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  • A spherical small ball of density  is gently released in a liquid of density <img alt="\mathrm{ \sigma(\rho&

A spherical small ball of density \rho is gently released in a liquid of density \mathrm{ \sigma(\rho>\sigma)}. The initial acceleration of the free fall of the ball will be

Option: 1

\mathrm{\left(\frac{\rho+\sigma}{\rho}\right) g}


Option: 2

\mathrm{\left(\frac{\rho-\sigma}{\sigma}\right) g}


Option: 3

\mathrm{\left(\frac{\rho-\sigma}{\rho}\right) g}


Option: 4

\mathrm{g}


Answers (1)

best_answer

Let m be the mass of the ball and V its volume. Its mass \mathrm{m=\rho V}. The weight of the ball is

\mathrm{ W=m g=\rho V g }
The volume of the liquid displaced \mathrm{=V}. If \mathrm{\sigma} is the density of the liquid, the weight of the liquid displaced is the upthrust U it experiences.

\mathrm{ U=V \sigma g }
\mathrm{\therefore \quad}  The net downward force acting on the body is

\mathrm{ F=W-U=(\rho-\sigma) V g }
The initial acceleration is

\mathrm{ a=\frac{F}{m}=\frac{(\rho-\sigma) V g}{\rho V}=\left(\frac{\rho-\sigma}{\rho}\right) g }

Posted by

sudhir kumar

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