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Dependence of Resistance on Temperature

The resistanc...

Question

The resistance in the two arms of the meter bridge are $5 Ω$ and $R Ω$ respectively, balance point is at $l_{1}$ from left end. When the resistance $R$ is shunted with an equal resistance the new balance point is $1.6 l_{1}$ . The resistance $R$ is

10 Ω

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15 Ω

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20 Ω

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25 Ω

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Solution

The correct option is B $15 Ω$
For first case:
$5 (100 - l_{1})$ = $R l_{1}$
$\Rightarrow 500 - 5 l_{1}$ = $R l_{1} \dots (1)$
For second case:
Resistance in second arm will become $\frac{R}{2}$
so, $5 (100 - 1.6 l_{1})$ = $\frac{R}{2} (1.6 l_{1})$
$\Rightarrow 500 - 8 l_{1} = 0.8 R l_{1} \dots (2)$
Subtracting equations $(1) - (2)$
$- 5 l_{1} + 8 l_{1} = R l_{1} - 0.8 R l_{1}$
$3 = 0.2 R$
$\Rightarrow R = 15 Ω$

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The resistance in the two arms of the meter bridge are 5 Ω and R Ω respectively, balance point is at l1 from left end. When the resistance R is shunted with an equal resistance the new balance point is 1.6l1. The resistance R is

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