Question

The resistance 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, then 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

The correct option is B 15 cm to the left

Using, $\frac{R_1}{l}=\frac{R_2}{100-l}$ .............(1)

Put $R_1=20\Omega$ and $R_2=30\Omega$ in (1),

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

$⟹l=40$ $cm$

Let the new balance point lies at a distance $l^{\prime }$.

Put $R_1=10\Omega$ and $R_2=30\Omega$ in (1),

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

$⟹l^{\prime }=25$ $cm$

$l^{\prime }<l$ $⟹$ The balance point shifts towards left.

$\therefore$ Shift of the balance point, $\Delta l=l-l^{\prime }=15$ cm towards left.