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
IIT 2011, Diffcult
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Using the condition for balanced Wheatstone bridge, we get
$\frac{X}{10}=\frac{(52+1) cm }{(100-52+2) cm }=\frac{53}{50}$
or $X=\frac{53 \times 10}{50}=10.6 \ \Omega$
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