A galvanometer of internal resistance 50\(\Omega\) and full scale deflection of 20 mA is converted into a voltmeter by connecting a resistor of resistance 1...

The maximum voltage that can be read by the voltmeter can be calculated using the formula: V = I(G + S) Where V is the voltage, I is the current passing through the circuit, G is the galvanometer resistance, and S is the shunt resistance. In this case, the full scale deflection current of the galvanometer is 20 mA, which is 0.02 A. The galvanometer resistance is given as 50 Ω, and the shunt resistance is 1950 Ω. To calculate the current passing through the circuit, we can use Ohm's Law: V = IR Where R is the total resistance in the circuit. In this case, the total resistance is the sum of the galvanometer resistance and the shunt resistance: R = G + S = 50 Ω + 1950 Ω = 2000 Ω Substituting the values into Ohm's Law, we get: V = IR = 0.02 A × 2000 Ω = 40 V Therefore, the maximum voltage that can be read by the voltmeter is 40V. Option D is correct.