2021 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (5) $4.0\ \text{cm}$

Given information

We know:

In the upright tube the water rises to a height $h = 2.0\ \text{cm}$.

The tube is then tilted $60^\circ$ from the vertical.

What the question asks

We want: the length $l$ of the water column along the tilted tube.

Relevant theory

The capillary rise increases when the capillary gets thinner.

Required equations

$h = \dfrac{2T\cos\theta}{r\rho g}$

$p = h\rho g$

$l\cos\alpha = h$

Applying it to this question

Tilting the tube changes nothing in $T$, $\theta$, $r$, $\rho$ or $g$.

So the vertical height $h$ of the water above the outside level is the same, $2.0\ \text{cm}$.

Only the length along the tube changes.

Both lengths are in centimetres, so $l$ comes out in centimetres.

$l\cos 60^\circ = 2.0$

$0.5\,l = 2.0$

$4.0$

$l = \dfrac{2.0}{0.5}$

Reasoning and calculation

The column is $4.0\ \text{cm}$ long.

Its vertical height is $4.0 \times 0.5 = 2.0\ \text{cm}$, as before.

The answer is (5).

Checks

$h / c$

Does it make sense?

The tube leans over. A leaning column has to be longer than its vertical height. So $l > 2.0\ \text{cm}$.

At $0^\circ$ the tube is upright and $l = 2.0\ \text{cm}$. At $60^\circ$ the length is double. The closer the tube gets to flat, the longer the column becomes.

Conclusion

$l = \dfrac{h}{\cos 60^\circ} = \dfrac{2.0}{0.5} = 4.0\ \text{cm}$

The vertical height stays $2.0\ \text{cm}$.

$l\cos 60^\circ = 2.0$, so $l = 4.0\ \text{cm}$. The answer is (5).

In a tilted capillary tube the vertical height of the liquid stays the same.

Draw the right-angled triangle. The column is the long side. Its vertical part is $l\cos\alpha$ with $\alpha$ taken from the vertical.

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