A 50Ω resistance Galvanometer has 25 divisions on its scale. On sending a current of 4×10^-4A in it, the pointer gives a deflection of 1 division. How much resistance should be connected with it so that this galvanometer becomes a voltmeter of 2.5 volt range?

To convert a galvanometer into a voltmeter, we need to connect a resistor in series with it. The value of the resistor can be calculated using the formula:

$R=\frac{G}{{\displaystyle \frac{I_f}{n}}}$

where R is the resistance to be connected in series with the galvanometer, G is the resistance of the galvanometer, If is the current required to produce a full-scale deflection on the galvanometer, and n is the number of divisions on the scale of the galvanometer corresponding to the full-scale deflection.
In this case, the galvanometer has a resistance of 50Ω and 25 divisions on its scale. On sending a current of $4\times {{10}^{-}}^{4}A$ in it, the pointer gives a deflection of 1 division. Therefore, the current required to produce a full-scale deflection is:

$I_f=\frac{4\times {10}^{-4}A}{1division}\times 25division=0.01A$

Substituting the given values in the formula, we get:

$R=\frac{50\Omega }{{\displaystyle \frac{0.01A}{25}}}=1200\Omega$

However, we need to find the resistance required for a voltmeter of 2.5 volts range. The voltage range of a voltmeter is given by:

$V_{range}=I_f\times (G+R)$

where G is the resistance of the galvanometer and R is the resistance connected in series with it.
We want the voltmeter to have a range of 2.5 volts, so we can write:

$2.5V=0.01A\times (50\Omega +R)$

$R=\frac{2.5V}{0.01A}-50\Omega$

$R=200\Omega$

Hence the correct answer is $200\Omega .$