With a resistance ' $X$ ' connected in series with a galvanometer of resistance $100 \Omega$, it acts as a voltmeter of range $0-15 \mathrm{~V}$. To double the range, a resistance of $1500 \Omega$ is to be connected in series with ' $X$ '. The value of ' X ' in ohm is

Converting a galvanometer to a voltmeter requires a series resistance $R_s$ such that the voltmeter range is $V = I_g (R_g + R_s)$, with $I_g$ as the full-scale deflection current, $R_g$ the galvanometer resistance, and $R_s$ the series resistance.

For a $15\text{ V}$ range with $R_g = 100\ \Omega$ and series resistance $X$,
$15 = I_g (100 + X)$

Doubling the range to $30\text{ V}$ requires an additional $1500\ \Omega$ in series, making the total series resistance $X + 1500$.
$30 = I_g (100 + X + 1500) = I_g (1600 + X)$

Dividing this equation by the first yields
$\frac{30}{15} = \frac{1600 + X}{100 + X}$
$2(100 + X) = 1600 + X$
$200 + 2X = 1600 + X$
$X = 1400\ \Omega$

Thus, $X = 1400\ \Omega$.