The Sun radiates energy in all directions. The average radiation received on the earth surface from the Sun is $1.4\raisebox{1ex}{$kW$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$. The average Earth-Sun distance is $1.5\times {10}^{11}m.$ The mass lost by the Sun per day is(1day=86400s)

The correct option is D

$3.8\times {10}^{14}kg$

Step 1: Given data

Average radiation received on the earth surface from the Sun, $P_{avg}=1.4\frac{kW}{m^2}=1.4\times {10}^3\frac{W}{m^2}$

Average distance between Earth-Sun, $R=1.5\times {10}^{11}m$

Time, $t=1day=86400s$

Step 2: Finding the total power

Total power emitted by the Sun, $P=P_{avg}\times surfaceoftheSun$

$\Rightarrow P=P_{avg}\times 4{\mathrm{\pi R}}^2$

$=1.4\times {10}^3\frac{W}{m^2}\times 4\pi {\left(1.5\times {10}^{11}m\right)}^2$

$=1.4\times {10}^3\times 4\times \frac{22}{7}\times 2.25\times {10}^{22}$$=39.6\times {10}^{25}W$

Step 3: Finding the mass lost

Power is defined as energy transferred per unit time, i.e., $P=\frac{E}{t}\Rightarrow E=Pt$…...(1)

Energy equivalence relation is given by $E=m{c}^2$…..….(2)

where E is the energy, m is the mass and c is the velocity of light, $=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Comparing equations (1) and (2), $m{c}^2=Pt$

$\Rightarrow m=\frac{Pt}{c^2}=\frac{39.6\times {10}^{25}W\times 86400s}{{\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)}^2}$

$=\frac{39.6\times {10}^{25}\times 86400}{9\times {10}^{16}}$

Thus the mass lost by the Sun per day is $m=3801.6\times {10}^{11}$$=3.8\times {10}^{14}kg$

Hence, the mass lost by the Sun per day is $3.8\times {10}^{14}kg$

Thus option D is correct.