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

A
$10.2 \ ohm$
B
$10.6 \ ohm$
C
$10.8 \ ohm$
D
$11.1 \ ohm$
(IIT-2011)
Solution
Using the condition for balanced Wheatstone bridge, we get
$\frac{X}{10}=\frac{(52+1) cm }{(100-52+2) cm }=\frac{53}{50}$
or $X=\frac{53 \times 10}{50}=10.6 \ \Omega$
Standard 12
Physics
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