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In a meter bridge, the gaps are enclosed by resistances of $2 \Omega$ and $3 \Omega$. The value of shunt to be added to $3 \Omega$ resistor to shift the balancing point by $22.5 \mathrm{~cm}$ is
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The correct answer is:
$2\ \Omega$
(Initial part)
$\frac{x}{100}-x=\frac{2}{3} \Rightarrow x=40 \mathrm{~cm}$
If there is shifting by $22.5 \mathrm{~cm}$. Then, to obtain the balance point in meter bridge
$\begin{aligned}
& \frac{2(3+x)}{3 x}=\frac{62.5}{37.5} \Rightarrow(6+2 x) 37.5=62.5 \times 3 x \\
& \Rightarrow \quad 225+75 x=187.5 x \\
& 187.5 x-75 x=225 \Rightarrow 112.5 x=225 \Rightarrow x=2
\end{aligned}$
$\frac{x}{100}-x=\frac{2}{3} \Rightarrow x=40 \mathrm{~cm}$
If there is shifting by $22.5 \mathrm{~cm}$. Then, to obtain the balance point in meter bridge
$\begin{aligned}
& \frac{2(3+x)}{3 x}=\frac{62.5}{37.5} \Rightarrow(6+2 x) 37.5=62.5 \times 3 x \\
& \Rightarrow \quad 225+75 x=187.5 x \\
& 187.5 x-75 x=225 \Rightarrow 112.5 x=225 \Rightarrow x=2
\end{aligned}$
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