Q53 / 69 · JEE Advanced 2011 (Paper 1) · Physics
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
- A$10.2 \Omega$
- Correct answerB$10.6 \Omega$
- C$10.8 \Omega$
- D$11.1 \Omega$
Solution
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.
Hence, the correct answer is option: 'B'.