The resistances in the two arms of the meter bridge are 5 Ω and R Ω… | NEET 2014 (AIPMT)

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Wheatstone Bridge & Metre Bridge

The resistances in the two arms of the meter bridge are 5 $\Omega$ and R $\Omega$ respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 $l$ 1. The resistance R is

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Answer: (b)

Metre bridge balance: $\dfrac{5}{R} = \dfrac{l_1}{100-l_1}$ .
Shunting $R$ with an equal resistance makes it $\dfrac{R}{2}$ , so $\dfrac{10}{R} = \dfrac{1.6l_1}{100-1.6l_1}$ .
Dividing: $2 = \dfrac{1.6(100-l_1)}{100-1.6l_1} \Rightarrow 200-3.2l_1 = 160-1.6l_1 \Rightarrow l_1 = 25\text{ cm}$ .
$\dfrac{5}{R} = \dfrac{25}{75} \Rightarrow R = 15\ \Omega$ .
So the answer is (b) $15\ \Omega$ .

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