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  • A small cork ball of density is immersed in water of density <img alt="\mathrm{\rho(>\

A small cork ball of density \mathrm{\sigma} is immersed in water of density \mathrm{\rho(>\sigma)} to a depth h and released. If \mathrm{\rho=4 \sigma, } the height up to which the ball will rise above the surface of water is

Option: 1

h


Option: 2

2h


Option: 3

3h


Option: 4

4h


Answers (1)

best_answer

Let V be the volume of the ball. The upward force when the ball is completely immersed in water is

                                    \mathrm{ U=\rho V g }
Downward force acting on the ball is its weight

                                         \mathrm{ W=\sigma V g }
Net upward force is

                           \mathrm{ F=U-W=(\rho-\sigma) V g }
Mass of the ball is \mathrm{m=\sigma V.} Therefore, upward acceleration is

                    \mathrm{ a=\frac{F}{m}=\frac{(\rho-\sigma) V g}{\sigma V}=\left(\frac{\rho-\sigma}{\sigma}\right) g }                       (1)
If u is the velocity of the ball when it emerges from water, then

                                \mathrm{ \Rightarrow \quad \begin{aligned} 0-u^2 & =-2 a h \\ u \quad u & =\sqrt{2 a h} \end{aligned} }
If H is the maximum height attained,
 

                             \mathrm{ \begin{aligned} 0-u^2 & =2 \times(-g) H \\ \Rightarrow \quad H & =\frac{u^2}{2 g}=\frac{2 a h}{2 g}=\frac{a h}{g} \end{aligned} }                                       (2)
Using (1) in (2) we get

                                          \mathrm{ H=\left(\frac{\rho-\sigma}{\sigma}\right) h }

\mathrm{\text { Putting } \rho=4 \sigma \text {, we get } H=3 h \text {, which is choice (c). }}

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SANGALDEEP SINGH

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