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A wooden ball of density \mathrm{\sigma} is released from the bottom of a tank which is filled with a liquid of density \mathrm{\rho ;(\rho>\sigma)} up to a height \mathrm{h_1}. The ball rises in the liquid, emerges from its surface and attains a height \mathrm{h_2} in air. If viscous effects are neglected, the ratio \mathrm{h_2 / h_1} is

Option: 1

\mathrm{\left(\frac{\rho}{\sigma}+1\right)}


Option: 2

\mathrm{\left(\frac{\rho}{\sigma}-1\right)}


Option: 3

\mathrm{\frac{\rho}{\sigma}}


Option: 4

\mathrm{\frac{\sigma}{\rho}}


Answers (1)

best_answer

Weight of the ball \mathrm{W=m g=\sigma V g}. Upthrust \mathrm{U=\rho V g}. Therefore, the net upward force acting on the ball is

                          \mathrm{ F=U-W=(\rho-\sigma) V g }
Now, mass of the ball is \mathrm{m=\sigma V}. Therefore, upward acceleration of the ball while it is rising in the liquid is

                   \mathrm{ a=\frac{F}{m}=\frac{(\rho-\sigma) V g}{\sigma V}=\left(\frac{\rho-\sigma}{\sigma}\right) g }
Velocity of the ball on reaching the surface of water is

                              \mathrm{ v=\sqrt{2 a h_1} }                                              (i)
This is the initial upward velocity of the ball in air. If it rises to a height \mathrm{h_2} in air, we have

                              \mathrm{ v=\sqrt{2 g h_2} }                                               (ii)
Equating (i) and (ii), we have \mathrm{a h_1=g h_2}

            or       \mathrm{ \frac{h_2}{h_1}=\frac{a}{g}=\frac{\rho-\sigma}{\sigma}=\left(\frac{\rho}{\sigma}-1\right) }

Posted by

Kuldeep Maurya

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