Question

Assume that you are sitting in the sun for $2.5$ hours. The area of the body exposed normally to sun rays is $1.3\,{m^2}$. The intensity of sun rays is $1.1\,kilowatt/{m^2}$. If your body completely absorbs the sun rays then the momentum transferred to your body will be(in kg-m/s).
A. 0.043 kg-m/s
B. 0.037 kg-m/s
C. 0.61 kg-m/s
D. -0.91 kg-m/s

Answer 1

Hint: to find momentum we need to find energy. And to find energy we need to determine the power. In the above question intensity and area have been mentioned. Hence we can start by finding power.In this question we need to convert time that is given in hours into seconds to get the answer in kg-m/s .

Formula used:
Power=intensity \[ \times \] area
Energy=power \[ \times \] time
Momentum=energy/c.

Complete step by step answer:
As mentioned in the above hint we will start by finding power first.
Given: intensity=$1.1\,kilowatt/{m^2}$, area= $1.3\,{m^2}$, time=2.5hours
To convert time in seconds:
\[T = 2.15 \times 60 \times 60\operatorname{s} \]
\[\Rightarrow T = 2.15 \times 3600\operatorname{s} .........(1)\]
Therefore power(P) is product of intensity(i) and area(a)
\[P = i \times a\]
\[\Rightarrow P = 1.1 \times 1.3W\]
\[\Rightarrow P = 1.1 \times 1.3 \times 1000\,KW........(2)\]

Now we find energy: energy is given by the product of time and power.
\[E = P \times T\]
Substituting values of equation 1 and 2 we get:
\[E = 1.1 \times 1300 \times 2.15 \times 3600J\]
\[\Rightarrow E = 11068200\,J......(3)\]
Now we can find momentum using the relation:
\[m = \dfrac{E}{c}\]
Where $E$=energy, $c$=speed in vacuum and $m$=momentum.
\[m = \dfrac{{11068200}}{{3 \times {{10}^8}}}\]
\[\Rightarrow m = \dfrac{{11068200}}{3} \times {10^{ - 8}}\]
\[\Rightarrow m = 3689400 \times {10^{ - 8}}\,kgm{s^{ - 1}}\]
\[\therefore m = 0.037\,kgm{s^{ - 1}}\]

Hence the correct answer is option B.

Note: Remember that you have to convert the units so as to answer the question in kgm/s. Remember to convert time from hours into seconds by multiplying it with 3600. Also remember to convert power from W to KW by multiplying it by 1000.