JEE Main 2014 · 11 Apr · Shift 1 · Physics Q30

← JEE Main 2014 · 11 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 30 of 90

In the circuit diagrams (A, B, C and D) shown below, R is a high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as :

Circuits A-C, each with a cell and key K₁. (A) S in series with G; R with key K₂ connected across G. (B) R in series with G and S; key K₂ across S. (C) S in series with G and R; key K₂ across R.

(D) R in series with G, a cell and key K₁; S with key K₂ connected across G.

Circuit A with

G=\dfrac{RS}{(R-S)}

Circuit B with G = S

Circuit C with G = S

Circuit D with

G=\dfrac{RS}{R-S}

Answer and worked solution

Official answer: D

In the half-deflection method a high resistance $R$ is in series with G. A shunt $S$ (switched by $K_2$ ) is placed across G and adjusted to halve the deflection. With $R\gg G$ the total current hardly changes, so G and S share it equally and $G\approx S$ .

Without that approximation, keeping the battery current fixed in the analysis gives $G=\dfrac{RS}{R-S}$ . Circuit D has this arrangement, and the key gives (4).

Topics:Galvanometer experiments

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 30