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)

$15cm$ to the left

Given : $P=20\Omega$ and $Q=30\Omega$

Let initial balancing point to be at $l$.

Using balanced wheatstone bridge, $\frac{l}{100-l}=\frac{P}{Q}$

$\therefore$ $\frac{l}{100-l}=\frac{20}{30}$

$⟹$ $l=40$ cm

Now the resistance in the left gap is halved i.e. $P=10\Omega$

So new balancing length becomes $l^{\prime }$.

$\therefore$ $\frac{l^{\prime }}{100-l^{\prime }}=\frac{10}{30}$

$⟹$ $l^{\prime }=25$ cm

Thus shift in the balancing length $\Delta l=40-25=15$ cm to the left.