Question 38: Properties of Matter

Question

A spherical ball of radius $$1 	imes 10^{-4} \mathrm{~m}$$ and density $$10^5 \mathrm{~kg} / \mathrm{m}^3$$ falls freely under gravity through a distance $$h$$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $$h$$ is approximately:
(The coefficient of viscosity of water is $$9.8 	imes 10^{-6} \mathrm{~N} \mathrm{~s} / \mathrm{m}^2$$)

Accepted Answer

To solve this problem, we can use the concepts of terminal velocity and the forces acting on the spherical ball. First, let's analyze the situation step-by-step.
When the ball falls freely under gravity, it achieves a terminal velocity $$v_t$$ in water. This terminal velocity is reached when the gravitational force is balanced by the drag force and the buoyant force in the water.
The forces acting on the ball are:
1. Gravitational Force: $$F_g = mg$$
2. Buoyant Force: $$F_b = ho_{	ext{water}} V g$$
3. Drag Force: $$F_d = 6 \pi \eta r v_t$$
Where,
$$m$$ is the mass of the ball.
$$g$$ is the acceleration due to gravity ($$9.8 \, \mathrm{m/s^2}$$).
$$ho_{	ext{water}}$$ is the density of water ($$1000 \, \mathrm{kg/m^3}$$).
$$V$$ is the volume of the ball ($$\frac{4}{3} \pi r^3$$).
$$\eta$$ is the coefficient of viscosity of water ($$9.8 	imes 10^{-6} \, \mathrm{Ns/m^2}$$).
$$r$$ is the radius of the ball ($$1 	imes 10^{-4} \, \mathrm{m}$$).
$$v_t$$ is the terminal velocity.
Using the equilibrium condition at terminal velocity:
$$F_g = F_b + F_d$$
$$mg = ho_{	ext{water}} V g + 6 \pi \eta r v_t$$
First, compute the mass of the ball:
$$m = ho_{	ext{ball}} 	imes V = ho_{	ext{ball}} 	imes \frac{4}{3} \pi r^3$$
$$m = 10^5 \, \mathrm{kg/m^3} 	imes \frac{4}{3} \pi (1 	imes 10^{-4} \, \mathrm{m})^3$$
$$m = 10^5 \,\mathrm{kg/m^3} 	imes \frac{4}{3} \pi 	imes 10^{-12} \,\mathrm{m^3}$$
$$m = \frac{4}{3} \pi 	imes 10^{-7} \, \mathrm{kg}$$
Next, solve for the terminal velocity $$v_t$$ using the equilibrium equation:
$$mg = ho_{	ext{water}} \frac{4}{3} \pi r^3 g + 6 \pi \eta r v_t$$
$$v_t = \frac{mg - ho_{	ext{water}} \frac{4}{3} \pi r^3 g}{6 \pi \eta r}$$
$$v_t = \frac{\frac{4}{3} \pi 	imes 10^{-7} 	imes 9.8 - 1000 	imes \frac{4}{3} \pi (1 	imes 10^{-4})^3 	imes 9.8}{6 \pi 	imes 9.8 	imes 10^{-6} 	imes 10^{-4}}$$
$$v_t = \frac{\frac{4}{3} \pi 	imes 10^{-7} 	imes 9.8 - 1000 	imes \frac{4}{3} \pi 	imes 10^{-12} 	imes 9.8}{6 \pi 	imes 9.8 	imes 10^{-10}}$$
$$v_t = \frac{\frac{4}{3} \pi 	imes 9.8 	imes 10^{-7} (1 - 10^{-5})}{6 \pi 	imes 9.8 	imes 10^{-10}}$$
$$v_t = \frac{\frac{4}{3} 	imes 10^{-7}}{6 	imes 10^{-10}}$$
$$v_t = \frac{4}{18} 	imes 10^3 \, \mathrm{m/s}$$
$$v_t \approx 222.22 \, \mathrm{m/s}$$
The height $$h$$ required to reach this terminal velocity while the ball falls freely under gravity can be found using the kinematic equation:
$$v_t = \sqrt{2gh}$$
$$h = \frac{v_t^2}{2g}$$
$$h = \frac{(222.22)^2}{2 	imes 9.8}$$
$$h = \frac{49328.88}{19.6}$$
$$h \approx 2517.8 \, \mathrm{m}$$
So, the closest value of $$h$$ is approximately:
Option A: 2518 m
.

Answer

2518 m

Answer

2396 m

Answer

2249 m

Answer

2296 m

Properties of Matter

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