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(A) Let the density of the ball be $\rho_b = 500 \ kg/m^3$ and the density of the liquid be $\rho_l = 1000 \ kg/m^3$. Let $V$ be the volume of the bal

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let the density of the ball be $ho_b = 500 \ kg/m^3$ and the density of the liquid be $ho_l = 1000 \ kg/m^3$. Let $V$ be the volume of the ball.When the ball is inside the liquid,the forces acting on it are the buoyant force $F_B = V ho_l g$ (upwards) and the weight $W = V ho_b g$ (downwards).The net force $F_{net} = F_B - W = V ho_l g - V ho_b g = V g (ho_l - ho_b)$.The acceleration $a$ of the ball inside the liquid is $a = \frac{F_{net}}{m} = \frac{V g (ho_l - ho_b)}{V ho_b} = g \frac{(ho_l - ho_b)}{ho_b}$.Substituting the values: $a = 10 	imes \frac{(1000 - 500)}{500} = 10 	imes \frac{500}{500} = 10 \ m/s^2$ (upwards).Let $v_0$ be the velocity of the ball when it strikes the liquid surface. Using the equation of motion $v = u + at$ for the motion inside the liquid,where the final velocity is $0$ at $t = 2 \ s$:$0 = v_0 - a t \Rightarrow v_0 = a t = 10 	imes 2 = 20 \ m/s$.For the free fall from height $h$,$v_0^2 = 2gh$.$20^2 = 2 	imes 10 	imes h \Rightarrow 400 = 20h \Rightarrow h = 20 \ m$.

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