MHT CET · Physics · Current Electricity
In a meter bridge experiment, the balance point is obtained if the gaps are closed by \(2 \Omega\) and \(3 \Omega\). A shunt of \(X \Omega\) is added to \(3 \Omega\) resistor to shift the null point by 22.5 cm . The value of ' \(x\) ' is
- A \(1 \Omega\)
- B \(2 \Omega\)
- C \(3 \Omega\)
- D \(4 \Omega\)
Answer & Solution
Correct Answer
(B) \(2 \Omega\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \text { In case } 1, \\
& \frac{2}{3}=\frac{l}{100-l} \\
\therefore \quad & 200-2 l=3 l \\
\therefore \quad & l=40 \mathrm{~cm}
\end{array}\)
In case 2,
\(\begin{aligned} & \frac{2}{\frac{3 x}{3+x}}=\frac{40+22.5}{100-(40+22.5)} \\ \therefore \quad & \frac{2(3+x)}{3 x}=\frac{62.5}{37.5}\end{aligned}\)
\(\begin{array}{ll}\therefore & (6+2 x) \times 37.5=3 x(62.5) \\ \therefore & 225+75 x=187.5 x \\ \therefore & x=\frac{225}{112.5}=2 \Omega\end{array}\)
& \text { In case } 1, \\
& \frac{2}{3}=\frac{l}{100-l} \\
\therefore \quad & 200-2 l=3 l \\
\therefore \quad & l=40 \mathrm{~cm}
\end{array}\)
In case 2,
\(\begin{aligned} & \frac{2}{\frac{3 x}{3+x}}=\frac{40+22.5}{100-(40+22.5)} \\ \therefore \quad & \frac{2(3+x)}{3 x}=\frac{62.5}{37.5}\end{aligned}\)
\(\begin{array}{ll}\therefore & (6+2 x) \times 37.5=3 x(62.5) \\ \therefore & 225+75 x=187.5 x \\ \therefore & x=\frac{225}{112.5}=2 \Omega\end{array}\)
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