33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Current Electricity
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32. (c) : Power = 100 W; Voltage of bulb = 200 V and
supply voltage (V s ) = 160 V
Therefore resistance of bulb (R)
2 2
V ( 200)
= = = 400 Ω
P 100
and power consumption (P)
Vs 2 ( 160)
2
= = = 64 W.
R 400
33. (c) : 1 kWh = 1000 Wh
= (1000 W) × (3600 s) = 36 × 10 5 J
34. (d) : Capacitance (C) = 4 mF = 4 × 10 –6 F; Voltage
(V) = 400 volts and resistance (R) = 2 kW = 2 × 10 3 W
Heat produced = Electrical energy stored = 1 2
CV
2
1 −6 2
= × ( 4× 10 ) × ( 400) = 032 . J.
2
35. (d) : Power (P) = 60 W and voltage (V ) = 220 V
Resistance of the filament,
R
V 2 ( 220)
2
= = = 807 Ω
P 60
36. (b) : 2 H 80
H = I Rt or R = = = 2 Ω
2 2
37. (d)
( It) ( 2 × 10)
38. (c) :
Let R be the resistance of the each bulb.
Case (i)
R R 2R
Net resistance of the circuit, R1
= + =
3 3 3
Power consumption by the bulbs = Power supply by the
sources
2 2
E 3E
⇒ P1
= =
( 2R
/ 3)
2R
Case (ii)
Net resistance of the circuit,
R 3R
R2
= + R =
2 2
Power consumption by the bulbs,
E
2
2 E
2
P
P
1 3 3 9
2 = = ⎛ ⎞
3R
2 3 ⎝ ⎜ ⎟ ; = × =
( / ) R ⎠ P2
2 2 4
39. (c) : Resistance of bulb,
V
2 RB = P
= ( 100)
2
500
= 20 Ω
Power of the bulb in the circuit,
P
P = VI; I = ; I = 500
100
= 5A
V B
V R = IR ⇒ (230 – 100) = 5 × R
\ R = 26 W
40. (c) : As both metal wires are of identical dimensions,
so their length and area of cross-section will be same. Let
them be l and A respectively. Then the resistance of the first
wire is
l
R1
= σ 1A
... (i)
l
and that of the second wire is R2
= σ A
... (ii)
As they are connected in series, so their effective
resistance is
l l
R s = R 1 + R 2 = + (using (i) and (ii))
σ1A
σ2A
l ⎛ 1 1 ⎞
= +
... (iii)
A ⎝
⎜ σ ⎠
⎟
1 σ2
If s eff is the effective conductivity of the combination, then
l
Rs = 2 ... (iv)
σ eff A
Equating eqns. (iii) and (iv), we get
2l
l ⎛ 1 1 ⎞
= +
σeff
A A ⎝
⎜ σ1 σ2
⎠
⎟
2 σ2 + σ1
2σσ
1 2
= or σeff
=
σeff
σσ 1 2
σ1+
σ2
41. (c) : The circuit is shown in the figure.
Resistance of the ammeter is
( 480 Ω)( 20 Ω)
R A =
= 19.
2 Ω
( 480 Ω+
20 Ω)
(As 480 W and 20 W are in
parallel)
As ammeter is in series with
40.8 W,
\ Total resistance of the
circuit is
R = 40.8 W + R A = 40.8 W + 19.2 W = 60 W
By Ohm’s law, current in the circuit is
V
I = R
= 30 V
60
= 1
A
2
= 05 . A
Ω
Thus the reading in the ammeter will be 0.5 A.
42. (c) : The current flowing in the different branches of
circuit is indicated in the figure.
2