current-electricity Question 48
Question: Q. 1. A resistance $R$ is to be measured using a meter bridge. Student chooses the standard resistance $S$ to be $100 \Omega$. He finds the null 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?
(a) He should measure $l_{1}$ more accurately.
(b) He should change $S$ to $1000 \Omega$ and repeat the experiment.
(c) He should change $S$ to $3 \Omega$ and repeat the experiment.
(d) He should give up hope of a more accurate measurement with a meter bridge.
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Solution:
Ans. Correct option : (c)
Explanation : To calculate resistance, $R$
$$ \begin{aligned} R & =S\left[\frac{l_{1}}{\left(100-l_{1}\right)}\right] \ & =100\left[\frac{2.9}{97.1}\right] \ & =2.98 \Omega \end{aligned} $$
So to get balance point near to $50 \mathrm{~cm}$ (middle) we have to take $S=3 \Omega$, as here $\mathrm{R}: S=2.9: 97.1$ implies that $S$ is nearly 33 times to $R$. In order to make ratio $R$ and $S=1: 1$, we must take the resistance $S=3 \Omega$.
