Rao IIT Academy/ 2014/ HSC - Board Physics (54) / Solutions
KOTA || MUMBAI || SILVASSA || VAPI || BOKARO
HSC -BOARD - 2014
Physics (54)
solutions
SECTION -I
Q.1 (i)
Explanation of capillary action: Suppose that a capillary tube is dipped in a liquid like water which wets the
capillary. Let us consider the situation before the movement of water inside the capillary. The shape of the
surface in the capillary is concave. Let us consider four points A, B, C, D such that (i) A be just above the
curved surface inside the capillary. (ii) B is just be low inside the capillary. (iii) C is just above the plane surface
outside the capillary. (iv) D is just below the plane surface outside the capillary PA , PB , PC , PD be the
pressures at these points.
(1/2)
A


B
C

D
(1/2)
As the pressure on the concave side of the liquid surface is greater than convex sides PA  PB
As the pressure is smae on the both side of a plane surface PC  PD
Now PA  PC = atmospheric pressure
....... (1)
........... (2)
................ (3)
(1/2)
From relation (1), (2) and (3) PD  PB .
But B and D are same horizonal level in liquid, where pressure at D is greater than pressure at B. Therefore
liquid cannot remain in equilibrium and it flows in to the capillary and rises above B till the pressure at B and
D becomes equal. This is the reason why where is rise of liquid inside the capillary tube.
(1/2)
(ii)
E be emissive power of blakc body for wavelength  .
E d  be energy density between wavelength  and   d 
The graph of E d  vs  is called black body spectrum.
E d 
T4
T3
T2
T1
T4
 T3  T2  T1
(1)

labelling
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(iii)
Escape velocity of body from the surface of the earth = 11.2 km/s
ve  2 v0
1 1 .2  1 0 3
= 1.414 v0
v0  7.920 103 m / s
 7.92 km / s
(iv)
n
v
2    2e 
where e  0.3d
for open end correction  2e
 2  0.3d
e  0.6  5  3cm
n
(v)
v
348

 348 Hz.
2  0.47  0.03 2  0.5
1
1
2
2
PV  mnvrms
 Mvrms
3
3
Vrms 
3PV
M
Vrms 
3RT
M
Vrms  T
(vi)
(vii)
{ PV = RT}
Hence proved
  
v   r
 

dv d    dr

 r  
dt
dt dt


d   dr  dv 


,

v
&
a
where
dt
dt
dt
    

a    r   v
(1)
(1/2)
(1/2)
m  1kg
O
T cos 30  mg
mg
T
cos 30
Fc  T sin 30


T
h
B
r
L
T co s 
A
C
T sin 
mg
Fc 
 sin 30
cos 30
mg
 mg  tan 30
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(viii)
m  50 gm , L  12cm
R  2cm
 ml 2 mR 2 
 l 2 R2 
I 

  m 

4 
 12
 12 4 

(1/2)
12  12 4 
I  50 
 
4
 12
(1/2)
 50 13
 650 dyne  cm 2
Q.2
(1)
(i)
Let M be mass of the earth, R be radius of the earth.
g d be gravitational acceleration at depth ' d ' from the earth surface.
g be gravitational acceleration on the earth surfaces.
 be the density of the earth.
S
P
C
(1/2)
‘P’ be the point inside the eaarth at depth ' d ' from earth surfaces.
........... (1)
(since CS =R)
 CS  CP  d ,  CP  R  d
4
G  R3 
GM
4G R 
g 2 , g 3
g
......... (2)

R
3
R2
(1/2)
(1/2)
g d = acceleration due to gravity at depth ' d '
=
G  Mass of the sphere with radius CP
CP 2
4
G  CP 3 
= 3
CP 2
(1/2)
4G CP 
........... (3)
3
Dividing eq. (3) by eq. (2)
g d CP R  d


g
R
R
 gd 
(1/2)
 d
 g d  g 1  R 


(1/2)
(ii) Given
  14  1010 N / m
F  4.2 108 N
A  1 1  1 m2
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Fh
Ax
Fh
x
A


(1/2)
(1/2)
4.2 108 1m
114 1010
(1)
4.2 108

14 1010
 0.3 102 m  3mm
(iii)
(1)
1
m 2 ( A2  X 2 )
2
1
2
2
P.E.  m X
2
 K.E. =
(1/2)
(1/2)
(1)
K .E.  P.E.
1
1
K A2  X 2  KX 2
2
2
2
2

A  2X

(iv)


X2 

X
(1/2)
A2
2
A
(1/2)
2
x

y  0.05sin   20t  
6

5

 0.05sin   20  0.1  
6

(1/2)
 5
 0.05sin   2  
 6
5 

 0.05sin  2  
6 

(1/2)
 7 
 0.05sin  
 6 
(1/2)
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

 0.05sin    
6



 0.05sin    
(1/2)
6

1
 0.05   0.025 m
(1)
2
Q.3 Statement:- According to the principle of parallel axes theorm of moment of inertia, the moment of inertia of
abody about an axes is equal to the sum of (i) it moment of inertia about a parallel axes through its centre of
mass and (ii) the product of the mass of the body and the square of the distances between two axes. (1)
Mathematical form: Let
I 0 be the M.I. of the body about an axis passing through any point O.
I c be the M.I. of the body about a parallel axis passing through its centre of mass C.
M be mass of the body, h be the distance between two axes.
Then according to the theorem of parallel axes of moment of inertia I 0  I c Mh2 .
Proof: Consider a mass element dm situated at the point ‘P’ of the body. ‘C’ be the centre of mass of the
body and ‘O’ be the any point. Joing PO, PC and PQ perpendicular to OC produced as shown in figure. The
axes passing through O and C are perpendicular to the plane of the figure.
(1/2)
dm P
Q
h
O
(1/2)
C
The M.I. of the element about the axis passing through the point O is OP 2 dm
2
 M.I of the body about the axis through the point O will be I 0   OP dm
parallel M.I. of the body about a parallel axis passing through the point C I 0   CP 2 dm
(1/2)
From figure. OP 2  OQ 2  PQ 2  (OC  CQ )2  PQ 2  h 2  2h CQ  CQ 2  PQ 2
[ CQ 2  PQ 2  CP 2 ]
= h 2  2h CQ  CP 2
2
2
2
 OP dm   h dm   2h CQ dm  CP dm
Since  dm = mass of the body = M
And it can be show that  CQ dm  0 as C is the centre of mass. I

Given
r
(1/2)
(1/2)
0
 I c Mh 2
(1/2)
0.25
 0.125mm
2
 0.125 103 m
T  0.0245N / m
h  4 102 m
  28 , g  9.8
T
hrg
2 cos 
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

2T cos 
hrg
(1/2)
2  0.0245cos 28
0.125 103  4 102  9.8
(1)
 882.9 kg / m3
(1)
OR
Consider a string of lenght ‘L’. Let ‘m’ be the mass per unit length of the stirng ‘T’ be tension in the string.
If transveres wave is produced in the string, the velocity of the wave is given by v 
T
.....(1)
m
(1/2)
(1/2)
If the string is plucked in middle, two incident and reflected wave will produce stationary wave. The string will
vibrate in different mode which is called mode of vibrations.
The simplest mode of vibration is as shown in figure, which is called fundamental mode of vibration. Here two
nodes and one antinode is formed. Let  and n be corresponding wavelength and frequency
(1/2)

L  /2

From equation (i) n  2 L  

T
m
1 T
2L m
This is called fundamental frequency

n
Young’s modulus
(1/2)
 v  n  n 2 L
  2L
.....(2)
Y
TL
Al
T
YA
M V

, m
L
L
L

(1)
(1)
AL
 A
L
1
 from equation (2) n 
2L
n
YAl
L
A
Yl
AL
1
2L
(1)
L2  L1  0.20
T2  T1  10%T1
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L
g
T  2

T L

T1

T2
(1/2)
L1
L2
T1
L1

0.9T1
L1  0.2
Squaring both side
1
L1

0.81 L1  0.2
(1/2)
(1/2)
L1  0.2  0.81 L1

L1  0.81L1  0.2
0.19 L1  0.2
0.20
0.19
L1  1.053 m
L1 
Q.4. (i)
(ii)
(iii)
(iv)
(1/2)
(b) Centrifugal force
2E
Y
(b) From 7700 Å to 4 × 106 Å
n1n2
(a) 2n  n
2
1
(c)
(v)
(b)  rad
(vi)
1
(d) 2  n  n 
1
2
(vii)
(b)
7
4
SECTION - II
Q.5 (i) In an atom, electron revolves around the nucleus in circular motion. ‘e’ be the charge on electron. The T be
the time required to complete one revolution.
(1/2)
Circumferece
2 r
T 
(1/2)
 Period of revolution  velocity
v
Circulating current I 
e
ev
I 
T
2 r
(1/2)
Magnetic moment associated with electron circular loop = M = IA  M 
M 
ev
 r2
2 r
ev r
2
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The direction of this magnetic moment is in to the plane of paper. Negatively charge electron is moving in
anticlockwise direction leading current in clockwise direction.
(ii) A photoelectric cell is device which converts light energy into electrical energy. It works on the principle
of photoelectric effect.
>
>
>
>
e
e
e
e
(1)
A
Construction : A photoelectric cell consist a small evacuated bulb. A thin layer of an alkali metal is deposited on inner surface of the bulb. The bulb is made of quartz, if cell is used with ultraviolet light. If the cell is
to be used with visible light only. the bulb is made of ordinary glass. A small portion of the surface of bulb is
left uncoated and serves as a window for incoming light. The coated surface of the bulb acts as cathode. The
anode is in shape of sphere.
(1)
(iii)
M  1.633
M  tan i p

(1/2)
i p  tan 1 M

 tan 1 1.633 
(1/2)
 58º 311
(iv)
(1)
T1  300 K
x1  2.4  105
x2  3.6  105
T2  ?
(1/2)
xT  costant
x1T1  x2T2
xT
T2  1 1
x2



(1/2)
2.4 105  300
3.6 10 5
(1/2)
T2  200 K
(1)
E
(v)
Filament
(cathode)

B
(1)
Incident beam
Reflected
beam
+

Anode
Electron
beam

Detector


d
Atomic
plate
Nickel crystal
Any threee labelling
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(vi)
Transmitter: A transmitter converts a signal produced by the source of information into a form suitable for
transmission through a channel and subsequent reception.
(1)
Receiver : A receiver extracts the desired message signals from the received signals at the channel output.
It consists of a pickup antenna to pick up signal, demodulator, an amplifier and transducer the receiver
reconstruct a recognizable from of the original message signal for delivering it to the user of information. (1)
(vii)
L
1
m  radius

B  4 103 T
e  16 mv
 16 10 3 v
Area covered in one revolution
A   L2
 1 
 

 
2
 1m 2
eB f A
 f 

(1/2)
e
BA
16 103
4 103 1
f  4 Hz
(viii)
(1/2)
E  hv

v
E
h
C

 Wave number

v
1 v

 c

E
hc

2.072 1.6 1019
6.63 1034  3 108

1
 1.667  106 m 1

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(1)
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Q.6.
(i)

Statement: The line integral of the magnetic field B around any closed path is equal to 0 times
 
the total current (I) passing through that closed path
 
  B . dI   0 I
(1)
Magnetic induction along the axis of toroid:
Toroid is a solenoid bent into a shape of hollow doughnut.
Consider a toroidal solenoid of average radius ‘r’ having centre carrying current I. In order to find magnetic
field produced at the centre along axis of toroid due to current flowing through coil, imagine an Amperial
loop of radius ‘r’ and traverse it in clock wise direction.
(1/2)
I current flowing
thro’ coil wound on
toroid
r radius of Amperian
loop
r
0
(1/2)
I
I

According to Ampere’s circuital law,

 B  dL
= µ0 I
Here current I flows through the ring as many times time as there are number of turns. Thus the total current
flowing through toroid is N I , where N is total number of turns.


  B  dL
= µ0 N I —————— (1)


Now, and are in same direction   B  dL

= B  dL

  B  dL
= B ( 2  r)
——— (2)
(1/2)
Comparing equation (1) and equation (2 )
µ0 N I = B ( 2  r)  B =
0 N I
2r
———— (3)
If ‘n’ is the number of turns per unit length of toroid then n =
N
2 r
Substituting this value in equation No (3) we get B = µ0 n I
(ii)
(1/2)
 h 20  2
rn  
n
2 

me


(1/2)
 h 2 0  2
r

 2   me 2   2 


(1/2)
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34 2

r2
 6.63 10 


3.14  9.1 1031  1.6 1019

(iii)
 8.85 1012   2 

2
(1/2)
2
43.96 1068  8.85  1012  4
3.14  9.1 1031  2.56  1038
(1/2)
 2.127 1010 m
(1)
 2.127 A
If a germanium crystal or silicon crystal is doped during manufacture in such a way that half of it is
p-type and other half is n-type, we get p-n junction.
P
N
N
P
(1)
P-N junction with
foward biased
P-N junction with
reversed bais
(a) Forward biased: A battery is connected across p-n junction diode such that, p-type is connected to
positive terminal and n-type is connected to negative terminal, then it is called forward biased. The potential
difference applied should be more than 0.3 V for germanium and more than 0.7 V for silicon. Then holes from
p-type region and electrons from n-type region moves towards barrier and it decreases width and height of
barrier. Hence an electric current flows through circuit. When applied potential is zero, then current also equal
to zero. When P.D. Increases the current is also increases but very slowly. When applied P.D. is more than
potential barrier, current in creases rapidly. This voltage is called knee voltage. The forward biased voltage
at which the current through diode increases rapidly is called knee voltage
(1)
(b) Reversed biased : A battery is connected across p-n junction diode such that, p-type is connected to
negative terminal and n-type is connected to positive terminal, then it is called reversed biased.
The holes from p-type region and electrons from n-type region moves away from junction and it increases
width and height of barrier. Hence there is no flow of an electric current in ideal case. In actualcase current is
very small.  in  A  . The width of potential barrier increases and diode offer very high resistance. The very
small current flows through circuit is called as reverse current.
If reversed bias voltage increased then the kinetic energy of electrons increases. At certain reversed bias
voltage the K.E. of electrons increases enough and they knock out the electrons from semiconductor atoms
Therefore current suddenly increase. That certain reversed bias voltage is called as breakdown voltage. At
breakdown voltage the current increases suddenly and destroy the junction permanently. The reversed bas
voltage at which P-N junction breaks and current suddenly increases is called as breakdown
voltage.
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(iv)
C2 , C2 and C3 are in series.
1
1
1
1
 C C C C
s
1
2
3

(1/2)
1
1
1


6
6
6 10
6 10
6 106
1
3

Cs 6  106

Cs  2 106 F
(1/2)
Cs and C4 are parallel
equivalent resistance

C  C p  C s  C4
(1/2)
 2 106  6 106
C  8 106

v  v1  v2  v3
240  3v1

(1/2)
v1  v2  v3
(1/2)
v1  80 volt
Also charge on C1 , C2 , C3 are same.
Q1  Q2  Q3  Q

Q  C1v1
 6 106  80
Q  480  10 6 C and Q4  6  106  240  1440C
Q.7.
(1/2)
Fresnel’s biprism experiment can be used to determine the wavelength of a monochromatic light. The fig.
shows the experimental arrangement and ray diagram.
(1)
Apparatus : An optical bench is used in the biprism expeiment. It consists of a heavy metal platform about
2m in length and carrying four vertical stands to a slits, the biprism, the lens and eyepiece. The stands can be
moved along the bench as well as perpendicular to the bench.
Adjustment : The slit S is adjusted to be vertical and narrow. It is illuminated by light from a monochromatic
source placed behind it. The light emerging from the slit is made incident on the biprism. The eyepiece stand
is arranged at about 1 m from the slit. The refracting edge of the biprism and the vertical cross-wire of the
eyepiece are arranged parallel to the slit and along a straight line. The sit is observed through the eyepiece
and the biprism is slowly rotated aobut the horizontal axis. When its refracting edge becomes exactly parallel to the slit, the interference pattern consisting of alternate bright and dark bands is seen through the
eyepiece. The slit must be suffciently narrow so that the bands are sharp and clear.
(1)
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Xd
, where X is the band width
D
or the fringe width d is the distance between the coherent sources and D is the distance between the sources
and the eyepiece. The distance D can be measured directly with the help of the scale marked on the optical
bench. To measure the band width X , the micrometer screw fitted to the eyepiece is adjusted such that the
vertical cross-wire is made coincide with one of the bright bands. The micromater reading X1 is noted By
rotating the screw in the same sense, the vertical cross-wire is made coincide with successive bright bands
and the corresponding reading X2,X3, X4.....etc. are noted. The mean value of (X2–X1),(X3–X2),(X4–X3)...
etc are noted. The mean value of (X2–X1),(X3–X2),(X4–X3).... etc gives the mean band width X . To measured the distance between two coherent sources, a convex lens is mounted on the stand between the
biprism and the eyepiece. Without disturbing the slit and the biprism, the eyepiece stand is moved along the
bench so that the distance the slit and the eyepiece is more than four times the focal length of the lens. The
lens stand is moved towards the biprism and its position (L1) is so adjusted that the two magnified images of
the slit are seen through the eyepiece by rotating the micrometer screw. The vertical cross-wire in the
eyepiece is made to coincide with each image and the corresponding reading is noted. The difference
between these two reading gives the distance d1 between the two magnified images. The lens is moved
towards the eyepiece and its position (L2) is so adjusted that two diminished images of the slit are seen
through the eyepiece. The distance d2 between these images is measure as in case of d1 Then the distance
Measurement : The wavelength of a monochromatic light is given by  
between the two coherent sources is given by d  d1 d 2
The wavelength of the monochromatic light is determined by formula i.e  
Xd
.
D
(2)
A
S1
S1
D1
S2
u



D2
S2
b

u

C
v
C
vg 
g
(1/2)
C
w
C C

w  g
 3 2
2.7  107  C   
 4 3
vw  vg 
 9 8 
2.7  107  C 

 12 

C  2.7 12 107
 3.24 108 m / s
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(1)
B

vw 

a
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OR
Transformer :
A transformer is a device with the help of which, a given alternating voltage can be increased or decreased to
any desired value. The first type of transformer which delivers an output voltage smaller than the input voltage
is called a step down transformer. The second type of transformer which delivers an output voltage larger than
the input voltage is called a step up transformer.
(1/2)
Principle : An e.m.f. is induced in the secondary coil due to the changes of current in a neighboring coil.
(1/2)
Construction: A simple transformer is as shown in figure. It consists of a closed, laminated soft iron core on
which two coils having different number of turns are wound. The coils are made of insulated wire. The coil to
which the A.C. input voltage is applied is called the primary coil and the coil across which the A.C. output
voltage is obtained is called the secondary coil.
(1/2)
0
i2
i1
0
N1
E1
P
E2
S
(1/2)
N2
0
0
Working ( Theory ): When an A.C. voltage is applied across the primary, the current through the primary
goes on changing. Due to this, the magnetic flux passing through the core goes on changing. As this changing
flux is linked with both the coils, an e.m.f. is induced in each coil.
Let N1 & N 2 be number of turns of the primary and secondary coils respectively, e1 & e2 be e.m.f. induced
of the primary and secondary coils respectively,  be the magnetic flux linked with each turns of the coils.
(1)
d
d
d
d
e1    N1    N1
& e2    N 2    N 2
(1/2)
dt
dt
dt
dt
e2 N2
N2

The
ratio
e1 N1
N1 is called the turns ratio of the transformer..
R  11 , x  45 cm

 R  100   x  55cm
(1/2)
Let x be theresistance of metal ring
 Equivalent resistance in left gap
1
1
1


Rp x / 2 x / 2

Rp 
(1/2)
x
4
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(1/2)
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From balancing condition
Rp
x
100   x
R

x / 4 45

11 55
x  9 4
x  36 


(ii)
(iii)
(iv)
1
r
(c) Missing electron in valence bond.
(d) Ionosphere
(c) increasing the length of wire
(v)
(a)
(i)
(vi)
(vii)
(1/2)
x 45
  11
4 55

Q.8


(1)
(a)
27

32
(c) Infinite resistance
(a) 0.149 × 10+7
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