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A meter bridge is set up as shown in figure, to determine an unknown resistance $X$ using a standard $10 \Omega$ resistor. The galvanometer shows null point when tapping key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is
Options:
Solution:
2597 Upvotes
Verified Answer
The correct answer is:
$10.6 \Omega$
$10.6 \Omega$
Using the concept of balanced Wheatstone bridge, we have
$$
\begin{array}{rlrl}
& \frac{P}{Q} & =\frac{R}{S} \\
\therefore & & \frac{X}{(52+1)} & =\frac{10}{(48+2)} \\
\therefore & & X & =\frac{10 \times 53}{50}=10.6 \Omega
\end{array}
$$
$\therefore$ Correct option is (b).
Analysis of Question
Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.
$$
\begin{array}{rlrl}
& \frac{P}{Q} & =\frac{R}{S} \\
\therefore & & \frac{X}{(52+1)} & =\frac{10}{(48+2)} \\
\therefore & & X & =\frac{10 \times 53}{50}=10.6 \Omega
\end{array}
$$
$\therefore$ Correct option is (b).
Analysis of Question
Question is moderately tough, because normally end corrections are not taught in the coaching/schools in this type of problem.
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