In meter bridge when P is kept in left gap, Q is kept in right gap, the balancing length is 40 cm. If Q is shunted by $10\Omega$ , the balance point shifts by 10 cm. resistance Q is

Solution:

Initially, Balanced condition of meter bridge $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{40}{100-40}=\frac{2}{3}..\left(\mathrm{i}\right)$

when Q is shunted with $10\mathrm{\Omega }$, the resistance of the right gap will be $\frac{10.\mathrm{Q}}{10+\mathrm{Q}}$

Since the balancing length is 50cm, $\frac{\mathrm{P}}{{\displaystyle \frac{10.\mathrm{Q}}{10+\mathrm{Q}}}}=\frac{50}{100-50}$

from (i) we get $\frac{2}{3}\mathrm{Q}=\frac{10.\mathrm{Q}}{10+\mathrm{Q}}\Rightarrow \mathrm{Q}=5\mathrm{\Omega }$