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Standard X

Physics

Resistivity

An unknown re...

Question

An unknown resistance $R_{1}$ is connected in series with a resistance of $10 Ω$ . This combination is connected to one gap of a metre bridge while a resistance $R_{2}$ is connected in the other gap. The balance
point is at $50$ cm. Now, when the $10 Ω$ resistance is removed the balance point shifts to 40 cm. The value of $R_{1}$ is (in ohms) :-

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Solution

The correct option is A 20
The null condition for a meter bridge is $\frac{R_{1}}{R_{2}} = \frac{l}{100 - l}$
for first case: $\frac{R_{1} + 10}{R_{2}} = \frac{50}{100 - 50} = 1$
or $R_{1} + 10 = R_{2} . . . (1)$
for 2nd case: $\frac{R_{1}}{R_{2}} = \frac{40}{100 - 40} = \frac{2}{3}$ or $R_{2} = 1.5 R_{1}$
now (1) becomes, $R_{1} + 10 = 1.5 R_{1}$
or $0.5 R_{1} = 10 \Rightarrow R_{1} = 20$

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The correct option is A 20The null condition for a meter bridge is R1R2=l100−l for first case: R1+10R2=50100−50=1or R1+10=R2...(1)for 2nd case: R1R2=40100−40=23 or R2=1.5R1now (1) becomes, R1+10=1.5R1or 0.5R1=10⇒R1=20