Question

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

• A. $15\, cm$ to the right
• B. $15 \,cm$ to the left
• C. $20 \,cm$ to the right
• D. $20 \,cm$ 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, $\,\,\, \frac{20}{30}=\frac{l}{100-l} \Rightarrow l=40 \,cm$
Now, $\,\,\, P'=\frac{20}{2}=10 cm$
$\therefore \,\,\, \frac{10}{30}=\frac{l'}{100-l'} \Rightarrow l'=25 \,cm$
So, balance point shifts by $(40-25) \,cm ,$ i.e. $15\, cm$ to the left.