QJEEM18Phc3e9: JEE Main 2018 Physics - Class 12 JEEM18Phc3e9 Solution: In a circuit for finding the resistance of a galvanometer by…

Question

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In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k

Ω

are used. The figure of merit of the galvanometer is 60

μ A /

division. In the absence of shunt resistance, the galvanometer produces a deflection of

θ

= 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of

θ / 2,

is closest to :

Accepted Answer

Current required by unit deflection is 60 $μ$ A.

For, $θ$ = 9 current is I = 9 $\times$ 60 $μ$ A

$\Rightarrow$ I = 540 $μ$ A = 540 $\times$ 10^{$-$ 6} A

Let G is resistance of galvanometer. Then,

$540 \times 10^{- 6} = \frac{6}{(11000 + G)}$

[11000 + G] 90 $\times$ 10^{$-$ 6} = 1

99000 + 9G = 10⁵

9G = 100000 $-$ 99000

9G = 1000

$G = \frac{1000}{9} Ω$

Also, in half deflection method,

$G = \frac{R S}{R - S} \Rightarrow \frac{1000}{9} = \frac{11000 S}{11000 - S}$

$\frac{1}{9} = \frac{11 S}{11000 - S} \Rightarrow 11000 - S = 99 S$

100 S = 11000 $\Rightarrow$ S = 110 $Ω$

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