In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found thatbalancing length is = 25 cm. If R is replaced by a wire of half length and half diameter that of R ofsame material, then the balancing distance is (in cm). What is the value of .

JEE MainPhysicsModerateCurrent Electricity

In a meter bridge experiment S is a standard resistance. R is a
resistance wire. It is found thatbalancing length is $\mathcal{l}$ = 25
cm. If R is replaced by a wire of half length and half diameter that of
R ofsame material, then the balancing distance is $\mathcal{l}'$ (in cm).
What is the value of $\mathcal{l}^{'}/10$ .

Show answer and solution

\frac{x}{R} = \frac{75}{25} = 3

R = \frac{\rho\mathcal{l}}{A} = \frac{4\rho\mathcal{l}}{\pi d^{2}}

R^{'} = \frac{4\rho\left( \frac{\mathcal{l}}{2} \right)}{\pi\left( \frac{d}{2} \right)^{2}} = 2R

then

\frac{X}{R^{'}} = \left( \frac{100 - \mathcal{l}}{\mathcal{l}} \right)

\frac{100 - \mathcal{l}}{\mathcal{l}} = \frac{X}{2R} = \frac{3}{2}

\mathcal{l =}40.00cm

Source: Grass-G / P-12 Physics paper 2 Nov.docx

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