Question

If $150J$of energy is incident on the area$2m^2$. If $Q_r=15J$, the coefficient of absorption is $0.6,$ then the amount of energy transmitted is

• A. $50J$
• B. $45J$
• C. $40J$
• D. $30J$

Answer 1

The correct option is B

$45J$

Solution:

Step 1: Given data:

$150J$ of energy is incident on a surface.

$15J$ of energy is reflected by it.

The coefficient of absorption is $r=0.6$.

Step 2: Formula used and find the amount of energy transmitted.

Understand that some thermal radiation is reflected, some are absorbed, and some is transferred when it strikes a body.

Therefore the thermal radiation $Q=Q_a+Q_r+Q_t$

Here, $Q$are the charge,$a$ area,$r$ radius, and $t$ time.

And, divide $Q$ both sides

$$\begin{gathered} \frac{Q}{Q}=\frac{Q_a}{Q}+\frac{Q_r}{Q}+\frac{Q_t}{Q} \\ 1=a+r+t \end{gathered}$$

Now substitute the value of $a,r,t$

$$\begin{gathered} \frac{15}{150}+0.6+x=1 \\ 0.1+0.6+x=1 \\ x=1-0.7 \\ x=0.3 \end{gathered}$$

Step 3: Now the transmitted power =$\frac{Q_t}{Q}$

$$\begin{gathered} 0.3=\frac{Q_t}{150J} \\ {Q}_t=150J\times \frac{3}{10} \\ {Q}_t=45J \end{gathered}$$

Therefore the transmitted energy is $45J$.

Hence, the correct option is (B).