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Q.

In the circuit shown $R_{2} - R_{1} = 10 Ω$ and resistance per unit length of wire $P Q = 1 Ω / c m$ and $P Q = 10 c m$ . If $R_{2}$ is made $20 Ω$ the to get zero deflection in galvanometer. S is midpoint of wire PQ?

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a

The jockey at Q can be moved towards right 2cm.

b

The jockey at S can be moved towards left a distance 5/3 cm.

c

The jockey at P can be moved towards right 2cm.

d

The jockey at all positions fixed and $R_{1}$ should be made 20

answer is A, C.

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Detailed Solution

Equivalent diagram is as shown in P is moved 2 cm right them $R_{1} = 12, R_{3} = 3 \frac{R_{1}}{R_{2}} = \frac{R_{2}}{R_{4}}$ (Hence wheat stone will be balanced)
If s is moved left $\frac{5}{3} c m$ then $R_{4} = \frac{20}{3}$ hence $\frac{R_{1}}{R_{2}} = \frac{R_{2}}{R_{4}}$ (hence wheat stone will be balanced)

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In the circuit shown R2−R1=10Ω and resistance per unit length of wirePQ=1Ω/cm and PQ=10cm. If R2 is made 20Ω the to get zero deflection in galvanometer. S is midpoint of wire PQ?