2022 A/L Physics Past Paper · Question 36 (English Medium)

Answer: (2) $500\ \text{cm}^3$

Given information

What we know:

The wooden cube has side $10\ \text{cm}$. Its density is $800\ \text{kg m}^{-3}$.

The water has density $1000\ \text{kg m}^{-3}$.

The cube hangs from a spring balance by a light string. The balance reads $3\ \text{N}$.

The figure shows the cube partly under the water.

The question does not give $g$. We use $g = 10\ \text{m s}^{-2}$, which makes the numbers come out in whole values.

What the question asks

We want: the volume of the wood that is under water.

What each letter means:

$V$ ($\text{m}^3$): the volume of the whole cube.

$V'$ ($\text{m}^3$): the volume of the cube that is under water.

$\rho_c$ ($\text{kg m}^{-3}$): the density of wood.

$\rho$ ($\text{kg m}^{-3}$): the density of water.

$W$ ($\text{N}$): the weight of the cube.

$T$ ($\text{N}$): the pull of the string, which is the balance reading.

$U$ ($\text{N}$): the upthrust of the water on the cube.

Relevant theory

What is happening:

Gravity pulls the cube down with its weight $W$.

The string pulls it up with $T$, the balance reading.

The water pushes up on the part that is under water. This push is the upthrust $U$.

The cube hangs still, so the up forces balance the down force.

$$T + U = W$$

In words: the string pull plus the upthrust equals the weight.

When a body is completely or partially immersed in a fluid at rest, it experiences an upthrust equal to the weight of the fluid displaced by the body.

If a volume V of a body is immersed in a fluid of density $\rho$ , the upthrust according to Archimedes principle, is $$U = V\rho g$$

In the formula $U = V\rho g$, the $V$ is the volume that is under water. We write it $V'$.

For example, push an empty plastic bottle down into a bucket of water. The more of the bottle goes under, the harder the water pushes it up. The push depends on how much water the bottle moves out of the way.

The cube is like that bottle. The part under water moves water out of the way, and that water's weight is the upthrust.

Required equations

$T + U = W$

$U = V'\rho g$

$W = V\rho_c g$

$V' = \dfrac{W - T}{\rho g}$

Applying it to this question

$\dfrac{\text{N}}{\text{kg m}^{-3} \times \text{m s}^{-2}} = \text{m}^3$, which is a volume.

$0.001$

$V = 0.1 \times 0.1 \times 0.1$

$8$

$W = V\rho_c g = 0.001 \times 800 \times 10$

$5$

$U = W - T = 8 - 3$

$0.0005$

$V' = \dfrac{U}{\rho g} = \dfrac{5}{1000 \times 10}$

$500$

$V' = 0.0005 \times 10^6$

Reasoning and calculation

The volume of wood under water is $500\ \text{cm}^3$.

That is option (2).

Checks

$(rho_w * v * g - t) / (rho * g)$

Does it make sense?

The whole cube is $1000\ \text{cm}^3$, and $500\ \text{cm}^3$ is less than that. So the cube can be partly under water, as in the figure.

Check the forces: $3 + 5 = 8\ \text{N}$. The string pull plus the upthrust equals the weight.

Check the upthrust: $500\ \text{cm}^3$ of water has mass $0.5\ \text{kg}$ and weight $5\ \text{N}$. That is the upthrust we found.

The question gives no $g$. With $g = 9.8\ \text{m s}^{-2}$ the volume would be about $494\ \text{cm}^3$, which is still closest to option (2).

Conclusion

$V' = 500\ \text{cm}^3$

The volume of wood under water is $500\ \text{cm}^3$.

The answer is (2).

When something hangs still, list the forces up and the forces down and make them equal.

The upthrust is the weight of the water the body pushes aside. So divide the upthrust by $\rho g$ to get the volume that is under water.

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