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# A problem from Hydraulic

“a stone of [wiki]density [/wiki] 4gm/cm3 is dropped freely in a liquid of density 0.8 gm/cm3 .what will be the acceleration of the sinking stone?”

Ans:

the force acting are

weight =mg= $\frac{4}{3}\pi r^{3}\rho g$ downwards (where $\rho$ is the density of the body)

[wiki]Buoyant force[/wiki] =$\frac{4}{3}\pi r^{3}\sigma g$

SO, THE NET DOWNWARD [wiki]FORCE [/wiki]IS
$\frac{4}{3}\pi r^{3}(\rho -\sigma ) g$

Therefore, acceleration = net downward force / mass

=$\frac{\frac{4}{3}\pi r^{3}(\rho -\sigma ) g}{\frac{4}{3}\pi r^{3}\rho }$
=$\left (1-\frac{\sigma }{\rho } \right )g$

Now substitute the values and calculate

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