2018 A/L Physics Past Paper · Question 47 (English Medium)

Wooden cubes with metal, floating — depth immersed

Figure shows three objects $(A), (B)$ and $(C)$ which are made using three uniform wooden cubes of identical dimensions made out of the same material, and three identical uniform metal cubes. In $(A)$ and $(B)$ the metal cubes are glued onto the top and the bottom of the wooden cubes respectively. In $(C)$ , the metal cube is embedded in the wooden cube, as shown in the figure. The three objects $(A), (B)$ and $(C)$ are now slowly lowered without changing their orientation and made to float vertically in a pool of water. If the depths to which the wooden cubes are immersed in water are $H_A, H_B$ and $H_C$ respectively, which of the following relationships is true?

(1) $H_A > H_B > H_C$
(2) $H_A = H_B > H_C$
(3) $H_A = H_B = H_C$
(4) $H_C > H_B > H_A$
(5) $H_A > H_C > H_B$

Answer and explanation

Answer: (5) $H_A > H_C > H_B$

What the question asks

We want: in which of the three does the wooden cube sink deepest?

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.

A floating body is balanced.

Its weight equals the upthrust.

The upthrust is the weight of the water it pushes aside: $V_{\text{water}}\,\rho_w g$ .

Required equations

$W = V_{submerged}\,\rho_w g$

Reading the options

The letters:

The bottom of the wooden cube has area $S$ . The wood has volume $V$ and density $\rho$ .

The metal has volume $v$ and density $\rho_m$ .

Each option is an order of the depths $H$ .

Reasoning and calculation

The same in all three:

the same wooden cube, the same metal cube, the same water.

Only one thing changes: where is the metal?

Ask two questions.

1. How heavy is the object?

2. Does anything other than the wood push water aside?

(A) Metal on top

$H_A = \dfrac{V\rho + v\rho_m}{S\rho_w}$

The metal is out of the water.

So only the wood pushes water aside.

$S H_A \rho_w = V\rho + v\rho_m$

(B) Metal underneath

$H_B = H_A - \dfrac{v}{S}$

The weight is the same as (A).

But now the metal is in the water.

It pushes water aside too.

So the wood does not need to sink as far.

$S H_B \rho_w + v\rho_w = V\rho + v\rho_m$

$H_C = H_A - \dfrac{v\rho}{S\rho_w}$

Some wood was cut out to make room for the metal.

So (C) is a little lighter.

The metal is out of the water. Only the wood pushes water aside.

$S H_C \rho_w = (V - v)\rho + v\rho_m$

Now compare.

$H_A - H_C = \frac{v}{S}\cdot\frac{\rho}{\rho_w} > 0$

$H_C - H_B = \frac{v}{S}\left(1 - \frac{\rho}{\rho_w}\right) > 0$

The second one is positive because wood is lighter than water ( $\rho < \rho_w$ ).

H_A

H_C

H_B

(A) sinks the deepest.

(B) sinks the least.

$H_A > H_C > H_B$

Checks

$(1 - 0.6) * 1$

Check the extremes.

If wood were as heavy as water ( $\rho \to \rho_w$ ): then $H_C \to H_B$ .

If wood weighed almost nothing ( $\rho \to 0$ ): then $H_C \to H_A$ .

Both agree with our order.

Conclusion

$H_A > H_C > H_B$

The answer is option (5).

Explore

The three objects are lowered slowly into the water and left to float.

Each sinks until the water it pushes aside weighs as much as it does.

In $B$ the metal under the water pushes water aside too, so its wood sinks less.

$C$ has less wood, so it is lighter than $A$ : $H_A > H_C > H_B$ .

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