Q200 / 225 · JEE Main 2004 · Physics
Time taken by a $836 \mathrm{~W}$ heater to heat one litre of water from $10^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is
- A$50 \mathrm{~s}$
- B$100 \mathrm{~s}$
- Correct answerC$150 \mathrm{~s}$
- D$200 \mathrm{~s}$
Solution
Let $\mathrm{t}$ be the time taken, then
$$
\begin{aligned}
& \frac{836 \times t}{4.2}=1000 \times 1 \times(40-10)[\text { using } Q=m s t] \\
& \Rightarrow t=150 \text { sec. }
\end{aligned}
$$
Hence, the correct answer is option: 'C'.