Pressure in Fluids and Atmospheric Pressure — Question 3

(a) As we know,

Pressure due to water column of height h = h ρ g

Given,

h = 1.5 m

ρ = 1000 kg m ^-3

g = 9.8 m s ²

Substituting the values in the formula above, we get,

$\text P = 1.5 \times 1000 \times 9.8 \\[0.5em] \text P = 14.7 \times 10^{3}\ \text {N m}^{-2} \\[0.5em]$

Hence, pressure = 1.47 x 10⁴ N m^-2

(b) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

A = 100 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 100 cm ² = $\dfrac{1 \times 100}{10000}$

Therefore, A = 10^-2 m ²

Substituting the values in the formula above, we get,

$1.47 \times 10^{4} = \dfrac{\text {Thrust (F)}}{10^{-2}} \\[0.5em] \Rightarrow \text {Thrust (F)} = 1.47 \times 10^{2} \\[0.5em] \Rightarrow \text {Thrust (F)} = 147\ \text N \\[0.5em]$

Hence, Thrust = 147 N