A ball whose density is $0.4 \times 10^3\,kg/m^3$ falls into water from a height of $9\,cm$ . To what depth does the ball sink ? ....... $cm$
Diffcult
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energy of the ball at the surface of the water

$=\mathrm{mgh}=\mathrm{V} \rho_{\mathrm{b}} \mathrm{gh}$

upthrust on $\mathrm{ball}=\mathrm{V} \rho_{\mathrm{w}} \mathrm{g}$

net force on the ball in water $=\left(\rho_{w}-\rho_{b}\right) V g$

 it is constant and upwards

 let it sink upto depth d then

$\left(\rho_{m}-\rho_{b}\right) V_{\rho}d=V \rho_{b} g h$

$\Rightarrow \mathrm{d}=\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{w}}-\rho_{\mathrm{b}}} \mathrm{h}$

$\Rightarrow \mathrm{d}=\frac{0.4 \times 10^{3}}{1 \times 10^{3}-0.4 \times 10^{3}} \times 9 \mathrm{cm}=\frac{4}{6} \times 9=6 \mathrm{cm}$

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