A uniformly tapering vessel shown in the figure is filled with a liquid of density $900 {kg/m}^{3}$ . The force that acts on the base of the vessel due to the liquid is (take $g = 10 {m/s}^{2}$ )

3.6 N

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7.2 N

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9.0 N

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12.6 N

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Solution

The correct option is B $7.2 N$
Given, $h = 0.4 m, ρ = 900 {kg/m}^{3}$
Area at the bottom $A = 2 \times 10^{- 3}$
Let $P$ be pressure at the bottom of the cylinder
Then, force at the bottom of cylinder is
$F = P A = ρ g h A$
$= 900 \times 10 \times 0.4 \times 2 \times 10^{- 3} = 7.2 N$

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A uniformly tapering vessel shown in the figure is filled with a liquid of density 900 kg/m3. The force that acts on the base of the vessel due to the liquid is (take g=10 m/s2)

The correct option is B 7.2 NGiven, h=0.4 m,ρ=900 kg/m3 Area at the bottom A=2×10−3 Let P be pressure at the bottom of the cylinder Then, force at the bottom of cylinder is F=PA=ρghA =900×10×0.4×2×10−3=7.2 N

A uniformly tapering vessel shown in the figure is filled with a liquid of density $900 {kg/m}^{3}$ . The force that acts on the base of the vessel due to the liquid is (take $g = 10 {m/s}^{2}$ )

The correct option is B $7.2 N$
Given, $h = 0.4 m, ρ = 900 {kg/m}^{3}$
Area at the bottom $A = 2 \times 10^{- 3}$
Let $P$ be pressure at the bottom of the cylinder
Then, force at the bottom of cylinder is
$F = P A = ρ g h A$
$= 900 \times 10 \times 0.4 \times 2 \times 10^{- 3} = 7.2 N$