Q155 / 180 · COMEDK 2017 · Physics

Two resistances are connected in two gaps of a meterbridge. The balance point is $20 \mathrm{~cm}$ from the zero en(d) A resistance of $15 \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \mathrm{~cm}$. The value of the smaller resistance $($ in $\Omega)$ is

  • A
    3
  • B
    6
  • Correct answer
    C
    9
  • D
    12

Solution

In first case, at balance point, $\frac{P}{Q}=\frac{l}{100-l}=\frac{20}{100-20}=\frac{1}{4} \Rightarrow Q=4 P...(i)$ Let $P$ be the smaller resistance, then in second case at balance point, $\begin{aligned} \frac{P+15}{Q} &=\frac{l}{100-l^{\prime}}=\frac{40}{60}=\frac{2}{3} \\ \Rightarrow \frac{P+15}{4 P} &=\frac{2}{3} \\ \Rightarrow 3 P+45 &=8 P \Rightarrow P=9 \Omega \end{aligned}$

Hence, the correct answer is option: 'C'.

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