33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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34 NEET-AIPMT Chapterwise Topicwise Solutions Physics
77. (a) : Terminal potential difference is 2.2 V when
circuit is open.
\ e.m.f. of the cell = E = 2.2 volt
Now, when the cell is connected to
the external resistance, current in
the circuit I is given by
E
I = R r = 22 .
ampere, where r is the internal resistance
+ 5 + r
of the cell.
Potential difference across the cell = IR
22 .
or, = × 5=
1.
8 or, 5 + r = 11/1.8
5 + r
11 110 − 90 10
∴ r = − 5 = = Ω
18 . 18 9
78. (c) : V − E V −12
= I ⇒ = 60
r
−2
5×
10
⇒ V = 15 V
2
79. (a) : i = = 05 . Ampere
4
V = e – ir = 2 – 0.5 × 0.1 = 1.95 V
80. (a) : The output power of a cell is given by
2
V
P = 2
( r + R)
R
Maximum power is delivered to the load only when
the internal resistance of the source is equal to the load
resistance (R). Then
2 2
V V
Pmax = = (where r = R)
4R
4r
81. (a) : Current drawn from the cell is
nε
ε
I = =
nr r
So, I is independent of n and I is constant.
82. (a) : Let r be resistivity of the material of the wire
and r be radius of the wire.
Therefore, resistance of 1 m wire is
ρ()
1 ρ
⎛ ρl
⎞
R = = R =
r r
⎝
⎜Q
2 2
A ⎠
⎟
π π
Let e be emf of each cell. In
first case, 10 cells each of emf
e are connected in series to
heat the wire of length 1 m by
DT(= 10°C) in time t.
( 10ε)
2
∴ =
R t ms∆T
…(i)
In second case,
resistance of same
wire of length 2 m is
ρ()
2 2ρ
R′ = = = 2R
2 2
πr
πr
Let n cells each of emf e are connected in series to heat
the same wire of length 2 m, by the same temperature DT
(= 10°C) in the same time t.
( nε)
2 t
∴ = ( 2ms ) ∆ T
…(ii)
2R
Divide (ii) by (i), we get
2
n
2
= 2 ⇒ n = 400 ∴ n=
20
200
83. (b) : Current in the circuit,
2E
i =
( r1+ r2
+ R) .
As per question, E – ir 1 = 0
E 2E
E
i = or,
r r + r + R
=
1 1 2 r1
or, 2r 1 = r 1 + r 2 + R or, R = r 1 – r 2
84. (d) : Current in the circuit,
V
I = R
= 6
3
= 2 A
Voltage drop across 2 W,
V 1 = 2 × 2 = 4 V
Voltmeter reading = 18 – 4 = 14 V
85. (b) : For maximum current, the two batteries should
be connected in series. The current will be maximum
when external resistance is equal to the total internal
resistance of cells i.e. 2 W. Hence power developed across
the resistance R will be
2 2
2 ⎛ 2E
⎞ ⎛ 2×
2⎞
I R=
⎜ R
2 2
⎝ R+
2r
⎟ = ⎜
⎠ ⎝ 2+
2
⎟
⎠
× = W
8−
4
86. (a) : I =
+ + = 4
1 2 9 12
= 1
3 A
87. (d) :
VP
− VQ
= 4 − 1 × =
3 3 3volt
V AB = V A – V B = 2 × 2 + 3 + 1 × 2 = 9 V
88. (b) : Since the galvanometer shows no deflection so
current will flow as shown in the figure.
VA
12 V 12
Current,
I = =
=
R1
+ R ( 500 + 100)
Ω 600
VB = IR =⎛ ⎝ ⎜ 12 ⎞
A
⎠
⎟ ( 100 Ω)
= 2 V
600
A