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In a meter bridge experiment, when a nichrome wire is in the right gap, the balancing length is $60 \mathrm{~cm}$. When the nichrome wire is uniformly stretched to increase its length by $20 \%$ and again connected in the right gap, the new balancing length is nearly
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$51\ cm$
Since $\mathrm{R} \propto l$
$\begin{aligned}
& \therefore \frac{R_L}{R_R}=\frac{60}{100-60}=\frac{60}{40}=\frac{6}{4}=\frac{3}{2}... (i) \\
& \Rightarrow \frac{R_L}{144 R_R}=\frac{l}{100-l}... (ii)
\end{aligned}$
From eqs. (i) and (ii)
$\begin{aligned}
& \frac{144 R_R}{R_R}=\frac{3 / 2}{l / 100-l}=\frac{3}{2} \times \frac{100-l}{l} \\
& \Rightarrow \quad \frac{100-l}{l}=\frac{2}{3} \times 144=2 \times 0.48=0.96 \\
& \therefore 100-i=0.96 l
\end{aligned}$
or, $1.96 l=100$
or, $l=\frac{100}{196}=51.02 \mathrm{~cm}=51 \mathrm{~cm}$
$\begin{aligned}
& \therefore \frac{R_L}{R_R}=\frac{60}{100-60}=\frac{60}{40}=\frac{6}{4}=\frac{3}{2}... (i) \\
& \Rightarrow \frac{R_L}{144 R_R}=\frac{l}{100-l}... (ii)
\end{aligned}$
From eqs. (i) and (ii)
$\begin{aligned}
& \frac{144 R_R}{R_R}=\frac{3 / 2}{l / 100-l}=\frac{3}{2} \times \frac{100-l}{l} \\
& \Rightarrow \quad \frac{100-l}{l}=\frac{2}{3} \times 144=2 \times 0.48=0.96 \\
& \therefore 100-i=0.96 l
\end{aligned}$
or, $1.96 l=100$
or, $l=\frac{100}{196}=51.02 \mathrm{~cm}=51 \mathrm{~cm}$
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