33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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66 NEET-AIPMT Chapterwise Topicwise Solutions Physics
Hints & Explanations
1. (c) :
T
V = V0 for 0 ≤t
≤
2
T
V = 0 for ≤t ≤ T
2
12 /
12 /
⎡ T ⎤ ⎡ T/
2 T
⎢ 2
∫V dt ⎥
∫ V0 2
⎤
⎢ dt+
∫ () 0 dt ⎥
T
Vrms = ⎢ 0 ⎥ ⎢ 0
/ 2 ⎥
⎢ ⎥ = ⎢
⎥
T
T
⎢ ⎥ ⎢
⎥
⎢ ∫ dt ⎥ ⎢ ∫ dt ⎥
⎣⎢
0 ⎦⎥
⎣⎢
0 ⎦⎥
= ⎡ ⎣ ⎢ V ⎤ ⎡
⎥ = ⎛ ⎞
⎤
⎢
⎦ ⎝
⎜
⎠
⎟⎥ = ⎡
⎣ ⎦ ⎣ ⎢ ⎤
0 2 12 /
⎥
T t T V T V
0 2 0 2 12 /
0 2 12 /
[] / T 2 2 ⎦
V0
∴ Vrms
=
2
I
2. (d) : Irms = 0 2
3. (a) : The major portion of the A.C. flows on the
surface of the wire. So where a thick wire is required,
a number of thin wires are joined together to give an
equivalent effect of a thick wire. Therefore multiple
strands are suitable for transporting A.C. Similarly
multiple strands can also be used for D.C.
4. (b) : The situation is as shown
in the figure.
As the iron rod is inserted, the
magnetic field inside the coil
magnetizes the iron, increasing the
magnetic field inside it. Hence,the inductance of the coil
increases. Consequently, the inductive reactance of the
coil increases. As a result, a larger fraction of the applied
AC voltage appears across the inductor, leaving less
voltage across the bulb. Therefore, the brightness of the
light bulb decreases.
5. (a) : In an A.C. circuit with inductance coil, the
voltage V leads the current I by a phase difference of
90° or the current I lags behind the voltage V by a phase
difference of 90°. Thus the voltage goes on decreasing
with the increase in time as shown in the graph (a).
6. (c) : Here, C = 40mF = 40 × 10 –6 F
V rms = 200 V; u = 50 Hz
The value of the current, I rms = ε rms
= εrmsωC
1
ωC
or I rms = 200 (2p × 50) × (40 × 10 –6 ) = 2.51 A
( w = 2pu)
7. (d) : When an ideal capacitor is connected with an ac
voltage source, current leads voltage by 90°. Since, energy
stored in capacitor during charging is spent in maintaining
charge on the capacitor during discharging. Hence over a
full cycle the capacitor does not consume any energy from
the voltage source.
8. (d) : The given equation of alternating voltage is
e= 200 2 sin 100t
...(i)
The standard equation of alternating voltage is
e = e 0 sinwt
Comparing (i) and (ii), we get
−1
e 0 = 200 2 V, ω = 100 rad s
The capacitive reactance is
XC = 1
C
= 1
ω 100 × 1× 10 −6
Ω
The r.m.s. value of the current in the circuit is
( )
vr.m.s.
e0
/ 2 200 2 / 2
ir.m.s.
= = =
XC
1/
ωC
−6
( 1/
100 × 10 )
= 200 × 100 × 10 –6 A = 2 × 10 –2 A = 20 mA
...(ii)
X
9. (c) : X = 1 and X′ = 1 ∴ X′ =
C ω
4Cω
4
10. (c) : When L is removed
XC
XC
tan f = ⇒ tan π = ...(i)
R 3 R
When C is removed,
XL
X
tan f =
L
⇒ tan π = ...(ii)
R 3 R
From (i) and (ii), X C = X L .
Since, X L = X C , the circuit is in resonance.
Z = R
Power factor, cos f = Z R
= =1
R R
11. (a) : When circuit is connected to an AC source of
12 V, gives a current of 0.2 A.
12
\ Impedance, Z = = 60 Ω
02 .
When the same circuit is connected to a DC source of
12 V, gives a current of 0.4 A.
12
\ Resistance, R = = 30 Ω
04 .