The pressure exerted by a mass of ‘x’ mg resting on the area of 1.00cm2 is 1.00 Pa, then ‘x’ is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Given data:

P=1Pa

$A = 10^{- 4} m^{2}$

$g = 9.8 m / s^{2}$

$m = x \times 10^{- 6} Kg$

$P = \frac{ma}{A}$

$= \frac{mg}{A}$

$1 = \frac{x \times 10^{- 6} \times 9.8}{10^{- 4}}$

$10^{- 4} = x \times 10^{- 6} \times 9.8$

$\frac{10^{- 4} \times 10^{6}}{9.8} = x$

$x = \frac{10^{3}}{98}$

$x = \frac{1}{98} \times 10^{3}$

$x = 0.0102 \times 10^{3}$

=10.2

The pressure exerted by a mass of ‘x’ mg resting on the area of 1.00 ${cm}^{2}$ is 1.00 Pa, then ‘x’ is