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CBSE Class 12-science Answered

a solid ball of density half of that of water falls freely under the action of gravity from a height of 19.6 and then water enters into tje water upto depth of？
Asked by sbtech357 | 22 Jul, 2018, 09:32: AM
answered-by-expert Expert Answer
when the ball is falling from a height 19.6 m, it is accelerated due to gravity.
Once the ball enters water, it is subjected to buoyancy force due to its low density.
The ball is subjected to retardation in water due to this buyoncy force.
 
Initial velocity u of the ball when it enters into water, begin mathsize 12px style u space equals square root of 2 g h end root space equals space square root of 2 cross times 9.8 cross times 19.6 end root space equals space 19.6 space m divided by s end style
Retardation r = Force / mass = begin mathsize 12px style fraction numerator V cross times open parentheses rho subscript w minus rho subscript b close parentheses cross times g over denominator V cross times rho subscript b end fraction space equals space g space space b e c a u s e space rho subscript b space equals 1 half rho subscript w end style
It is assumed that V is volume of ball, g is acceleration due to gravity. ρw and ρb are densities of water and ball respectively.
 
depth S travelled by ball in water is given by,   S = u2 / (2×r) = (19.6×19.6)/(2×9.8) = 19.6 m
Answered by Thiyagarajan K | 22 Jul, 2018, 11:51: AM

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