A uniformly tapering vessel is filled to the top with a liquid of density $900 k g m^{- 3}$ . The force that acts on the base of the vessel due to the liquid is $(g = 10 m s^{- 2})$

3.6 N

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72 N

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9.0 N

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14.4 N

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Solution

The correct option is A $72 N$
pressure $P$ at the base $= h ρ g$
$P = 0.4 \times 900 \times 10 \Rightarrow P = 3.6 \times 10^{3} N / m^{2}$
$∵$ Force = pressure $\times$ area
$\Rightarrow F = 3.6 \times 10^{3} \times 2 \times 10^{- 3} = 7.2 N$

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A uniformly tapering vessel is filled to the top with a liquid of density 900kgm−3. The force that acts on the base of the vessel due to the liquid is (g=10ms−2)

The correct option is A 72 Npressure P at the base =hρgP=0.4×900×10⇒P=3.6×103N/m2∵ Force = pressure × area⇒F=3.6×103×2×10−3=7.2N

A uniformly tapering vessel is filled to the top with a liquid of density $900 k g m^{- 3}$ . The force that acts on the base of the vessel due to the liquid is $(g = 10 m s^{- 2})$

The correct option is A $72 N$
pressure $P$ at the base $= h ρ g$
$P = 0.4 \times 900 \times 10 \Rightarrow P = 3.6 \times 10^{3} N / m^{2}$
$∵$ Force = pressure $\times$ area
$\Rightarrow F = 3.6 \times 10^{3} \times 2 \times 10^{- 3} = 7.2 N$