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In meter-bridge experiment a resistance of $18 \Omega$ is connected in left gap and an
unknown resistance $\mathrm{R}$ is connected in right gap. The null point is obtained at ' $\ell_{1}{ }^{\circ}$
from left end. If unknown resistance is replaced by $\left(\frac{R}{3}\right) \Omega$, the null point is obtained
at $1 \cdot 5 \ell_{1}$. The unknown resistance is
Options:
unknown resistance $\mathrm{R}$ is connected in right gap. The null point is obtained at ' $\ell_{1}{ }^{\circ}$
from left end. If unknown resistance is replaced by $\left(\frac{R}{3}\right) \Omega$, the null point is obtained
at $1 \cdot 5 \ell_{1}$. The unknown resistance is
Solution:
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Verified Answer
The correct answer is:
$18 \Omega$
(A)
$\frac{18}{\ell_{1}}=\frac{\mathrm{R}}{100-\ell_{1}}$...(1)
$\frac{18}{1.5 \ell_{1}}=\frac{\mathrm{R} / 3}{100-1.5 \ell_{1}}=\frac{\mathrm{R}}{300-4.5 \ell_{1}}$...(2)
Dividing Eq. (1) by Eq. (2) and solving, $\ell_{1}=50 \mathrm{~cm}$
Substituting this value of $\ell_{1}$ in Eq. (1)
we get $\mathrm{R}=18 \Omega$
$\frac{18}{\ell_{1}}=\frac{\mathrm{R}}{100-\ell_{1}}$...(1)
$\frac{18}{1.5 \ell_{1}}=\frac{\mathrm{R} / 3}{100-1.5 \ell_{1}}=\frac{\mathrm{R}}{300-4.5 \ell_{1}}$...(2)
Dividing Eq. (1) by Eq. (2) and solving, $\ell_{1}=50 \mathrm{~cm}$
Substituting this value of $\ell_{1}$ in Eq. (1)
we get $\mathrm{R}=18 \Omega$
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