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Q.

One atmosphere is numerically equal to approximately

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a

$10^{4} dynes {cm}^{- 2}$

b

$10^{8} dynes {cm}^{- 2}$

c

$10^{2} dynes {cm}^{- 2}$

d

$10^{6} dynes {cm}^{- 2}$

answer is A.

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Detailed Solution

$1 atm = 1.03125 \times 10^{5} pa$

$= 1.03125 \times 10^{6} gm / cm - s^{2}$

$= 1.03125 \times 10^{6} dynes / {cm}^{2}$

$1 \times 10^{6} dynes / {cm}^{2}$

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One atmosphere is numerically equal to approximately