A spherical marble of radius 1 , cm is stuck | vedclass

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(D) The force exerted by the water on the marble is due to the hydrostatic pressure at the depth of the marble. The pressure $P$ at the bottom of the

Vedclass · Accepted Answer

$0.31 \, N$ downwards

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(D) The force exerted by the water on the marble is due to the hydrostatic pressure at the depth of the marble. The pressure $P$ at the bottom of the bucket is given by $P = ho g h$,where $ho = 1000 \, kg/m^3$ is the density of water,$g = 10 \, m/s^2$ is the acceleration due to gravity,and $h = 10 \, cm = 0.1 \, m$ is the height of the water column.The force $F$ acting downwards on the marble due to the water pressure is $F = P 	imes A$,where $A$ is the cross-sectional area of the hole (which is equal to the cross-sectional area of the marble at its equator).The area $A = \pi r^2$,where $r = 1 \, cm = 0.01 \, m$.Substituting the values:$F = (1000 \, kg/m^3) 	imes (10 \, m/s^2) 	imes (0.1 \, m) 	imes \pi 	imes (0.01 \, m)^2$$F = 1000 	imes 10 	imes 0.1 	imes 3.14159 	imes 0.0001$$F = 1000 	imes 0.000314159 \approx 0.314 \, N$.Since the pressure acts downwards on the top surface of the marble,the net force due to water is $0.31 \, N$ downwards.

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