A galvanometer of 50 ohm resistance has 25 divisions. A current of $4 \times 10^{-4}$ ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of :-

Full-scale deflection current: $I_g = 25 \times 4 \times 10^{-4} = 10^{-2}$ A
To make a voltmeter, a resistance R is connected in series with the galvanometer.
According to the question: $25 = I_g(R + R_g)$
$25 = (4 \times 10^{-4} \times 25)(R + 50)$
$\Rightarrow R + 50 = \dfrac{25}{10^{-2}} = 2500$
$\Rightarrow R = 2450\ \Omega$
So a resistance of 2450 $\Omega$ should be connected in series.