When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is:

We are told that, when the diode conducts in the forward direction, the potential difference appearing across it is $$V_D = 0.5\ \text{V}.$$ At the same time, the maximum safe current that may pass through the diode is specified as $$I_{\text{max}} = 10\ \text{mA} = 0.01\ \text{A}.$$ A battery of emf $$E = 1.5\ \text{V}$$ is to be used, and we must choose a series resistance that keeps the diode current at or below this safe limit.

The battery’s emf will be shared between the diode and the series resistor. So the potential difference available across the resistor is obtained by simple subtraction:

$$V_R = E - V_D = 1.5\ \text{V} - 0.5\ \text{V} = 1.0\ \text{V}.$$

Now we invoke Ohm’s law, which states $$V = I R.$$ Solving for the resistance gives

$$R = \frac{V}{I}.$$

Substituting the voltage that must appear across the resistor and the maximum permissible current, we get

$$R_{\text{min}} = \frac{V_R}{I_{\text{max}}} = \frac{1.0\ \text{V}}{0.01\ \text{A}} = 100\ \Omega.$$

This value ensures that, even under maximum forward bias, the current cannot exceed the safe limit of $$10\ \text{mA}.$$ Any smaller resistance would allow more current than permitted, whereas a larger resistance would only reduce the current further, remaining safe.

Hence, the correct answer is Option C.