A ball rises to the surface at a constant vel | vedclass

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(A) Let the density of the ball be $\rho$ and its volume be $V$. The density of the liquid is $4\rho$.The weight of the ball is $F_G = V\rho g$ (actin

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$3 : 1$

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(A) Let the density of the ball be $\rho$ and its volume be $V$. The density of the liquid is $4\rho$.The weight of the ball is $F_G = V\rho g$ (acting downwards).The buoyant force acting on the ball is $F_B = V(4\rho)g = 4V\rho g$ (acting upwards).Since the ball rises at a constant velocity,the net force on it is zero. The forces acting on the ball are the buoyant force upwards,and the weight and the viscous friction force $F_v$ downwards.Therefore,$F_B = F_G + F_v$.Substituting the values: $4V\rho g = V\rho g + F_v$.Thus,$F_v = 3V\rho g$.The ratio of the force of friction to the weight of the ball is $\frac{F_v}{F_G} = \frac{3V\rho g}{V\rho g} = \frac{3}{1}$.So,the ratio is $3 : 1$.

$A$ ball rises to the surface at a constant velocity in a liquid whose density is $4$ times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball and its weight is

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