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?

Given Data:

• Resistance of the galvanometer (R_g) = 50Ω
• Current per division = 4 × 10^-4 A
• Number of divisions = 25
• Desired voltage range (V) = 2.5 V

Maximum deflection current is given by: 

I_max = current per division × number of divisions

I_max = (4 × 10^-4) × 25 = 10 × 10^-3 = 0.01 A

The total resistance of the voltmeter is: 

R_total = R_g + R

Using Ohm's Law: 

V = I_max × R_total

Substituting R_total = R_g + R: 

V = I_max × (R_g + R)

Rearrange to solve for R: 

R = V / I_max - R_g

Substituting V = 2.5 V, I_max = 0.01 A, and R_g = 50Ω

R = 2.5 / 0.01 - 50

R = 250 - 50

R = 200Ω

Final Answer:

The resistance that should be connected in series with the galvanometer to make it a voltmeter of 2.5 V range is: 200Ω