In a meter bridge experiment, the null point is at a distance of 30 cm from A. If a resistance of 16 Ω is conn

Question

In a meter bridge experiment, the null point is at a distance of 30 cm from A. If a resistance of 16 Ω is connected in parallel with resistance Y, the null point occurs at 50 cm from A. The value of the resistance Y is:

Accepted Answer

For a meter bridge, the ratio of resistances in the gaps is equal to the ratio of the balancing lengths: $\frac{X}{Y} = \frac{l}{100 - l}$.

**Case 1:** Null point at $l_1 = 30$ cm.
$\frac{X}{Y} = \frac{30}{100 - 30} = \frac{30}{70} = \frac{3}{7} \implies X = \frac{3}{7}Y$.

**Case 2:** A resistance of $16\ \Omega$ is connected in parallel with $Y$. The new equivalent resistance $Y'$ is:
$Y' = \frac{16Y}{16 + Y}$.
The new null point is at $l_2 = 50$ cm.
$\frac{X}{Y'} = \frac{50}{100 - 50} = 1 \implies X = Y'$.

**Solving:** substitute $X$ from Case 1 into the equation from Case 2:
$\frac{3}{7}Y = \frac{16Y}{16 + Y}$.
Dividing both sides by $Y$ (since $Y 
eq 0$):
$\frac{3}{7} = \frac{16}{16 + Y}$
$3(16 + Y) = 112$
$48 + 3Y = 112$
$3Y = 64 \implies Y = \frac{64}{3}\ \Omega$.

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