GSEB Class 11 Physics Solutions Chapter 9 Mechanical Properties of Solids

Question 1. A steel wire of length 4.7 m and cross-sectional area \( 3.0 \times 10^{-5} \text{m}^2 \) stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area \( 4.0 \times 10^{-5} \text{m}^2 \) under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Answer:
Here, for steel wire:
Length of wire, \( l_1 = 4.7 \text{ m} \)
Area of cross-section, \( a_1 = 3.0 \times 10^{-5} \text{ m}^2 \)
Stretching, \( \Delta l_1 = \Delta l \) (let's assume)
Stretching force on steel, \( F_1 = F \)
For copper wire:
Length of wire, \( l_2 = 3.5 \text{ m} \)
Area of cross-section, \( a_2 = 4.0 \times 10^{-5} \text{ m}^2 \)
Stretching, \( \Delta l_2 = \Delta l \) (given)
Force on copper, \( F_2 = F \)
Let \( Y_1 \) and \( Y_2 \) be the Young's modulus of the steel and copper wires, respectively.
We know that Young's modulus \( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{F \times l}{A \times \Delta l} \)
For steel wire:
\( Y_1 = \frac{F_1 \times l_1}{a_1 \times \Delta l_1} = \frac{F \times 4.7}{3 \times 10^{-5} \times \Delta l} \quad \text{....(i)} \)
For copper wire:
\( Y_2 = \frac{F_2 \times l_2}{a_2 \times \Delta l_2} = \frac{F \times 3.5}{4 \times 10^{-5} \times \Delta l} \quad \text{....(ii)} \)
To find the ratio \( Y_1 / Y_2 \), we divide equation (i) by equation (ii):
\( \frac{Y_1}{Y_2} = \frac{\frac{F \times 4.7}{3 \times 10^{-5} \times \Delta l}}{\frac{F \times 3.5}{4 \times 10^{-5} \times \Delta l}} \)
\( \frac{Y_1}{Y_2} = \frac{F \times 4.7}{3 \times 10^{-5} \times \Delta l} \times \frac{4 \times 10^{-5} \times \Delta l}{F \times 3.5} \)
\( \frac{Y_1}{Y_2} = \frac{4.7 \times 4}{3 \times 3.5} = \frac{18.8}{10.5} \)
\( \frac{Y_1}{Y_2} \approx 1.7904 \approx 1.8 \)
The ratio of the Young's modulus of steel to that of copper is approximately 1.8.
In simple words: We calculated how stiff each wire is using its length, area, and how much it stretched. Since both wires stretched the same amount under the same load, we could find the ratio of their Young's moduli by dividing their formulas. The steel wire's stiffness is about 1.8 times that of the copper wire.

Exam Tip: Remember the formula for Young's modulus and ensure consistent units. When comparing two materials under similar conditions, setting up a ratio of their Young's moduli simplifies the calculation significantly.

Question 2. Figure shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for his material?
Answer:
From the given graph, we can find the values needed.
For a stress of \( 150 \times 10^6 \text{ Nm}^{-2} \), the corresponding strain is \( 0.002 \).
(a) Young's modulus of the material \( (Y) \) is given by the formula:
\( Y = \frac{\text{Stress}}{\text{Strain}} \)
\( Y = \frac{150 \times 10^6 \text{ Nm}^{-2}}{0.002} \)
\( Y = \frac{150 \times 10^6}{2 \times 10^{-3}} = 75 \times 10^9 \text{ Nm}^{-2} \)
\( Y = 7.5 \times 10^8 \text{ Nm}^{-2} \)
(b) Yield strength of a material is defined as the maximum stress it can sustain before it permanently deforms.
From the graph, the approximate yield strength of the given material is observed near the peak of the curve, which is approximately \( 300 \times 10^6 \text{ Nm}^{-2} \). The actual value is slightly less than this peak, indicating the point where it starts to yield. So, it is about \( 3 \times 10^8 \text{ Nm}^{-2} \).
In simple words: We looked at the graph. Young's modulus, which shows how stretchy the material is, was found by dividing a stress value by its strain value. The yield strength, which is how much force the material can take before it bends permanently, was found by looking at the highest point on the curve before it starts to deform more easily.

Exam Tip: To find Young's modulus from a stress-strain curve, pick any point in the linear elastic region and divide the stress by the strain. The yield strength is the maximum stress the material can endure before plastic (permanent) deformation begins, often identified at the highest point of the elastic region or the onset of non-linearity.

Question 3. The stress-strain graphs for materials A and B are shown in figure (a) and (b) (a) Which of the materials has greater Young's modulus? (b) Which of the two is the stronger material?
Answer:
The provided stress-strain graphs illustrate the behavior of two different materials, A and B.
(a) Young's modulus is represented by the slope of the stress-strain curve in the elastic region. A steeper slope indicates a higher Young's modulus. By comparing the two graphs, we observe that for the same amount of strain, material A experiences more stress than material B. Therefore, material A has a steeper slope, which means its Young's modulus \( \left(= \frac{\text{Stress}}{\text{Strain}}\right) \) is greater than that of material B.
(b) The strength of a material is determined by the total amount of stress (load) it can withstand before it breaks or fractures. This is represented by the breaking point on the stress-strain curve. Looking at the graphs, material A can endure more stress without breaking compared to material B, whose breaking point is indicated by point D. This indicates that material A is stronger than material B because it can handle a larger load before failing.
In simple words: (a) Material A has a bigger Young's modulus because its graph line goes up more steeply, meaning it needs more force to stretch the same amount. (b) Material A is stronger because it can handle more force before it breaks compared to material B, which breaks at point D.

Exam Tip: Remember that a steeper slope on a stress-strain graph indicates a higher Young's modulus (greater stiffness), while a higher breaking point on the stress axis signifies greater strength (ability to withstand more load before fracture).

Question 4. Examine the following 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 rubber is significantly less than that of steel. This is evident because if both steel and rubber wires have the same length and cross-sectional area and are subjected to the same deforming force, the steel wire will show much less extension compared to the rubber wire. In simpler terms, \( Y_{\text{steel}} > Y_{\text{rubber}} \). This means that to achieve the same amount of strain, steel requires a much greater stress than rubber.
(b) True. The stretching of a coil spring is primarily determined by its shear modulus. When a coil spring stretches, there is no significant change in the overall length of the wire that forms the coil, nor is there a change in its volume. Instead, the deformation primarily involves a change in the shape of the coil (twisting of the wire). Since shear modulus relates to changes in shape, it accurately describes the stretching behavior of the coil spring.
In simple words: (a) This statement is false. Steel is much stiffer than rubber, so it has a higher Young's modulus. Steel resists stretching more than rubber. (b) This statement is true. When you pull a spring, it changes its shape by twisting, not by changing its length or volume much. The shear modulus measures how a material resists shape change, so it's the right way to describe how a spring stretches.

Exam Tip: Understand that Young's modulus measures resistance to length change (stiffness), while shear modulus measures resistance to shape change. Steel is much stiffer than rubber, and a coil spring's stretching is mainly due to the twisting (shear) of its wire.

Question 5. Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Young's modulus of steel is \( 2.0 \times 10^{11} \text{Pa} \). Compute the elongations of steel and brass wires. (1 Pa = 1 Nm\(^{-2}\)).
Answer:
The setup shows a 4.0 kg mass hanging from the steel wire, and a 6.0 kg mass hanging from the brass wire, which is itself suspended from the 4.0 kg mass. This means the steel wire supports both masses, while the brass wire only supports the 6.0 kg mass.
For the steel wire:
Total force supported, \( F_1 = (4.0 \text{ kg} + 6.0 \text{ kg}) \times 9.8 \text{ m/s}^2 = 10 \times 9.8 \text{ N} = 98 \text{ N} \)
Length of steel wire, \( l_1 = 1.5 \text{ m} \)
Diameter of wire, \( D = 0.25 \text{ cm} = 0.25 \times 10^{-2} \text{ m} \)
Radius, \( r_1 = \frac{D}{2} = \frac{0.25 \times 10^{-2} \text{ m}}{2} = 0.125 \times 10^{-2} \text{ m} \)
Area, \( A_1 = \pi r_1^2 = \pi (0.125 \times 10^{-2})^2 \text{ m}^2 \)
Young's modulus of steel, \( Y_1 = 2.0 \times 10^{11} \text{ Pa} \)
The elongation \( \Delta l_1 \) is given by \( \Delta l_1 = \frac{F_1 \times l_1}{A_1 \times Y_1} \)
\( \Delta l_1 = \frac{98 \text{ N} \times 1.5 \text{ m}}{\pi (0.125 \times 10^{-2})^2 \text{ m}^2 \times 2.0 \times 10^{11} \text{ Pa}} \)
Using \( \pi \approx \frac{22}{7} \):
\( \Delta l_1 = \frac{98 \times 1.5}{ \frac{22}{7} \times (0.125 \times 10^{-2})^2 \times 2 \times 10^{11}} \)
\( \Delta l_1 = \frac{147}{ \frac{22}{7} \times 0.015625 \times 10^{-4} \times 2 \times 10^{11}} \)
\( \Delta l_1 = \frac{147 \times 7}{ 22 \times 0.015625 \times 2 \times 10^{7}} \)
\( \Delta l_1 = \frac{1029}{ 0.6875 \times 10^{7}} \approx 1.496 \times 10^{-4} \text{ m} \)
Approximately, \( \Delta l_1 = 1.5 \times 10^{-4} \text{ m} \)
For the brass wire:
Force supported, \( F_2 = 6.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 58.8 \text{ N} \)
Length of brass wire, \( l_2 = 1.0 \text{ m} \)
Diameter is the same, so radius \( r_2 = 0.125 \times 10^{-2} \text{ m} \)
Area, \( A_2 = \pi r_2^2 = \pi (0.125 \times 10^{-2})^2 \text{ m}^2 \)
Young's modulus of brass, \( Y_2 = 0.91 \times 10^{11} \text{ Pa} \) (as implicitly used in the source calculation, this value is given from typical brass properties for similar problems).
The elongation \( \Delta l_2 \) is given by \( \Delta l_2 = \frac{F_2 \times l_2}{A_2 \times Y_2} \)
\( \Delta l_2 = \frac{58.8 \text{ N} \times 1.0 \text{ m}}{\pi (0.125 \times 10^{-2})^2 \text{ m}^2 \times 0.91 \times 10^{11} \text{ Pa}} \)
Using \( \pi \approx \frac{22}{7} \):
\( \Delta l_2 = \frac{58.8 \times 1}{ \frac{22}{7} \times (0.125 \times 10^{-2})^2 \times 0.91 \times 10^{11}} \)
\( \Delta l_2 = \frac{58.8 \times 7}{ 22 \times 0.015625 \times 0.91 \times 10^7} \)
\( \Delta l_2 = \frac{411.6}{ 0.313 \times 10^7} \approx 1.31 \times 10^{-4} \text{ m} \)
Approximately, \( \Delta l_2 = 1.3 \times 10^{-4} \text{ m} \).
In simple words: We first calculated the total force on each wire. The steel wire held both weights, while the brass wire held only the lower weight. Using the wire's length, cross-sectional area, and how stiff the material is (Young's modulus), we then found out how much each wire would stretch. Steel stretched about \( 1.5 \times 10^{-4} \text{ m} \) and brass stretched about \( 1.3 \times 10^{-4} \text{ m} \).

Exam Tip: Be careful to identify the total load (force) acting on each wire correctly. The formula for elongation \( \Delta l = \frac{F l}{AY} \) is fundamental; ensure all units are consistent (SI units are preferred).

Question 6. 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:
Here, we have an aluminium cube with a side length of \( L = 10 \text{ cm} = 0.1 \text{ m} \).
The area of the face to which the mass is attached (and where the force is applied) is \( A = L \times L = L^2 \).
\( A = (0.1 \text{ m})^2 = 0.01 \text{ m}^2 \).
The mass attached to this face is \( M = 100 \text{ kg} \).
The tangential force \( F \) acting on the face due to this mass is \( F = M \times g \).
Assuming \( g = 9.8 \text{ m/s}^2 \), \( F = 100 \text{ kg} \times 9.8 \text{ m/s}^2 = 980 \text{ N} \).
The shear stress on the face is \( \text{Shear Stress} = \frac{F}{A} = \frac{980 \text{ N}}{0.01 \text{ m}^2} = 9.8 \times 10^4 \text{ Nm}^{-2} \).
The shear modulus of aluminium is \( \eta = 25 \text{ GPa} = 25 \times 10^9 \text{ Nm}^{-2} \).
We know the relation for shear modulus:
\( \eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} \)
The shear strain is defined as \( \frac{\Delta y}{L} \), where \( \Delta y \) is the vertical displacement of the face.
So, \( \eta = \frac{\text{Shear Stress}}{\Delta y / L} \)
Rearranging to find \( \Delta y \):
\( \Delta y = \frac{\text{Shear Stress} \times L}{\eta} \)
\( \Delta y = \frac{9.8 \times 10^4 \text{ Nm}^{-2} \times 0.1 \text{ m}}{25 \times 10^9 \text{ Nm}^{-2}} \)
\( \Delta y = \frac{9.8 \times 10^3}{25 \times 10^9} \text{ m} \)
\( \Delta y = 0.392 \times 10^{-6} \text{ m} \)
\( \Delta y = 3.92 \times 10^{-7} \text{ m} \)
In centimeters, \( \Delta y = 3.92 \times 10^{-5} \text{ cm} \).
The vertical deflection of the face is approximately \( 0.392 \times 10^{-6} \text{ m} \).
In simple words: We calculated the force acting on the cube's face due to the attached mass. Then, we found the shear stress, which is force per unit area. Using the shear modulus (which tells us how much the material resists shape change) and the cube's side length, we calculated how much the face would move downwards.

Exam Tip: For shear modulus problems, clearly identify the tangential force and the area it acts upon to calculate shear stress. Remember that shear strain is the ratio of displacement to the original length perpendicular to the force.

Question 7. 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 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. Young's modulus, Y = \( 2.0 \times 10^{11} \text{ Pa} \)?
Answer:
Here, the total mass to be supported by the structure is \( M = 50,000 \text{ kg} \).
The total weight of the structure to be supported by the four columns is \( W = M \times g \).
Assuming \( g = 9.8 \text{ m/s}^2 \), \( W = 50,000 \times 9.8 \text{ N} = 490,000 \text{ N} \).
Since this weight is evenly supported by 4 columns, the compressional force on each column \( (F) \) is:
\( F = \frac{W}{4} = \frac{490,000 \text{ N}}{4} = 122,500 \text{ N} \).
For each column:
Inner radius, \( r_1 = 30 \text{ cm} = 0.3 \text{ m} \)
Outer radius, \( r_2 = 60 \text{ cm} = 0.6 \text{ m} \)
The area of cross-section of each hollow cylindrical column \( A \) is given by \( A = \pi (r_2^2 - r_1^2) \).
\( A = \pi ((0.6 \text{ m})^2 - (0.3 \text{ m})^2) = \pi (0.36 - 0.09) \text{ m}^2 = 0.27 \pi \text{ m}^2 \).
Young's modulus for mild steel, \( Y = 2.0 \times 10^{11} \text{ Pa} \).
We need to calculate the compressional strain of each column.
The compressional stress is \( \text{Stress} = \frac{F}{A} \).
The compressional strain is given by the formula:
\( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{F / A}{Y} = \frac{F}{A \times Y} \)
Substitute the values:
\( \text{Strain} = \frac{122,500 \text{ N}}{0.27 \pi \text{ m}^2 \times 2.0 \times 10^{11} \text{ Pa}} \)
Using \( \pi \approx \frac{22}{7} \):
\( \text{Strain} = \frac{122,500}{0.27 \times \frac{22}{7} \times 2.0 \times 10^{11}} \)
\( \text{Strain} = \frac{122,500 \times 7}{0.27 \times 22 \times 2.0 \times 10^{11}} \)
\( \text{Strain} = \frac{857,500}{11.88 \times 10^{11}} \approx 7.218 \times 10^{-6} \)
So, the compressional strain of each column is approximately \( 7.22 \times 10^{-6} \).
The given solution has a slight variation in the force and values, arriving at \( 2.88 \times 10^{-6} \). Let's re-evaluate using the values from the provided solution if it leads to a different result.
Following the solution's calculation which leads to \( 0.722 \times 10^{-6} \text{ (per column)} \) by using \( F = 50,000 \times 9.8 / 4 \) and \( A = \pi (r_2^2 - r_1^2) = 0.27 \pi \):
\( \text{Compressional Strain} = \frac{F}{A \times Y} = \frac{50000 \times 9.8 / 4}{0.27 \pi \times 2 \times 10^{11}} = \frac{122500}{0.27 \times \frac{22}{7} \times 2 \times 10^{11}} \approx 7.22 \times 10^{-6} \)
The solution shows an intermediate step of `0.722 × 10^-6` which is then multiplied by 4 to get `2.88 × 10^-6`. This implies that `0.722 × 10^-6` is the *total* strain, not per column, or there's a misunderstanding. The question asks for the strain *of each column*. Let's stick with the individual column strain. The calculation shown in the provided solution gives: \( \text{Strain} = \frac{50,000 \times 9.8 \times 7}{4 \times 0.27 \times 22 \times 2 \times 10^{11}} \approx 0.722 \times 10^{-6} \)
This `0.722 x 10^-6` is for *one* column. The next line `0.722 x 10^-6 x 4` is a bit confusing because the question asks for strain of *each* column. Strain is an intensive property and does not multiply by the number of columns. The final answer of `2.88 x 10^-6` might be a total effective strain for the entire system, or simply a miscalculation. Given the wording, the strain *of each column* should be `0.722 x 10^-6`. I will present this value. Therefore, the compressional strain of each column is approximately \( 0.722 \times 10^{-6} \).
In simple words: First, we found the total weight of the structure and divided it by four to get the force on each column. Then, we calculated the area of the hollow part of each column. Finally, using the column's Young's modulus, we figured out how much each column compresses relative to its original size, which is its compressional strain.

Exam Tip: When dealing with multiple supporting structures, ensure you correctly distribute the total load to find the force on each individual component. Remember that strain is a dimensionless quantity, representing the fractional change in length or volume.

Question 8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N force, producing only elastic deformation. Calculate the resulting strain? Y for copper = \( 1.1 \times 10^{11} \text{Nm}^{-2} \).
Answer:
Here, the Young's modulus for copper is \( Y = 1.1 \times 10^{11} \text{ Nm}^{-2} \).
The cross-sectional dimensions are \( 15.2 \text{ mm} \times 19.1 \text{ mm} \).
Let's convert these dimensions to meters:
\( 15.2 \text{ mm} = 15.2 \times 10^{-3} \text{ m} \)
\( 19.1 \text{ mm} = 19.1 \times 10^{-3} \text{ m} \)
The area of cross-section \( A \) is:
\( A = (15.2 \times 10^{-3} \text{ m}) \times (19.1 \times 10^{-3} \text{ m}) \)
\( A = 290.32 \times 10^{-6} \text{ m}^2 \)
The applied force is \( F = 44500 \text{ N} \).
The longitudinal stress is given by \( \text{Stress} = \frac{F}{A} \).
The resulting strain (longitudinal strain) is given by the formula:
\( \text{Strain} = \frac{\text{Stress}}{Y} = \frac{F / A}{Y} = \frac{F}{A \times Y} \)
Substitute the values into the formula:
\( \text{Strain} = \frac{44500 \text{ N}}{(290.32 \times 10^{-6} \text{ m}^2) \times (1.1 \times 10^{11} \text{ Nm}^{-2})} \)
\( \text{Strain} = \frac{44500}{290.32 \times 1.1 \times 10^5} \)
\( \text{Strain} = \frac{44500}{319.352 \times 10^5} \)
\( \text{Strain} = \frac{4.45 \times 10^4}{3.19352 \times 10^7} \)
\( \text{Strain} \approx 0.00013934 \)
\( \text{Strain} \approx 1.3934 \times 10^{-4} \)
Rounded to three significant figures, the resulting strain is \( 0.139 \). The calculation provided in the source `139.34 × 10^-3` is incorrect; it should be `1.3934 × 10^-4`. I will use the correct calculation result.
The resulting strain is approximately \( 1.39 \times 10^{-4} \).
In simple words: We first calculated the area of the copper piece's cross-section. Then, using the given force and the Young's modulus of copper, we found the material's strain. This strain shows how much the material stretches relative to its original size when pulled.

Exam Tip: Always ensure unit consistency, converting all dimensions to meters (SI units) before calculation. For rectangular cross-sections, the area is simply length times width. Strain is a dimensionless quantity.

Question 9. A steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress is not to exceed \( 10^8 \text{ Nm}^{-2} \), what is the maximum load the cable can support?
Answer:
Here, the radius of the steel cable is \( r = 1.5 \text{ cm} \).
Let's convert the radius to meters: \( r = 1.5 \times 10^{-2} \text{ m} \).
The maximum stress the cable should not exceed is \( \text{Max. Stress} = 10^8 \text{ Nm}^{-2} \).
The area of the cross-section of the cable \( A \) is given by \( A = \pi r^2 \).
\( A = \pi (1.5 \times 10^{-2} \text{ m})^2 = \pi (2.25 \times 10^{-4}) \text{ m}^2 \).
We know that Stress = Force / Area.
Therefore, the maximum load (force) the cable can support is \( F = \text{Max. Stress} \times A \).
\( F = 10^8 \text{ Nm}^{-2} \times \pi (2.25 \times 10^{-4}) \text{ m}^2 \)
\( F = 10^8 \times \pi \times 2.25 \times 10^{-4} \text{ N} \)
\( F = 2.25 \pi \times 10^4 \text{ N} \)
Using \( \pi \approx 3.142 \):
\( F = 2.25 \times 3.142 \times 10^4 \text{ N} \)
\( F = 7.0695 \times 10^4 \text{ N} \)
The maximum load the cable can support is approximately \( 7.1 \times 10^4 \text{ N} \).
In simple words: We took the cable's radius to find its cross-sectional area. Then, knowing the highest stress it can handle, we multiplied that stress by the area to find the total force, or load, the cable can safely hold without breaking or deforming too much.

Exam Tip: Remember the basic definition of stress as force per unit area. For circular cables, the area is \( \pi r^2 \). Ensure radius is in meters for calculations in SI units.

Question 10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2 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:
Let \( a_1 \) and \( a_2 \) be the areas of cross-section of the copper and iron wires, respectively. Let \( d_1 \) and \( d_2 \) be their respective diameters.
The area of a wire with diameter \( d \) is \( A = \frac{\pi d^2}{4} \).
So, \( a_1 = \frac{\pi d_1^2}{4} \) and \( a_2 = \frac{\pi d_2^2}{4} \).
Therefore, \( \frac{a_1}{a_2} = \frac{\pi d_1^2 / 4}{\pi d_2^2 / 4} = \frac{d_1^2}{d_2^2} = \left(\frac{d_1}{d_2}\right)^2 \).
The length of each wire is \( L = 2 \text{ m} \).
Let \( \Delta l \) be the extension produced in each wire. Since the bar is supported symmetrically and each wire has the same tension, the extension \( \Delta l \) will be the same for all wires.
Let \( F \) be the tension produced in each wire. Since the total mass is 15 kg and it is supported by three wires, and each wire is stated to have the "same tension," we take \( F \) as a generic force per wire.
From the relation for Young's modulus, \( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/L} \).
Rearranging for strain: \( \text{Strain} = \frac{F}{A \times Y} \).
Since the strains are equal for both copper and iron wires:
\( \frac{F_1}{a_1 Y_1} = \frac{F_2}{a_2 Y_2} \)
Given that each wire has the same tension, \( F_1 = F_2 = F \).
So, \( \frac{F}{a_1 Y_1} = \frac{F}{a_2 Y_2} \)
This simplifies to \( a_1 Y_1 = a_2 Y_2 \).
Therefore, \( \frac{a_1}{a_2} = \frac{Y_2}{Y_1} \).
Substituting the expression for the ratio of areas in terms of diameters:
\( \left(\frac{d_1}{d_2}\right)^2 = \frac{Y_2}{Y_1} \)
So, \( \frac{d_1}{d_2} = \sqrt{\frac{Y_2}{Y_1}} \).
For this type of problem, typical Young's moduli values are:
\( Y_{\text{copper}} = Y_1 \approx 1.1 \times 10^{11} \text{ Pa} \)
\( Y_{\text{iron}} = Y_2 \approx 1.9 \times 10^{11} \text{ Pa} \)
Using these values as implied by the calculation in the source:
\( \frac{d_1}{d_2} = \sqrt{\frac{1.9 \times 10^{11}}{1.1 \times 10^{11}}} = \sqrt{\frac{19}{11}} \)
\( \frac{d_1}{d_2} = \sqrt{1.727} \approx 1.314 \)
Thus, the ratio of the diameters is approximately \( d_1 : d_2 = 1.31 : 1 \).
In simple words: Because the bar is supported evenly and each wire pulls with the same force, all wires stretch by the same amount. By relating this stretching to each material's stiffness (Young's modulus) and their cross-sectional areas (which depend on diameter), we found that the ratio of the diameters depends on the square root of the inverse ratio of their Young's moduli.

Exam Tip: When problems involve symmetrical support and equal tension, assume equal extension for all supporting wires. The key is to equate the strains (since length is also equal) and then use the Young's modulus formula to find the relationship between areas or diameters.

Question 11. A 14.5 kg mass, fastened to one end of a steel wire of unstretched length 1 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 cm². Calculate the elongation of the wire when, the mass is at the lowest point of its path?
Answer:
Here, the mass attached to one end of the steel wire is \( m = 14.5 \text{ kg} \).
The length of the steel wire is \( l = 1 \text{ m} \).
The frequency of whirling is \( \nu = 2 \text{ rev/s} = 2 \text{ rps} \).
The angular frequency \( \omega \) is \( \omega = 2 \pi \nu = 2 \pi \times 2 = 4 \pi \text{ rad/s} \).
The cross-sectional area of the wire is \( A = 0.065 \text{ cm}^2 \).
Converting to \( \text{m}^2 \): \( A = 0.065 \times (10^{-2} \text{ m})^2 = 0.065 \times 10^{-4} \text{ m}^2 = 6.5 \times 10^{-6} \text{ m}^2 \).
The Young's modulus for steel is \( Y = 2 \times 10^{11} \text{ Pa} \).
We need to calculate the elongation \( \Delta l \) of the wire.
When the mass is at the lowest point of its path in a vertical circle, the tension (stretching force) \( T \) in the wire is the sum of the gravitational force and the centripetal force:
\( T = mg + m l \omega^2 \)
Assuming \( g = 9.8 \text{ m/s}^2 \):
\( T = (14.5 \text{ kg} \times 9.8 \text{ m/s}^2) + (14.5 \text{ kg} \times 1 \text{ m} \times (4 \pi \text{ rad/s})^2) \)
\( T = 142.1 \text{ N} + (14.5 \times 16 \pi^2) \text{ N} \)
Using \( \pi^2 \approx 9.87 \) (as specified in the source solution for this problem):
\( T = 142.1 \text{ N} + (14.5 \times 16 \times 9.87) \text{ N} \)
\( T = 142.1 \text{ N} + (14.5 \times 157.92) \text{ N} \)
\( T = 142.1 \text{ N} + 2289.84 \text{ N} = 2431.94 \text{ N} \)
Now, we calculate the elongation \( \Delta l \) using the formula:
\( \Delta l = \frac{T \times l}{A \times Y} \)
\( \Delta l = \frac{2431.94 \text{ N} \times 1 \text{ m}}{(6.5 \times 10^{-6} \text{ m}^2) \times (2 \times 10^{11} \text{ Pa})} \)
\( \Delta l = \frac{2431.94}{13 \times 10^5} \text{ m} \)
\( \Delta l = \frac{2431.94}{1.3 \times 10^6} \text{ m} \)
\( \Delta l \approx 1.8707 \times 10^{-3} \text{ m} \)
\( \Delta l \approx 1.87 \text{ mm} \).
The elongation of the wire is approximately \( 1.87 \times 10^{-3} \text{ m} \).
In simple words: At the lowest point of the circle, the wire pulls on the mass with a force that includes both gravity and the force needed to keep it moving in a circle. We calculated this total pulling force. Then, using the wire's length, its thickness (area), and how stiff it is (Young's modulus), we found out how much the wire would stretch.

Exam Tip: For vertical circular motion, remember that the tension in the wire is highest at the bottom, where it must support both gravity and the centripetal force. Ensure you use the correct formula for tension and accurately calculate the angular frequency.

Question 12. Compute the bulk modulus of water from the following data: initial volume = 100.0 litre, pressure increase = 100.0 atm (1 atm = \( 1.013 \times 10^5 \text{ Pa} \)), 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:
Given data:
Initial volume, \( V_1 = 100.0 \text{ litre} \).
Final volume, \( V_2 = 100.5 \text{ litre} \).
Pressure increase, \( P = 100.0 \text{ atm} \).
First, convert units to SI:
\( 1 \text{ litre} = 10^{-3} \text{ m}^3 \).
So, \( V_1 = 100.0 \times 10^{-3} \text{ m}^3 \).
\( V_2 = 100.5 \times 10^{-3} \text{ m}^3 \).
Change in volume, \( \Delta V = V_2 - V_1 = (100.5 - 100.0) \times 10^{-3} \text{ m}^3 = 0.5 \times 10^{-3} \text{ m}^3 \).
Convert pressure to Pascals:
\( P = 100.0 \times (1.013 \times 10^5 \text{ Pa}) = 1.013 \times 10^7 \text{ Pa} \).
The bulk modulus \( K_w \) of water is given by the formula:
\( K_w = \frac{P \times V_1}{\Delta V} \) (taking \( V_1 \) as the reference volume for fractional change).
\( K_w = \frac{(1.013 \times 10^7 \text{ Pa}) \times (100.0 \times 10^{-3} \text{ m}^3)}{0.5 \times 10^{-3} \text{ m}^3} \)
\( K_w = \frac{1.013 \times 10^7 \times 100}{0.5} \)
\( K_w = \frac{101.3 \times 10^7}{0.5} = 202.6 \times 10^7 \text{ Pa} = 2.026 \times 10^9 \text{ Pa} \).
So, the bulk modulus of water is \( 2.026 \times 10^9 \text{ Pa} \).
Now, let's compare it with the bulk modulus of air at S.T.P. (Standard Temperature and Pressure).
Given \( K_{\text{air}} = 1.0 \times 10^{-4} \text{ GPa} \).
Convert \( K_{\text{air}} \) to Pascals: \( 1 \text{ GPa} = 10^9 \text{ Pa} \).
\( K_{\text{air}} = 1.0 \times 10^{-4} \times 10^9 \text{ Pa} = 10^5 \text{ Pa} \).
Now, let's find the ratio:
\( \frac{K_w}{K_{\text{air}}} = \frac{2.026 \times 10^9 \text{ Pa}}{10^5 \text{ Pa}} = 2.026 \times 10^4 = 20260 \).
The ratio is very large, approximately 20260.
This large ratio exists because liquids (like water) are much less compressible than gases (like air). In simple terms, the molecules in liquids are already quite close together, and the intermolecular forces are strong, making it very difficult to push them even closer. In contrast, gas molecules are far apart with weak intermolecular forces, so they can be compressed much more easily. Therefore, a given pressure change causes a much smaller fractional volume change in water than in air.
In simple words: We used the change in volume and pressure to figure out how resistant water is to being squeezed (its bulk modulus). Then we compared this to air's bulk modulus. The main difference is that water molecules are already packed tightly, making it hard to compress, while air molecules are far apart and easy to squeeze together. That's why water's bulk modulus is so much higher.

Exam Tip: Remember to convert all given values to consistent SI units (Pascals for pressure, cubic meters for volume) before calculating bulk modulus. The significantly higher bulk modulus for liquids compared to gases highlights their incompressibility due to strong intermolecular forces and close molecular packing.

Question 13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is \( 1.03 \times 10^3 \text{ kgm}^{-3} \), compressibility of water is \( 45.8 \times 10^{-11} \text{ Pa}^{-1} \)?
Answer:
Given data:
Pressure, \( P = 80.0 \text{ atm} \).
Converting to Pascals: \( P = 80.0 \times 1.013 \times 10^5 \text{ Pa} = 8.104 \times 10^6 \text{ Pa} \).
Compressibility of water, \( \frac{1}{K} = 45.8 \times 10^{-11} \text{ Pa}^{-1} \).
Density of water at the surface, \( \rho = 1.03 \times 10^3 \text{ kgm}^{-3} \).
Let \( \rho' \) be the density of water at the given depth.
If \( V \) and \( V' \) are the volumes of a certain mass \( M \) of water at the surface and at the given depth, respectively:
\( V = \frac{M}{\rho} \) and \( V' = \frac{M}{\rho'} \).
The change in volume is \( \Delta V = V - V' = M \left( \frac{1}{\rho} - \frac{1}{\rho'} \right) \).
The volumetric strain is \( \frac{\Delta V}{V} = \frac{M \left( \frac{1}{\rho} - \frac{1}{\rho'} \right)}{M/\rho} = \rho \left( \frac{1}{\rho} - \frac{1}{\rho'} \right) = 1 - \frac{\rho}{\rho'} \).
We know that compressibility is \( \frac{1}{K} = \frac{\text{Volumetric Strain}}{\text{Pressure}} = \frac{\Delta V/V}{P} \).
So, \( \frac{1}{K} = \frac{1 - \rho/\rho'}{P} \).
Rearranging for \( 1 - \frac{\rho}{\rho'} \):
\( 1 - \frac{\rho}{\rho'} = P \times \frac{1}{K} \)
Substitute the values:
\( 1 - \frac{1.03 \times 10^3}{\rho'} = (8.104 \times 10^6 \text{ Pa}) \times (45.8 \times 10^{-11} \text{ Pa}^{-1}) \)
\( 1 - \frac{1.03 \times 10^3}{\rho'} = 371.1392 \times 10^{-5} \)
\( 1 - \frac{1.03 \times 10^3}{\rho'} = 3.711392 \times 10^{-3} = 0.003711392 \)
\( \frac{1.03 \times 10^3}{\rho'} = 1 - 0.003711392 \)
\( \frac{1.03 \times 10^3}{\rho'} = 0.996288608 \)
\( \rho' = \frac{1.03 \times 10^3}{0.996288608} \)
\( \rho' \approx 1033.8 \text{ kgm}^{-3} \)
The density of water at this depth is approximately \( 1.034 \times 10^3 \text{ kgm}^{-3} \).
In simple words: We used the given pressure, the water's normal density, and its compressibility (how much it can be squeezed) to find its new density at a greater depth. When water is under high pressure, its volume slightly reduces, making its density slightly higher.

Exam Tip: For problems involving compressibility and density, remember the relationship between volumetric strain and density change: \( \frac{\Delta V}{V} = 1 - \frac{\rho}{\rho'} \). Convert all units to SI (Pascals for pressure, kg/m\(^3\) for density) for consistent calculations.

Question 14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm?
Answer:
Given data:
Hydraulic pressure, \( P = 10 \text{ atm} \).
Convert pressure to Pascals: \( P = 10 \times (1.013 \times 10^5 \text{ Pa}) = 1.013 \times 10^6 \text{ Pa} \).
The bulk modulus \( K \) for glass is \( 37 \times 10^9 \text{ Nm}^{-2} \) (as stated to be from a reference table).
We need to compute the fractional change in volume \( \frac{\Delta V}{V} \).
The bulk modulus is defined as \( K = \frac{P}{\Delta V/V} \).
Rearranging for fractional change in volume:
\( \frac{\Delta V}{V} = \frac{P}{K} \)
Substitute the values:
\( \frac{\Delta V}{V} = \frac{1.013 \times 10^6 \text{ Pa}}{37 \times 10^9 \text{ Pa}} \)
\( \frac{\Delta V}{V} = \frac{1.013}{37 \times 10^3} = \frac{1.013}{37000} \)
\( \frac{\Delta V}{V} \approx 0.000027378 \)
\( \frac{\Delta V}{V} \approx 2.74 \times 10^{-5} \).
This can also be expressed as a percentage: \( 2.74 \times 10^{-3} \% \).
The fractional change in volume of the glass slab is approximately \( 2.74 \times 10^{-5} \).
In simple words: We used the given pressure and the known stiffness of glass (its bulk modulus) to find how much its volume would shrink. The fractional change in volume simply tells us the ratio of how much the volume changed compared to its original volume.

Exam Tip: Remember that bulk modulus measures resistance to volume change. The formula \( \frac{\Delta V}{V} = \frac{P}{K} \) is key. Always ensure that pressure and bulk modulus are in consistent units (Pascals) for accurate results.

Question 15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of \( 7 \times 10^6 \text{Pa} \). K for copper = \( 140 \times 10^9 \text{Pa} \).
Answer:
Given data:
Edge length of the copper cube, \( L = 10 \text{ cm} = 0.1 \text{ m} \).
Hydraulic pressure, \( P = 7 \times 10^6 \text{ Pa} \).
Bulk modulus for copper, \( K = 140 \times 10^9 \text{ Pa} \).
First, calculate the initial volume of the cube, \( V \):
\( V = L^3 = (0.1 \text{ m})^3 = 0.001 \text{ m}^3 \).
We need to determine the volume contraction \( \Delta V \).
The formula for bulk modulus is \( K = \frac{P}{-\Delta V/V} \), where the negative sign indicates volume contraction for a positive pressure.
Rearranging to find \( \Delta V \):
\( \Delta V = -\frac{P \times V}{K} \)
Substitute the given values:
\( \Delta V = -\frac{(7 \times 10^6 \text{ Pa}) \times (0.001 \text{ m}^3)}{140 \times 10^9 \text{ Pa}} \)
\( \Delta V = -\frac{7 \times 10^6 \times 10^{-3}}{140 \times 10^9} \text{ m}^3 \)
\( \Delta V = -\frac{7 \times 10^3}{140 \times 10^9} \text{ m}^3 \)
\( \Delta V = -\frac{1}{20} \times 10^{-6} \text{ m}^3 \)
\( \Delta V = -0.05 \times 10^{-6} \text{ m}^3 \).
Converting to \( \text{cm}^3 \): \( 1 \text{ m}^3 = 10^6 \text{ cm}^3 \).
\( \Delta V = -0.05 \text{ cm}^3 \).
The negative sign confirms that it is a volume contraction.
The volume contraction of the copper cube is \( 0.05 \text{ cm}^3 \).
In simple words: We first found the original volume of the copper cube. Then, using the applied pressure and how much copper resists squeezing (its bulk modulus), we calculated how much its volume would decrease. The negative sign simply tells us that the volume is getting smaller.

Exam Tip: Remember to calculate the initial volume correctly from the given dimensions. When using the bulk modulus formula, the negative sign for \( \Delta V \) correctly indicates a contraction under positive pressure. Ensure all units are in the SI system.

Question 16. How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer:
Given data:
Initial volume of water, \( V = 1 \text{ litre} \).
The water is to be compressed by \( 0.10\% \). This means the change in volume \( \Delta V \) is negative.
\( \Delta V = -0.10\% \text{ of } V = -\frac{0.10}{100} \times V = -\frac{1}{1000} \times V \).
So, \( \frac{\Delta V}{V} = -\frac{1}{1000} \).
The bulk modulus of water is \( K = 2.2 \times 10^9 \text{ Nm}^{-2} \) (or Pascals).
We need to find the change in pressure \( \Delta P \).
The bulk modulus formula is \( K = \frac{\Delta P}{-\Delta V/V} \).
Rearranging for \( \Delta P \):
\( \Delta P = -K \times \frac{\Delta V}{V} \)
Substitute the values:
\( \Delta P = -(2.2 \times 10^9 \text{ Pa}) \times \left(-\frac{1}{1000}\right) \)
\( \Delta P = (2.2 \times 10^9 \text{ Pa}) \times (1 \times 10^{-3}) \)
\( \Delta P = 2.2 \times 10^6 \text{ Pa} \).
Therefore, the pressure on the water should be increased by \( 2.2 \times 10^6 \text{ Pa} \) to compress it by 0.10%.
In simple words: We wanted to squeeze water by a tiny amount, so we needed to find out how much to increase the pressure. Using water's stiffness (bulk modulus) and the desired volume change, we calculated the extra pressure needed to achieve that compression.

Exam Tip: Ensure the fractional change in volume is correctly represented as a negative value for compression. The bulk modulus formula relates pressure change to fractional volume change; paying attention to signs will lead to the correct positive pressure increase.

Question 16. How much should the pressure on a litre of water be changed to compress it by 0.10%?
Answer: Here, the initial volume \( V = 1 \) litre. The change in volume \( \Delta V = -0.10\% \) of \( V \).
So, \( \Delta V = - \frac{0.10}{100} \times 1 = - \frac{1}{1000} \) litre.
Let \( \Delta P \) be the change in pressure needed to compress 1 litre of water.
The bulk modulus of water \( K = 2.2 \times 10^9 \, \text{Nm}^{-2} \).
Using the relation, \( K = \frac{\Delta P}{(\Delta V/V)} \), we get
\( \Delta P = -K \times \frac{\Delta V}{V} \)
\( \Delta P = 2.2 \times 10^9 \times \frac{(1/1000)}{1} \)
\( \Delta P = \frac{2.2 \times 10^9}{1000} \)
\( \Delta P = 2.2 \times 10^6 \, \text{Nm}^{-2} \)
In simple words: To compress one litre of water by 0.10%, you need to apply a pressure of \( 2.2 \times 10^6 \) Newtons per square metre. The negative sign for volume change shows that it's a compression.

Exam Tip: Remember to express percentage changes as a fraction or decimal, and be mindful of units consistency when performing calculations in physics problems.