Question

The resistances in left and right gap of a meter bridge are $20\Omega$ and $30\Omega$ respectively. When the resistance in the left gap is reduced to half its value, the balance point shifts by

• A. $15cm$ to the right
• B. $15cm$ to the left
• C. $20cm$ to the right
• D. $20cm$ to the left

Answer 1

Answer: (B)

Balance condition in meter bridge is given by
$\frac{P}{Q}=\frac{l}{100-l}$
where, $P$ is the resistance in left gap and $Q$ is the resistance in right gap, and $l$ is length of wire from one end (i.e. left end) where null or balance point is obtained.
Here, $\frac{20}{30}=\frac{l}{100-l}\Rightarrow l=40cm$
Now, $P^{\prime }=\frac{20}{2}=10cm$
$\therefore \frac{10}{30}=\frac{l^{\prime }}{100-l^{\prime }}\Rightarrow l^{\prime }=25cm$
So, balance point shifts by $(40-25)cm,$ i.e. $15cm$ to the left.