Question

In a metre bridge, the gaps are closed by resistances $P$ and $Q$, $P$ being less than $Q$. A balance is obtained when the jockey makes contact at a point of bridge wire $40\text{cm}$ from one end. On shunting the coil $Q$ with a resistance of $50\Omega$, the balance point is moved through $10\text{cm}$. The resistance $P$ and $Q$ are

• A. $25\Omega ,16.67\Omega$
• B. $16.67\Omega ,25\Omega$
• C. $10\Omega ,25\Omega$
• D. $8\Omega ,12\Omega$

Answer 1

The correct option is B $16.67\Omega ,25\Omega$

Balanced condition of meter bridge in $1st$ case:
$\frac{P}{Q}=\frac{40}{60}=\frac{2}{3}$
Upon shunting $Q$,
$\frac{P}{Q||50\Omega }=\frac{40+10}{60-10}=1$
or $\frac{P(50+Q)}{50Q}=1$
$\Rightarrow \frac{50Q}{50+Q}=\frac{2}{3}Q$
$\Rightarrow Q=25\Omega$
$\therefore P=\frac{2}{3}Q=\frac{50}{3}=16.67\Omega$