Q1: 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…

Given: For steel wire: Length, L_s = 4.7 m Cross-sectional area, A_s = 3.0 × 10^-5 m^2 For copper wire: Length, L_c = 3.5 m Cross-sectional area, A_c = 4.0 × 10^-5 m^2 It is given that the wires stretch by the same amount under the same load. Let the load be F and the elongation be Δ L. So, F_s =…

Q8: A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic…

Given: Cross-sectional dimensions = 15.2 mm × 19.1 mm Applied force (tension), F = 44,500 N Material is copper. From Table 8.1, Young's modulus for copper, Y = 110 × 10^9 N m^-2 = 1.1 × 10^11 Pa To Find: The resulting strain. Formula: Young's modulus, Y = Stress/Strain Therefore, Strain = Stress/Y…

Mechanical Properties Of Solids MECHANICAL PROPERTIES OF SOLIDS Class 11 Physics NCERT Solutions

Q1Exercises

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 of $4.0 \times 10^{-5} \text{ m}^2$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Solution

Given: For steel wire: Length,

L_s = 4.7 \text{ m}

Cross-sectional area,

A_s = 3.0 \times 10^{-5} \text{ m}^2

For copper wire: Length,

L_c = 3.5 \text{ m}

Cross-sectional area,

A_c = 4.0 \times 10^{-5} \text{ m}^2

It is given that the wires stretch by the same amount under the same load. Let the load be

F

and the elongation be

\Delta L

. So,

F_s = F_c = F

and

\Delta L_s = \Delta L_c = \Delta L

.

To Find: The ratio of the Young's modulus of steel to that of copper,

\frac{Y_s}{Y_c}

.

Formula: The Young's modulus

Y

is given by:

Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}

From this, the elongation

\Delta L

can be written as:

\Delta L = \frac{F \cdot L}{A \cdot Y}

Calculation: For the steel wire:

\Delta L_s = \frac{F \cdot L_s}{A_s \cdot Y_s}

For the copper wire:

\Delta L_c = \frac{F \cdot L_c}{A_c \cdot Y_c}

Since

\Delta L_s = \Delta L_c

:

\frac{F \cdot L_s}{A_s \cdot Y_s} = \frac{F \cdot L_c}{A_c \cdot Y_c}

Canceling

F

from both sides and rearranging to find the ratio

\frac{Y_s}{Y_c}

:

\frac{Y_s}{Y_c} = \frac{L_s}{A_s} \times \frac{A_c}{L_c}

Substituting the given values:

\frac{Y_s}{Y_c} = \frac{4.7}{3.0 \times 10^{-5}} \times \frac{4.0 \times 10^{-5}}{3.5}

\frac{Y_s}{Y_c} = \frac{4.7 \times 4.0}{3.0 \times 3.5} = \frac{18.8}{10.5} \approx 1.79

Final Answer: The ratio of the Young's modulus of steel to that of copper is approximately

1.79 : 1

.

Q2Exercises

Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?

Solution

(a) Young's Modulus

Concept: Young's modulus (

Y

) is the ratio of stress to strain in the region where the stress-strain curve is linear (obeys Hooke's law). It is the slope of the linear portion of the graph.

Formula:

Y = \frac{\text{Stress}}{\text{Strain}}

Calculation: From the graph (Figure 8.9), we can pick a point in the linear region. Let's choose the point where the stress is

150 \times 10^6 \text{ N m}^{-2}

. The corresponding strain at this point is

0.002

.

Y = \frac{150 \times 10^6 \text{ N m}^{-2}}{0.002}

Y = \frac{150 \times 10^6}{2 \times 10^{-3}} \text{ N m}^{-2}

Y = 75 \times 10^9 \text{ N m}^{-2} = 75 \text{ GPa}

Final Answer for (a): The Young's modulus for the material is

75 \times 10^9 \text{ N m}^{-2}

or

75

GPa.

(b) Approximate Yield Strength

Concept: The yield strength is the stress at the elastic limit (yield point), beyond which the material undergoes plastic deformation. On the stress-strain curve, this is the point where the graph starts to deviate from a straight line.

Observation from Graph: Looking at Figure 8.9, the curve is linear up to a stress of about

300 \times 10^6 \text{ N m}^{-2}

. After this point, the curve starts to bend, indicating the onset of plastic deformation.

Final Answer for (b): The approximate yield strength for this material is

300 \times 10^6 \text{ N m}^{-2}

or

300

MPa.

Q3Exercises

The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale.

(a)

Which of the materials has the greater Young's modulus?

(b)

Which of the two is the stronger material?

Solution

(a) Greater Young's Modulus

Reasoning: Young's modulus (

Y

) is defined as the ratio of stress to strain within the elastic limit. On a stress-strain graph, this corresponds to the slope of the linear portion of the curve.

Y = \frac{\text{Stress}}{\text{Strain}} = \text{Slope of the curve}

A steeper slope indicates a larger Young's modulus, which means a larger stress is required to produce the same amount of strain.

From Figure 8.10, the slope of the graph for material A is greater than the slope of the graph for material B. Therefore, material A has a greater Young's modulus.

Answer (a): Material A has the greater Young's modulus.

(b) Stronger Material

Reasoning: The strength of a material is determined by the maximum stress it can withstand before fracturing. This is known as the ultimate tensile strength. The fracture point is the point on the graph where the material breaks.

From Figure 8.10, the graph for material A reaches a higher stress value before the fracture point compared to material B. This means material A can withstand a greater stress before breaking.

Answer (b): Material A is the stronger material.

Q4Exercises

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.

Solution

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

Answer: False.

Reason: Young's modulus is a measure of a material's resistance to elastic deformation under load. A material with a higher Young's modulus is more rigid and requires a large force to produce a small change in length. Steel is much more rigid than rubber. For a given applied force, steel stretches very little, while rubber stretches significantly. This means that for the same stress, the strain in steel is much less than the strain in rubber. Since

Y = \frac{\text{Stress}}{\text{Strain}}

, a smaller strain for the same stress results in a larger Young's modulus. Therefore, the Young's modulus of steel (approx.

200 \times 10^9 \text{ Pa}

) is much greater than that of rubber (approx.

0.01 \times 10^9 \text{ Pa}

). The statement is false.

(b) The stretching of a coil is determined by its shear modulus.

Answer: True.

Reason: When a coil spring is stretched or compressed, the wire of the spring does not simply elongate along its length. Instead, the wire undergoes twisting or torsion. This twisting action is a type of shearing deformation. The restoring force in the spring arises from the material's resistance to this shear. The elastic property that relates shearing stress to shearing strain is the shear modulus (or modulus of rigidity,

G

). Therefore, the stretching of a coil is determined by the shear modulus of the material of the wire.

Q5Exercises

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution

Given: Diameter of both wires,

d = 0.25 \text{ cm} = 0.25 \times 10^{-2} \text{ m}

Radius of both wires,

r = \frac{d}{2} = 0.125 \times 10^{-2} \text{ m}

For steel wire: Unloaded length,

L_s = 1.5 \text{ m}

Young's modulus of steel,

Y_s = 2.0 \times 10^{11} \text{ Pa}

(from Table 8.1)

For brass wire: Unloaded length,

L_b = 1.0 \text{ m}

Young's modulus of brass,

Y_b \approx 91 \times 10^9 \text{ Pa}

(a standard value, as brass is an alloy of copper and zinc. From Table 8.2, shear modulus G is 36 GPa and G is approx Y/3, so Y is approx 108 GPa. We'll use the value for copper from the book as a close approximation,

Y_b \approx Y_{copper} = 1.1 \times 10^{11} \text{ Pa}

)

Loads: Mass attached to steel wire,

m_s = 4.0 \text{ kg}

Mass attached to brass wire,

m_b = 6.0 \text{ kg}

Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

To Find: Elongation of steel wire,

\Delta L_s

Elongation of brass wire,

\Delta L_b

Formula: Elongation,

\Delta L = \frac{F \cdot L}{A \cdot Y}

, where

A = \pi r^2

.

Calculation: First, calculate the cross-sectional area,

A

:

A = \pi r^2 = 3.14 \times (0.125 \times 10^{-2})^2 \text{ m}^2

A = 3.14 \times 1.5625 \times 10^{-6} \text{ m}^2 \approx 4.9 \times 10^{-6} \text{ m}^2

Next, calculate the forces (tensions) in the wires: The brass wire supports the

6.0 \text{ kg}

mass. Force on brass wire,

F_b = m_b \times g = 6.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 58.8 \text{ N}

The steel wire supports both the

4.0 \text{ kg}

and

6.0 \text{ kg}

masses. Total mass on steel wire,

M = m_s + m_b = 4.0 + 6.0 = 10.0 \text{ kg}

Force on steel wire,

F_s = M \times g = 10.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \text{ N}

Now, compute the elongations:

Elongation of the steel wire ( $\Delta L_s$ ):

\Delta L_s = \frac{F_s \cdot L_s}{A \cdot Y_s} = \frac{98 \text{ N} \times 1.5 \text{ m}}{(4.9 \times 10^{-6} \text{ m}^2) \times (2.0 \times 10^{11} \text{ Pa})}

\Delta L_s = \frac{147}{9.8 \times 10^5} \text{ m} = 15 \times 10^{-5} \text{ m} = 1.5 \times 10^{-4} \text{ m}

Elongation of the brass wire ( $\Delta L_b$ ): (Using

Y_b = 1.1 \times 10^{11} \text{ Pa}

)

\Delta L_b = \frac{F_b \cdot L_b}{A \cdot Y_b} = \frac{58.8 \text{ N} \times 1.0 \text{ m}}{(4.9 \times 10^{-6} \text{ m}^2) \times (1.1 \times 10^{11} \text{ Pa})}

\Delta L_b = \frac{58.8}{5.39 \times 10^5} \text{ m} \approx 10.9 \times 10^{-5} \text{ m} = 1.09 \times 10^{-4} \text{ m}

Final Answer: The elongation of the steel wire is

1.5 \times 10^{-4} \text{ m}

. The elongation of the brass wire is approximately

1.1 \times 10^{-4} \text{ m}

.

Q6Exercises

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?

Solution

Given: Edge length of the aluminium cube,

L = 10 \text{ cm} = 0.1 \text{ m}

Mass attached,

m = 100 \text{ kg}

Shear modulus of aluminium,

G = 25 \text{ GPa} = 25 \times 10^9 \text{ Pa}

Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

To Find: The vertical deflection of the face,

\Delta x

.

Formula: The shear modulus

G

is given by:

G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{\Delta x/L}

where

F

is the tangential force,

A

is the area of the face on which the force is applied,

L

is the length of the side perpendicular to the face, and

\Delta x

is the deflection.

Calculation: The force applied is the weight of the mass, which acts tangentially to the top face:

F = m \times g = 100 \text{ kg} \times 9.8 \text{ m/s}^2 = 980 \text{ N}

The area of the face on which this force acts is:

A = L^2 = (0.1 \text{ m})^2 = 0.01 \text{ m}^2

Rearranging the formula for shear modulus to solve for

\Delta x

:

\Delta x = \frac{F \cdot L}{A \cdot G}

Substituting the values:

\Delta x = \frac{980 \text{ N} \times 0.1 \text{ m}}{(0.01 \text{ m}^2) \times (25 \times 10^9 \text{ Pa})}

\Delta x = \frac{98}{0.25 \times 10^9} \text{ m} = \frac{98}{25 \times 10^7} \text{ m}

\Delta x = 3.92 \times 10^{-7} \text{ m}

Final Answer: The vertical deflection of this face is

3.92 \times 10^{-7} \text{ m}

.

Q7Exercises

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.

Solution

Given: Total mass supported,

M = 50,000 \text{ kg}

Number of columns,

n = 4

Inner radius of each column,

r_i = 30 \text{ cm} = 0.3 \text{ m}

Outer radius of each column,

r_o = 60 \text{ cm} = 0.6 \text{ m}

Material is mild steel. From Table 8.1, Young's modulus for steel,

Y = 200 \times 10^9 \text{ N m}^{-2} = 2.0 \times 10^{11} \text{ Pa}

Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

To Find: The compressional strain of each column.

Formula: Young's modulus,

Y = \frac{\text{Compressional Stress}}{\text{Compressional Strain}}

Therefore, Compressional Strain =

\frac{\text{Compressional Stress}}{Y}

Compressional Stress =

\frac{F}{A}

, where

F

is the force on one column and

A

is its cross-sectional area.

Calculation: Total load (weight) of the structure:

W = M \times g = 50,000 \text{ kg} \times 9.8 \text{ m/s}^2 = 490,000 \text{ N} = 4.9 \times 10^5 \text{ N}

Since the load is distributed uniformly among 4 columns, the compressive force on each column is:

F = \frac{W}{4} = \frac{4.9 \times 10^5 \text{ N}}{4} = 1.225 \times 10^5 \text{ N}

The cross-sectional area of each hollow cylindrical column is:

A = \pi (r_o^2 - r_i^2)

A = \pi ((0.6)^2 - (0.3)^2) \text{ m}^2

A = \pi (0.36 - 0.09) \text{ m}^2 = 0.27 \pi \text{ m}^2

A \approx 0.27 \times 3.14 \text{ m}^2 \approx 0.848 \text{ m}^2

Now, calculate the compressional stress on each column:

\text{Stress} = \frac{F}{A} = \frac{1.225 \times 10^5 \text{ N}}{0.848 \text{ m}^2} \approx 1.444 \times 10^5 \text{ N m}^{-2}

Finally, calculate the compressional strain:

\text{Strain} = \frac{\text{Stress}}{Y} = \frac{1.444 \times 10^5 \text{ N m}^{-2}}{2.0 \times 10^{11} \text{ N m}^{-2}}

\text{Strain} \approx 0.722 \times 10^{-6} = 7.22 \times 10^{-7}

Final Answer: The compressional strain of each column is approximately

7.22 \times 10^{-7}

.

Q8Exercises

A piece of copper having a rectangular cross-section of $15.2 \text{ mm} \times 19.1 \text{ mm}$ is pulled in tension with $44,500 \text{ N}$ force, producing only elastic deformation. Calculate the resulting strain?

Solution

Given: Cross-sectional dimensions =

15.2 \text{ mm} \times 19.1 \text{ mm}

Applied force (tension),

F = 44,500 \text{ N}

Material is copper. From Table 8.1, Young's modulus for copper,

Y = 110 \times 10^9 \text{ N m}^{-2} = 1.1 \times 10^{11} \text{ Pa}

To Find: The resulting strain.

Formula: Young's modulus,

Y = \frac{\text{Stress}}{\text{Strain}}

Therefore, Strain =

\frac{\text{Stress}}{Y}

Stress =

\frac{F}{A}

Calculation: First, convert the dimensions to meters and calculate the cross-sectional area,

A

: Width,

b = 15.2 \text{ mm} = 15.2 \times 10^{-3} \text{ m}

Thickness,

d = 19.1 \text{ mm} = 19.1 \times 10^{-3} \text{ m}

A = b \times d = (15.2 \times 10^{-3} \text{ m}) \times (19.1 \times 10^{-3} \text{ m})

A = 290.32 \times 10^{-6} \text{ m}^2 \approx 2.9 \times 10^{-4} \text{ m}^2

Next, calculate the tensile stress:

\text{Stress} = \frac{F}{A} = \frac{44,500 \text{ N}}{2.9032 \times 10^{-4} \text{ m}^2} \approx 1.533 \times 10^8 \text{ N m}^{-2}

Finally, calculate the resulting strain:

\text{Strain} = \frac{\text{Stress}}{Y} = \frac{1.533 \times 10^8 \text{ N m}^{-2}}{1.1 \times 10^{11} \text{ N m}^{-2}}

\text{Strain} \approx 1.39 \times 10^{-3}

Final Answer: The resulting strain is approximately

1.39 \times 10^{-3}

.

Q9Exercises

A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed $10^8 \text{ N m}^{-2}$ , what is the maximum load the cable can support ?

Solution

Given: Radius of the steel cable,

r = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}

Maximum allowable stress,

\sigma_{max} = 10^8 \text{ N m}^{-2}

To Find: The maximum load (force) the cable can support,

F_{max}

.

Formula: Stress is defined as force per unit area:

\sigma = \frac{F}{A}

Therefore, the maximum force is:

F_{max} = \sigma_{max} \times A

where the cross-sectional area

A = \pi r^2

.

Calculation: First, calculate the cross-sectional area of the cable:

A = \pi r^2 = 3.14 \times (1.5 \times 10^{-2} \text{ m})^2

A = 3.14 \times 2.25 \times 10^{-4} \text{ m}^2

A = 7.065 \times 10^{-4} \text{ m}^2

Now, calculate the maximum load:

F_{max} = (10^8 \text{ N m}^{-2}) \times (7.065 \times 10^{-4} \text{ m}^2)

F_{max} = 7.065 \times 10^4 \text{ N}

Final Answer: The maximum load the cable can support is

7.065 \times 10^4 \text{ N}

.

Q10Exercises

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.

Solution

Given: Length of each wire,

L_{Cu} = L_{Fe} = L = 2.0 \text{ m}

Tension in each wire is the same,

F_{Cu} = F_{Fe} = F

. The bar is rigid and supported symmetrically, which implies that the elongation in each wire must be the same for the bar to remain horizontal.

\Delta L_{Cu} = \Delta L_{Fe} = \Delta L

Young's modulus for copper (from Table 8.1),

Y_{Cu} = 110 \times 10^9 \text{ Pa}

Young's modulus for iron (from Table 8.1),

Y_{Fe} = 190 \times 10^9 \text{ Pa}

To Find: The ratio of the diameter of the copper wire to the iron wire,

\frac{d_{Cu}}{d_{Fe}}

.

Formula: The elongation

\Delta L

is given by:

\Delta L = \frac{F \cdot L}{A \cdot Y}

where the cross-sectional area

A = \pi (d/2)^2 = \frac{\pi d^2}{4}

.

Calculation: Since

\Delta L_{Cu} = \Delta L_{Fe}

:

\frac{F_{Cu} \cdot L_{Cu}}{A_{Cu} \cdot Y_{Cu}} = \frac{F_{Fe} \cdot L_{Fe}}{A_{Fe} \cdot Y_{Fe}}

Given that

F_{Cu} = F_{Fe}

and

L_{Cu} = L_{Fe}

, the equation simplifies to:

\frac{1}{A_{Cu} \cdot Y_{Cu}} = \frac{1}{A_{Fe} \cdot Y_{Fe}}

A_{Cu} \cdot Y_{Cu} = A_{Fe} \cdot Y_{Fe}

Substitute the expression for area,

A = \frac{\pi d^2}{4}

:

\left(\frac{\pi d_{Cu}^2}{4}\right) \cdot Y_{Cu} = \left(\frac{\pi d_{Fe}^2}{4}\right) \cdot Y_{Fe}

Cancel

\frac{\pi}{4}

from both sides:

d_{Cu}^2 \cdot Y_{Cu} = d_{Fe}^2 \cdot Y_{Fe}

Rearrange to find the ratio of the diameters squared:

\frac{d_{Cu}^2}{d_{Fe}^2} = \frac{Y_{Fe}}{Y_{Cu}}

Take the square root of both sides:

\frac{d_{Cu}}{d_{Fe}} = \sqrt{\frac{Y_{Fe}}{Y_{Cu}}}

Substitute the values for Young's moduli:

\frac{d_{Cu}}{d_{Fe}} = \sqrt{\frac{190 \times 10^9}{110 \times 10^9}} = \sqrt{\frac{19}{11}} \approx \sqrt{1.727}

\frac{d_{Cu}}{d_{Fe}} \approx 1.314

Final Answer: The ratio of the diameter of the copper wire to that of the iron wire is approximately

1.31

.

Q11Exercises

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 $0.065 \text{ cm}^2$ . Calculate the elongation of the wire when the mass is at the lowest point of its path.

Solution

Given: Mass,

m = 14.5 \text{ kg}

Unstretched length of steel wire,

L = 1.0 \text{ m}

Angular velocity,

\omega = 2 \text{ rev/s}

Cross-sectional area,

A = 0.065 \text{ cm}^2 = 0.065 \times 10^{-4} \text{ m}^2

Young's modulus for steel,

Y_s = 2.0 \times 10^{11} \text{ Pa}

(from Table 8.1) Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

To Find: The elongation of the wire,

\Delta L

.

Formula: Elongation,

\Delta L = \frac{F \cdot L}{A \cdot Y}

. At the lowest point of the vertical circle, the total tension (

F

) in the wire is the sum of the weight of the mass (

mg

) and the centripetal force (

mr\omega^2

).

F = mg + mr\omega^2

Here, the radius of the circle

r

is the length of the wire,

L

.

Calculation: First, convert the angular velocity from rev/s to rad/s:

\omega = 2 \frac{\text{rev}}{\text{s}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 4\pi \text{ rad/s}

Next, calculate the total tension (

F

) in the wire at the lowest point:

F = mg + mL\omega^2 = m(g + L\omega^2)

F = 14.5 \text{ kg} \times (9.8 \text{ m/s}^2 + (1.0 \text{ m}) \times (4\pi \text{ rad/s})^2)

F = 14.5 \times (9.8 + 16\pi^2) \text{ N}

Using

\pi^2 \approx 9.87

:

F = 14.5 \times (9.8 + 16 \times 9.87) \text{ N}

F = 14.5 \times (9.8 + 157.92) \text{ N} = 14.5 \times 167.72 \text{ N} \approx 2432 \text{ N}

Now, calculate the elongation

\Delta L

:

\Delta L = \frac{F \cdot L}{A \cdot Y_s}

\Delta L = \frac{2432 \text{ N} \times 1.0 \text{ m}}{(0.065 \times 10^{-4} \text{ m}^2) \times (2.0 \times 10^{11} \text{ Pa})}

\Delta L = \frac{2432}{0.13 \times 10^7} \text{ m} = \frac{2432}{1.3 \times 10^6} \text{ m}

\Delta L \approx 1870.8 \times 10^{-6} \text{ m} = 1.87 \times 10^{-3} \text{ m} = 1.87 \text{ mm}

Final Answer: The elongation of the wire at the lowest point of its path is approximately

1.87 \text{ mm}

.

Q12Exercises

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm ( $1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$ ), 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.

Solution

Given: Initial volume,

V = 100.0 \text{ litre}

Final volume,

V_{final} = 100.5 \text{ litre}

Pressure increase,

\Delta p = 100.0 \text{ atm}

1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}

Note: There is a logical inconsistency in the question. An increase in pressure should cause a decrease in volume (compression). The final volume should be less than the initial volume. We will assume the change in volume is a decrease of 0.5 litre, i.e.,

\Delta V = -0.5

litre.

Change in volume,

\Delta V = V_{final} - V = (100.0 - 0.5) - 100.0 = -0.5 \text{ litre}

To Find:

Bulk modulus of water, $B_{water}$ .
Compare $B_{water}$ with $B_{air}$ .
Explain why the ratio is large.

Formula: Bulk modulus,

B = -\frac{\Delta p}{\Delta V / V}

Calculation:

Compute Bulk Modulus of Water Convert pressure increase to Pascals: $\Delta p = 100.0 \text{ atm} \times (1.013 \times 10^5 \text{ Pa/atm}) = 1.013 \times 10^7 \text{ Pa}$ Calculate the fractional change in volume: $\frac{\Delta V}{V} = \frac{-0.5 \text{ litre}}{100.0 \text{ litre}} = -0.005$ Now, compute the bulk modulus: $B_{water} = -\frac{1.013 \times 10^7 \text{ Pa}}{-0.005} = \frac{1.013 \times 10^7}{5 \times 10^{-3}} \text{ Pa}$ $B_{water} = 0.2026 \times 10^{10} \text{ Pa} = 2.026 \times 10^9 \text{ Pa}$ This value is close to the standard value given in Table 8.3 ( $2.2 \times 10^9 \text{ Pa}$ ).
Compare with Bulk Modulus of Air From Table 8.3, the bulk modulus of air at STP is $B_{air} = 1.0 \times 10^5 \text{ Pa}$ . Ratio = $\frac{B_{water}}{B_{air}} = \frac{2.026 \times 10^9 \text{ Pa}}{1.0 \times 10^5 \text{ Pa}} = 2.026 \times 10^4 \approx 20260$
Explanation for the Large Ratio The bulk modulus is a measure of a substance's incompressibility. The large ratio of $B_{water}$ to $B_{air}$ indicates that water is about 20,000 times less compressible than air. The reason for this lies in the intermolecular structure. In a liquid like water, the molecules are very close to each other and are held by strong intermolecular forces. This tight packing strongly resists any attempt to push them closer together (compression). In a gas like air, the molecules are very far apart and the intermolecular forces are negligible. There is a large amount of empty space between the molecules, making it very easy to reduce the volume by pushing the molecules closer together. Therefore, gases are highly compressible, while liquids are nearly incompressible.

Final Answer: The bulk modulus of water is approximately

2.03 \times 10^9 \text{ Pa}

. The bulk modulus of water is about 20,260 times greater than that of air. This is because the molecules in water are closely packed and resist compression, whereas the molecules in air are far apart and easily compressed.

Q13Exercises

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03 \times 10^3 \text{ kg m}^{-3}$ ?

Solution

Given: Pressure at depth,

P = 80.0 \text{ atm}

Density at the surface,

\rho_0 = 1.03 \times 10^3 \text{ kg m}^{-3}

Pressure at the surface,

P_0 = 1 \text{ atm}

(atmospheric pressure) Bulk modulus of water,

B = 2.2 \times 10^9 \text{ Pa}

(from Table 8.3)

To Find: The density of water at the given depth,

\rho

.

Formula: Bulk modulus,

B = -\frac{\Delta p}{\Delta V / V_0}

Density,

\rho = \frac{m}{V}

Calculation: Let a mass

m

of water have volume

V_0

and density

\rho_0

at the surface. At the depth, the same mass

m

has volume

V

and density

\rho

.

m = \rho_0 V_0 = \rho V \implies V = V_0 \frac{\rho_0}{\rho}

The change in volume is

\Delta V = V - V_0 = V_0 \frac{\rho_0}{\rho} - V_0 = V_0 \left(\frac{\rho_0}{\rho} - 1\right)

. The fractional change in volume is

\frac{\Delta V}{V_0} = \frac{\rho_0}{\rho} - 1

.

The increase in pressure is

\Delta p = P - P_0 = 80.0 \text{ atm} - 1 \text{ atm} = 79.0 \text{ atm}

. Convert

\Delta p

to Pascals:

\Delta p = 79.0 \times (1.013 \times 10^5 \text{ Pa}) \approx 8.0 \times 10^6 \text{ Pa}

From the bulk modulus formula:

\frac{\Delta V}{V_0} = -\frac{\Delta p}{B}

Substitute the expression for fractional volume change:

\frac{\rho_0}{\rho} - 1 = -\frac{\Delta p}{B}

\frac{\rho_0}{\rho} = 1 - \frac{\Delta p}{B}

\rho = \frac{\rho_0}{1 - \frac{\Delta p}{B}}

Now, substitute the numerical values:

\rho = \frac{1.03 \times 10^3 \text{ kg m}^{-3}}{1 - \frac{8.0 \times 10^6 \text{ Pa}}{2.2 \times 10^9 \text{ Pa}}}

\rho = \frac{1.03 \times 10^3}{1 - 0.003636} = \frac{1.03 \times 10^3}{0.996364}

\rho \approx 1.03376 \times 10^3 \text{ kg m}^{-3}

Final Answer: The density of water at the given depth is approximately

1.034 \times 10^3 \text{ kg m}^{-3}

.

Q14Exercises

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

Solution

Given: Hydraulic pressure,

p = 10 \text{ atm}

Material is glass. From Table 8.3, the bulk modulus of glass,

B = 37 \times 10^9 \text{ Pa} = 37 \text{ GPa}

.

1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}

To Find: The fractional change in volume,

\frac{\Delta V}{V}

.

Formula: The bulk modulus

B

is defined as:

B = -\frac{p}{(\Delta V / V)}

where the negative sign indicates that an increase in pressure causes a decrease in volume.

Calculation: First, convert the pressure to Pascals:

p = 10 \text{ atm} \times (1.013 \times 10^5 \text{ Pa/atm}) = 1.013 \times 10^6 \text{ Pa}

Rearrange the formula to solve for the fractional change in volume:

\frac{\Delta V}{V} = -\frac{p}{B}

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} \approx -0.02738 \times 10^{-3}

\frac{\Delta V}{V} \approx -2.74 \times 10^{-5}

Final Answer: The fractional change in volume of the glass slab is approximately

-2.74 \times 10^{-5}

. The negative sign indicates a decrease in volume.

Q15Exercises

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^6 \text{ Pa}$ .

Solution

Given: Edge length of the copper cube,

L = 10 \text{ cm} = 0.1 \text{ m}

Hydraulic pressure,

p = 7.0 \times 10^6 \text{ Pa}

Material is copper. From Table 8.3, the bulk modulus of copper,

B = 140 \times 10^9 \text{ Pa} = 140 \text{ GPa}

.

To Find: The volume contraction (decrease in volume),

\Delta V

.

Formula: The bulk modulus

B

is defined as:

B = -\frac{p}{(\Delta V / V)}

Calculation: First, calculate the initial volume of the cube:

V = L^3 = (0.1 \text{ m})^3 = 0.001 \text{ m}^3 = 10^{-3} \text{ m}^3

Rearrange the formula for bulk modulus to solve for the volume contraction,

\Delta V

:

\Delta V = -\frac{p \cdot V}{B}

Substitute the given values:

\Delta V = -\frac{(7.0 \times 10^6 \text{ Pa}) \times (10^{-3} \text{ m}^3)}{140 \times 10^9 \text{ Pa}}

\Delta V = -\frac{7.0 \times 10^3}{140 \times 10^9} \text{ m}^3

\Delta V = -\frac{7.0}{140} \times 10^{-6} \text{ m}^3 = -0.05 \times 10^{-6} \text{ m}^3

\Delta V = -5.0 \times 10^{-8} \text{ m}^3

The volume contraction is the magnitude of

\Delta V

, which is

5.0 \times 10^{-8} \text{ m}^3

. To express this in cubic centimeters (

cm^3

):

5.0 \times 10^{-8} \text{ m}^3 \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^3 = 5.0 \times 10^{-8} \times 10^6 \text{ cm}^3 = 0.05 \text{ cm}^3

Final Answer: The volume contraction of the solid copper cube is

5.0 \times 10^{-8} \text{ m}^3

(or

0.05 \text{ cm}^3

).

Q16Exercises

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Solution

Given: Initial volume,

V = 1 \text{ litre}

Percentage compression =

0.10\%

This means the fractional change in volume is negative.

\frac{\Delta V}{V} = -\frac{0.10}{100} = -0.001 = -10^{-3}

Material is water. From Table 8.3, the bulk modulus of water,

B = 2.2 \times 10^9 \text{ Pa}

.

To Find: The change in pressure,

\Delta p

, required.

Formula: The bulk modulus

B

is defined as:

B = -\frac{\Delta p}{(\Delta V / V)}

Calculation: Rearrange the formula to solve for the change in pressure,

\Delta p

:

\Delta p = -B \times \left(\frac{\Delta V}{V}\right)

Substitute the given values:

\Delta p = -(2.2 \times 10^9 \text{ Pa}) \times (-10^{-3})

\Delta p = 2.2 \times 10^6 \text{ Pa}

Final Answer: The pressure on a litre of water should be changed (increased) by

2.2 \times 10^6 \text{ Pa}

to compress it by 0.10%.

Mechanical Properties Of Solids MECHANICAL PROPERTIES OF SOLIDSClass 11 Physics NCERT Solutions

Solution

Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?

Solution

The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale.

Solution

Read the following two statements below carefully and state, with reasons, if it is true or false.

Solution

Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.

Solution

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?

Solution

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.

Solution

A piece of copper having a rectangular cross-section of 15.2 mm×19.1 mm15.2 \text{ mm} \times 19.1 \text{ mm}15.2 mm×19.1 mm is pulled in tension with 44,500 N44,500 \text{ N}44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Solution

A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m−210^8 \text{ N m}^{-2}108 N m−2, what is the maximum load the cable can support ?

Solution

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.

Solution

Solution

Solution

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03×103 kg m−31.03 \times 10^3 \text{ kg m}^{-3}1.03×103 kg m−3?

Solution

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

Solution

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0×106 Pa7.0 \times 10^6 \text{ Pa}7.0×106 Pa.

Solution

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Solution

A piece of copper having a rectangular cross-section of $15.2 \text{ mm} \times 19.1 \text{ mm}$ is pulled in tension with $44,500 \text{ N}$ force, producing only elastic deformation. Calculate the resulting strain?

A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed $10^8 \text{ N m}^{-2}$ , what is the maximum load the cable can support ?

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03 \times 10^3 \text{ kg m}^{-3}$ ?

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^6 \text{ Pa}$ .