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An unknown resistance $R_{1}$ is connected in series with a resistance of $10 ~\Omega$. This combination is connected to one gap of a meter bridge, while a resistance $R_{2}$ is connected in the other gap. The balance point is at $50 \mathrm{~cm}$. Now, when the $10 ~\Omega$ resistance is removed, the balance point shifts to $40 \mathrm{~cm}$. The value of $R_{1}$ (in $\Omega$ ) is
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Verified Answer
The correct answer is:
$20$
In first case, $P=R_{1}+10, Q=R_{2}, l=50 \mathrm{~cm}$
For balanced condition,
$\frac{P}{Q}=\frac{l}{100-l}$
$\Rightarrow \quad \frac{R_{1}+10}{R_{2}}=\frac{50}{50}$
$\Rightarrow \quad R_{1}+10=R_{2}$
Similarly, for second case,
$P=R_{1}, Q=R_{2}, l=40 \mathrm{~cm}$
$\Rightarrow \quad \frac{R_{1}}{R_{2}}=\frac{40}{100-40}=\frac{40}{60}$
$\Rightarrow \quad R_{2}=\frac{3}{2} R_{1}$
Putting value of $R_{2}$ from $\mathrm{Eq}$. (ii) in $\mathrm{Eq}$ (i), we get
$$
\begin{aligned}
R_{1}+10 &=\frac{3}{2} R_{1} \\
\Rightarrow \quad R_{1} &=2 \times 10=20 \Omega
\end{aligned}
$$
For balanced condition,
$\frac{P}{Q}=\frac{l}{100-l}$
$\Rightarrow \quad \frac{R_{1}+10}{R_{2}}=\frac{50}{50}$
$\Rightarrow \quad R_{1}+10=R_{2}$
Similarly, for second case,
$P=R_{1}, Q=R_{2}, l=40 \mathrm{~cm}$
$\Rightarrow \quad \frac{R_{1}}{R_{2}}=\frac{40}{100-40}=\frac{40}{60}$
$\Rightarrow \quad R_{2}=\frac{3}{2} R_{1}$
Putting value of $R_{2}$ from $\mathrm{Eq}$. (ii) in $\mathrm{Eq}$ (i), we get
$$
\begin{aligned}
R_{1}+10 &=\frac{3}{2} R_{1} \\
\Rightarrow \quad R_{1} &=2 \times 10=20 \Omega
\end{aligned}
$$
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