Physics - Modern Physics Question with Solution | TestHub

From Einstein's mass energy relation the energy released is

$∆E = ∆m{c}^2$

where$∆m$is mass and c is speed is light.

Given

$\Delta \text{m}=\text{1mg}=1\times 10^{-6}kg,\; c=3\times 10^8\text{m}/\text{s}$

$\therefore ∆E = 1 \times {10}^{-6} \times {\left(3 \times {10}^8\right)}^2$

$∆E = 9\times {10}^{10} \mathrm{J}$

The rate at which energy is dissipated is known as power,$ie,$

$P=\frac{∆ E}{t}\mathrm{ }\mathrm{ }=\frac{9 \times {10}^{10}}{1}=9\times {10}^{10}\mathrm{W}$

Since,${10}^3 \mathrm{W} = 1\mathrm{k}\mathrm{ }\mathrm{W}$

$\therefore P= 9 \times {10}^7 \mathrm{k}\mathrm{W}$