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