Q 12-13-057JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
The mass defect in a particular reaction is $0.4\ \text{g}$. The amount of energy liberated is $n\times10^7\ \text{kW h}$, where $n = $ ______. (speed of light $= 3\times10^8\ \text{m s}^{-1}$)
Numerical value type. Enter your answer.
Answer: 1
$$E = \Delta mc^2 = 0.4\times10^{-3}\times9\times10^{16} = 3.6\times10^{13}\ \text{J}$$
$1\ \text{kW h} = 3.6\times10^6\ \text{J}$, so $E = 10^7\ \text{kW h}$ and $n = 1$.
Solution by Sreeraj P, M.Sc Physics