A table tennis ball has radius (3/2) x 10^-2 m and mass (22/7) x 10^-3 kg. It is slowly pushed down

Q: Why is the work done by the viscous force zero when pushing the ball down?

The problem states the ball is pushed down *slowly*. Viscous force (drag) is directly proportional to velocity ($F_v = 6\pi\eta r v$). If the velocity is nearly zero, the viscous force is negligible, and thus the work done by it is effectively zero.

Q: Why is the external work positive? Shouldn't we be pushing in the direction of gravity?

Calculate the forces! The buoyant force on the ping-pong ball is much greater than its weight (gravity). If you just let it go, it floats. To submerge it, you must actively push *downward* against the strong upward buoyant force. Since your force is downward and the displacement is downward, the work you do is positive.

Q: How did we know the work done during ascent is exactly $0.077\text{ J}$?

During the downward push, $W_{\text{ext}} + W_g + W_B = 0$, which means $W_{\text{ext}} = -(W_g + W_B)_{\text{downward}}$. During the upward ascent, the displacement reverses direction, so the signs of the work done by gravity and buoyancy reverse. Therefore, $(W_g + W_B)_{\text{upward}} = -(W_g + W_B)_{\text{downward}} = W_{\text{ext}} = 0.077\text{ J}$.

Q: Why do we use $v = 7\text{ m/s}$ to calculate the maximum viscous force?

Stokes' Law states that viscous force increases linearly with speed. As the ball accelerates upwards through the water, its speed increases. It reaches its absolute maximum speed right at the moment it breaks the water surface, which we calculated as $7\text{ m/s}$.

Q: Does the mass of the ball affect the maximum height $H$ it reaches in the air?

Indirectly, yes! While the kinematic formula $H = v^2 / 2g$ doesn't explicitly show mass, the launch velocity $v$ heavily depends on the mass. A lighter ball experiences a greater net upward acceleration in the water, leading to a higher exit velocity $v$, and consequently, a greater height $H$.

Question

A table tennis ball has radius $(3/2) 	imes 10^{-2}	ext{ m}$ and mass $(22/7) 	imes 10^{-3}	ext{ kg}$. It is slowly pushed down into a swimming pool to a depth of $d = 0.7	ext{ m}$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is(are) correct?

[Given: $\pi = 22/7$, $g = 10	ext{ ms}^{-2}$, density of water $= 1 	imes 10^3	ext{ kgm}^{-3}$, viscosity of water $= 1 	imes 10^{-3}	ext{ Pa-s}$.]

Options:

(A) The work done in pushing the ball to the depth $d$ is $0.077	ext{ J}$.

(B) If we neglect the viscous force in water, then the speed $v = 7	ext{ m/s}$.

(C) If we neglect the viscous force in water, then the height $H = 1.4	ext{ m}$.

(D) The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500/9$.

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of Mechanics and Fluid Dynamics. It tests your ability to seamlessly transition between the Work-Energy Theorem, Archimedes' Principle, Kinematics, and Stokes' Law. Let's dive into the physics of this table tennis ball's journey.

### Analyzing the Setup
Imagine a table tennis ball floating on a swimming pool. We slowly push it down to a depth of $d = 0.7	ext{ m}$. Then, we let it go! It shoots up, pops out of the water, and flies into the air. We need to analyze the work done, its speed, and the forces acting on it.

### The Work-Energy Theorem (Option A)
When we push the ball down *slowly*, the change in kinetic energy is zero ($\Delta K = 0$). According to the Work-Energy Theorem, the total work done by all forces—our external push, gravity, and buoyancy—must add up to zero. Since it's moving very slowly, we can safely ignore the viscous drag for this part of the motion.

The buoyant force pushes up, while gravity and our hand push down. The net upward force is buoyancy minus gravity:
$$F_{	ext{net}} = F_B - mg = ho_w V g - mg$$

Let's calculate the volume of the ball:
$$V = \frac{4}{3}\pi r^3 = \frac{4}{3} 	imes \frac{22}{7} 	imes (1.5 	imes 10^{-2})^3 = \frac{99}{7} 	imes 10^{-6}	ext{ m}^3$$

Now, substituting this into our net force equation:
$$F_{	ext{net}} = (10^3) \left(\frac{99}{7} 	imes 10^{-6}ight) (10) - \left(\frac{22}{7} 	imes 10^{-3}ight) (10) = \frac{0.99 - 0.22}{7} = 0.11	ext{ N}$$

The work done by our external force to push it down by $0.7	ext{ m}$ is:
$$W_{	ext{ext}} = F_{	ext{net}} 	imes d = 0.11 	imes 0.7 = 0.077	ext{ J}$$
**Option A is absolutely correct!**

### Speed at the Surface (Option B)
Now, we release the ball. It accelerates upwards. If we neglect the viscous force, the work done by the net upward force (which is buoyancy minus gravity) will completely convert into the ball's kinetic energy when it reaches the surface. Notice that this work is exactly equal to the work we did to push it down!

$$W_{	ext{net}} = \Delta K \implies 0.077 = \frac{1}{2}mv^2$$

Plugging in the mass of the ball:
$$0.077 = \frac{1}{2} \left( \frac{22}{7} 	imes 10^{-3} ight) v^2$$
$$0.077 = \frac{11}{7} 	imes 10^{-3} 	imes v^2 \implies v^2 = 49$$

Taking the square root, the speed $v$ is exactly **$7	ext{ m/s}$**. **Option B is also correct!**

### Maximum Height (Option C)
Once the ball leaves the water, buoyancy disappears. Only gravity acts on it, slowing it down. The maximum height $H$ it reaches is given by standard kinematics:
$$H = \frac{v^2}{2g} = \frac{49}{2 	imes 10} = 2.45	ext{ m}$$

Option C claims it's $1.4	ext{ m}$, so **Option C is incorrect.**

### The Forces Ratio (Option D)
Finally, let's evaluate Option D. We need the ratio of the net force (excluding viscosity) to the maximum viscous force. We already found the net force is $0.11	ext{ N}$.

The viscous force is given by Stokes' Law: $F_v = 6\pi \eta r v$. It reaches its maximum value when the speed is maximum, which is $7	ext{ m/s}$ just before it exits the water.

Let's calculate this maximum viscous force:
$$F_{v,	ext{max}} = 6 	imes \frac{22}{7} 	imes 10^{-3} 	imes 1.5 	imes 10^{-2} 	imes 7 = 1.98 	imes 10^{-3}	ext{ N}$$

Dividing the net force by this viscous force gives us:
$$	ext{Ratio} = \frac{0.11}{1.98 	imes 10^{-3}} = \frac{110}{1.98} = \frac{500}{9}$$

**Option D is correct!** The correct options are A, B, and D.

Select Answer:

Visualized Solution

\text{Analyzing the Setup}

\text{Work-Energy Theorem (Option A)}

\text{Work Done by External Force}

\text{Speed at Surface (Option B)}

\text{Calculating } v

\text{Maximum Height (Option C)}

\text{Forces Ratio (Option D)}

\text{Calculating the Ratio}

\text{Conclusion}

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup Imagine a table tennis ball floating on a swimming pool

The Work-Energy Theorem (Option A) When we push the ball down slowly, the change in kinetic energy is zero ( $Δ K = 0$ )

Speed at the Surface (Option B) Now, we release the ball

Maximum Height (Option C) Once the ball leaves the water, buoyancy disappears

The Forces Ratio (Option D) Finally, let's evaluate Option D

Similar Questions

Comprehension Passage

The value of $X$ is _______.

The value of $Y$ is _______.

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As the bubble moves upwards, besides the buoyancy force the following forces are acting on it.

When the gas bubble is at a height y from the bottom, its temperature is

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Select Answer:

Visualized Solution

\text{Analyzing the Setup}

\text{Work-Energy Theorem (Option A)}

\text{Work Done by External Force}

\text{Speed at Surface (Option B)}

\text{Calculating } v

\text{Maximum Height (Option C)}

\text{Forces Ratio (Option D)}

\text{Calculating the Ratio}

\text{Conclusion}

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup Imagine a table tennis ball floating on a swimming pool

The Work-Energy Theorem (Option A) When we push the ball down slowly, the change in kinetic energy is zero ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔK=0)

Speed at the Surface (Option B) Now, we release the ball

Maximum Height (Option C) Once the ball leaves the water, buoyancy disappears

The Forces Ratio (Option D) Finally, let's evaluate Option D

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of X is _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of Y is _______.

Comprehension Passage

As the bubble moves upwards, besides the buoyancy force the following forces are acting on it.

When the gas bubble is at a height y from the bottom, its temperature is

The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)

The Work-Energy Theorem (Option A) When we push the ball down slowly, the change in kinetic energy is zero ( $Δ K = 0$ )

The value of $X$ is _______.

The value of $Y$ is _______.