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A body of specific heat $0.2 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}$ is heated through $100^{\circ} \mathrm{C}$. The percentage increase in its mass is
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The correct answer is:
$9.3 \times 10^{-11} \%$
Heat taken by the given body $=m c \Delta \theta$
{specific heat of body $=0.2 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}$, final temperature of the body $\left.=100^{\circ} \mathrm{C}\right\}$
$\begin{aligned} \Delta E & =m \times 0.2 \times 100=20 \mathrm{mkcal} \\ & =20 \mathrm{~m} \times 4.2 \times 10^3 \mathrm{~J}\end{aligned}$
Now gain in mass is given by
$\Delta m=\frac{\Delta E}{C^2}$ $\left(\because E=m c^2\right)$
$=\frac{20 \mathrm{~m} \times 4.2 \times 10^3}{\left(3 \times 10^8\right)^2}$
Therefore, percentage increase in mass is given by
$\begin{aligned} & =\frac{\Delta m}{m} \times 100 \\ & =\frac{20 m \times 4.2 \times 10^3}{\left(3 \times 10^8\right)^2 \times m} \times 100 \\ & =9.3 \times 10^{-11} \%\end{aligned}$
{specific heat of body $=0.2 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}$, final temperature of the body $\left.=100^{\circ} \mathrm{C}\right\}$
$\begin{aligned} \Delta E & =m \times 0.2 \times 100=20 \mathrm{mkcal} \\ & =20 \mathrm{~m} \times 4.2 \times 10^3 \mathrm{~J}\end{aligned}$
Now gain in mass is given by
$\Delta m=\frac{\Delta E}{C^2}$ $\left(\because E=m c^2\right)$
$=\frac{20 \mathrm{~m} \times 4.2 \times 10^3}{\left(3 \times 10^8\right)^2}$
Therefore, percentage increase in mass is given by
$\begin{aligned} & =\frac{\Delta m}{m} \times 100 \\ & =\frac{20 m \times 4.2 \times 10^3}{\left(3 \times 10^8\right)^2 \times m} \times 100 \\ & =9.3 \times 10^{-11} \%\end{aligned}$
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