"Properties of Solids and Liquids" — Practice

Question 1

A tank is filled with water and accelerated horizontally at $a=g/2$. Find the angle the free surface makes with the horizontal, and state which way a submerged cork tethered to the floor leans.

Accepted Answer

$	an	heta=a/g=0.5$, so $	heta=26.6^{\circ}$, with the surface dropping towards the front.

The cork leans **forwards**, through the same $26.6^{\circ}$ from the vertical. Buoyancy acts opposite to $\vec g_{eff}$, and since $\vec g_{eff}$ points down and backwards, buoyancy points up and forwards.

A pendulum bob in the same tank leans backwards by $26.6^{\circ}$ — the cork and the bob tilt through equal angles on opposite sides of the vertical.

Question 2

A closed rectangular tank of length $3$ m and height $1.5$ m is completely filled with water and accelerated at $2$ m s$^{-2}$. Find the pressure difference between the bottom rear corner and the top front corner. Take $g=10$ m s$^{-2}$.

Accepted Answer

Moving from the top front corner to the bottom rear corner, $\Delta x=-3$ m and $\Delta y=-1.5$ m.

$\Delta P=ho aL+ho gH=1000\left(2	imes3+10	imes1.5ight)$

$\Delta P=1000	imes21=2.1	imes10^{4}$ Pa

The acceleration contributes $6000$ Pa, equivalent to an extra $0.6$ m of water depth measured backwards along the tank.

Question 3

A cylindrical vessel of radius $0.2$ m is filled with water to a depth of $0.25$ m and rotated about its axis at $\omega=14$ rad s$^{-1}$. Find how far the water rises at the rim above its original level, and check that it does not spill from a vessel $0.4$ m tall.

Accepted Answer

$\dfrac{\omega^{2}R^{2}}{4g}=\dfrac{196	imes0.04}{40}=0.196$ m

The rim rises $0.196$ m to a height of $0.446$ m, which **exceeds** the vessel height of $0.4$ m, so water does spill.

The limiting angular speed follows from $\omega^{2}R^{2}/4g=0.15$, giving $\omega=\dfrac{2}{0.2}\sqrt{10	imes0.15}=12.2$ rad s$^{-1}$.

Question 4

Water stands to a depth $h$ against a vertical gate of width $b$, hinged along its top edge. Find the horizontal force that must be applied at the bottom edge to hold the gate shut.

Accepted Answer

Total thrust $F=	frac12ho gbh^{2}$, acting at depth $2h/3$, which is $2h/3$ below the hinge.

Taking moments about the top hinge, with the applied force $P$ at depth $h$:

$Ph=\left(	frac12ho gbh^{2}ight)\left(	frac{2h}{3}ight)$

$P=	frac13ho gbh^{2}$

The hinge therefore carries the remaining $	frac16ho gbh^{2}$.

Question 5

A uniform cylinder of density $800$ kg m$^{-3}$ and length $0.4$ m floats upright in water. Find the period of its small vertical oscillations, and state how the answer changes if the whole beaker is placed in a lift accelerating upward at $g/4$.

Accepted Answer

$h_{sub}=\dfrac{800}{1000}	imes0.4=0.32$ m, so $T=2\pi\sqrt{0.32/10}=1.12$ s.

In the lift the **draught is unchanged**, because $g_{eff}$ cancels from the floating condition. But the restoring force is stiffer, and the period becomes

$T'=2\pi\sqrt{\dfrac{h_{sub}}{g_{eff}}}=2\pi\sqrt{\dfrac{0.32}{12.5}}=1.01$ s

So the block floats just as deep but bobs faster.

Question 6

A block of wood floats in a beaker of water resting on a balance. A student presses the block fully under water with a finger. Does the balance reading change, and by how much?

Accepted Answer

The reading **increases** by the extra downward force from the finger.

Before: the block displaced its own weight, so the pan carried the weight of beaker plus water plus block.

After: the block displaces its full volume $V$, so the buoyant force is $ho_l Vg$, exceeding the block's weight $ho_b Vg$. The finger must push down with $\left(ho_l-ho_bight)Vg$, and that force is transmitted to the pan.

Increase in reading $=\left(ho_l-ho_bight)Vg$.

Question 7

A uniform steel rod of length $L$ and density $\rho$ hangs vertically from its upper end. Find its extension, and compare it with the extension produced by hanging an equal mass from the lower end of a massless rod.

Accepted Answer

Tension at height $x$ above the bottom is $ho Agx$, so

$\Delta L=\displaystyle\int_0^{L}\frac{ho Agx}{AY}dx=\frac{ho gL^{2}}{2Y}=\frac{MgL}{2AY}$

An equal mass hung at the end gives $MgL/AY$.

The self-weight extension is therefore exactly **half**, because the average tension along the rod is $Mg/2$ rather than $Mg$.

Question 8

A rod of length $L$, area $A$, density $\rho$ and Young's modulus $Y$ is rotated in a horizontal plane about one end at angular speed $\omega$. Find the tension at the midpoint and the total extension.

Accepted Answer

For the part beyond $r$, of mass $ho A(L-r)$ with centre of mass at $(L+r)/2$:

$T(r)=ho A(L-r)\omega^{2}\dfrac{L+r}{2}=	frac12ho A\omega^{2}\left(L^{2}-r^{2}ight)$

At $r=L/2$: $T=	frac38ho A\omega^{2}L^{2}$.

$\Delta L=\displaystyle\int_0^{L}\frac{T(r)}{AY}dr=\frac{ho\omega^{2}}{2Y}\left[L^{3}-\frac{L^{3}}{3}ight]=\frac{ho\omega^{2}L^{3}}{3Y}$

Question 9

A steel rail $10$ m long is clamped rigidly at both ends at $15^{\circ}$C. Find the stress developed when it is heated to $45^{\circ}$C, and state how the answer would change for a $20$ m rail. Take $Y=2	imes10^{11}$ Pa and $\alpha=1.2	imes10^{-5}$ K$^{-1}$.

Accepted Answer

$\sigma=Y\alpha\Delta	heta=\left(2	imes10^{11}ight)\left(1.2	imes10^{-5}ight)(30)$

$\sigma=7.2	imes10^{7}$ Pa $=72$ MPa, compressive.

For the $20$ m rail the stress is **identical**. The length cancels because both the free expansion and the elastic compression needed to undo it scale with $L$.

What does change is the force required at the clamps only through the area, not the span.

Question 10

A soap bubble of radius $3$ cm and another of radius $5$ cm are joined by a fine tube. Describe what happens, and find the radius of curvature of the film across the junction. Take $T=0.03$ N m$^{-1}$.

Accepted Answer

Excess pressures: $4T/0.03=4$ Pa and $4T/0.05=2.4$ Pa.

The **smaller** bubble is at higher pressure, so air flows from it into the larger one. The small bubble shrinks and the large one grows.

The common film bulges into the larger bubble, with

$r=\dfrac{r_1r_2}{r_2-r_1}=\dfrac{3	imes5}{5-3}=7.5$ cm

Question 11

A tank of cross-section $2$ m$^{2}$ is filled with water to a height of $5$ m and drains through a hole of area $10$ cm$^{2}$ in its base. Find the time to fall to half depth, and the reaction force on the tank at that instant.

Accepted Answer

$t=\dfrac{A}{a}\sqrt{\dfrac{2}{g}}\left(\sqrt{H_1}-\sqrt{H_2}ight)=\dfrac{2}{10^{-3}}\sqrt{0.2}\left(\sqrt5-\sqrt{2.5}ight)$

$t=2000	imes0.447	imes\left(2.236-1.581ight)=586$ s

At $H=2.5$ m the thrust is

$F=2ho gHa=2(1000)(10)(2.5)\left(10^{-3}ight)=50$ N

directed backwards, away from the jet.

Question 12

A tank is filled to height $H$. Show that holes at depths $h$ and $H-h$ give jets landing at the same distance, and find the depth giving the greatest range.

Accepted Answer

Speed $v=\sqrt{2gh}$; fall time from height $H-h$ is $\sqrt{2(H-h)/g}$.

$x=\sqrt{2gh}\sqrt{\dfrac{2(H-h)}{g}}=2\sqrt{h(H-h)}$

This expression is unchanged when $h$ and $H-h$ are swapped, so the two holes land together.

Maximising $h(H-h)$ gives $h=H/2$ and $x_{max}=H$ — the greatest range equals the depth of the tank.

Practice — "Properties of Solids and Liquids"

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