JEE Main 2017 · 8 Apr · Shift 1 · Physics Q30

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PhysicsSingle correct+4 / -1Question 30 of 90

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key $K_3$ is open. The balance point is at a length $l_1$ cm from P when two way key $K_3$ is plugged in between 2 and 1, while the balance point is at a length $l_2$ cm from P when key $K_3$ is plugged in between 3 and 1. The ratio of two resistances $\dfrac{R_1}{R_2}$ , is found to be :

Potentiometer wire PQ driven by cell E₁ through rheostat Rh₁ and key K₁. A second circuit of cell E₂, rheostat Rh₂, key K₂ and ammeter A passes current through R₁ and R₂ in series; the galvanometer G and two-way key K₃ (terminals 1, 2, 3) connect the junctions to a jockey on PQ.

\dfrac{l_1}{l_1+l_2}

\dfrac{l_2}{l_2-l_1}

\dfrac{l_1}{l_1-l_2}

\dfrac{l_1}{l_2-l_1}

Answer and worked solution

Official answer: D

The current through $R_1$ and $R_2$ stays the same. With $K_3$ between 2 and 1 the potentiometer balances the drop across $R_1$ : $IR_1=kl_1$ . Between 3 and 1 it balances $R_1+R_2$ : $I(R_1+R_2)=kl_2$ . So
$\frac{R_1}{R_2}=\frac{l_1}{l_2-l_1}.$

Topics:Potentiometer

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 30