A radiation of 200 W is incident on a surface which is 60 % reflecting and 40 % absorbing. The total force on the surface is

Question

A radiation of 200 W is incident on a surface which is 60 % reflecting and 40 % absorbing. The total force on the surface is (A) 1.07 × 10-6 N (B) 1.3 × 10-6 N (C) 1.07 × 10-7 N (D) 1.03 × 10-7 N

Tardigrade Admin · Accepted Answer

Here, F text total =F text re +Fa b s =(1.2 P/C)+(0.4 P/C)=(1.6 P/C) =(1.6 × 200/3 × 108)=1.07 × 10-6 N

Q. A radiation of $200 W$ is incident on a surface which is $60%$ reflecting and $40%$ absorbing. The total force on the surface is

$1.07 \times 1 0^{- 6} N$

$1.3 \times 1 0^{- 6} N$

$1.07 \times 1 0^{- 7} N$

$1.03 \times 1 0^{- 7} N$

Here, $F_{total} = F_{re} + F_{ab s}$
$= \frac{1.2 P}{C} + \frac{0.4 P}{C} = \frac{1.6 P}{C}$
$= \frac{1.6 \times 200}{3 \times 1 0 ^{8}} = 1.07 \times 1 0^{- 6} N$

Q. A radiation of 200W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is

1.07×10−6N

1.3×10−6N

1.07×10−7N

1.03×10−7N