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A meter bridge with two resistances connected in the two gaps has a balance point at 40 cm. If a resistance 10 $Ω$ is connected in series with the smaller resistor, the balance point shifts by 20 cm. Neglecting any corrections, the larger resistance

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$12 Ω$

$40 Ω$

$60 Ω$

$8 Ω$

answer is D.

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Detailed Solution

$\frac{R_{1}}{R_{2}} = \frac{40}{100 - 40}$
and $\frac{R_{1} + 10}{R_{2}} = \frac{60}{100 - 60}$ $\Rightarrow$ $R_{2}$ = 12 $Ω$

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