A voltmeter has a resistance of $50\Omega$ is connected across a cell of e.m.f. 2V and internal resistance $10\Omega$. The reading of voltmeter is a. 1.667V b. 16.7V c. 167V d. 0.167V

Sure, here are the step by step solutions: Given: - Voltage of the battery, E = 2V - Internal resistance of the battery, r = 10Ω - Resistance of the load, R = 50Ω Formula used: - Ohm's Law: V = IR - Equation for total resistance in a circuit with a battery and a resistor in series: R_total = R_load + r Solution: 1. Calculate the total resistance of the circuit: $R_{total}=R_{load}+r=50\Omega +10\Omega =60\Omega$ 2. Use Ohm's Law to calculate the current in the circuit: $I=\frac{E}{R_{total}}=\frac{2V}{60\Omega }=0.0333A$ 3. Use Ohm's Law again to calculate the voltage across the load resistor: $V_{load}=I{R}_{load}=0.0333A\cdot 50\Omega =1.667V$ Therefore, the voltage across the load resistor is 1.667V.