A composite solid cylinder consisting parts A and B is made from materials whose densities are 600 kg m⁻³ and 2000 kg m⁻³ respectively. The height of part A is 9 cm and that of B is 1 cm. The cylinder floats in water of density 1000 kg m⁻³ as shown in the figure. What is the height (h) of the cylinder under water?

2021 A/L Physics Past Paper · Question 21 (English Medium)

Depth of a floating composite cylinder

A composite solid cylinder consisting parts $A$ and $B$ is made from materials whose densities are $600\ \text{kg m}^{-3}$ and $2000\ \text{kg m}^{-3}$ respectively. The height of part $A$ is $9\ \text{cm}$ and that of $B$ is $1\ \text{cm}$ . The cylinder floats in water of density $1000\ \text{kg m}^{-3}$ as shown in the figure. What is the height ( $h$ ) of the cylinder under water?

(1) $2.6\ \text{cm}$
(2) $5.4\ \text{cm}$
(3) $7.4\ \text{cm}$
(4) $8.0\ \text{cm}$
(5) $9.0\ \text{cm}$

Answer and explanation

Answer: (3) $7.4\ \text{cm}$

Given information

What we know:

Part $A$ : density $600\ \text{kg m}^{-3}$ , height $9\ \text{cm}$ .

Part $B$ : density $2000\ \text{kg m}^{-3}$ , height $1\ \text{cm}$ .

Water: density $1000\ \text{kg m}^{-3}$ .

Both parts have the same cross-section area $S$ .

What the question asks

Find the height $h$ of the cylinder that is under water.

The plan has two steps.

First, find the total weight of the cylinder.

Then make the upthrust equal to that weight.

Relevant theory

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 body is to float in a liquid, the forces on it should be in equilibrium. That is, the upthrust should be equal to the weight of the body.

The mass of a unit volume of a substance is termed its density.

Required equations

$m = \rho V$

$U = \rho_w\, S h\, g$

$U = mg$

Applying it to this question

The cylinder is at rest on the water.

Only two forces act on it: its weight down and the upthrust up.

They are equal.

$54$

$m_A/S = \rho_A h_A = 600 \times 0.09$

$20$

$m_B/S = \rho_B h_B = 2000 \times 0.01$

$74$

$m/S = 54 + 20$

$mg = \rho_w S h g$

$0.074$

$h = \dfrac{m/S}{\rho_w} = \dfrac{74}{1000}$

The heights are in metres and the densities in $\text{kg m}^{-3}$ .

$\text{kg m}^{-2}$ divided by $\text{kg m}^{-3}$ gives metres.

$0.074\ \text{m} = 7.4\ \text{cm}$ .

Reasoning and calculation

The cylinder is lighter than the same volume of water. It floats.

Its water-line is at $0.074\ \text{m}$ from the bottom.

That is $7.4\ \text{cm}$ .

The answer is (3).

Checks

$(dA * hA + dB * hB) / dw$

Does it make sense?

The depth $7.4\ \text{cm}$ is less than the height of $A$ , which is $9\ \text{cm}$ . So part $B$ stays above the water, as the figure shows.

The cylinder is $10\ \text{cm}$ tall. Its mean density is $\dfrac{74}{0.10} = 740\ \text{kg m}^{-3}$ , less than water. So it floats, with $74\%$ of its height under water.

Conclusion

$h = \dfrac{600 \times 0.09 + 2000 \times 0.01}{1000} = 0.074\ \text{m} = 7.4\ \text{cm}$

The total mass for each unit of area is $54 + 20 = 74\ \text{kg m}^{-2}$ .

Upthrust equals weight: $\rho_w h = 74$ , so $h = 0.074\ \text{m}$ .

The answer is (3).

A floating body: $upthrust = weight$

Write both sides with the same area $S$ . It cancels, and you are left with the heights and densities.

Exam tip

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