The area of base of a cylindrical vessel is 300 cm ². Water (density = 1000 kg m ^-3) is poured into it up to a depth of 6 cm. Calculate (a) the pressure and (b) the thrust of water on the base. (g = 10 m s^-2).

(a) As we know,

Pressure due to water column of height h = hρg

Given,

ρ = 1000 kg m ^-3

g = 10 m s^-2

h = 6 cm

Converting cm to m

100 cm = 1 m

So, 6 cm = $\dfrac{1}{100}$ x 6 = 0.06 m

Hence, h = 0.06 m

Substituting the values in the formula above we get,

$P = 0.06 \times 1000 \times 10 \\[0.5em] P = 600 \text {Pa} \\[0.5em]$

Hence, Pressure = 600 Pa

(b) As we know,

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

A = 300 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 300 cm ² = $\dfrac{1 \times 300}{10000}$

Therefore, A = 3 x 10^-2 m ²

Substituting the values in the formula we get,

$600 = \dfrac{\text {Thrust (F)}}{3 \times 10^{-2}} \\[0.5em] \Rightarrow \text {Thrust (F)} = 600 \times 3 \times 10^{-2}\\[0.5em] \Rightarrow \text {Thrust (F)} = 18 N \\[0.5em]$

Hence, thrust = 18 N