The base of a cylindrical vessel is $300 {c m}^{2}$ . Water is poured into it upto a depth of $6 c m$ . What will be the thrust acting on the base (in Newtons) ? Take density of water $= 1000 k g$ $m^{- 3}$ and $g = 10 m$ $s^{- 2}$

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Solution

hydrostatic pressure = $\frac{T h r u s t}{A r e a} = ϱ g h$
Thrust = $10^{3} \times 10 \times 0.06 \times 0.03$
= 18 N.

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Similar questions

300 c m^{2}

6 c m

ρ_{w} = 1000 k g / m^{3}

g = 10 m / s^{2}

300 c m^{2}

m / s^{2}

k g / m^{3}

300 c m^{2}

1000 k g m^{- 3}

6 c m

g = 10 m s^{- 2}

The area of base of a cylindrical vessel is 300cmsquare water (density = 1000kg m-3 ) is poured into it up to a depth of 6cm.

Calculate

The pressure

The thrust of water on the base.(g=10ms-2)

c m^{2}

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