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If $150 \mathrm{~J}$ of energy is incident on area $2 \mathrm{~m}^{2}$. If $Q_{r}=15 \mathrm{~J}$, coefficient of absorption is $0.6$, then amount of energy transmitted is
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The correct answer is:
$45 \mathrm{~J}$
When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted.
$$
Q=Q_{a}+Q_{r}+Q_{t}
$$
and $\quad \frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}=a+r+t=1$
$$
\Rightarrow \quad \frac{15}{150}+0.6+x=1
$$
or $0.1+0.6+x=1$
or $x=0.3$
Transmitting power, $t=\frac{Q_{t}}{Q}$
or $0.3=\frac{Q_{t}}{150}$
$$
\Rightarrow \quad Q_{t}=45 \mathrm{~J}
$$
$$
Q=Q_{a}+Q_{r}+Q_{t}
$$
and $\quad \frac{Q_{a}}{Q}+\frac{Q_{r}}{Q}+\frac{Q_{t}}{Q}=a+r+t=1$
$$
\Rightarrow \quad \frac{15}{150}+0.6+x=1
$$
or $0.1+0.6+x=1$
or $x=0.3$
Transmitting power, $t=\frac{Q_{t}}{Q}$
or $0.3=\frac{Q_{t}}{150}$
$$
\Rightarrow \quad Q_{t}=45 \mathrm{~J}
$$
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