A rubber ball of mass 10 g and radius 2 cm is submerged in water up to a depth of 5 cm and released. Find the height up to which the ball pops up above the surface of water, neglecting the resistance of water and air. The density of water = `1g*cm^(-3)`.
Text Solution
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Volume of the rubber ball, `V=4/3xx22/7xx(2)^(3)cm^(3)=704/21cm^(3)` Resultant upward force = buoyant force - weight of the rubber ball or, `10a=Vxx1xxg-10xxg" "`[a = upward acceleration] or, `10a=(704g)/21-10gor,a=494/210gcm*s^(-2)` Let the velocity of the ball when it reaches the surface of water be v, then `v=sqrt(2as)=sqrt((2xx494)/210xx980xx5)` `=sqrt((2xx494xx14xx5)/3)=151.83cm*s^(-1)` If the ball pops up to a height x above the surface of water, then `v^(2)=2gxor,x=v^(2)/(2g)=((151.83)^(2))/(2xx980)=11.76cm`.
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