Question

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 _________.

Answer 1

In balancing
$\frac{R}{S} = \frac{25}{75}$
New resistance $R' = \frac{\rho\ell}{A}$
$= \frac{\rho\times\frac{\ell}{2}}{\frac{A}{4}} = \frac{\rho \ell }{2}\times4A$
$R' = 2R$
$\frac{2R}{S} = \frac{\ell'}{100-\ell'}$
$2\times \frac{1}{3} = \frac{\ell '}{100-\ell '} = 3\ell' = 200 - 2\ell'$
$5\ell ' = 200$
$\ell' = 40$
$\therefore $ Correct answer $40$