This MCQ module is based on: NCERT Exercises and Solutions: Mechanical Properties of Solids
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NCERT Exercises and Solutions: Mechanical Properties of Solids
🎓 Class 11
Physics
CBSE
Theory
Ch 8 – Mechanical Properties of Solids
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Mechanical Properties of Solids
Chapter Summary — Key Formulas
| Quantity | Formula | Unit |
|---|---|---|
| Stress | \(\sigma = F/A\) | Pa (N/m²) |
| Longitudinal Strain | \(\varepsilon = \Delta L / L\) | Dimensionless |
| Volumetric Strain | \(\varepsilon_V = \Delta V / V\) | Dimensionless |
| Shearing Strain | \(\gamma = \Delta x / L = \tan\theta \approx \theta\) | Dimensionless (rad) |
| Young's Modulus | \(Y = \frac{FL}{A \cdot \Delta L}\) | Pa (solids only) |
| Bulk Modulus | \(B = \frac{-pV}{\Delta V}\) | Pa (all states) |
| Shear Modulus | \(G = \frac{F/A}{\Delta x / L}\) | Pa (solids only) |
| Compressibility | \(k = 1/B\) | Pa&supmin;¹ |
| Poisson's Ratio | \(\sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}}\) | Dimensionless (0 to 0.5) |
| Elastic Energy | \(U = \frac{1}{2} F \cdot \Delta L = \frac{1}{2} \sigma \varepsilon V\) | J |
| Energy Density | \(u = \frac{1}{2} \sigma \varepsilon = \frac{1}{2} Y \varepsilon^2\) | J/m³ |
Keywords
Elasticity
Plasticity
Stress
Strain
Hooke's Law
Young's Modulus
Bulk Modulus
Shear Modulus
Poisson's Ratio
Elastic Limit
Yield Point
Fracture Point
Ductile
Brittle
Compressibility
Elastic Energy
Energy Density
I-Beam
NCERT Exercises
Exercise 8.1
A steel wire of length 4.7 m and cross-sectional area \(3.0 \times 10^{-5}\) m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area \(4.0 \times 10^{-5}\) m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Given:
Steel: \(L_s = 4.7\) m, \(A_s = 3.0 \times 10^{-5}\) m²
Copper: \(L_c = 3.5\) m, \(A_c = 4.0 \times 10^{-5}\) m²
Same force \(F\) and same extension \(\Delta L\).
Solution:
Since \(\Delta L = \frac{FL}{AY}\), and \(\Delta L\) is the same for both: \[\frac{FL_s}{A_s Y_s} = \frac{FL_c}{A_c Y_c}\] \[\frac{L_s}{A_s Y_s} = \frac{L_c}{A_c Y_c}\] \[\frac{Y_s}{Y_c} = \frac{L_s \cdot A_c}{L_c \cdot A_s} = \frac{4.7 \times 4.0 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}} = \frac{18.8}{10.5} = 1.79\] Answer: \(\boxed{\frac{Y_s}{Y_c} \approx 1.79}\). Steel has about 1.8 times the Young's modulus of copper.
Steel: \(L_s = 4.7\) m, \(A_s = 3.0 \times 10^{-5}\) m²
Copper: \(L_c = 3.5\) m, \(A_c = 4.0 \times 10^{-5}\) m²
Same force \(F\) and same extension \(\Delta L\).
Solution:
Since \(\Delta L = \frac{FL}{AY}\), and \(\Delta L\) is the same for both: \[\frac{FL_s}{A_s Y_s} = \frac{FL_c}{A_c Y_c}\] \[\frac{L_s}{A_s Y_s} = \frac{L_c}{A_c Y_c}\] \[\frac{Y_s}{Y_c} = \frac{L_s \cdot A_c}{L_c \cdot A_s} = \frac{4.7 \times 4.0 \times 10^{-5}}{3.5 \times 3.0 \times 10^{-5}} = \frac{18.8}{10.5} = 1.79\] Answer: \(\boxed{\frac{Y_s}{Y_c} \approx 1.79}\). Steel has about 1.8 times the Young's modulus of copper.
Exercise 8.2
A wire is stretched by a force of 100 N. If the original diameter is 0.25 cm and the extension is 0.025 cm over a length of 1.0 m, find Young's modulus of the material.
Given:
\(F = 100\) N, \(d = 0.25\) cm \(= 2.5 \times 10^{-3}\) m, \(r = 1.25 \times 10^{-3}\) m
\(\Delta L = 0.025\) cm \(= 2.5 \times 10^{-4}\) m, \(L = 1.0\) m
Step 1: Area \[A = \pi r^2 = \pi \times (1.25 \times 10^{-3})^2 = \pi \times 1.5625 \times 10^{-6} = 4.909 \times 10^{-6} \text{ m}^2\] Step 2: Young's Modulus \[Y = \frac{FL}{A \cdot \Delta L} = \frac{100 \times 1.0}{4.909 \times 10^{-6} \times 2.5 \times 10^{-4}}\] \[= \frac{100}{1.227 \times 10^{-9}} = 8.15 \times 10^{10} \text{ Pa}\] Answer: \(\boxed{Y \approx 8.15 \times 10^{10} \text{ Pa} \approx 81.5 \text{ GPa}}\)
\(F = 100\) N, \(d = 0.25\) cm \(= 2.5 \times 10^{-3}\) m, \(r = 1.25 \times 10^{-3}\) m
\(\Delta L = 0.025\) cm \(= 2.5 \times 10^{-4}\) m, \(L = 1.0\) m
Step 1: Area \[A = \pi r^2 = \pi \times (1.25 \times 10^{-3})^2 = \pi \times 1.5625 \times 10^{-6} = 4.909 \times 10^{-6} \text{ m}^2\] Step 2: Young's Modulus \[Y = \frac{FL}{A \cdot \Delta L} = \frac{100 \times 1.0}{4.909 \times 10^{-6} \times 2.5 \times 10^{-4}}\] \[= \frac{100}{1.227 \times 10^{-9}} = 8.15 \times 10^{10} \text{ Pa}\] Answer: \(\boxed{Y \approx 8.15 \times 10^{10} \text{ Pa} \approx 81.5 \text{ GPa}}\)
Exercise 8.3
The stress-strain graphs for materials A and B are shown below. The graphs are drawn to the same scale.
(a) Which material has the greater Young's modulus?
(b) Which material is stronger?
(a) Greater Young's modulus:
Young's modulus is the slope of the initial linear portion of the stress-strain curve. Material A has a steeper initial slope than B (for the same strain, A has higher stress). Therefore, Material A has the greater Young's modulus.
(b) Stronger material:
Strength is measured by the ultimate tensile stress — the maximum stress the material can withstand before fracture. From the graphs, Material A reaches a higher maximum stress before fracture than Material B. Therefore, Material A is stronger.
However, Material B has a larger fracture strain — it can undergo more deformation before breaking. So B is more ductile while A is stiffer and stronger.
Young's modulus is the slope of the initial linear portion of the stress-strain curve. Material A has a steeper initial slope than B (for the same strain, A has higher stress). Therefore, Material A has the greater Young's modulus.
(b) Stronger material:
Strength is measured by the ultimate tensile stress — the maximum stress the material can withstand before fracture. From the graphs, Material A reaches a higher maximum stress before fracture than Material B. Therefore, Material A is stronger.
However, Material B has a larger fracture strain — it can undergo more deformation before breaking. So B is more ductile while A is stiffer and stronger.
Exercise 8.4
Read the following two statements below carefully and state, with reasons, if each 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.
(a) False. Young's modulus of steel (≈ 200 GPa) is far greater than that of rubber (≈ 0.01 GPa). Steel resists longitudinal deformation much more than rubber. The common misconception arises because rubber stretches more easily — but this means it has a lower modulus, not a higher one.
(b) True. When a helical coil spring is stretched, the wire of the coil is actually subjected to a torsional (twisting) deformation, not simple elongation. Torsion is a form of shearing, so the relevant modulus is the shear modulus (modulus of rigidity), not Young's modulus.
(b) True. When a helical coil spring is stretched, the wire of the coil is actually subjected to a torsional (twisting) deformation, not simple elongation. Torsion is a form of shearing, so the relevant modulus is the shear modulus (modulus of rigidity), not Young's modulus.
Exercise 8.5
Two wires of diameter 0.25 cm, one made of steel and the other of brass, are loaded as shown in the figure. The unloaded length of the steel wire is 1.5 m and that of brass is 1.0 m. Compute the elongation of the steel and brass wires. Given: \(Y_{\text{steel}} = 2.0 \times 10^{11}\) Pa, \(Y_{\text{brass}} = 0.91 \times 10^{11}\) Pa.
Given:
Both wires: \(d = 0.25\) cm \(= 2.5 \times 10^{-3}\) m, \(r = 1.25 \times 10^{-3}\) m
\(A = \pi r^2 = \pi \times (1.25 \times 10^{-3})^2 = 4.909 \times 10^{-6}\) m²
Forces on each wire:
The brass wire supports only the 6 kg load below it: \(F_{\text{brass}} = 6 \times 9.8 = 58.8\) N
The steel wire supports both the 4 kg load at the junction AND the entire brass wire + 6 kg load:
\(F_{\text{steel}} = (4 + 6) \times 9.8 = 10 \times 9.8 = 98\) N
Elongation of steel wire: \[\Delta L_s = \frac{F_s L_s}{A Y_s} = \frac{98 \times 1.5}{4.909 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{147}{9.818 \times 10^{5}} = 1.497 \times 10^{-4} \text{ m}\] \[\boxed{\Delta L_s \approx 1.50 \times 10^{-4} \text{ m} = 0.15 \text{ mm}}\] Elongation of brass wire: \[\Delta L_b = \frac{F_b L_b}{A Y_b} = \frac{58.8 \times 1.0}{4.909 \times 10^{-6} \times 0.91 \times 10^{11}} = \frac{58.8}{4.467 \times 10^{5}} = 1.316 \times 10^{-4} \text{ m}\] \[\boxed{\Delta L_b \approx 1.32 \times 10^{-4} \text{ m} = 0.13 \text{ mm}}\]
Both wires: \(d = 0.25\) cm \(= 2.5 \times 10^{-3}\) m, \(r = 1.25 \times 10^{-3}\) m
\(A = \pi r^2 = \pi \times (1.25 \times 10^{-3})^2 = 4.909 \times 10^{-6}\) m²
Forces on each wire:
The brass wire supports only the 6 kg load below it: \(F_{\text{brass}} = 6 \times 9.8 = 58.8\) N
The steel wire supports both the 4 kg load at the junction AND the entire brass wire + 6 kg load:
\(F_{\text{steel}} = (4 + 6) \times 9.8 = 10 \times 9.8 = 98\) N
Elongation of steel wire: \[\Delta L_s = \frac{F_s L_s}{A Y_s} = \frac{98 \times 1.5}{4.909 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{147}{9.818 \times 10^{5}} = 1.497 \times 10^{-4} \text{ m}\] \[\boxed{\Delta L_s \approx 1.50 \times 10^{-4} \text{ m} = 0.15 \text{ mm}}\] Elongation of brass wire: \[\Delta L_b = \frac{F_b L_b}{A Y_b} = \frac{58.8 \times 1.0}{4.909 \times 10^{-6} \times 0.91 \times 10^{11}} = \frac{58.8}{4.467 \times 10^{5}} = 1.316 \times 10^{-4} \text{ m}\] \[\boxed{\Delta L_b \approx 1.32 \times 10^{-4} \text{ m} = 0.13 \text{ mm}}\]
Exercise 8.6
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 applied to the opposite face. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Given:
Edge \(L = 10\) cm \(= 0.10\) m
Area of face: \(A = L^2 = (0.10)^2 = 0.01\) m²
\(m = 100\) kg, \(F = mg = 100 \times 9.8 = 980\) N
\(G = 25 \times 10^9\) Pa
Using: \(G = \frac{F/A}{\Delta x / L}\), so \(\Delta x = \frac{FL}{AG}\) \[\Delta x = \frac{980 \times 0.10}{0.01 \times 25 \times 10^9} = \frac{98}{2.5 \times 10^8} = 3.92 \times 10^{-7} \text{ m}\] Answer: \(\boxed{\Delta x = 3.92 \times 10^{-7} \text{ m} = 0.392 \,\mu\text{m}}\).
The deflection is extremely small — aluminium is quite rigid against shear.
Edge \(L = 10\) cm \(= 0.10\) m
Area of face: \(A = L^2 = (0.10)^2 = 0.01\) m²
\(m = 100\) kg, \(F = mg = 100 \times 9.8 = 980\) N
\(G = 25 \times 10^9\) Pa
Using: \(G = \frac{F/A}{\Delta x / L}\), so \(\Delta x = \frac{FL}{AG}\) \[\Delta x = \frac{980 \times 0.10}{0.01 \times 25 \times 10^9} = \frac{98}{2.5 \times 10^8} = 3.92 \times 10^{-7} \text{ m}\] Answer: \(\boxed{\Delta x = 3.92 \times 10^{-7} \text{ m} = 0.392 \,\mu\text{m}}\).
The deflection is extremely small — aluminium is quite rigid against shear.
Exercise 8.7
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 to be distributed equally, calculate the compressive strain of each column. Given: \(Y_{\text{steel}} = 2.0 \times 10^{11}\) Pa.
Given:
Total mass: \(M = 50{,}000\) kg. Load per column: \(m = 50{,}000/4 = 12{,}500\) kg
\(F = mg = 12{,}500 \times 9.8 = 1.225 \times 10^5\) N
Inner radius: \(r_1 = 0.30\) m, Outer radius: \(r_2 = 0.60\) m
Cross-sectional area (hollow cylinder): \[A = \pi(r_2^2 - r_1^2) = \pi(0.36 - 0.09) = \pi \times 0.27 = 0.8482 \text{ m}^2\] Compressive stress: \[\sigma = \frac{F}{A} = \frac{1.225 \times 10^5}{0.8482} = 1.444 \times 10^5 \text{ Pa}\] Compressive strain: \[\varepsilon = \frac{\sigma}{Y} = \frac{1.444 \times 10^5}{2.0 \times 10^{11}} = 7.22 \times 10^{-7}\] Answer: \(\boxed{\varepsilon = 7.22 \times 10^{-7}}\). This is an extremely small strain, confirming that steel columns can easily support large loads with negligible deformation.
Total mass: \(M = 50{,}000\) kg. Load per column: \(m = 50{,}000/4 = 12{,}500\) kg
\(F = mg = 12{,}500 \times 9.8 = 1.225 \times 10^5\) N
Inner radius: \(r_1 = 0.30\) m, Outer radius: \(r_2 = 0.60\) m
Cross-sectional area (hollow cylinder): \[A = \pi(r_2^2 - r_1^2) = \pi(0.36 - 0.09) = \pi \times 0.27 = 0.8482 \text{ m}^2\] Compressive stress: \[\sigma = \frac{F}{A} = \frac{1.225 \times 10^5}{0.8482} = 1.444 \times 10^5 \text{ Pa}\] Compressive strain: \[\varepsilon = \frac{\sigma}{Y} = \frac{1.444 \times 10^5}{2.0 \times 10^{11}} = 7.22 \times 10^{-7}\] Answer: \(\boxed{\varepsilon = 7.22 \times 10^{-7}}\). This is an extremely small strain, confirming that steel columns can easily support large loads with negligible deformation.
Exercise 8.8
A piece of copper having a rectangular cross-section of \(15.2\) mm × \(19.1\) mm is pulled in tension with a force of \(44{,}500\) N, producing only elastic deformation. Calculate the resulting strain. Given: \(Y_{\text{Cu}} = 1.1 \times 10^{11}\) Pa.
Given:
Cross-section: \(15.2 \times 10^{-3}\) m \(\times\) \(19.1 \times 10^{-3}\) m
\(A = 15.2 \times 10^{-3} \times 19.1 \times 10^{-3} = 2.903 \times 10^{-4}\) m²
\(F = 44{,}500\) N, \(Y = 1.1 \times 10^{11}\) Pa
Stress: \[\sigma = \frac{F}{A} = \frac{44{,}500}{2.903 \times 10^{-4}} = 1.533 \times 10^{8} \text{ Pa}\] Strain: \[\varepsilon = \frac{\sigma}{Y} = \frac{1.533 \times 10^{8}}{1.1 \times 10^{11}} = 1.39 \times 10^{-3}\] Answer: \(\boxed{\varepsilon = 1.39 \times 10^{-3}}\)
Cross-section: \(15.2 \times 10^{-3}\) m \(\times\) \(19.1 \times 10^{-3}\) m
\(A = 15.2 \times 10^{-3} \times 19.1 \times 10^{-3} = 2.903 \times 10^{-4}\) m²
\(F = 44{,}500\) N, \(Y = 1.1 \times 10^{11}\) Pa
Stress: \[\sigma = \frac{F}{A} = \frac{44{,}500}{2.903 \times 10^{-4}} = 1.533 \times 10^{8} \text{ Pa}\] Strain: \[\varepsilon = \frac{\sigma}{Y} = \frac{1.533 \times 10^{8}}{1.1 \times 10^{11}} = 1.39 \times 10^{-3}\] Answer: \(\boxed{\varepsilon = 1.39 \times 10^{-3}}\)
Exercise 8.9
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 \(10^8\) Pa, what is the maximum load the cable can support?
Given:
\(r = 1.5\) cm \(= 0.015\) m
Maximum stress: \(\sigma_{\max} = 10^8\) Pa
Area: \[A = \pi r^2 = \pi \times (0.015)^2 = \pi \times 2.25 \times 10^{-4} = 7.069 \times 10^{-4} \text{ m}^2\] Maximum load: \[F = \sigma_{\max} \times A = 10^8 \times 7.069 \times 10^{-4} = 7.069 \times 10^{4} \text{ N}\] Answer: \(\boxed{F_{\max} = 7.07 \times 10^4 \text{ N} \approx 70.7 \text{ kN}}\).
This can support a mass of approximately \(m = F/g = 70{,}700/9.8 \approx 7{,}214\) kg.
\(r = 1.5\) cm \(= 0.015\) m
Maximum stress: \(\sigma_{\max} = 10^8\) Pa
Area: \[A = \pi r^2 = \pi \times (0.015)^2 = \pi \times 2.25 \times 10^{-4} = 7.069 \times 10^{-4} \text{ m}^2\] Maximum load: \[F = \sigma_{\max} \times A = 10^8 \times 7.069 \times 10^{-4} = 7.069 \times 10^{4} \text{ N}\] Answer: \(\boxed{F_{\max} = 7.07 \times 10^4 \text{ N} \approx 70.7 \text{ kN}}\).
This can support a mass of approximately \(m = F/g = 70{,}700/9.8 \approx 7{,}214\) kg.
Exercise 8.10
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are copper and the middle one is iron. Determine the ratios of their diameters if each is to have the same tension. Given: \(Y_{\text{Cu}} = 1.2 \times 10^{11}\) Pa, \(Y_{\text{Fe}} = 1.9 \times 10^{11}\) Pa.
Given: All three wires have same length \(L = 2.0\) m and same tension \(T\).
Since the bar is rigid and supported symmetrically, for each wire to carry the same tension, they must also have the same extension \(\Delta L\) (otherwise the rigid bar would tilt).
Condition: Same \(F\) (tension) and same \(\Delta L\) for all wires.
\[\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y} = \frac{4FL}{\pi d^2 Y}\] For equal \(\Delta L\) with same \(F\) and \(L\): \[\frac{4FL}{\pi d_{\text{Cu}}^2 Y_{\text{Cu}}} = \frac{4FL}{\pi d_{\text{Fe}}^2 Y_{\text{Fe}}}\] \[d_{\text{Cu}}^2 Y_{\text{Cu}} = d_{\text{Fe}}^2 Y_{\text{Fe}}\] \[\frac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\frac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} = \sqrt{\frac{1.9 \times 10^{11}}{1.2 \times 10^{11}}} = \sqrt{1.583} = 1.258\] Answer: \(\boxed{\frac{d_{\text{Cu}}}{d_{\text{Fe}}} = 1.26}\). The copper wires must be about 26% thicker in diameter than the iron wire so that all three have equal extensions under equal tension.
Since the bar is rigid and supported symmetrically, for each wire to carry the same tension, they must also have the same extension \(\Delta L\) (otherwise the rigid bar would tilt).
Condition: Same \(F\) (tension) and same \(\Delta L\) for all wires.
\[\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y} = \frac{4FL}{\pi d^2 Y}\] For equal \(\Delta L\) with same \(F\) and \(L\): \[\frac{4FL}{\pi d_{\text{Cu}}^2 Y_{\text{Cu}}} = \frac{4FL}{\pi d_{\text{Fe}}^2 Y_{\text{Fe}}}\] \[d_{\text{Cu}}^2 Y_{\text{Cu}} = d_{\text{Fe}}^2 Y_{\text{Fe}}\] \[\frac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\frac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} = \sqrt{\frac{1.9 \times 10^{11}}{1.2 \times 10^{11}}} = \sqrt{1.583} = 1.258\] Answer: \(\boxed{\frac{d_{\text{Cu}}}{d_{\text{Fe}}} = 1.26}\). The copper wires must be about 26% thicker in diameter than the iron wire so that all three have equal extensions under equal tension.
Exercise 8.11
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 \(0.065\) cm². Calculate the elongation of the wire when the mass is at the lowest point of the path. Given: \(Y_{\text{steel}} = 2 \times 10^{11}\) Pa.
Given:
\(m = 14.5\) kg, \(L = 1.0\) m, \(A = 0.065\) cm² \(= 0.065 \times 10^{-4}\) m² \(= 6.5 \times 10^{-6}\) m²
\(\omega = 2\) rev/s \(= 2 \times 2\pi\) rad/s \(= 4\pi\) rad/s
\(Y = 2 \times 10^{11}\) Pa
At the lowest point of the vertical circle:
The wire must support both the weight and provide centripetal force (both directed upward along the wire toward the centre):
\[T = mg + m\omega^2 L\] \[T = 14.5 \times 9.8 + 14.5 \times (4\pi)^2 \times 1.0\] \[T = 142.1 + 14.5 \times 16\pi^2\] \[T = 142.1 + 14.5 \times 157.91\] \[T = 142.1 + 2289.7 = 2431.8 \text{ N}\] Elongation: \[\Delta L = \frac{TL}{AY} = \frac{2431.8 \times 1.0}{6.5 \times 10^{-6} \times 2 \times 10^{11}} = \frac{2431.8}{1.3 \times 10^{6}}\] \[\Delta L = 1.87 \times 10^{-3} \text{ m} = 1.87 \text{ mm}\] Answer: \(\boxed{\Delta L \approx 1.87 \text{ mm}}\).
Note: The centripetal force (2289.7 N) is much larger than the weight (142.1 N), which is why the wire elongation is significant. We should verify that the total stress (2431.8 / 6.5×10&supmin;&sup6; = 3.74×10&sup8; Pa) is below the elastic limit of steel (∼250 MPa for mild steel), and indeed 374 MPa exceeds this, suggesting the wire may undergo permanent deformation. In practice, high-tensile steel would be used.
\(m = 14.5\) kg, \(L = 1.0\) m, \(A = 0.065\) cm² \(= 0.065 \times 10^{-4}\) m² \(= 6.5 \times 10^{-6}\) m²
\(\omega = 2\) rev/s \(= 2 \times 2\pi\) rad/s \(= 4\pi\) rad/s
\(Y = 2 \times 10^{11}\) Pa
At the lowest point of the vertical circle:
The wire must support both the weight and provide centripetal force (both directed upward along the wire toward the centre):
\[T = mg + m\omega^2 L\] \[T = 14.5 \times 9.8 + 14.5 \times (4\pi)^2 \times 1.0\] \[T = 142.1 + 14.5 \times 16\pi^2\] \[T = 142.1 + 14.5 \times 157.91\] \[T = 142.1 + 2289.7 = 2431.8 \text{ N}\] Elongation: \[\Delta L = \frac{TL}{AY} = \frac{2431.8 \times 1.0}{6.5 \times 10^{-6} \times 2 \times 10^{11}} = \frac{2431.8}{1.3 \times 10^{6}}\] \[\Delta L = 1.87 \times 10^{-3} \text{ m} = 1.87 \text{ mm}\] Answer: \(\boxed{\Delta L \approx 1.87 \text{ mm}}\).
Note: The centripetal force (2289.7 N) is much larger than the weight (142.1 N), which is why the wire elongation is significant. We should verify that the total stress (2431.8 / 6.5×10&supmin;&sup6; = 3.74×10&sup8; Pa) is below the elastic limit of steel (∼250 MPa for mild steel), and indeed 374 MPa exceeds this, suggesting the wire may undergo permanent deformation. In practice, high-tensile steel would be used.
Frequently Asked Questions - NCERT Exercises and Solutions: Mechanical Properties of Solids
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MyAiSchool Class 11 Physics Chapter 8 Mechanical Properties of Solids solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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Physics Class 11 Part II – NCERT (2025-26)
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