A capillary tube dipped vertically in a vessel of water with a large cross-sectional area is shown in the figure. This set-up is fixed in a lift which is at rest. The capillary rise is 8 mm when the open end of the tube is 40 mm above the water level in the vessel. If the lift (I) moves downward with an acceleration of 5 m s⁻², (II) falls freely, what would the capillary rise be, respectively?

2020 A/L Physics Past Paper · Question 45 (English Medium)

Capillary rise in a lift — downward acceleration and free fall

A capillary tube dipped vertically in a vessel of water with a large cross-sectional area is shown in the figure. This set-up is fixed in a lift which is at rest. The capillary rise is $8\ \text{mm}$ when the open end of the tube is $40\ \text{mm}$ above the water level in the vessel. If the lift

(I) moves downward with an acceleration of $5\ \text{m s}^{-2}$ ,

(II) falls freely,

what would the capillary rise be, respectively?

(1) $4\ \text{mm}$ , $0$
(2) $16\ \text{mm}$ , $0$
(3) $4\ \text{mm}$ , $8\ \text{mm}$
(4) $16\ \text{mm}$ , $32\ \text{mm}$
(5) $16\ \text{mm}$ , $40\ \text{mm}$

Answer and explanation

Answer: (5) $16\ \text{mm}$ , $40\ \text{mm}$

What the question asks

We must find: the capillary rise when (I) the lift accelerates downward at $5\ \text{m s}^{-2}$ , and (II) the lift falls freely; at rest $h_0 = 8$ mm, the tube's open end is $40$ mm above the water level.

Relevant theory

This is

known as capillary rise.

Considering the equilibrium of forces,

$\therefore F = ma$

F - Resultant force in newtons

Required equations

$2\pi rT\cos\theta = \pi r^2 h_0\rho g$

$2\pi rT\cos\theta - \pi r^2 h\rho g = \pi r^2 h\rho(-a)$

$2\pi rT\cos\theta = \pi r^2 h\rho(g - a)$

Reasoning and calculation

What changes, what stays the same: the surface tension $T$ , the contact angle, the tube's radius, and the water's density stay the same — so the upward pulling force $2\pi rT\cos\theta$ does not change.

Only the lift's acceleration $a$ changes.

So the amount of column that can be supported is $m(g - a)$ .

Proportion: since the pulling force is unchanged, $h(g - a)$ is unchanged: $h \propto \dfrac{1}{g - a}$ .

$2$

$\frac{h_1}{h_0} = \frac{g}{g - a} = \frac{10}{10 - 5}$

$16$

$h_1 = 2 \times 8$

$40$

$g - a = 0 \;\Rightarrow\; \text{no support is needed for the column}$

(II) explained: if $g - a = 0$ then $h(g - a) =$ a constant is satisfied by no finite $h$ — the pulling force keeps drawing the water up.

Once the water reaches the tube's open end ( $40$ mm), the surface curvature there simply changes; it does not overflow.

Reading the options

Comparing with the choices: (I) $16$ mm, (II) $40$ mm — option (5).

Checks

$h0 * g / (g - a)$

Dimension check: $\dfrac{g}{g - a}$ is dimensionless; $h_1$ has the same unit as $h_0$ (mm) ✓.

Boundary check: at $a = 0$ , $h = 8$ mm (rest) ✓.

As $a \to g$ , $h \to \infty$ — the tube's own length is the limit ✓.

With an upward acceleration ( $a < 0$ ), $h < 8$ mm ✓.

Conclusion

$h = h_0\,\frac{g}{g - a}$

$h \propto \dfrac{1}{g - a}$ : (I) $8 \times \dfrac{10}{5} = 16$ mm; (II) $g - a = 0$ , the water rises to the top of the tube, $40$ mm — the answer is (5).

Assuming that in free fall surface tension also vanishes because of "weightlessness".

Surface tension is a molecular force — it does not change; what vanishes is only the need to support the column's weight.

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