Question

In a meter bridge experiment, resistances are connected as shown in figure. The balancing length $l_1$ is $55\text{cm}$. Now an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $75\text{cm}$. The value of $x$ is

• A. $\frac{54}{13}\Omega$
• B. $\frac{20}{11}\Omega$
• C. $\frac{48}{11}\Omega$
• D. $\frac{11}{48}\Omega$

Answer 1

The correct option is C $\frac{48}{11}\Omega$

For the meter bridge shown in figure the balancing length is $l_1$ $=$ $55$ $\text{cm}$.

Applying balanced condition for meter bridge

$\frac{R_P}{R_Q}=\frac{l_1}{100-l_1}$

$\frac{3}{R_Q}=\frac{55}{45}$

$\therefore R_Q=\frac{45\times 3}{55}=\frac{27}{11}\Omega$

Now an unknown resistance is connected in series with $\text{P}$ and the balancing length is found to be $l_2=75\text{cm}$

Applying the balancing condition for 2nd case,

$\frac{3+x}{R_Q}=\frac{l_2}{100-l_2}$

$(\because R_Q=27/11\Omega$ & $l_2=75\text{cm})$

$\frac{(3+x)11}{27}=\frac{75}{25}$

$(3+x)11=81$

$\therefore x=\frac{48}{11}\Omega$

Ans : $\left(c\right)$

Why this question ? Tip: In such problems of meter bridge whenever an unknown resistance is connected in series or parallel with any of the arm resistances, first find out net resistance for that arm and then apply balanced bridge condition for new null point.