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Q.

The resistance in the two arms of a meter bridge are $5 Ω$ and $R Ω$ respectively and balancing length is l₁. 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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a

$10 Ω$

b

$15 Ω$

c

$20 Ω$

d

$25 Ω$

answer is B.

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Detailed Solution

$\frac{5}{R} = \frac{ℓ_{1}}{100 - ℓ_{1}} and \frac{5}{R / 2} = \frac{1 .6 ℓ_{1}}{100 - 1 .6 ℓ_{1}}$

$\Rightarrow R = 15 Ω$

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