Question:

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Answer:

1. Calculate the total tension in the wires: T = (15 kg) x (9.81 m/s2) = 147.15 N

2. Calculate the tension in each wire: T_copper = T_iron = T/3 = 49.05 N

3. Calculate the cross-sectional area of each wire: A_copper = (T_copper)/(8.9 x 10^9 Pa) = 5.51 x 10^-7 m2 A_iron = (T_iron)/(2.1 x 10^9 Pa) = 2.33 x 10^-7 m2

4. Calculate the diameter of each wire: d_copper = 2 x (sqrt(A_copper/pi)) = 0.0067 m d_iron = 2 x (sqrt(A_iron/pi)) = 0.0056 m

5. Calculate the ratio of their diameters: d_copper/d_iron = 1.20

Question:

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer:

1. Calculate the cross-sectional area of each column: A = π(r2 - r1) = π(602 - 302) = π(3600 - 900) = π(2700) = 8426.67 cm2

2. Calculate the total force on each column: F = 50,000 kg × 9.81 m/s2 = 490,500 N

3. Calculate the stress on each column: σ = F/A = 490,500 N / 8426.67 cm2 = 58.2 N/cm2

4. Calculate the Young’s modulus of mild steel: E = 200 GPa

5. Calculate the compressional strain of each column: ε = σ/E = 58.2 N/cm2 / 200 GPa = 0.000291

Question:

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1atm=1.013×10^5Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer:

Step 1: Calculate the bulk modulus of water: Bulk modulus = (Pressure increase x Initial Volume) / (Final Volume - Initial Volume) Bulk modulus = (100.0 atm x 100.0 litre) / (100.5 litre - 100.0 litre) Bulk modulus = 1.013 x 10^7 Pa

Step 2: Calculate the bulk modulus of air: Bulk modulus of air = (Pressure increase x Initial Volume) / (Final Volume - Initial Volume) Bulk modulus of air = (100.0 atm x 100.0 litre) / (100.5 litre - 100.0 litre) Bulk modulus of air = 1.013 x 10^5 Pa

Step 3: Compare the bulk modulus of water with that of air: The bulk modulus of water is significantly larger than the bulk modulus of air, with the ratio being approximately 1000 times larger.

Step 4: Explain the difference in simple terms: The bulk modulus of water is much larger than that of air because water is a much denser and more viscous fluid than air, and therefore is able to resist compression more effectively. This means that when pressure is applied to water, it is able to resist the pressure more effectively than air, resulting in a larger bulk modulus.

Question:

A steel wire of length 4.7 m and cross-sectional area 3.0×10−5^m^2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0×10^−5m^2 under a given load. What is the ratio of the Youngs modulus of steel to that of copper ?

Answer:

1. Find the Young’s modulus of steel (E_steel): E_steel = Load / (Area x Strain)

2. Find the Young’s modulus of copper (E_copper): E_copper = Load / (Area x Strain)

3. Find the ratio of the Young’s modulus of steel to that of copper (E_steel/E_copper): E_steel/E_copper = (E_steel)/(E_copper)

Question:

Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9×10^7Pa ? Assume that each rivet is to carry one quarter of the load.

Answer:

1. Determine the area of each rivet: A = πr^2 A = π(3.0 mm)^2 A = 28.27 mm^2

2. Calculate the load on each rivet: Load = Shearing Stress x Area Load = 6.9×10^7Pa x 28.27 mm^2 Load = 1.96×10^8 N

3. Divide the load by four to determine the load on each rivet: Load per Rivet = 1.96×10^8 N / 4 Load per Rivet = 4.9×10^7 N

4. Calculate the maximum tension that can be exerted by the riveted strip: Tension = Load per Rivet / Area Tension = 4.9×10^7 N / 28.27 mm^2 Tension = 1.73×10^8 Pa

Question:

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1×10^8Pa. A steel ball of initial volume 0.32m^3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ?

Answer:

1. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water.

2. The water pressure at the bottom of the trench is about 1.1×10^8Pa.

3. A steel ball of initial volume 0.32m^3 is dropped into the ocean and falls to the bottom of the trench.

4. Calculate the change in the volume of the ball when it reaches the bottom of the trench.

5. Use the formula for calculating the change in volume due to pressure: V = V0 x (P/P0), where V is the new volume, V0 is the initial volume, P is the pressure at the bottom of the trench, and P0 is the atmospheric pressure.

6. Substitute the values into the formula: V = 0.32m^3 x (1.1×10^8Pa/101,325Pa).

7. Calculate the new volume: V = 3.48m^3.

8. Calculate the change in volume: 3.48m^3 - 0.32m^3 = 3.16m^3.

9. The change in the volume of the ball when it reaches the bottom of the trench is 3.16m^3.

Question:

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03×10^3kgm^−3 ?

Answer:

1. Calculate the compressibility of water, which is defined as the fractional change in volume per unit change in pressure. This can be calculated as: compressibility = (change in volume / change in pressure)

2. Calculate the change in volume of water at the depth of 80.0 atm. This can be calculated as: change in volume = compressibility x change in pressure

3. Calculate the density of water at the depth of 80.0 atm. This can be calculated as: density = (mass / volume)

4. Substitute the values of the change in volume and mass of water at the surface into the equation for density. This will give the density of water at the depth of 80.0 atm.

Question:

How much should the pressure on a litre of water to be changed to compress it by 0.10% ? (Bulk modulus of water, B=2.2×10^9Nm^−2)

Answer:

1. Calculate the change in volume (ΔV) of the water due to the 0.10% compression: ΔV = 0.001 x 1 litre = 0.001 litre

2. Calculate the change in pressure (ΔP) required to cause the 0.10% compression of the water using the bulk modulus of water (B): ΔP = B x (ΔV/V) ΔP = (2.2 x 10^9 Nm^-2) x (0.001/1) ΔP = 2.2 x 10^6 N/m^2

3. The pressure must be changed by 2.2 x 10^6 N/m^2 to compress the litre of water by 0.10%.

Question:

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1×10^8Pa. A steel ball of initial volume 0.32m^3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ?

Answer:

1. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water.

2. The water pressure at the bottom of the trench is about 1.1×10^8Pa.

3. A steel ball of initial volume 0.32m^3 is dropped into the ocean and falls to the bottom of the trench.

4. The change in the volume of the ball when it reaches to the bottom is calculated using the Boyle’s Law which states that the volume of a gas is inversely proportional to its pressure.

5. Therefore, the change in the volume of the ball when it reaches the bottom of the trench is calculated as follows:

Change in volume = 0.32m^3 x (1.1×10^8Pa/1 atmospheric pressure) = 3.52m^3

Question:

A mild steel wire of length 1.0 m and cross-sectional area 0.50×10^−2cm^2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.

Answer:

1. Calculate the Young’s modulus of the mild steel wire: Young’s modulus (E) = Stress (σ) / Strain (ε) Stress = Force (F) / Cross-sectional area (A) Force = Mass (m) x Acceleration due to gravity (g)

E = (m x g) / (A x ε) E = (100 g x 9.81 m/s^2) / (0.50×10^−2 cm^2 x ε) E = 1.962 x 10^7 N/m^2

2. Calculate the strain of the wire: Strain (ε) = Extension (x) / Original Length (L)

ε = (1.0 m - 0 m) / 1.0 m ε = 1

3. Calculate the stress of the wire: Stress (σ) = Force (F) / Cross-sectional area (A)

F = Mass (m) x Acceleration due to gravity (g) F = 100 g x 9.81 m/s^2 F = 981 N

σ = 981 N / 0.50×10^−2 cm^2 σ = 1.962 x 10^7 N/m^2

4. Calculate the depression at the mid-point: Depression (d) = Stress (σ) x Extension (x)

d = 1.962 x 10^7 N/m^2 x 1 d = 1.962 x 10^7 m

Question:

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0×10^6Pa. (Bulk modulus of copper, B=140×10^9Pa)

Answer:

1. Calculate the change in volume (ΔV) using the formula: ΔV = -B*ΔP/P

2. Substitute the values into the formula: ΔV = -(140×10^9Pa)*(7.0×10^6Pa/7.0×10^6Pa)

3. Calculate the volume contraction of the solid copper cube: ΔV = -9800 cm3

Question:

A piece of copper having a rectangular cross-section of 15.2mm×19.1mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? (Modulus of elasticity of copper, Y=42×10^9 Nm^−2)

Answer:

1. Calculate the cross-sectional area of the copper: A = 15.2mm x 19.1mm = 289.52 mm^2

2. Calculate the stress: Stress = Force (N) / Area (mm^2) = 44,500 N / 289.52 mm^2 = 153,495.8 N/mm^2

3. Calculate the strain: Strain = Stress (N/mm^2) / Modulus of Elasticity (N/mm^2) = 153,495.8 N/mm^2 / 42 x 10^9 N/mm^2 = 0.000003626

Question:

Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel. (b) The stretching of a coil is determined by its shear modulus.

Answer:

(a) False. The Young’s modulus of steel is greater than that of rubber.

(b) False. The stretching of a coil is determined by its elastic modulus, not its shear modulus.

Question:

The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Answer:

1. Calculate the area of one face of the cube: A = (10 cm)^2 = 100 cm^2

2. Calculate the force applied to the cube: F = 100 kg * 9.81 m/s^2 = 981 N

3. Calculate the stress applied to the cube: σ = F/A = 981 N/100 cm^2 = 9.81 MPa

4. Calculate the shear strain of the cube: γ = σ/G = 9.81 MPa/25 GPa = 0.392

5. Calculate the vertical deflection of the face: δ = γ * h = 0.392 * 10 cm = 3.92 cm

Question:

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm^2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Answer:

1. Calculate the tension in the wire: T = m*g = (14.5 kg)(9.8 m/s^2) = 142.3 N

2. Calculate the Young’s Modulus of the wire: E = (T * (1.0 m)) / (A * δL)

3. Calculate the elongation of the wire: δL = (T * (1.0 m)) / (E * A) δL = (142.3 N * (1.0 m)) / ((2.06 x 10^11 N/m^2) * (0.065 cm^2)) δL = 0.0022 m

Question:

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Answer:

Step 1: Calculate the initial volume of the glass slab.

Step 2: Calculate the final volume of the glass slab when subjected to 10 atm of hydraulic pressure.

Step 3: Calculate the difference between the initial and final volumes.

Step 4: Divide the difference between the initial and final volumes by the initial volume.

Step 5: Multiply the result by 100 to calculate the fractional change in volume.