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A resistance $\mathrm{R}$ is to be measured using a meter bridge, student chooses the standard resistance $S$ to be $100 \Omega$. He findsthenull point at $l_1=2.9 \mathrm{~cm}$. He is told to attempt to improve the accuracy.
Which of the following is a useful way?
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Which of the following is a useful way?
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The correct answer is:
He should change $\mathrm{S}$ to $3 \Omega$ and repeat the experiment
He should change $\mathrm{S}$ to $3 \Omega$ and repeat the experiment
Adjusting the blance point near the middle of the bridge, i.e.. when $l_1$ is close to $50 \mathrm{~cm}$. requires a suitable choice of $S, R$ is unknown resistance :
Since, $\quad \frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{R} l_1}{\mathrm{R}\left(100-l_1\right)}$
$\frac{\mathrm{R}}{\mathrm{S}}=\frac{l_1}{100-l_1} \quad$ or $\quad \mathrm{R}=\mathrm{S}\left[\frac{l_1}{100-l_1}\right]$
$\mathrm{R}=\mathrm{S}\left[\frac{2.9}{97.1}\right]$
So, here, $\mathrm{R}: \mathrm{S}=2.9: 97.1$ implies that the $\mathrm{S}$ is nearly 33 times to that of $\mathrm{R}$. In orded to make this ratio $1: 1$ it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$,
Since, $\quad \frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{R} l_1}{\mathrm{R}\left(100-l_1\right)}$
$\frac{\mathrm{R}}{\mathrm{S}}=\frac{l_1}{100-l_1} \quad$ or $\quad \mathrm{R}=\mathrm{S}\left[\frac{l_1}{100-l_1}\right]$
$\mathrm{R}=\mathrm{S}\left[\frac{2.9}{97.1}\right]$
So, here, $\mathrm{R}: \mathrm{S}=2.9: 97.1$ implies that the $\mathrm{S}$ is nearly 33 times to that of $\mathrm{R}$. In orded to make this ratio $1: 1$ it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$,
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