A barometer having a cross - sectional area of $1.00 c m^{2}$ at sea level measures a pressure of 76.0 cm of Hg. The pressure exerted by this column of mercury is equal to the pressure exerted by the air on $1 c m^{2}$ of the earth's surface. The total mass of earth's atmosphere is___. (Average radius of earth is p km and density of mercury is 13.6 g/ml)

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Solution

The correct option is A

$P r e s s u r e = \frac{F o r c e}{A r e a} = \frac{m a s s \times a c c e l e r a t i o n}{a r e a}$

$h d g = \frac{m \times g}{4 π r^{2}} \Rightarrow m = h d 4 π r^{2}$

$\Rightarrow m = 0.76 \times 13.6 \times 10^{3} \times 4 π ⟮ P \times 10^{3} ⟯^{2}$

= $4 π P^{2} \times 0.76 \times 13.6 \times 10^{9} k g$

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The correct option is A Pressure=ForceArea=mass×accelerationarea hdg=m×g4πr2⇒m=hd4πr2 ⇒m=0.76×13.6×103×4π⟮P×103⟯2 = 4πP2×0.76×13.6×109 kg