A spherical ball of radius r and relative den | vedclass

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(A) The volume of the ball is $V = \frac{4}{3}\pi r^3$. The density of the ball is $\sigma = 0.5\rho$.Initially,the ball floats with half its volume s

Vedclass · Accepted Answer

$\frac{5}{12}\pi r^4 \rho g$

Vedclass · Answer

(A) The volume of the ball is $V = \frac{4}{3}\pi r^3$. The density of the ball is $\sigma = 0.5\rho$.Initially,the ball floats with half its volume submerged. The buoyant force $F_B = V_{submerged} \rho g = (V/2)\rho g = (2/3)\pi r^3 \rho g$. Since it is in equilibrium,the weight of the ball $W = V \sigma g = V(0.5\rho)g = (2/3)\pi r^3 \rho g$. Thus,$F_B = W$.To push the ball down until it is fully submerged,we must apply an external force to overcome the net buoyant force. The work done is equal to the change in potential energy of the system.When the ball is pushed down by a distance $x$,the additional volume submerged is $A(x)dx$. The work done $W = \int_{0}^{r} F_{ext} dx$. The net force required at any depth $y$ (where $y$ is the depth of the center below the equilibrium position) is $F_{ext} = F_{buoyant} - Weight = (V_{submerged}(y) - V_{submerged, initial})\rho g$.Integrating the force over the displacement,the work done is $W = \frac{5}{12}\pi r^4 \rho g$.

$A$ spherical ball of radius $r$ and relative density $0.5$ is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that the whole of it is just immersed in water is: (where $\rho$ is the density of water)

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