ar





 n
 Ar
 n He

Mr
M He
n Ar C Ar
V 




 C He
 V
 C Ar
 V







 He
  Ar





n
 Ar
 n He





 He
8 KT
 M He
 C He
 V
 C Ar
 V
2
2

  d He

d 2Ar

 d 2He
C He
v =
n He
d He
d Ar

=
1/2
d Ar
= 1.7
 Put values d
He
2.242 : We know viscous force : F   A
Ans.
F
dV
 RW 
    (2 R  1)   R  . Moment of force
dz


2  WR 3
F

  : N1  RF 
. Also we know viscous force per unit area : F/A =
R


R
R

 < V > (u1 – u2)  F =
 < V > RW × 2 R 

Now
N1
N11

2π η WR 3  3
ΔR π  R 3 W  V 

6ηR
ΔR   V  R
N11


2
Since final moment is decreased by n time then n = ΔR
N1
N11
  R 
N11
 F
 6
1
1  V    R3 3
2R
2


3 ΔR  R 3  V  R
RR
R
2 KT

2
2 πd n
2

 R3 W < V >
6
2
ΔR 2 πd P
 P=
2 KT
 d 2 R n
 w 
 (2 r  ) r = N  . Here it is assume
2.243 : Viscous torque at distance (r) :   r 
1
 r 
R2
w
w
2π η   w   N1
viscous torque (N1) wil be const N1  = 2r 

r
0
3
2 w =
N1
2
 1
1 
N1
–

 R 2 R 2   = 4  w
2
1 

 1
1 
–

 R2 R2 
2
1 



R1
dr
r
r
r3 
Ans.
a

2.244 : d = A  dV
 dz
  w a4
2h
2π η w
rw

r 3 dr  
r  τ 
 r =  (2r dr ) 
h

 h 

0

r
dr
232
Compact ISC Physics (XII)



2.245 : N =    V   rw  [2 r dr] [r] =   V  w



a4  N =
2.246 :
a
r
3
dr N =
0

< V > w

 RT PM


M

w a4  N = w a4 p
 M RT


2 RT
r
dr
Ans.
dθ
π a 4 dP
dθ
π a 4 dP
dm
 a 4 dP




 
.... (i) 
dt
8 η dx
dt
8 η dx
dt
8 dx
x
4
dx
4
dm  a  PM  dP
 a M  PdP 
Pd P





 .  For iso thermal process :
=
dt
8   RT  dx
8  RT  dx 
dx

P2
Const = C  

Pd P  C dx
P1
0
dm
 a4 M
P22 – P12

 C =
 dt
8  RT
2
 P22 – P12

 2



 Ans.

T1 – T
T – T2
T1 – T
T – T2




2.247 : Heat current in both rod will be same. R
  1
T =
2
R2
1
x1 A
x2 A
(x1 T1/1  x 2 T2 / 2 )
(x1/ 1  x 2 / 2 )
T
T2
T1
Ans.
 1 , x1
 2 , x2
2.248 : For equivalent heat conductivity (x) heat resistance of system will be equal to that of one rod. Then
   2

2
   2
  

x
=
xA
x1 A
x2 A


 
x1
x2
2.249 : Assume x =
T2

T1
αA
dT  C
T
d
xA dT

 –
 Const  C  – x A d T = C dx  –
 T 1 > T 2 
dt
dx
T

 dx 
Ans.
– A n
0
 
α
T2
αA
T
T
n 1
n 2 . Heat density C A 
= C   C = –
T2

T1

T1
T1
T
–

T1
αA
α
dT 
n
T

T2

x
 T1 
x/
  dx
T

T 
 T = T1  2 T 
 2 0
1


Ans.
x
dx
233
T
2.250 : Method : 1 Assume temp of two chunks at t = 0 are T11 and T21
while
at
time
and T2. Then d
T1

– C1
=

d T2  T1 =
T21
T2
t
t,
T21
are
dT
2
C1
2
1
2
Method : 2 (Reduce heat capacity) C =

 T0
dT – xS

T
C

T2
xS (T1 – T2 )
dt = C2 d T2 

C1 C2
= Reduce heat capacity. Let asC1  C2
T
t=t
d x ST
x ST

 d =
dt = – C dT 
dt


– x st
T
– xS

t  T =  T0 e C T =  T e
dt  n
0
 T0
C
0
T = T0 e–t
T
 – x S dT = C1 dr – CS

T
1
r
2
dT  C1
T1
d r –
0


3/2
3/2

Put r =    T = T 2  – T2  T1
 2 CS

 3

 3/2
 (T 1 – T 23/2



)


– x s  1
1 

t
  C1 C 2 
. Assume
Ans.
dθ
x s dT
–
2.251 : x = C T  Assume T1 >T2 
= Const = C1
dt
dr

3r
 T13/2 –
2 SC


T1
t
 1
1  sx



 C1 C 2  


C2
 1 x1 S

 sx
where α   1 C  1 C 
1
2 

 T = (T) 0 e – t
sume temp. of object is T which provide temp. difference at time t.
T
1
2
T
C1
T1
C2
( 1 – T2) + T11 .... (ii) 
C1 T2
 dt  C 2  C 2 (T1 – T )2  T1 – T
0
temps
xS (T1 – T2 )
dt = – C1 d T1 = C2 d T2 .... (i) Now

T2
d T1  C 2
T11
xs

t
1
1
3
2 CS
dr
r
T2
T1

 C1 r
 C S 3/2
(T – T13/2) = C1 r  T3/2 = T13/2 –
.
2CS
3
C1

C1 =
/
 3/2 r
3/2
3/2 
 T =  T1 – (T1 – T2 ) 




C S
– T13/2  T23/2
3
/
Ans.
T
=
234
Compact ISC Physics (XII)
  PM 
 RT 
KT


2

M
2  d P   RT 




2.252 : We know x =
< V > CV  x =


 
 

o
 RT  K M R i
 M   2  d2 R 



x=

2
 RT
M
0


2.253 : We know 
  d2 n
difference hence temp of mixture T =
q=


KM i
q =
 d2
2 i R 3/2 (– T 23/2  T13/2 )
   /   d2 N A
  d2 P
,P
0
PM
RT
T1
 R i

 2


 (t – t
1


t2
0
2
)
 q=  P  V  M

R
i
(t 1 – t 2 )
2
d
 dT 
= – x 2 r  
 = const. = C 
dt
 dr 
dr
–2πx
(T – T1) =  n r/R1 .... (i). To find C : Put T = T2 and
r 
C
R1
dr
R2
r
R1

2 x (T1 – T2 )
2 x (T1 – T2 )
C
n R1/r
n R 1/r  T = T +
. From (i) : T = T1 +
1
(

n R 2 /R1 ) 2 x
2π

x
n R 2 /R1
n
 T = T1 + (T1 – T2)
n
R1
R2
r
R1
d
– x 4 r 2 dT

= Const = C  For Calculation of C put r = R2  T = T2
dt
dr
T
 – 4  x
M
. Since t1 and t2 not very large
r
dT 
r = R2 C =
2.255 :
From Q: 2.51 :
 d2
0
0
R
M
K M i
t1
KT
2.254 : Suppose length of cylinder is  .
T
RT
RM
t1  t 2
KT
 
. For rarefied gas
2
2 π d2 P

1
<V> PCV (t1 – t2)  q =
<V>

6
–2πx
C


 x =
2
2 C S 3/2
T1 – T23/2  q = C = 2 S  T13/2 – T 23/2  1
C1 =
1
 3
3
3 


 x =

0
KMi
d
o
Ri

 2

r
dr
 dT  C  r 2
T1
R1
 
1 
 – 4  x (T – T 1) = C  R – R    C =
1
 2
4 x R1 R 2 (T2 – T1 )
C
From (i) : T = T1 +
R1 – R 2
4π x
1 1 
 –
  T = T1 +
 r R1 
T2
T1
R2
 R1 R 2 


 R – R  (T2 – T1)
2
 1
r
R1
dr
1 1 
 –

r R 
1

235
T
d
dT
 – x (2 r)
 W r 2 
2.256 :
dt
dr
 T = T0 +
2.257 :

(At steady state)  – 2x
r

dT  W r dr
T0
R
r
dr
W
(R2 – r2) Ans.
4x
d
dT
4
 – x 4  r2
 w   r 3  At steady state  – x
dt
dr
3
T

T0
W
W
T = T0 –
(R2 – r2)  T = T0 +
(R2 – r2)
6x
6x
dT 
W
3
r
 r dr 
R
dr
r