Question

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

Answer 1

Answer: 40.00

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