Physics of Solids Numerical Problems - Chapter 17 of FSc/ICS Physics

Physics of Solids Numerical Problems - Chapter 17 | Physics Notes

Physics of Solids Numerical Problems - Chapter 17

Problem 17.1: A 1.25 cm diameter cylinder is subjected to a load of 2500 kg. Calculate the stress on the bar in mega pascals.

Given Data:

Diameter of Cylinder \( d = 1.25 \, \text{cm} = 0.0125 \, \text{m} \)
Load \( m = 2500 \, \text{kg} \)
Gravitational Acceleration \( g = 9.8 \, \text{m/s}^2 \)

To Determine:

Stress \( \sigma = ? \) (in MPa)

Calculations:

\[ \sigma = \frac{F}{A} = \frac{mg}{\left( \frac{\pi d^2}{4} \right)} = \frac{4mg}{\pi d^2} \]

\[ = \frac{4 \times 2500 \times 9.8}{3.14 \times (0.0125)^2} \]

\[ = \frac{4 \times 2500 \times 9.8}{3.14 \times 0.00015625} \]

\[ = \frac{98000}{0.000490625} \]

\[ = 199.7 \times 10^6 \, \text{Pa} \]

\[ = 199.7 \, \text{MPa} \approx 200 \, \text{MPa} \]

Problem 17.2: A 1.0 m long copper wire is subjected to stretching force and its length increases by 20 cm. Calculate the tensile strain and the percent elongation which the wire undergoes.

Given Data:

Original Length \( l = 1.0 \, \text{m} \)
Elongation \( \Delta l = 20 \, \text{cm} = 0.2 \, \text{m} \)

To Determine:

Tensile Strain = ?
Percent Elongation = ?

Calculations:

(a) Tensile Strain:

\[ \text{Tensile Strain} = \frac{\Delta l}{l} \]

\[ = \frac{0.2}{1} \]

\[ = 0.2 \]

(b) Percent Elongation:

\[ \% \text{Elongation} = \frac{\Delta l}{l} \times 100 \]

\[ = \frac{0.2}{1} \times 100 \]

\[ = 20\% \]

Problem 17.3: A wire 2.5 m long and cross-section area \( 10^{-5} \, m^2 \) is stretched 1.5 mm by a force of 100 N in the elastic region. Calculate (i) the strain (ii) Young's modulus (iii) the energy stored in the wire.

Given Data:

Length \( l = 2.5 \, \text{m} \)
Cross-Section Area \( A = 10^{-5} \, \text{m}^2 \)
Force \( F = 100 \, \text{N} \)
Elongation \( \Delta l = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \)

To Determine:

Strain = ?
Young's Modulus \( Y = ? \)
Energy stored in the wire = ?

Calculations:

(i) Strain:

\[ \text{Strain} = \frac{\Delta l}{l} \]

\[ = \frac{1.5 \times 10^{-3}}{2.5} \]

\[ = 6 \times 10^{-4} \]

(ii) Young's Modulus:

\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} \]

\[ = \frac{100 / 10^{-5}}{6 \times 10^{-4}} \]

\[ = \frac{10^7}{6 \times 10^{-4}} \]

\[ = 1.666 \times 10^{10} \, \text{Pa} \]

\[ \approx 1.67 \times 10^{10} \, \text{Pa} \]

(iii) Energy stored in the wire:

\[ \text{Energy} = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume} \]

\[ = \frac{1}{2} \times \frac{F}{A} \times \frac{\Delta l}{l} \times A \times l \]

\[ = \frac{1}{2} F \Delta l \]

\[ = \frac{1}{2} \times 100 \times 1.5 \times 10^{-3} \]

\[ = 0.075 \, \text{J} \]

\[ = 7.5 \times 10^{-2} \, \text{J} \]

Problem 17.4: What stress would cause a wire to increase in length by 0.01% if the Young's modulus of the wire is \( 12 \times 10^{10} \, \text{Pa} \)? What force would produce this stress if the diameter of the wire is 0.56 mm?

Given Data:

Percent Elongation = 0.01%
Young's Modulus \( Y = 12 \times 10^{10} \, \text{Pa} \)
Diameter \( d = 0.56 \, \text{mm} = 0.56 \times 10^{-3} \, \text{m} \)

To Determine:

Stress = ?
Force \( F = ? \)

Calculations:

First, calculate the strain:

\[ \% \text{Elongation} = 0.01\% \]

\[ \Rightarrow \frac{\Delta l}{l} \times 100 = 0.01 \]

\[ \Rightarrow \frac{\Delta l}{l} = \frac{0.01}{100} = 10^{-4} \]

Now calculate the stress:

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

\[ \Rightarrow \text{Stress} = Y \times \text{Strain} \]

\[ = 12 \times 10^{10} \times 10^{-4} \]

\[ = 12 \times 10^6 \, \text{Pa} \]

Now calculate the force:

\[ \text{Stress} = \frac{F}{A} \]

\[ \Rightarrow F = \text{Stress} \times A \]

\[ = \text{Stress} \times \frac{\pi d^2}{4} \]

\[ = 12 \times 10^6 \times \frac{3.14 \times (0.56 \times 10^{-3})^2}{4} \]

\[ = 12 \times 10^6 \times \frac{3.14 \times 3.136 \times 10^{-7}}{4} \]

\[ = 12 \times 10^6 \times \frac{9.847 \times 10^{-7}}{4} \]

\[ = 12 \times 10^6 \times 2.46175 \times 10^{-7} \]

\[ = 2.9541 \, \text{N} \]

\[ \approx 2.95 \, \text{N} \]

Problem 17.5: The length of a steel wire is 1.0 m and its cross-sectional area is \( 0.01 \times 10^{-4} \, m^2 \). Calculate the work done in stretching the wire when a force of 100 N is applied within the elastic region. Young's modulus of steel is \( 2.0 \times 10^{11} \, Pa \).

Given Data:

Length \( l = 1.0 \, \text{m} \)
Cross-Sectional Area \( A = 0.01 \times 10^{-4} \, \text{m}^2 = 10^{-6} \, \text{m}^2 \)
Force \( F = 100 \, \text{N} \)
Young's Modulus \( Y = 2.0 \times 10^{11} \, \text{Pa} \)

To Determine:

Work Done = ?

Calculations:

First, find the elongation \( \Delta l \):

\[ Y = \frac{F/A}{\Delta l/l} \]

\[ \Rightarrow \Delta l = \frac{F \cdot l}{Y \cdot A} \]

\[ = \frac{100 \times 1.0}{2.0 \times 10^{11} \times 10^{-6}} \]

\[ = \frac{100}{2.0 \times 10^5} \]

\[ = 5 \times 10^{-4} \, \text{m} \]

Now calculate the work done:

\[ \text{Work} = \frac{1}{2} F \Delta l \]

\[ = \frac{1}{2} \times 100 \times 5 \times 10^{-4} \]

\[ = 0.025 \, \text{J} \]

Problem 17.6: A cylindrical copper wire and a cylindrical steel wire each of length 1.5 m and diameter 2.0 mm are joined at one end to form a composite wire 3.0 m long. The wire is loaded until its length becomes 3.003 m. Calculate the strain in copper and steel wires and the force applied to the wire. (Young's modulus of copper is \( 1.2 \times 10^{11} \, \text{Pa} \) and for steel is \( 2.0 \times 10^{11} \, \text{Pa} \)).

Given Data:

Copper Wire: Length \( l_c = 1.5 \, \text{m} \), Diameter \( d = 2.0 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
Steel Wire: Length \( l_s = 1.5 \, \text{m} \), Diameter \( d = 2.0 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
Composite Length = 3.0 m
Final Length = 3.003 m
Total Elongation \( \Delta l_{\text{total}} = 0.003 \, \text{m} \)
Young's Modulus for Copper \( Y_c = 1.2 \times 10^{11} \, \text{Pa} \)
Young's Modulus for Steel \( Y_s = 2.0 \times 10^{11} \, \text{Pa} \)

To Determine:

Strain in Copper Wire = ?
Strain in Steel Wire = ?
Force Applied = ?

Calculations:

Let \( \Delta l_c \) = elongation in copper wire

Let \( \Delta l_s \) = elongation in steel wire

\[ \Delta l_c + \Delta l_s = 0.003 \, \text{m} \quad \text{(1)} \]

Since both wires have the same cross-sectional area and experience the same force:

\[ \text{Stress} = \frac{F}{A} \]

For copper: \( Y_c = \frac{F/A}{\Delta l_c / l_c} \Rightarrow \frac{F}{A} = Y_c \cdot \frac{\Delta l_c}{l_c} \)

For steel: \( Y_s = \frac{F/A}{\Delta l_s / l_s} \Rightarrow \frac{F}{A} = Y_s \cdot \frac{\Delta l_s}{l_s} \)

Since stress is equal: \( Y_c \cdot \frac{\Delta l_c}{l_c} = Y_s \cdot \frac{\Delta l_s}{l_s} \)

\[ \Rightarrow 1.2 \times 10^{11} \cdot \frac{\Delta l_c}{1.5} = 2.0 \times 10^{11} \cdot \frac{\Delta l_s}{1.5} \]

\[ \Rightarrow 1.2 \Delta l_c = 2.0 \Delta l_s \]

\[ \Rightarrow \Delta l_c = \frac{2.0}{1.2} \Delta l_s = \frac{5}{3} \Delta l_s \quad \text{(2)} \]

Substitute (2) into (1):

\[ \frac{5}{3} \Delta l_s + \Delta l_s = 0.003 \]

\[ \frac{8}{3} \Delta l_s = 0.003 \]

\[ \Delta l_s = 0.003 \times \frac{3}{8} = 0.001125 \, \text{m} \]

\[ \Delta l_c = \frac{5}{3} \times 0.001125 = 0.001875 \, \text{m} \]

Now calculate strains:

Strain in copper wire = \( \frac{\Delta l_c}{l_c} = \frac{0.001875}{1.5} = 0.00125 \)

Strain in steel wire = \( \frac{\Delta l_s}{l_s} = \frac{0.001125}{1.5} = 0.00075 \)

Calculate cross-sectional area:

\[ A = \frac{\pi d^2}{4} = \frac{3.14 \times (2 \times 10^{-3})^2}{4} = 3.14 \times 10^{-6} \, \text{m}^2 \]

Physics of Solids Numerical Problems - Chapter 17 | Physics Notes