A ball whose density is 0.4 × 10^3 kg/m | vedclass

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Question

(b) The velocity of ball before entering the water surface
$v = \sqrt {2gh} = \sqrt {2g 	imes 9} $
When ball enters into water, due to upthrust of

Vedclass · Accepted Answer

$6 $

Vedclass · Answer

(b) The velocity of ball before entering the water surface
$v = \sqrt {2gh} = \sqrt {2g 	imes 9} $
When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)
The retardation, a = $\frac{{{m{apparent weight}}}}{{{m{mass of ball}}}}$
$\frac{{ = V(ho - \sigma )g}}{{Vho }}$$ = \left( {\frac{{ho - \sigma }}{ho }} ight)g$$ = \left( {\frac{{0.4 - 1}}{{0.4}}} ight) 	imes g$$ = - \frac{3}{2}g$
If h be the depth upto which ball sink, then,
$0 - {v^2} = 2 	imes \left( { - \frac{3}{2}g} ight) 	imes h$==> $2g 	imes 9 = 3gh$

$h = 6 cm.$

A ball whose density is $0.4 × 10^3 kg/m^3$ falls into water from a height of $9 cm$ . To what depth does the ball sink........ $cm$

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