Question

One atmosphere is numerically equal to approximately:

• A. ${10}^{6}dyne{cm}^{-2}$
• B. ${10}^{2}dyne{cm}^{-2}$
• C. ${10}^{4}dyne{cm}^{-2}$
• D. ${10}^{8}dyne{cm}^{-2}$

Answer 1

The correct option is A ${10}^{6}dyne{cm}^{-2}$

$P_{dyne/c{m}^2}=P_{atm}\times 1.01325\times {10}^6$

$P_{atm}=1$

then $P_{dynec{m}^2}=1.01325\times {10}^6$

Option A is Correct.