Question

A conducting wire is connected in the right gap and known resistance in the left gap of a meter bridge of balancing length is $60\text{cm}$. The balancing length becomes $40\text{cm}$ when the wire is stretched. Length of the wire increases by

• A. $40\%$
• B. $30\%$
• C. $20\%$
• D. $50\%$

Answer 1

The correct option is D $50\%$

In $\text{CASE-I}:$

$\frac{R_l}{R_1}=\frac{l}{100-l}=\frac{60}{40}=\frac{3}{2}$

In $\text{CASE-II}:$

After stretching, the new balancing length is $40\text{cm}$.

$\frac{R_l}{R_2}=\frac{l_1}{100-l_1}=\frac{40}{60}=\frac{2}{3}$

The resistance and length relation in the case of stretching is $R\propto l^2\Rightarrow l\propto \sqrt{R}$

$\frac{l_2}{l_1}=\sqrt{\frac{R_2}{R_1}}=\sqrt{\frac{3}{2}\times \frac{3}{2}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

% increasing in length
$(\frac{l_2}{l_1}-1)100\text{\%}=(\sqrt{\frac{R_2}{R_1}}-1)100\text{\%}$

$=(\frac{3}{2}-1)100\text{\%}$

$=50\text{\%}$

Hence, option (d) is the correct answer.