In a meter bridge experiment S is a standard resistance.

Question:

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

Solution:

(40) For the given meter bridge

$\frac{R}{S}=\frac{\ell_{1}}{100-\ell_{1}}$ Where, $\ell_{1}=$ balancing length

$\Rightarrow \quad \frac{R}{S}=\frac{25}{75}=\frac{1}{3}$               ....(1)

New resistance,

$R^{\prime}=\frac{\rho \frac{\ell}{2}}{\frac{A}{4}}=\rho \frac{\ell \times 2}{A}$        $\left(\because \mathrm{R}=\rho \frac{\ell}{\mathrm{A}}\right)$

$\Rightarrow R^{\prime}=2 R$

$\frac{R^{\prime}}{S}=\frac{\ell_{2}}{100-\ell_{2}}$

$\Rightarrow \quad \frac{2 R}{S}=\frac{\ell_{2}}{100-\ell_{2}}$

$\Rightarrow \quad 2 \times \frac{1}{3}=\frac{\ell_{2}}{100-\ell_{2}}$            Using (i)

$\Rightarrow \quad \ell_{2}=40 \mathrm{~cm}$

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