The resistances in the two gaps of a balanced meter bridge are 'X' $\Omega$ and '3X' $\Omega$ respectively. If the resistances are interchanged the balance point shifts by

For a balanced meter bridge, the condition is $\dfrac{R_1}{R_2} = \dfrac{l}{100 - l}$

Initially, the resistances are $X$ and $3X$. Let the balancing length be $l_1$.

$\dfrac{X}{3X} = \dfrac{l_1}{100 - l_1}$

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

$100 - l_1 = 3l_1$

$4l_1 = 100 \Rightarrow l_1 = 25$ cm

When the resistances are interchanged, the new resistances in the gaps are $3X$ and $X$. Let the new balancing length be $l_2$.

$\dfrac{3X}{X} = \dfrac{l_2}{100 - l_2}$

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

$300 - 3l_2 = l_2$

$4l_2 = 300 \Rightarrow l_2 = 75$ cm

The shift in the balance point is $\Delta l = l_2 - l_1 = 75 - 25 = 50$ cm.

Answer: $50$ cm