EXERCISE # O-1
1.
Ans. (B)
–
– E0
–
–
–
–
–
–
–
O–
–
–
–
–
–
Î0
K
+
E0 +
+
+
+
+ x
+
+
+
+
3d
(1) Direction remain same
(2) Potential continuously increase
2.
Ans. (D)
C
2
C
C
C
C
A
Ceq =
3.
B
11C
7
Ans. (B)
A
Q
Q
as VA–VB = 0, VC–VA = IR1, VA–VD = IR2, VC – VB =
, VB–VA=
C1
C2
R1
I
C
G
C1
B
\ IR1 =
Q
C1 ....(i)
IR 2 =
Q
C 2 ...(ii)
Þ
1
R1 C 2
=
R 2 C1
I
R2
D
C2
K
4.
Ans. (B)
During time 't2' capacitor is discharging with the help of resistor 'R'
\ q = q0e–t/RC
V = V0 e–t/RC [ Q Q=CV]
As V0 =
2V
V
;V=
3
3
t2 = RC ln2
During time 't1' capacitor is charging with the help of battery.
\ q = q0 (1-e–t/RC) or V = V0 (1-e–t/RC)
2V
V
;V=
3
3
as V0 =
5.
t1 = RC ln2
T = t1 + t2 = 2RC ln2
Ans. (A)
i=
6.
V
V
VS
=
=
R æ r´ d ö rd
ç
÷
è S ø
Ans. (C)
q = 2 × 5 = 10 µC
Ceq = 1
2
Ceq =
5
10
7
Vof5 may be more than 2.
2
EXERCISE # O-2
1.
Ans. (B, C, D)
(B) First C increases
1
1 Q2
2
CV &
(C)
2
2 KC
V
in both case potential difference across plate is same.
d
Ans. (A, B, C, D)
Q – Remain same
Capacitance become KC
Ans. (A, B, C)
Potential across A & B should same.
(D) E =
2.
3.
t
4.
5.
6.
7.
8.
V0
= V0 (1 - e RC )
2
Ans. (ABD)
At t= 0, capacitor act a short circuit & current in resistor branch = 0.
At t= ¥, iR =2 mA; iC = 0.
Paragraph for question nos. 5 to 7
Ans. (A)
Electric field will be from A to B
Ans. (D)
Electric field can be either from A to B or from B to A depending on which is plate is positively
charged.
Ans. (C)
ò E·dl is negative of potential difference and it depend on initial and final point.
Ans. (A)
Current flow is defined with the flow of charge.
L®R
–
+
i
L®R
i
3
9.
Ans. (C)
r
t= Î
0
f = qE
Work energy Þ
1
d
mV2 = F.
2
2
T
2
V
R
T = mg +
mg
10.
mV2
Fd
= mg +
R
R
Ans. (B)
1 Q2
Uq + UR = U0 =
2 C
UC = UR
U0 1 Q2
=
2 2 C
Ans. (A) P, R, S; (B) P, R, S; (C) R; (D) Q, R]
UL =
11.
(A) C ®
C0
,
2
q=
C0 V0
2
AC 0
KC 0
Q V
=
, V = = 0 (K + 1)
2d - d + d / K (K + 1)
C 2K
(B) Same as A
(C) C' = KC0
Q = = KC0V0
C=
Cnew =
V=
KC 0
(K + 1)
KC 0 V0
(K + 1) = V (K + 1)
0
KC 0
(D) C' = KC0
Cnew =
KC 0
(K + 1)
Q = Cnew × V0 =
KC 0 V0
(K + 1)
4
EXERCISE # S
1.
2.
Ans. 9
Charge on 3µF = 3µC
Charge on 6µF = 6µC
by charge conservation = 9µC
Ans. 8
-s1
v2
s1
s2
v1
-s2
3.
V1 = V2 Þ
s1
s
s
8
5 ´ 10 -2 = 2 ´ 8 ´ 10 -2 Þ 1 =
e0
e0
s2 5
Ans. 30 V
(1)
(2) 3c
3c
3c
6v 2v
4.
Charge conservation on plates 1, 2 & 3
CV = 3C × 8 + C × 6
V = 30
Ans. 2 mA
Q
200 - Q
–3
–3
(d
–
2
×
10
t)
=
AÎ0
AÎ0 (d + 2 × 10 t)
diff.
i(d – 2Vt) + Q(–2V) = –i (d + 2V + t) + (200 – Q) (2V)
Þ 2id = 200 × 2V × 10–6
i = 2000 µA
5.
Ans. –3Q,
–3Q1
+ 20Q 20Q - 10Q 10Q
,
+
- 3Q
3
3
3
3
Q1
–Q1
(Q1–10Q)
–3Q1
(10Q – Q1)
æ 2Q1 ö
2(Q1 - 10Q)
VA – VC = 0 = ç 2A Î ÷ L +
= 2L
2A Î0
0 ø
è
5
Q1 =
20Q
3
–q
10
q
R
6.
Ans.
–iR –
20 i
–(100-q) 100–q
q 100 - q
+
=0
10
20
iR = 5 +
R
3q
20
dq 3 æ 100
ö
=
+q÷
ç
dt 20 è 3
ø
q = 200 (1 – e
7.
-
3t
20R
)
C 0 V0
Ans. V1 = C + C = common potential
0
1
C 0 V1
C 02 V1
=
V2 =
(C 0 + C1 ) (C 0 + C1 ) 2
C 0n
V0 = V
Vn =
(C 0 + C1 ) n
1/ n
C + C1 æ V0 ö
Þ 0
=ç ÷
C0
èVø
6
EXERCISE # O-1
1.
2.
3.
Ans. (D)
C ® Average velocity remain same.
Increase in temperature causes increase in velocity of electrons thus increasing the collision frequency.
Ans. (A)
Any such black box can be replaced by an effective emf and an effective
r
I
resistor connected in series.(see fig.)
\ E = I(R + r) where E is the effective emf and r is the effective
R
E
resistance.
Using the given data we find E = 12 volts and r = 2 ohms.
Therefore R should equal 118 ohms in order to get a current of
0.1 amp.
Ans. (A)
A
C
D
B
4.
I3
E
VA – VB = 2
I3 R3 = 1
BCDE, KVL Þ R2I2 = 2
Ans. (D)
2
æIö
P1 = P2 = P3 = ç ÷ R
è3ø
2
æ 2I ö
P4 = ç ÷ R
è 3ø
2
5.
æIö
P5 = ç ÷ R
è3ø
Ans. (B)
2V
2V
2V
2/3
2
4
=
3
3
Charge Þ q = CV = 4µC
Vacross 3 µF = 2 –
7
6.
7.
Ans. (D)
We have to take two readings one with resistance R and other without R.
A & B ® connection of E is wrong In C ® Key should be with R.
Ans. (A)
Rx =
dR x
( R 0 )( x )
dx dx
+
; ln Rx = ln R0 + ln x – ln (1 – x); R = 0 +
1- x
x 1- x
x
dR x
1
Systematic error is reduced as effect of cosmelting wire is......... R is minimum has x =
2
x
So random error is minimum
EXERCISE # O-2
1.
Ans. (A,B)
l
R = ò rx
0
dx
rx ö
æ
pç r + ÷
l ø
è
2
=
r
pr 2
é
ù
ê
ú
dx
1
r
ê
ú
x
1
r
=
2
pr 2 ê æ l1 ö ú
Now R1 = ò0
rx ö
æ
1+ ÷
pç r + ÷
êë èç
l ø úû
l ø
è
l1
2.
1
3
=
l
1+ 1 4
l
Ans. (B,D)
Direction of current from right of left
V
6W
3
18 + 12 + V
V=6
i
18 + 12 + V = 0
V = –30
3.
Ans. (B,C,D)
i ® will same
i = neA Vd
i 2
J=
, j sE
A
Current through the wire is same iA = iB = iC Q i=neAVd and J =
8
i
A
4.
Sol.
Ans. (A)
Reading of C = V {i in that branch = 0}
Reading of A =
V
R
Þ
R
C
A
5.
Sol.
V
=R
V/R
Ans. (D)
VB = iRammeter = 0 Þ Vcap. = 0
R
V
A
R
6.
All current will pass through ammeter
Ans. (C)
O
R
Sol.
2V
V
R
3V
2
i=
V
2R
V
R
L
R
2V
at t = 0 inductor is open circuit i =
V
Þ
V
2R
V/2
O
3V/2
2V
O
9
p.d. V –
7.
V 3V
=
2 2
Ans. (B)
10V
V
A
Initially V1 + V2 = 10V ...(i)
10V
V1
+ 2V2 = 10V ...(ii)
3
From equation (i) & (ii)
We get V1 = 6 volt, V2 = 4 volt
Finally
\ Final reading =
8.
V
A
V1
= 2 volt
3
Ans. (C)
V1 RA 4
=
=
V2 RV 6 & RA = 2W Þ RV = 3W
9.
Ans. (A)
RV R
V1
RV + R
1
3
= 3 = ÞR= W
RA
2V2 4
5
10.
11.
12.
13.
Sol.
14.
Sol.
Ans. (A)
Ans. (B)
Ans. (C)
dR d l1 d l 2
=
+
R
l1
l2
Ans. (B)
dQ
= k (T – T0)
dt
100 = k(40 – 20)
k = 5W/°C
400 = 5(T – 20)
T = 100°C
So water comes to Bioling point but entire energy is radiated. So it won't boil.
Ans. (B)
dQ
= k(T – T0) = 100 k (40 – 20)
dt
k = 5 W/°C
10
15.
Sol.
dQ 2
= k ( 45 - 20) = 25 × 5 = 125 W
dt
Ans. (D)
For 1st 60 sec, we can use approx method
dT
k
=
(Ta v - T0 )
dt ms + C 0
16.
1.
5
5
[ 75.5 - 20
=
60 ms + 100
Þ
ms + 100 = 3450
m × 4200 = 3350
Ans. (A) ®(P); (B) ®(P,R,S,T); (C) ®(Q,R,S,T); (D) ®(R,T)
EXERCISE # S
Ans. 5
All the elements of circuit are in parallel arrangement
1
1
1
1
1
1
1
4
2
=
+
+
+
+
+
2
+
R
=
Req 40 40 40 40 20 20 40 20 ; eq = 5W; Power = V /R = 5 W
3360
= 0.8 kg
4200
m 0.8 3
=
m = 0.8 l
vol. =
p 1000
Ans. 2
m=
2.
If IG=0 then it becomes a balanced whetstone bridge l1 =
3.
Ans. 6
R
9
V1
x
V2
R
9
R
2
200
2
cm
´ 1m = ´ 100 =
3
3
3
R
V3
R
R
18
V=27 volt
27
x - 9 x - 9 x - 18
+
+
= 0 Þ x = 12 \ V3 = 6 volt
R
R
R
27
4.
Ans. 3
5.
E1
3E
E ´ 40
E ´ 40
=1 Þ 1 = 3
and E2 =
Þ
E2
E2
100
100
Ans. 35
R1
x2
=
Þ x = 85
From the given circuit diagram
R2 (100 ) 2 - x 2
E1 =
R2
y2
Þ y = 50
Also R =
(100 ) 2 - y 2
1
D = (85 – 50) cm = 35 cm
11
EXERCISE # O-1
1.
Ans. (A)
\ Incident angle = reflected angle.
2.
3.
Ans. (D)
as u ® f
as u ® ¥
Ans. (D)
4.
Ans. (A)
m®¥
m®0
A
Optical path = òµdz =
ò (Az
2
+ 1)dz
0
Geometrical path = C
æ AC4
ö
+ C÷ – C
Extra = ç
è 4
ø
5.
6.
Ans. (A)
Due to refraction.
Ans. (C)
1m
7.
Ans. (C)
q
l
x
y
l cos q
=y
µ
l sin q = x
cos2q + sin2q = 1
12
Ans. (C)
9.
Ans. (A)
CORRECTION
Area will be less
10.
Ans. (B)
11.
A light ray strikes a hexagonal ice crystal ice crystal floating in the air as shown in the figure.
//
///
////
8.
//
///
///
///
///
The correct path of ray may be ?
(A)
(B)
incoming ray
(C)
(D)
Ans. (A)
CORRECTION ANS (C)
12. Ans. (A)
13. Ans. (A)
14. Ans. (C)
15. Ans. (A)
16. Ans. (B)
17. Ans. (A)
18. Ans. (B)
19. Ans. (C)
EXERCISE # O-2
1.
2.
3.
4.
5.
Ans. (A,D)
Ans. (B,C)
Ans. (D)
Ans. (C)
Ans. (BD)
13
6.
Ans. (ABC)
7.
Ans. (C)
8.
Ans. (B)
9.
10.
Ans. (D)
Ans. (A)
-1 æ sin i1 ö
m1 sin r1 = sin i1 Þ r1 = sin çè m ÷ø
1
11.
Ans. (A)
m1
m1 R1 sin i1
sin
i
=
sin
r
;
2
2
m2
m 2 R2 m1 = sin r2;
12.
13.
14.
15.
16.
Ans.
Ans.
Ans.
Ans.
Ans.
ö
-1 æ R1
r2 = sin çè m R sin i1 ÷ø
2 2
(BD)
(A,C,D)
(A) PQR (B) Q (C) S (D) ST
(A) PRST (B) Q (C) PQST (D) PRST
(A,B,C)
EXERCISE # S
1.
Ans. q < sin -1 ( 2sin15° )
2.
æ 16
ö
Ans. ç mm ÷
è 25
ø
3.
Ans. y = -
4.
5.
x
+ 0.1
1200
Ans : Real, below principal axis, anywhere b/w P & F1]
qö q
æh
Ans. d1 = tan–1 ç - tan ÷ +
2ø 2
èf
qö q
-1 æ h
d2 = tan ç + tan ÷ 2ø 2
èf
6.
qö
qö
-1 æ h
-1 æ h
& Angle between rays = tan ç + tan ÷ - tan ç - tan ÷
2ø
2ø
èf
èf
Ans. d = 6 cm
8.
4p
mm
9
Ans. (i) l0 = 600 nm, n = 1.5 (ii) i = sin-1 (0.75) = 48.59º
9.
Ans. r1 =
7.
10.
11.
Ans. (i) 2°, (ii)
a
a
a
, r2 =
, r3 =
3
11
50
4GM
Rc 2
Ans. (i) f = – 10 cm, (ii) (10, 2)
Ans.
14