Assume you are sitting in sun for 2.5 hours. The area of your body exposed normally to sun rays ${1.3 m}^{2}$ . 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):

0.043

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0.037

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0.61

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- 0.91

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Solution

The correct option is A $0.043$
$P o w e r = i n t e n s i t y \times a r e a$
Intensity and area are given ,
$P o w e r = 1.3 \times 1.1 K W$
$E n e r g y = p o w e r \times t i m e$
Time is 2.5hr
$E n e r g y = 1300 \times 1.1 \times 2.5 \times 3600 J$
$= 12870000 J$
$M o m e n t u m = \frac{e n e r g y}{c}$
$= \frac{12870000}{3 \times 10^{8}}$
$= 0.043$

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Assume you are sitting in sun for 2.5 hours. The area of your body exposed normally to sun rays 1.3m2. The intensity of sun rays is 1.1 Kilowatt/m2. If your body completely absorbs the sun rays then the momentum transferred to your body will be (in Kg-m/s):

The correct option is A 0.043Power=intensity×areaIntensity and area are given ,Power=1.3×1.1KWEnergy=power×timeTime is 2.5hrEnergy=1300×1.1×2.5×3600J =12870000J Momentum=energyc =128700003×108 =0.043

Assume you are sitting in sun for 2.5 hours. The area of your body exposed normally to sun rays ${1.3 m}^{2}$ . 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):