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A glass full of water has a bottom of area $20 {cm}^{2}$ , top of area $20 {cm}^{2}$ , height 20 cm and volume half a liter. (i) Find the force exerted by the water on the bottom (ii) Considering the equilibrium of the water; Find the resultant force exerted by the sides of the glass on the water. $[Atomspheric pressure = 1.0 \times 10^{5} {Nm}^{- 2}, Density of water = 1000 {kgm}^{- 3} and g = 10 {ms}^{- 2}]$

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(i) 404 N (ii) 5 N upward

(i) 204 N (ii) 1 N upward

(i) 104 N (ii) 5N downward

(i) 304 N (ii) 10 N downward

answer is C.

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Detailed Solution

P = P_{0} + h ρ g

$= 10^{5} + 0.2 \times 1000 \times 10$

$= 10^{5} + 2 \times 10^{3}$

$= 10^{5} + 0.02 \times 10^{5}$

$P = (1.02) \times 10^{5} {Nm}^{- 2}$

$(i) F = P . A = 1.02 \times 10^{5} \times 20 \times 10^{- 4}$

$= 2.04 \times 10^{2} = 204 N$

(ii) To find out the resultant force exerted by the sides of the glass from the F.B.D of water inside the glass

$P_{a} \times A + mg = Ah ρ_{w} g + F_{s} + P_{a} A$

$mg = Ah ρ_{w} {g+F}_{s}$

$F_{s} = mg - Ah ρ_{w} g$

$= 0 {.5×10-20×10}^{- 4} \times 20 \times 10^{- 2} \times 1000 \times 10$

$= 5 - 4 \times 10^{2} \times 10^{- 4} \times 10^{- 2} \times 10^{4}$

$F_{s} = 5 - 4 = 1N upward$

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