1. Given data

• Mass of metal ball: $m_m = 0.1\,\text{kg}$
• Initial temperature of metal ball: $T_m = 500^\circ\text{C}$
• Heat capacity of vessel: $C_v = 800\,\text{J K}^{-1}$
• Mass of water: $m_w = 0.5\,\text{kg}$
• Initial temperature of water and vessel: $T_i = 30^\circ\text{C}$
• Specific heat of water: $c_w = 4200\,\text{J kg}^{-1}\text{K}^{-1}$
• Specific heat of metal: $c_m = 400\,\text{J kg}^{-1}\text{K}^{-1}$

Let the final equilibrium temperature be $T$.

2. Heat lost by metal ball

$Q_{\text{lost}} = m_m c_m (500 - T)$

$Q_{\text{lost}} = 0.1 \times 400 \times (500 - T)$

$Q_{\text{lost}} = 40(500 - T)$

3. Heat gained by water and vessel

Water gains heat:

$Q_w = m_w c_w (T - 30) = 0.5 \times 4200 \times (T - 30) = 2100(T-30)$

Vessel gains heat:

$Q_v = C_v (T - 30) = 800(T-30)$

So total heat gained:

$Q_{\text{gained}} = [2100 + 800](T-30) = 2900(T-30)$

4. Apply principle of calorimetry

$Q_{\text{lost}} = Q_{\text{gained}}$

$40(500 - T) = 2900(T-30)$

$20000 - 40T = 2900T - 87000$

$107000 = 2940T$

$T \approx 36.4^\circ\text{C}$

5. Find increment in water temperature

Initial temperature of water = $30^\circ\text{C}$

Final temperature = $36.4^\circ\text{C}$

Increase in temperature:

$\Delta T = 36.4 - 30 = 6.4^\circ\text{C}$

Percentage increment in temperature of water:

$\%\text{ increment} = \frac{6.4}{30} \times 100 \approx 21.3\%$

This is approximately $20\%$.

6. Check options

• A: $20\%$ ✅
• B: $25\%$
• C: $15\%$
• D: $30\%$

So the correct option is A.