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A glass full of water has a bottom of area `20 cm^(2)`, top of area `20 cm`, height `20 cm` and volume half a litre. a. Find the force exerted by the water on the bottom. b. Considering the equilibrium of the `v`, after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`. Density of water `= 1000 kg//m^(-3)` and `g = 10 m//s^(2)` |
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Answer» Correct Answer - A::B::D `Pa=1.0xx10^5N/m^2`, `=Pw=10^3mg/m^3` g=10m/s^2` `v=500m/=500g=0.5kg` a. Force exerted at the bottom = Force due to cylindrical water column + atm. Force `=AxxhxxP_wxxg+P_axxA` `=A(hrho_wg+P_a)` `=204N` b. To find out the resultant force exerted by the sides of the glass, from the free body diagram of water inside the glass, `P_axxA+mg=Axxhxxrho_2xxg+F_s+P_axxA` `rarr mg=Axxhxxp_wxg+F_s` `rarr 0.5xx1=20xx10^-4xx20x10^-2xx10^-3xx10+F_s` `rarr F_s=5-4=1N` (upward) This force 1N is provided by the sides of the glass. |
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