In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is l = 25 cm. If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance l' (in cm) will now be \underline\hspace2cm .

Current Electricity PYQ — JEE Main 2020 · 9 Jan · Shift 2 · Q41 with Solution

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Correct answer: 40

For a meter bridge at balance point,

\frac{R}{S} = \frac{l}{100-l}

where:

Given $l = 25\,\text{cm}$ ,

\frac{R}{S} = \frac{25}{75} = \frac{1}{3}

So,

R = \frac{S}{3}

Resistance of a wire is

R = \rho \frac{L}{A}

Now the new wire has:

So its cross-sectional area becomes

A' = \pi \left(\frac{d'}{2}\right)^2 \propto (d')^2

Thus,

A' = \frac{A}{4}

Therefore new resistance $R'$ is

R' = \rho \frac{L'}{A'} = \rho \frac{L/2}{A/4} = 2\rho \frac{L}{A} = 2R

So,

R' = 2R = 2\cdot \frac{S}{3} = \frac{2S}{3}

Now,

\frac{R'}{S} = \frac{l'}{100-l'}

So,

\frac{2}{3} = \frac{l'}{100-l'}

Cross-multiplying,

3l' = 2(100-l')

3l' = 200 - 2l'

5l' = 200

l' = 40

The new balancing length is

\boxed{40\,\text{cm}}

Stored correct answer = $40$

This matches the derived answer.