In a meter bridge experiment, resistances are connected as shown in following figure. balancing leng

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3.Current Electricity

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In a meter bridge experiment, resistances are connected as shown in the following figure. The balancing length $l_1$ is $55\, cm$. Now, an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $75\, cm$. The value of $x$ is

A

$\frac{{54}}{{13}}\,\Omega $

B

$\frac{{20}}{{11}}\,\Omega $

C

$\frac{{48}}{{11}}\,\Omega $

D

$\frac{{11}}{{48}}\,\Omega $

Solution

For the given metre bridge,

$\frac{P}{Q}=\frac{\ell_{1}}{100-\ell_{1}}$

$\ell_{1}=55 \mathrm{\,cm}$

or $\quad 100-\ell_{1}=45 \mathrm{\,cm}$

$\because $ $\mathrm{P}=3\, \Omega$

or $\mathrm{Q}=3 \times \frac{45}{55}=3 \times \frac{9}{11}=\frac{27}{11}\, \Omega$

when $\mathrm{x}$ is connected in series with $P, \ell_{1}=75 \mathrm{\,cm}$

or $\frac{\mathrm{P}+\mathrm{x}}{\mathrm{Q}}=\frac{75 \mathrm{\,cm}}{25 \mathrm{\,cm}}$

$3+x=3 \times \frac{27}{11}$

$\mathrm{x}=\frac{81}{11}-3$ or $\quad x=\frac{48}{11} \,\Omega$

Standard 12

Physics

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