Question

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

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

Answer 1

Answer: (C)

$\frac{48}{11} O$

Answer 2

$\frac{P}{Q}=\frac{l_{1}}{100-l_{1}}$ or $\frac{3}{Q}=\frac{55}{100-55}=\frac{55}{45}$
$\therefore Q=3 \times \frac{45}{55}=\frac{27}{11}\, \Omega$
When $x$ is connected in series with $P, l_{1}=75\, cm$
$\therefore \frac{P+x}{Q}=\frac{75}{25}=3$ or $\frac{3+x}{(27 / 11)}=3$
or $x=\frac{81}{11}-3=\frac{48}{11}\, \Omega$