In the circuit shown R 1- R 2-10 - and resistance per unit length of wire PQ -1 - - cm and length PQ -10 cm- If R 2 is made 20 - then to get zero deflection in galvanometer- S is midpoint of wire PQ-A- The jockey at P can be moved towards right 2 cmB- The jockey at Q can be moved towards left 2 cmC- The jockey at S can be moved towards left a distance 5-3 mD- The jockey at all position fixed and R 1 should be made 20 -

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In the circui...

Question

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

The jockey at P can be moved towards right 2 cm

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The jockey at Q can be moved towards left 2 cm

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The jockey at S can be moved towards left a distance

\frac{5}{3} m

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The jockey at all position fixed and

R_{1}

should be made 20

Ω

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Solution

The correct options are
A The jockey at P can be moved towards right 2 cm
C The jockey at S can be moved towards left a distance $\frac{5}{3} m$
D The jockey at all position fixed and $R_{1}$ should be made 20 $Ω$

equivalent diagram is as shown is P is moved 2cm right them $R_{1} = 12, R_{3} = 3$
$\frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}$ (Hence wheat stone will be balanced)
If s is moved left $\frac{5}{3}$ cm then $R_{3} = \frac{10}{3}$ and $R_{4} = \frac{20}{3}$ hence $\frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}$ (hence wheat stone will be balanced)

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

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