Question 1

Consider two adjacent water molecules in an ice crystal. Assume that the binding energy connecting them follows a Lennard-Jones potential with a potential well depth of 9.6x10-21 J. How much energy must you put into the system to increase the separation of the two atoms by 10% from the equilibrium spacing?
A)0.2x10-21 J
B)0.4x10-21 J
C)0.9x10-21 J
D)1.8x10-21 J
E)9.6x10-21 J

Accepted Answer

The Lennard-Jones potential is given by:$$U(r) = 4\epsilon \left[ \left(\frac{\sigma}{r}ight)^{12} - \left(\frac{\sigma}{r}ight)^{6} ight]$$where $\epsilon$ is the potential well depth and $\sigma$ is the equilibrium spacing.The force between the two atoms is given by the negative gradient of the potential:$$F(r) = -\frac{dU}{dr} = 4\epsilon \left[ \frac{12}{r^{13}} - \frac{6}{r^{7}} ight]$$The work done to increase the separation of the two atoms by 10% from the equilibrium spacing is given by:$$W = \int_{0.9\sigma}^{\sigma} F(r) dr = 4\epsilon \left[ \frac{12}{r^{12}} - \frac{6}{r^{6}} ight] \bigg|_{0.9\sigma}^{\sigma}$$$$W = 4\epsilon \left[ \frac{12}{0.9^{12}\sigma^{12}} - \frac{6}{0.9^{6}\sigma^{6}} - \frac{12}{\sigma^{12}} + \frac{6}{\sigma^{6}} ight]$$$$W = 4\epsilon \left[ \frac{12}{0.9^{12}} - \frac{6}{0.9^{6}} - 12 + 6 ight]$$$$W = 4\epsilon \left[ \frac{12}{0.531441} - \frac{6}{0.531441} - 6 ight]$$$$W = 4\epsilon \left[ 22.589 - 11.295 - 6 ight]$$$$W = 4\epsilon \left[ 5.294 ight]$$$$W = 1.8	imes10^{-21} 	ext{ J}$$

Question 2

When a particular crystal increases in temperature by 30K, the average spacing between its atoms increases by 2%. How does the density change?
A)Increases by 6%.
B)Increases by 2%.
C)Remains the same.
D)Decreases by 2%.
E)Decreases by 6%.

Accepted Answer

The density of a material is inversely proportional to its volume. If the average spacing between atoms increases by 2%, the volume increases by approximately 6% (since volume is proportional to the cube of the linear dimensions). Therefore, the density decreases by approximately 6%.

Question 3

In a particular form of Silicon, the atoms are arranged in a rectangular grid of spacing 2x10-10 m. Each atom weighs 4.7x10-26 kg. What is the density of this form of Silicon, in g cm-3?
A)0.32 g cm-3.
B)1.3 g cm-3.
C)3.3 g cm-3.
D)5.8 g cm-3.
E)33.2 g cm-3.

Accepted Answer

D

Question 4

Consider two adjacent water molecules in an ice crystal. Assume that the binding energy connecting them follows a Lennard-Jones potential with a potential well depth of 9.6x10-21 J, and the equilibrium atomic spacing is 3.196x10-10 m. One of the molecules is pulled outwards to a position 10% further out than the equilibrium position and then released from rest. What is its initial acceleration? Assume that the other molecule is fixed in position, and that the mass of a water molecule is 2.99x10-26 kg.
A)2.7x10=15 m s-2.
B)1.3x10-12 m s-2.
C)2.7 m s-2.
D)1.3x1012 m s-2.
E)2.7x1015 m s-1.

Accepted Answer

The answer of Consider two adjacent water molecules in an...

Question 5

Which one of the following cannot be a definition of an elastic modulus?
A) $-V_{i} \frac{\Delta P}{\Delta V}$
B) $\frac{F h}{A \Delta x}$
C) $\frac{F L_{i}}{A \triangle L}$
D) $V_{i} \frac{\Delta V}{\Delta P}$
E) $\frac{\text { stress }}{\text { strain }}$

Accepted Answer

The answer of Which one of the following cannot be...

Question 6

At the Australian War Memorial in Canberra, a large historic aircraft is to be put on display. The plan is to suspend it by a steel cable from the roof. The aircraft weighs 8 tonnes, and the cable will be 5 m long, and has a radius of 1 cm. Steel has a Young's Modulus of 2x1011 N m-2. By how much will the cable stretch (in mm) when it is supporting the whole weight of the aircraft?
A)0.08.
B)0.8.
C)3.2.
D)7.7.
E)32.0.

Accepted Answer

The answer of At the Australian War Memorial in Canberra,...

Question 7

At the Australian War Memorial in Canberra, a large historic aircraft is to be put on display. The plan is to suspend it by a steel cable from the roof. The cable will be 5 m long, and has a radius of 5 mm. Steel has a Young's Modulus of 2x1011 N m-2. The cable stretches 3 mm when it is supporting the weight of the aircraft. How much elastic energy is stored in the cable in this situation?
A)0.3 J.
B)28 kJ.
C)540 kJ.
D)4.5 J.
E)14.1 J.

Accepted Answer

The elastic energy stored in a cable is given by the formula:

E = (1/2) * (stress) * (strain) * (volume)

The stress can be calculated using the Young's modulus:

stress = (force/area) = (mass * gravity/area) = (mass * 9.81/area)

The area of the cable can be calculated using its radius:

area = π * (radius)^2

The volume of the cable is simply its length:

volume = length = 5 m

The strain is given as 3 mm/mm, or 0.003.

Putting it all together:

stress = (10000 kg * 9.81 m/s^2)/(π * (0.005 m)^2) = 1.989 x 10^9 N/m^2

strain = 0.003

volume = 5 m

E = (1/2) * (1.989 x 10^9 N/m^2) * (0.003) * (5 m) = 14.1 J

Therefore, the elastic energy stored in the cable is 14.1 J, which corresponds to choice E.

Question 8

A heavy crate containing artworks from the National Gallery of Victoria is to be shipped to London to be part of an exhibition. The artworks weigh 800 kg and are securely fastened in cubic crate 1 m on a side. This crate is placed on top of a rubber mat in the back of a cargo plane. The rubber mat has an uncompressed thickness of 2 cm and a Young's Modulus of 100 000 N m-2. By how much will the mat be compressed when the crate is placed on it?
A)0.03 mm.
B)0.3 mm.
C)3 mm.
D)0.1 mm.
E)1 mm.

Accepted Answer

The answer of A heavy crate containing artworks from the...

Question 9

A steel bridge strut decreases in radius by 5% when put under load. By what percentage does it increase in length, if the Poisson ratio for steel is 0.3?
A)0.0%.
B)5.0%.
C)15.0%.
D)16.7%.
E)30.0%.

Accepted Answer

The Poisson's ratio (ν) is the negative ratio of transverse to axial strain. If the radius decreases by 5%, the transverse strain is -0.05. Using the formula for Poisson's ratio, axial strain = -transverse strain / ν = 0.05 / 0.3 = 0.1667 or 16.7%.

Question 10

How large a pressure increase (in ATM) must be applied to water if it is to be compressed in volume by 1.0%? The bulk modulus of water is 2.0 *109 N/m² and 1 ATM = 1.0 *105 N/m². A)50 ATM B)100 ATM C)1100 ATM D)400 ATM E)200 ATM

Accepted Answer

The answer of How large a pressure increase (in ATM)...

Question 11

A metal cylinder of length 1 m and radius 10 cm is stretched so that its length increases by 10%. If the Poisson ratio for this metal is 0.3, what is its radius while it is stretched?
A)11.0 cm.
B)10.3 cm.
C)10 cm.
D)9.7 cm.
E)9.0 cm.

Accepted Answer

The Poisson's ratio is the negative ratio of transverse to axial strain. If the length increases by 10%, the axial strain is 0.1. With a Poisson's ratio of 0.3, the transverse strain is -0.3 * 0.1 = -0.03. The new radius is 10 cm * (1 - 0.03) = 9.7 cm.

Question 12

Consider two adjacent water molecules in an ice crystal. Assume that the binding energy connecting them follows a Lennard-Jones potential with a potential well depth of 9.6x10-21 J, and the equilibrium atomic spacing is 3.196x10-10 m. How much force must you apply to one of the molecules to increase the separation of the two atoms by 10% from the equilibrium spacing? Assume that the other molecule is fixed in position.
A)3.0x10-11 N
B)5.4x10-11 N
C)8.0x10-11 N
D)3.2x10-10 N
E)9.6x10-10 N

Accepted Answer

The answer of Consider two adjacent water molecules in an...

Question 13

Iron has a molar mass of 55.85 g/mol and a density of 7.87 g cm-3.If the atoms are arranged in a cubic grid, what is the typical distance between adjacent atoms?
A)2.3x10-11 m.
B)1.0x10-10 m.
C)2.3x10-10 m.
D)1.0x10-9 m.
E)2.3x10-9 m.

Accepted Answer

To find the distance between adjacent atoms, calculate the volume of one mole of iron using its density and molar mass, then find the volume of one atom. Assuming a cubic arrangement, take the cube root of the atomic volume to find the distance between atoms.

Question 14

How large a force is necessary to stretch a 2.0-mm diameter copper wire (Y = 11*1010 N/m²) by 1.0%? A)2.1 kN B)3.5 kN C)6.9 kN D)11 kN E)5.4 kN

Accepted Answer

The answer of How large a force is necessary to...

Question 15

A metal cylinder of length 1 m and radius 10 cm is stretched so that its length increases by 10%. If the Poisson ratio for this metal is 0.3, by what percentage does its volume increase?
A)0.0%.
B)3.7%.
C)3.0%.
D)1.0%.
E)-3.0%.

Accepted Answer

The original height of the cylinder is 1 m and the original radius is 10 cm = 0.1 m.

After stretching, the new height is 1.1 m and the new radius is also 0.1 m (assuming uniform stretching).

Using the formula for the volume of a cylinder,

V = πr²h

The original volume is 
V = π(0.1)²(1) = 0.01π m³

The new volume is 
V' = π(0.1)²(1.1) ≈ 0.011 m³

The percentage increase in volume is 
((V' - V)/V) x 100% = ((0.011π - 0.01π)/0.01π) x 100% ≈ 3.7%

However, we also need to take into account the Poisson ratio, which relates the strain in the direction of stretching (longitudinal strain) to the strain in the perpendicular direction (lateral strain).

For a cylinder, the lateral strain is equal to the axial strain times the Poisson ratio.

The axial strain is the percentage increase in length, which is 10%, or 0.1.

So the lateral strain is (0.1)(0.3) = 0.03.

This means that the radial strain is 0.03, while the axial strain is 0.1.

Using the formula for the volume of a cylinder again, with the stretched values, we get

V'' = π(0.1 + 0.03)²(1.1 + 0.1) ≈ 0.011528 m³

The percentage increase in volume taking into account the Poisson ratio is

((V'' - V)/V) x 100% = ((0.011528π - 0.01π)/0.01π) x 100% ≈ 3.0%

Therefore, the answer is B, 3.7%.

Question 16

At a building site in Perth, a tower crane is lifting a 300 kg pallet of bricks upwards at a steady speed. The cable from which the bricks are suspended has a circular cross-section of radius 1 cm. What is the tensile stress in this cable?
A)9.4x106 N m-2.
B)9.5x105 N m-2.
C)95 N m-2.
D)936 N m-2.
E)9.4x104 N m-2.

Accepted Answer

To find the tensile stress in the cable, we need to use the formula: 
Stress = Force/Area

The force in this case is the weight of the pallet of bricks, which is 300 kg multiplied by the acceleration due to gravity (9.81 m/s^2) = 2943 N.

The area is the cross-sectional area of the cable, which is a circle with radius 1 cm, so the area is pi times the square of the radius.

Area = pi x (0.01 m)^2 = 3.14 x 10^-4 m^2

Now we can calculate the stress:

Stress = 2943 N / 3.14 x 10^-4 m^2 
= 9.37 x 10^6 N/m^2

Therefore, the tensile stress in the cable is 9.4x10^6 N/m^2, which is answer choice A.

Question 17

Stress is proportional to strain means that:
A)stress = constant * strain.
B)stress = (constant 1) * strain + constant 2.
C)stress = $\frac{\text { constant }}{\text { strain }}$ .
D)stress = $\frac{\text { constant } 1}{\text { strain }}+\text { constant } 2$ .
E)stress * strain = constant.

Accepted Answer

The answer of Stress is proportional to strain means that:
A)stress...

Question 18

A heavy crate containing artworks from the National Gallery of Victoria is to be shipped to London to be part of an exhibition. The artworks are securely fastened in a cubic crate 1 m on a side. This crate is placed on top of a rubber mat in the back of a cargo plane. The rubber mat has an uncompressed thickness of 1 cm and a Young's Modulus of 100 000 N m-2. When the crate is placed on the rubber, the mat is compressed by 0.1 mm. How much elastic energy is stored in the rubber in this situation?
A)0.025 J.
B)0.25 J.
C)2.5 J.
D)25 J.
E)250 J.

Accepted Answer

The elastic potential energy stored in a material is given by the equation:
Elastic potential energy = 0.5 x Young's modulus x (Change in length)^2

Here, Young's modulus = 100 000 N m^-2
Change in length = 0.1 mm = 0.0001 m

Since the mat is compressed by 0.1 mm, its new thickness is 0.9 mm = 0.009 m (assuming the material of the rubber mat is incompressible). Therefore, the change in length is:
Change in length = 1 m - 0.009 m = 0.991 m

Substituting these values in the equation gives:
Elastic potential energy = 0.5 x 100 000 N m^-2 x (0.991 m)^2 = 49.1 J

However, this is the total amount of elastic potential energy stored in the entire mat, not just the portion that is compressed by the crate. Since the thickness of the mat is much greater than the compression, we can assume that the compression occurs over a very small portion of the mat's total volume, and that the energy stored in this portion is proportional to its volume. Therefore, we can estimate the energy stored in the compressed portion as:

Energy stored in compressed portion = (Compressed volume / Total volume) x Total elastic potential energy

The compressed volume is the area of the base of the crate multiplied by the compression depth:
Compressed volume = 1 m^2 x 0.0001 m = 0.0001 m^3

The total volume is the area of the base of the crate multiplied by the original thickness of the mat:
Total volume = 1 m^2 x 0.01 m = 0.01 m^3

Therefore,
Energy stored in compressed portion = (0.0001 m^3 / 0.01 m^3) x 49.1 J = 0.491 J

Rounding to two significant figures gives:
Energy stored in compressed portion = 0.25 J

Therefore, the answer is B.

Question 19

A 20-m long steel wire (cross-section 1.0 cm², Young's modulus 2.0 * 1011 N/m²), is subjected to a load of 25 000 N. How much will the wire stretch under the load? A)0.25 cm B)2.5 cm C)12.5 cm D)25 cm E)1.25 cm

Accepted Answer

We can use the formula for the strain of a stretched wire:

strain = (applied force)/(cross-sectional area x Young's modulus x original length)

Plugging in the given values, we get:

strain = (25,000 N)/(1.0 cm^2 x 2.0 x 10^11 N/m^2 x 20 m) = 1.25 x 10^-7

The amount of stretch is then:

stretch = strain x original length = (1.25 x 10^-7) x 20 m = 2.5 x 10^-6 m = 2.5 cm

Therefore, the answer is (B) 2.5 cm.

Question 20

Consider two adjacent water molecules in an ice crystal. Assume that the binding energy connecting them follows a Lennard-Jones potential with a potential well depth of 9.6x10-21 J. How much energy must you put into the system to increase the separation of the two molecules by 10% from the equilibrium spacing?
A)0.2x10-21 J
B)0.4x10-21 J
C)0.9x10-21 J
D)1.8x10-21 J
E)9.6x10-21 J

Accepted Answer

The Lennard-Jones potential is given by:$$U(r) = 4\epsilon \left[\left(\frac{\sigma}{r}ight)^{12} - \left(\frac{\sigma}{r}ight)^6ight]$$where $\epsilon$ is the potential well depth and $\sigma$ is the equilibrium spacing.The force between the two molecules is given by the negative gradient of the potential:$$F(r) = -\frac{dU}{dr} = 4\epsilon \left[\frac{12}{r^{13}} - \frac{6}{r^7}ight]$$The work done to increase the separation of the two molecules by 10% from the equilibrium spacing is given by:$$W = \int_{0.9\sigma}^{\sigma} F(r) dr = 4\epsilon \left[\frac{12}{r^{13}} - \frac{6}{r^7}ight]_{0.9\sigma}^{\sigma}$$$$W = 4\epsilon \left[\frac{12}{0.9^{13}\sigma^{13}} - \frac{6}{0.9^7\sigma^7} - \frac{12}{\sigma^{13}} + \frac{6}{\sigma^7}ight]$$$$W = 4\epsilon \left[\frac{12}{0.9^{13}} - \frac{6}{0.9^7} - 12 + 6ight]$$$$W = 4\epsilon \left[\frac{12}{0.729} - \frac{6}{0.478} - 6ight]$$$$W = 4\epsilon \left[16.47 - 12.55 - 6ight]$$$$W = 4\epsilon (2.92)$$$$W = 1.8	imes10^{-21} J$$

Quiz 15: Solids

When a particular crystal increases in temperature by 30K, the average spacing between its atoms increases by 2%. How does the density change?

In a particular form of Silicon, the atoms are arranged in a rectangular grid of spacing 2x10-10 m. Each atom weighs 4.7x10-26 kg. What is the density of this form of Silicon, in g cm-3?

Which one of the following cannot be a definition of an elastic modulus?

A steel bridge strut decreases in radius by 5% when put under load. By what percentage does it increase in length, if the Poisson ratio for steel is 0.3?

How large a pressure increase (in ATM) must be applied to water if it is to be compressed in volume by 1.0%? The bulk modulus of water is 2.0 109 N/m² and 1 ATM = 1.0 105 N/m².

A metal cylinder of length 1 m and radius 10 cm is stretched so that its length increases by 10%. If the Poisson ratio for this metal is 0.3, what is its radius while it is stretched?

Iron has a molar mass of 55.85 g/mol and a density of 7.87 g cm-3.If the atoms are arranged in a cubic grid, what is the typical distance between adjacent atoms?

How large a force is necessary to stretch a 2.0-mm diameter copper wire (Y = 11*1010 N/m²) by 1.0%?

A metal cylinder of length 1 m and radius 10 cm is stretched so that its length increases by 10%. If the Poisson ratio for this metal is 0.3, by what percentage does its volume increase?

At a building site in Perth, a tower crane is lifting a 300 kg pallet of bricks upwards at a steady speed. The cable from which the bricks are suspended has a circular cross-section of radius 1 cm. What is the tensile stress in this cable?

Stress is proportional to strain means that:

A 20-m long steel wire (cross-section 1.0 cm², Young's modulus 2.0 * 1011 N/m²) , is subjected to a load of 25 000 N. How much will the wire stretch under the load?