Given data

• Battery emf: $V = 1.5\,\text{V}$
• Forward voltage drop across diode: $V_D = 0.5\,\text{V}$
• Maximum safe current: $I = 10\,\text{mA} = 0.01\,\text{A}$

Voltage across the series resistor When the diode is forward biased, it drops $0.5\,\text{V}$. So the remaining voltage across the resistor is $V_R = 1.5 - 0.5 = 1.0\,\text{V}$

Use Ohm’s law to find minimum resistance To keep current at or below the safe limit, $R_{\min} = \frac{V_R}{I} = \frac{1.0}{0.01} = 100\,\Omega$

Match with options

• A: $50\,\Omega$
• B: $200\,\Omega$
• C: $300\,\Omega$
• D: $100\,\Omega$

Therefore, the correct option is: $\boxed{\text{D: }100\,\Omega}$