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Question 332: [Matter > Solids]
Diagram shows the arrangement of atoms in a particular crystal.

Each atom is at corner of a cube.
Mass of each atom is 3.5 × 10^–25 kg. Density of the crystal is 9.2 × 10³ kg m^–3.
What is the shortest distance between centres of two adjacent atoms?
A 3.8 × 10^–29 m           B 6.2 × 10^–15 m           C 3.4 × 10^–10 m           D 3.0 × 10^–9 m

Reference: Past Exam Paper – June 2012 Paper 12 Q22 & June 2015 Paper 11 Q21



Solution 332:

Answer: C.

A cube has 8 corners (4 at the top and 4 at the bottom). Note that the centre of an atom is situated at a corner of a cube.

The shortest distance between the centres of two adjacent atoms is therefore the length of a side of the cube.

An atom can be considered to have a spherical shape.

{Now, you will need to imagine a cube in 3D. You really need to visualize this in 3D or else you won’t be able to understand the calculations. Try to draw what is being described below while you read.}

Consider a corner of a cube. Let it be one of the 4 top corners. Since the centre of an atom is placed at the corner, the upper half of the sphere (hemisphere) can be neglected. Now, only a section of the lower hemisphere would be inside the cube. This section is actually 1/8 of a whole sphere.

The volumes of atoms inside a cube would actually look like this:

At each corner, 1/8 of the volume of a sphere (that is, 1/8 of an atom) is inside the cube. Since there are 8 corners, the total number of atoms inside the cube is 8 (1/8) = 1 atom.

The density of any single cube in the lattice is the same as the density of the whole lattice.

As explained, a cube contains a total of 1 atom inside it. So, when considering the density of a cube, we only need to consider the mass of 1 atom.

Density = mass / volume

Volume of a cube = mass of 1 atom /density = (3.5x10^-25) / 9.2 x 10³ = 3.8x10^-29 m³

The shortest distance between the centres of 2 adjacent atoms is equal to the length of a side of the cube.

Length of cube = (3.8x10^-29)^1/3 = 3.40x10^-10m

Question 333: [Pressure > Liquids]
A tube is arranged as a siphon in a beaker of liquid. The point A is 60mm above the free surface if the liquid and B is 20mm below it. The tube is first filled with liquid and the end B is closed with a finger.

Find the amounts by which the pressures at A and B are above or below atmospheric pressure. When the finger is removed from B, liquid runs out. Why is this? When will the flow stop?
(Density of liquid = 0.9x10³ kgm^-3; take the acceleration of free fall, g as 10 ms^-2.)

Reference: “Pacific Physics A Level”, Volume 1, by POH LIONG YONG, pg 135 Example 7.3



Solution 333:
Let h_A and h_B be the heights of points A and B above and below the normal water surface, respectively.
{Remember that the tube is first filled with liquid and the end B is closed with a finger.}
If p_A = pressure at A, then
{Atmospheric pressure acts at the surface of the liquid in the beaker. Since end B of the tube is closed with a finger, the system is in a state of equilibrium.
The equation below gives us the pressure acting at the level of the liquid surface. The pressure inside the tube acting at the level of the liquid surface should be equal to the outside atmospheric pressure acting on the liquid surface in the beaker.
In addition to the pressure at p_A, the column of liquid inside the tube, below A, exerts some pressure at the level of the liquid surface.}
p_A + h_Aρg = atmospheric pressure, p_o
p_A = p_o – h_Aρg 
p_A = p_o – (6.0x10^-3) (0.9x10³) (10) = p_o – 5.4x10² Pa
Pressure at A = 5.4x10² Pa below atmospheric pressure

{The equation below now gives us the relationship of the pressure at the level of end B (unlike the equation above which gives us the relationship of pressures at the level of the liquid in the beaker).
The pressure at B acts on the finger. This is equal to the sum of the atmospheric pressure and the pressure due to the column of liquid from the level of the liquid above the level of end B.}
Pressure at B = p_o + h_Bρg
Pressure at B = p_o + (20x10^-3) (0.9x10³) (10) = p_o + 1.8x10² Pa
Pressure at B = 1.8x10² Pa above atmospheric pressure
{The finger also exerts the same pressure at B, causing the system to be in equilibrium.}


{When the finger is removed, the finger no longer exerts a pressure on the liquid inside the tube. Instead, the pressure acting against the liquid in the tube at end B is now the atmospheric pressure. The system is no longer in equilibrium}
When the finger is removed from B, liquid runs out because the pressure inside the tube at B is greater (which is the value calculated above) than the pressure outside, the atmospheric pressure (when the finger is removed, the outside atmospheric pressure now acts at end B of the tube while the liquid inside the tube at end B still has the pressure calculated above – which is greater than the atmospheric pressure).
{So, the resultant pressure goes outside. So, the liquid, which was already in the tube (as stated in the question) moves out of the tube at B.}

The flow will stop only when the liquid level in the beaker comes down to the level of B. {When the liquid comes down to the level of B, there is no longer any column of liquid from the level of the surface of the liquid in the beaker above end B. That is, h_B = 0. So, pressure at B = atmospheric pressure. Since the pressure at the surface of the liquid in the beaker is also the atmospheric pressure, the net difference in pressure is zero. So, the liquid stops flowing out.}








Question 334: [Matter > Young modulus]
Which property of metal wire depends on its Young modulus?
A ductility
B elastic limit
C spring constant
D ultimate tensile stress

Reference: Past Exam Paper – June 2012 Paper 11 Q23



Solution 334:

Answer: C.

Young modulus, E = stress / strain

Stress = Force / Cross-sectional area = F / A

Strain = extension / original length = e / L

E = (F/A) / (e/L) = FL / Ae

Hooke’s law: F = ke    where k is the spring constant

Since a metal wire obeys Hooke’s law up to some limit, the spring constant depends on the force, which can be found in the formula for the Young modulus.

E = (ke)L / Ae = kL / A

So, the spring constant, k depends on Young modulus E, original length L and cross-sectional area A.

For the above equation to hold, the metal wire should obey Hooke’s law. Hooke’s law is obeyed up to the limit of proportionality. Since the elastic limit and the ultimate tensile strength are beyond the limit of proportionality, these properties of the metal wire do not depend on its Young modulus. [B and D are incorrect]

Material that undergo plastic deformation before breaking are said to be ductile. Materials that obey Hooke’s law undergo elastic deformation – that is, they regain their original shape when the deforming force is removed. [A is incorrect]