Question 284 of 380

The resistance per centimeter of a meter bridge wire is $r$, with $X\Omega$ resistance in left gap. Balancing length from left end is at $40cm$ with $25\Omega$ resistance in right gap. Now the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for same settings will be at

A$20cm$B$10cm$C$80cm$D$40cm$

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Answer: (D) $40cm$

For first case:
$\dfrac{X}{r{l}_{1}}=\dfrac{25}{r{l}_{2}}$ ... (i)
For second case:
$\dfrac{X}{\left(2r\right){{l}^{'}}_{1}}=\dfrac{25}{\left(2r\right)l{'}_{2}}$ ... (ii)
From (i) and (ii), we get
$\dfrac{l{'}_{1}}{{l}_{1}}=\dfrac{l{'}_{2}}{{l}_{2}}$ also total length is constant.
Therefore, balance length doesn't change and
$l{'}_{1}={l}_{1}=40cm$