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In a meter bridge experiment $\mathrm{S}$ is a standard resistance. $\mathrm{R}$ 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 $\left.l^{\prime} \text { (in } \mathrm{cm}\right)$ will now be
$36$
$37$
$33$
$40$
Solution
In balancing
$\frac{R}{S}=\frac{25}{75}$
New resistance $\mathrm{R}^{\prime}=\frac{\rho \ell}{\mathrm{A}}$
$=\frac{\rho \times \frac{\ell}{2}}{\frac{\mathrm{A}}{4}}=\frac{\rho \ell}{2} \times 4 \mathrm{A}$
$\mathrm{R}^{\prime}=2 \mathrm{R}$
$\frac{2 \mathrm{R}}{\mathrm{S}}=\frac{\ell}{100-\ell^{\prime}}$
$2 \times \frac{1}{3}=\frac{\ell^{\prime}}{100-\ell^{\prime}}=3 \ell^{\prime}=200-2 \ell^{\prime}$
$5 \ell '=200$
$\ell^{\prime}=40$
