Q73 / 75 · JEE Main 2020 (09 Jan Shift 2) · Physics

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

Correct answer: 40

Solution

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

R = \frac{p l}{A} = \frac{4 p l}{π d^{2}}

R^{'} = \frac{4 p (\frac{l}{2})}{π {(\frac{d}{2})}^{2}} = 2 R

π {(\frac{d}{2})}^{2}

then

\frac{X}{R'} = (\frac{100 - l}{l})

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

l = 40.00 c m

Hence, the correct answer is 40.

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