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Problem Sets:
Problem
There is water in a container. A plastic test tube has a small iron block hanging underneath it, and the test tube floats on the water surface. Now the small iron block is removed and placed inside the test tube, and the test tube still floats on the water surface. Then:
- Select the correct answer. A. The water level in the container will rise B. The water level in the container will fall C. The volume of water displaced by the plastic test tube increases D. The volume of water displaced by the plastic test tube remains unchanged
- Instead of putting the iron block inside the tube, the block put at the bottom of the container. Which of the above is correct?
[Q1] The total displaced volume is $V_{disp} = \frac{m_t + m_i}{\rho_w}$ in both initial and final states, so the water level is unchanged. The volume displaced by the tube changes from $V_{disp,t,1} = V_{disp} - V_i$ to $V_{disp,t,2} = V_{disp}$. Thus, the volume displaced by the test tube increases. The correct answer is C.
[Q2] The total displaced volume changes from $V_{disp,1} = \frac{m_t + m_i}{\rho_w}$ to $V_{disp,3} = \frac{m_t}{\rho_w} + V_i$. Since $\rho_i > \rho_w$, it follows that $V_{disp,1} > V_{disp,3}$. The total volume of displaced water decreases. The correct answer is B.
Here is the solution based on the principles of buoyancy.
[Q1] Iron block placed inside the test tube
1. Analyze Total Displaced Volume Let $m_t$ and $m_i$ be the masses of the test tube and the iron block, respectively. Let $\rho_w$ be the density of water.
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Initial State (State 1): The test tube and the hanging iron block float as a single system. According to the principle of flotation, the total buoyant force equals the total weight of the system.
$$F_{B,1} = (m_t + m_i)g$$The total volume of water displaced, $V_{disp,1}$, is related to the buoyant force by Archimedes' principle: $F_{B,1} = \rho_w V_{disp,1} g$.
$$V_{disp,1} = \frac{m_t + m_i}{\rho_w}$$ -
Final State (State 2): The iron block is inside the test tube, and the system still floats. The total weight remains the same.
$$F_{B,2} = (m_t + m_i)g$$The new total displaced volume, $V_{disp,2}$, is:
$$V_{disp,2} = \frac{m_t + m_i}{\rho_w}$$ -
Conclusion on Water Level: Since $V_{disp,1} = V_{disp,2}$, the total volume of displaced water is unchanged. Therefore, the water level in the container remains the same. This eliminates options A and B.
2. Analyze Volume Displaced by the Test Tube Let $V_{disp,t,1}$ and $V_{disp,t,2}$ be the volumes of water displaced by the test tube in states 1 and 2. Let $V_i$ be the volume of the iron block.
- In State 1, the total displaced volume is the sum of the volume displaced by the tube and the volume of the fully submerged iron block: $$V_{disp,1} = V_{disp,t,1} + V_i$$
- In State 2, the iron block is inside the tube and does not displace water directly. The total displaced volume is solely due to the test tube: $$V_{disp,2} = V_{disp,t,2}$$
- Equating the expressions for total displaced volume ($V_{disp,1} = V_{disp,2}$), we get: $$V_{disp,t,1} + V_i = V_{disp,t,2}$$
- Since the volume of the iron block $V_i$ is positive, $V_{disp,t,2} > V_{disp,t,1}$. The volume of water displaced by the plastic test tube increases. This confirms option C is correct and D is incorrect.
[Q2] Iron block put at the bottom of the container
1. Analyze Total Displaced Volume
- Initial State (State 1): As derived before, the total displaced volume is: $$V_{disp,1} = \frac{m_t + m_i}{\rho_w}$$
- Final State (State 3): The test tube and iron block are separate.
- The test tube floats alone, displacing a volume $V_{disp,t,3} = \frac{m_t}{\rho_w}$.
- The iron block sinks to the bottom, displacing a volume equal to its own, $V_i$.
- The total displaced volume is the sum: $$V_{disp,3} = V_{disp,t,3} + V_i = \frac{m_t}{\rho_w} + V_i$$
2. Compare Displaced Volumes Let $\rho_i$ be the density of iron. We have $m_i = \rho_i V_i$.
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We compare $V_{disp,1}$ and $V_{disp,3}$:
$$V_{disp,1} = \frac{m_t + m_i}{\rho_w} = \frac{m_t}{\rho_w} + \frac{m_i}{\rho_w} = \frac{m_t}{\rho_w} + \frac{\rho_i V_i}{\rho_w}$$ $$V_{disp,3} = \frac{m_t}{\rho_w} + V_i$$ -
Since iron is denser than water, $\rho_i > \rho_w$, which implies $\frac{\rho_i}{\rho_w} > 1$.
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Therefore, $\frac{\rho_i V_i}{\rho_w} > V_i$.
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This leads to the conclusion that $V_{disp,1} > V_{disp,3}$.
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Conclusion on Water Level: The total volume of displaced water decreases. Therefore, the water level in the container will fall. This confirms option B is correct.