Resistances in two gaps of a metre bridge are 8 Ω and 24 Ω, respectively. The distance by which the balance point shifts by interchanging the resistances in the two gaps, is

50 cm

The correct answer is Option (2) → 50 cm

Given:

$R_1 = 8\,Ω$, $R_2 = 24\,Ω$, Total length of wire = $100\,cm$

Condition for balance:

$\frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$

Substitute:

$\frac{8}{24} = \frac{l_1}{100 - l_1}$

$\frac{1}{3} = \frac{l_1}{100 - l_1}$

$100 - l_1 = 3l_1$

$100 = 4l_1$

$l_1 = 25\,cm$

After interchanging resistances:

$\frac{R_1}{R_2} = \frac{24}{8} = \frac{l_2}{100 - l_2}$

$3 = \frac{l_2}{100 - l_2}$

$100 - l_2 = \frac{l_2}{3}$

$300 - 3l_2 = l_2$

$300 = 4l_2$

$l_2 = 75\,cm$

Shift in balance point:

$Δl = |l_2 - l_1| = |75 - 25| = 50\,cm$

Final Answer: Shift in balance point = 50 cm