2018 A/L Physics Past Paper · Question 41 (English Medium)

Answer: (4) Figure (4)

What the question asks

We want: the graph of the level $h$ against the temperature $\theta$, from $0\ ^\circ\text{C}$.

Relevant theory

In the temperature range 0 °C to 4 °C, the volume of water decreases with rise of temperature.

It is clear from the graph that the volume of water increases with rise of temperature only for temperatures greater than 4 °C.

The key idea.

The large cavity is full. When its water changes volume, the change shows only in the small sphere and the tube.

In a narrow place, the same change of volume gives a big change of level.

$$\frac{dh}{d\theta} = \frac{1}{A(h)}\,\frac{dV}{d\theta}$$

Here $A(h)$ is the cross-sectional area at height $h$.

Required equations

$\frac{dh}{d\theta} = \frac{1}{A(h)}\frac{dV}{d\theta}$

Reasoning and calculation

The axes.

Across: the temperature $\theta$ in °C.

Up: the level $h$ above $XY$.

Marks: $r$ is the middle of the sphere. $2r$ is its top. $2r + l$ is the end of the tube.

At the start $\theta = 0\ ^\circ\text{C}$.

The small sphere is half full. So $h = r$.

Three places where things change.

At $4\ ^\circ\text{C}$ the water stops shrinking and starts to expand.

At $h = r$ the sphere is widest.

At $h = 2r$ the sphere turns into the tube.

Now each part.

From 0 to $4\ ^\circ\text{C}$ the water shrinks. So $h$ falls. It is lowest at $4\ ^\circ\text{C}$.

After that $h$ rises. Near $h = r$ the sphere is wide. So $h$ rises slowly.

Above that the sphere gets narrower. So $h$ rises faster.

In the narrowing tube it rises faster still.

The values at the ends.

It starts at $h = r$.

The lowest $h$ is at $4\ ^\circ\text{C}$. It is below $r$, but not zero.

Then it rises up to $2r + l$.

The shape of the curve.

At $4\ ^\circ\text{C}$ the curve is flat for a moment.

Near $h = r$ the rise is slow.

At $2r$ there is no corner, because the sphere turns smoothly into the tube.

Near the top the curve bends upwards. The rise keeps getting faster.

Reading the options

It falls from $r$. It is lowest at $4\ ^\circ\text{C}$. It rises slowly near $r$. Then it rises fast, with no corner.

Only figure (4) has this shape.

Why the other options fail

This option says: figure (1). Here $h$ rises from the start.

But from 0 to $4\ ^\circ\text{C}$ the water shrinks. So $h$ must fall first.

So it is wrong.

This option says: figure (2). Here too $h$ rises from the start.

This option says: figure (3). Its curve has a sharp corner at $2r$.

But the sphere turns smoothly into the tube. So the slope changes smoothly too.

This option says: figure (5). Its rise slows down near $2r$.

But near the top of the sphere the cross-section is small. The same change of volume gives a bigger change of $h$. So $h$ rises faster there.

Conclusion

$h$ falls until $4\ ^\circ\text{C}$.

Then it rises with no corner. It is slow near $r$ and fast in the narrow tube.

The answer is option (4).

Explore

The water starts at $0\ ^\circ\text{C}$. The small sphere is half full.

Up to $4\ ^\circ\text{C}$ the water shrinks. The level falls.

Above $4\ ^\circ\text{C}$ the water expands. In the wide middle of the sphere the level rises slowly, then faster.

In the narrow tube it shoots up. That is figure (4).