33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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50 NEET-AIPMT Chapterwise Topicwise Solutions Physics
73. (c) : Here, resistance of the galvanometer = G
Current through the galvanometer,
IG = 02 I = 02 .
I = 1
. % of
I
100 500
\ Current through the shunt,
1 499
IS
= I− IG
= I− I = I
500 500
As shunt and galvanometer are in parallel
\ I G G = I S S
⎛ 1 ⎞ 499
G
I G S S
⎝
⎜
500 ⎠
⎟ = ⎛ ⎝ ⎜ ⎞
500 ⎠
⎟ or =
499
Resistance of the ammeter R A is
1 1 1 1 1 500
= + = + =
RA
G S G G G
499
RA = 1 G
500
Vg
74. (a) : S =
( I − Ig
)
Neglecting I g
V
−3 g 25 × 10 V
∴ S = =
= 0. 001Ω
I 25 A
75. (d) : Let resistance R is to be put in series with
galvanometer G to keep the main current in the circuit
unchanged.
GS
∴ + R=
G
G + S
2
2
GS G + GS−GS
G
R= G− ⇒ R =
=
G + S
G+
S G+
S
76. (a) : Here, Resistance of galvanometer,
G = 100 W
Current for full scale deflection, I g = 30 mA
= 30 × 10 –3 A
Range of voltmeter, V = 30 V
To convert the galvanometer into an voltmeter of a given
range, a resistance R is connected in series with it as shown
in the figure.
From figure, V = I g (G + R)
or R = V −G
Ig
30
= − 100 Ω= 1000 − 100 = 900 Ω
−3
30 × 10
77. (c) : iG = (I – i)S where G is the galvanometer
resistance and S is the shunt used with the ammeter.
1.0 × 60 = (5 – 1)S where S is the shunt used to read a 5 A
current when the galvanometer can stand by 1 A.
10 . × 60
S = = 15 Ω in parallel.
4
78. (b) : Total initial resistance
= R G + R 1 = (50 + 2950) W = 3000 W
e = 3 V
3V
∴ Current = = 1× 10 −3
A=1mA
3000 Ω
vvv
If the deflection has to be reduced to 20 divisions, current
2
i = 1 mA × as the full deflection scale for 1 mA = 30
3
divisions.
2
3 V= 3000 Ω× 1 mA =xΩ×
mA
3
⇒ x = 3000 × 1 × 3 = 4500 Ω
2
But the galvanometer resistance = 50 W
Therefore the resistance to be added
= (4500 – 50)W = 4450 W.
79. (a) : Let the shunt resistance be S.
Given: I = 750 A,
I g = 100 A, R G = 13 W
From the figure,
I g R G = (I – I g )S
or 100 × 13 = [750 – 100]S
or 1300 = 650 S
\ S = 1300/650 = 2 W
80. (d) : The total current shown by the galvanometer is
25 × 4 × 10 –4 A.
\ I g = 10 –2 A
The value of resistance connected in series to convert
galvanometer into voltmeter of 25 V is
R
V 25
= − G = − 50 = 2450 Ω.
I
−2
g 10
81. (a) : Voltmeter is used to measure the potential
difference across a resistance and it is connected in
parallel with the circuit. A high resistance is connected
to the galvanometer in series so that only a small fraction
(I g ) of the main circuit current (I) passes through it.
If a considerable amount of current is allowed to pass
through the voltmeter, then the reading obtained by
this voltmeter will not be close to the actual potential
difference between the same two points.
82. (b) : The shunt and galvanometer are in parallel.
1 1 1 18
Therefore, = + or R q =
R eq 9 2 e Ω
11
Using Ohm’s law, V = IR eq = 1 × 18 18
11
= 11
V.
\ Current through shunt = V R s
18 / 11 9
= =
2 11 08 . amp
83. (a) : To convert a galvanometer into ammeter, one
needs to connect a low resistance in parallel so that
maximum current passes through the shunt wire and
ammeter remains protected.