Agalvanometer with a resistance of 40.0 ω deflects full scale at a current of 2.0 ma. what resistance should be used with this galvanometer in order to construct a voltmeter that can read a maximum of 50 v?

The resistance needed to be added is R
The Current is 2 ma
The voltage reading is a maximum of 50 volts.
The ma meter has an internal resistance of 40 ohms.

Formula
E = I * R

Givens
E = 50
I = 2 ms
R = R + 40

Solution
E = I * R
I = 2 ma [ 1 amp / 1000 ma] = 0.002 amp
50 = 0.002 * (R + 40)  Divide by 0.002
50/0.002 = R + 40
25000 = R + 40 Subtract 40 from both sides.
R = 25000 - 40
R = 24960     Answer

The meter will still deflect full-scale when the current through it is 2 mA.
(That's 0.002 Amp.)
You want that to happen when the meter is connected to two points
with a potential difference of 50v between them.

The total resistance in the meter circuit has to be

R = (voltage) / (current) = (50v) / (0.002 A) = 25,000 Ohms.

You already have 40 ohms in the meter's mechanism.
So technically, if you want to nail it right-on, you'll want
a resistor of exactly 24,960 ohms in series with the galvanometer. 

(You can't buy a resistor with this kind of precision.  But there are
ways to take a resistor from the store, and increase its resistance
little by little, until you have exactly what you want.)

Less force and energy was applied

The correct answer is c. hope this