Q140 / 150 · MHT CET 2023 (12 May Shift 2) · Physics
Two resistance $\mathrm{X}$ and $\mathrm{Y}$ are connected in the two gaps of a meterbridge and the null points is obtained at $20 \mathrm{~cm}$ from zero end. When the resistance of $20 \Omega$ is connected in series with the smaller of the two resistance $\mathrm{X}$ and $\mathrm{Y}$, the null point shifts to $40 \mathrm{~cm}$ from left end. The value of smaller resistance in ohm is
- A$6$
- B$9$
- Correct answerC$12$
- D$15$
Solution
For a meterbridge, $\frac{\mathrm{X}}{\mathrm{Y}}=\frac{l}{100-l}$
In the first case, $l=20 \mathrm{~cm}$
$$
\begin{aligned}
\frac{\mathrm{X}}{\mathrm{Y}} & =\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4} \\
\therefore \quad 4 \mathrm{X} & =\mathrm{Y}
\end{aligned}
$$
In the second case, $l^{\prime}=40 \mathrm{~cm}$
$$
\begin{array}{ll}
\therefore & \frac{\mathrm{X}^{\prime}}{\mathrm{Y}^{\prime}}=\frac{40}{100-40}= \\
& \text { But, } \mathrm{X}^{\prime}=\mathrm{X}+2 \\
\therefore \quad & \frac{\mathrm{X}+20}{\mathrm{Y}}=\frac{2}{3} \\
\therefore \quad & \frac{\mathrm{X}+20}{4 \mathrm{X}}=\frac{2}{3} \\
\therefore \quad & 8 \mathrm{X}=3 \mathrm{X}+60 \\
\therefore \quad & \mathrm{X}=12 \Omega
\end{array}
$$
Hence, the correct answer is option: 'C'.