Question

In a meter bridge experiment, resistance are connected as shown in figure, The balancing length 1, is 55cm. Now an unknown resistance x is connected in series with P and new balancing length is found to be 75cm. 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

Answer: (C)

For the balance point of meterbridge.
Case - I, when balancing length is 55 cm
$\frac{P}{Q}=\frac{{\ell }_1}{100-{\ell }_1}$
$\Rightarrow \frac{3}{Q}=\frac{55}{45}$
$Q=\frac{45}{55}\times 3$
Case - II When an unknown resistance $x$ is involved
$\frac{P+x}{Q}=\frac{{\ell }_1}{100-{\ell }_1}$
$\frac{3+x}{Q}=\frac{75}{25}$
$\Rightarrow 3+x=\frac{75}{25}\times Q$...(ii)
By putting the value of $Q$ from equation (i) in equation (ii)
$3+x=\frac{75}{25}\times \frac{45\times 3}{55}$
$x=\frac{81}{11}-3=\frac{81-33}{11}=\frac{48}{11}\Omega$