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 $Y \neq 0$): $\frac{3}{7} = \frac{16}{16 + Y}$ $3(16 + Y) = 112$ $48 + 3Y = 112$ $3Y = 64 \implies Y = \frac{64}{3}\ \Omega$.