a beaker of circular cross section radius 4cm is filled with hg upto height 10cm. the force exerted by the mercury on the bottom of the beaker will be

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Sachin · Answer

P= F/AArea= πr2 = 3.14*4*4= 50.24cm2= 0.005024m2P=hdg= 0.1*13.6*10^3*9.8= 13.328kpaHere d is density of mercury.Now F= P*A= 13.328*0.005024 66.99= 67N

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a beaker of circular cross section radius 4cm is filled with hg upto height 10cm. the force exerted by the mercury on the bottom of the beaker will be

a beaker of circular cross section radius 4cm is filled with hg upto height 10cm. the force exerted by the mercury on the bottom of the beaker will be

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a beaker of circular cross section radius 4cm is filled with hg upto height 10cm. the force exerted by the mercury on the bottom of the beaker will be

a beaker of circular cross section radius 4cm is filled with hg upto height 10cm. the force exerted by the mercury on the bottom of the beaker will be

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