Discussion Forum

A meter bridge is set up as shown in the figure, to determine an unknown resistance X using a standard $10 \Omega$  resistor. The galvanometer shows a null point when the tapping key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for ends A and B. The determined value of X is 

Answer:

Explanation:

 Using the concept  of balanced 

 Wheatstone bridge , we have 

 $\frac{p}{Q}=\frac{R}{S}$

 $\therefore$   $\frac{ X}{(52+1)}=  \frac{10}{48+2}$

 $\therefore$    X= $\frac{10 \times 53}{50}$ =10.6 $\Omega$ 

 $\therefore$  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.