1. Given data

   • Power of heater: $P = 836\,\text{W} = 836\,\text{J s}^{-1}$
   • Volume of water: $1$ litre
   • Since density of water is $1\,\text{kg/L}$, mass of water: $m = 1\,\text{kg}$
   • Initial temperature: $10^\circ C$
   • Final temperature: $40^\circ C$
   • Rise in temperature: $\Delta T = 40 - 10 = 30^\circ C$
   • Specific heat capacity of water: $c \approx 4180\,\text{J kg}^{-1}\,^\circ\text{C}^{-1}$

2. Heat required to raise the temperature $Q = mc\Delta T$ Substituting values: $Q = (1)(4180)(30) = 125400\,\text{J}$

3. Use power relation $P = \frac{Q}{t} \implies t = \frac{Q}{P}$ $t = \frac{125400}{836} = 150\,\text{s}$

4. Match with options

   • A: $150\,\text{s}$ ✅
   • B: $100\,\text{s}$
   • C: $50\,\text{s}$
   • D: $200\,\text{s}$

Therefore, the correct option is A.