TS EAMCET · Physics · Current Electricity
In the meter bridge experiment, the length \(A B\) of the wire is \(1 \mathrm{~m}\). The resistors \(X\) and \(Y\) have values \(5 \Omega\) and \(2 \Omega\) respectively. When a shunt resistance \(S\) is connected to \(X\), the balancing point is found to be \(0.625 \mathrm{~m}\) from \(A\). Then, the resistance of the shunt is

- A \(5 \Omega\)
- B \(10 \Omega\)
- C \(7.5 \Omega\)
- D \(12.5 \Omega\)
Answer & Solution
Correct Answer
(B) \(10 \Omega\)
Step-by-step Solution
Detailed explanation
Here in given condition, we have \[ \begin{aligned} \frac{b x}{\frac{b+x}{2}} & =\frac{0.625}{0.375} \\ \frac{b x}{(b+x) 2} & =\frac{25}{15} \\ \frac{5 b}{(b+5) 2} & =\frac{5}{3} \\ \frac{b}{2 b+10} & =\frac{1}{3} \\ 3 b-2 b & =10 \\ b & =10 \Omega \end{aligned} \]
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