JEE Advance - Physics (2011 - Paper 1 Offline - No. 13)
A meter bridge is set up as shown, to determine an unknown resistance X using a standard 10 $$\Omega$$ 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
10.2 $$\Omega$$
10.6 $$\Omega$$
10.8 $$\Omega$$
11.1 $$\Omega$$
Explanation
Corrected length L1 (AJ) = 52 + 1 = 53 cm
Corrected length L2 (BJ) = (100 $$-$$ 52) + 2 = 50 cm
For a balanced Wheatstone bridge,
$${X \over {10}} = {{{L_1}} \over {{L_2}}} = {{53} \over {50}}$$
$$\Rightarrow$$ X = 10.6 $$\Omega$$
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