MHT CET · Physics · Thermal Properties of Matter
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
- A \(50 \mathrm{~J}\)
- B \(45 \mathrm{~J}\)
- C \(40 \mathrm{~J}\)
- D \(30 \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(45 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
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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