Concept Card: Hooke’s Law & Young’s Modulus of Elasticity
1. Foundational Definition of Hooke’s Law:
Within the proportional limit of an elastic material, the internal restoring stress developed across any section is strictly directly proportional to the longitudinal strain produced:
$$\sigma \propto \varepsilon \implies \mathbf{\sigma = Y \cdot \varepsilon}$$
where $Y$ is Young’s Modulus of Elasticity.
In terms of applied tensile force $F$, original length $L$, cross-sectional area $A$, and elongation $\Delta L$:
$$\frac{F}{A} = Y \left(\frac{\Delta L}{L}\right) \implies \mathbf{F = \left(\frac{Y A}{L}\right) \Delta L}$$
This establishes an exact mathematical equivalence to an ideal linear spring:
$$\mathbf{F = k_{\text{wire}} \Delta L \quad \text{where} \quad k_{\text{wire}} = \frac{Y A}{L}}$$
The effective stiffness $k_{\text{wire}}$ depends on geometry ($k \propto A/L$), whereas $Y$ is an invariant material constant.
2. Young’s Modulus: Dimensions, Physical Units & Microscopic Origin
- Mathematical Definition:
$$\mathbf{Y = \frac{\text{Tensile / Compressive Stress}}{\text{Longitudinal Strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}}$$ - SI Units & Dimensions:
- SI Unit: $\text{N/m}^2$ or $\text{Pascal (Pa)}$.
- Dimensional Formula: $[Y] = \frac{[F]}{[A]} = \mathbf{[M L^{-1} T^{-2}]}$.
- Microscopic Interatomic Model:
At the atomic scale, a solid behaves as a 3D lattice of point masses connected by interatomic springs of stiffness $k_0$. If $r_0$ is the equilibrium interatomic separation:
$$\mathbf{Y = \frac{k_0}{r_0} \implies k_0 = Y r_0}$$
For structural metals ($Y \approx 2 \times 10^{11}\text{ Pa}, r_0 \approx 2.5 \times 10^{-10}\text{ m}$), the interatomic bond stiffness is $k_0 \approx 50\text{ N/m}$.
3. Combination of Elastic Wires (Spring Analogies)
(A) Wires Connected in Series (End-to-End):
When two wires are joined end-to-end and pulled with tensile force $F$:
- The tension $F$ is identical in both wires.
- The total elongation is the sum: $\Delta L_{\text{total}} = \Delta L_1 + \Delta L_2$.
- Equivalent stiffness:
$$\frac{F}{k_{\text{eq}}} = \frac{F}{k_1} + \frac{F}{k_2} \implies \mathbf{\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{L_1}{Y_1 A_1} + \frac{L_2}{Y_2 A_2}}$$
(B) Wires Connected in Parallel (Side-by-Side):
When two wires are hung side-by-side supporting a rigid load:
- Both wires undergo identical elongation: $\Delta L_1 = \Delta L_2 = \Delta L$.
- The total force is shared: $F_{\text{total}} = F_1 + F_2$.
- Equivalent stiffness:
$$\mathbf{k_{\text{eq}} = k_1 + k_2 = \frac{Y_1 A_1}{L_1} + \frac{Y_2 A_2}{L_2}}$$
4. Energy Considerations in Elastic Deformation
- Total Elastic Strain Energy ($U$):
$$U = \int_0^{\Delta L} F(x) dx = \int_0^{\Delta L} \left(\frac{Y A}{L} x\right) dx = \mathbf{\frac{1}{2} \frac{Y A (\Delta L)^2}{L} = \frac{1}{2} F \Delta L = \frac{1}{2} k \Delta L^2}$$ - Elastic Energy Density ($u$):
$$\mathbf{u = \frac{U}{\text{Volume}} = \frac{1}{2} \frac{F \Delta L}{A L} = \frac{1}{2} \sigma \varepsilon = \frac{\sigma^2}{2 Y} = \frac{1}{2} Y \varepsilon^2}$$ - The $50\%$ Energy Dissipation Theorem in Gradual Loading:
When a mass $M$ is attached to a vertical wire and lowered slowly to equilibrium, the loss in gravitational potential energy is $\Delta U_g = M g \Delta L$.
However, the elastic energy stored in the wire is only $U_{\text{elastic}} = \frac{1}{2} M g \Delta L$.
Conclusion: Exactly **half** ($\frac{1}{2} M g \Delta L$) of the gravitational energy lost is dissipated as heat and internal damping!
5. Critical Insights & Exam Traps in JEE:
- Trap 1 (Material Invariance of $Y$): Cutting a wire in half doubles its stiffness ($k’ = \frac{Y A}{L/2} = 2k$), but its Young’s modulus $Y$ remains completely unchanged. $Y$ depends solely on the material and temperature.
- Trap 2 (Sudden vs. Gradual Loading):
- Gradual Loading: Load descends through quasi-static equilibrium $\implies \Delta L_0 = \frac{M g L}{Y A}$.
- Sudden Release from Rest: Mass oscillates with amplitude $\Delta L_0$, attaining a maximum elongation:
$$\mathbf{\Delta L_{\max} = 2 \Delta L_0 = \frac{2 M g L}{Y A}}$$
Sudden loading produces **twice** the elongation and **twice** the peak stress of gradual loading!
- Trap 3 (Elongation of Tapered Wires): For a wire whose radius tapers linearly from $r_1$ to $r_2$ over length $L$, the elongation under load $F$ is:
$$\mathbf{\Delta L = \frac{F L}{\pi r_1 r_2 Y}}$$ - Trap 4 (Rotating Rod Elongation): For a rod of mass $M$ and length $L$ rotating about one end with angular speed $\omega$:
$$\mathbf{\Delta L = \frac{M \omega^2 L^2}{3 A Y}}$$
Solved Examples
Example 1 (Composite Series Wire – Stress, Strain & Stiffness Analysis):
A composite structural wire consists of a steel segment of length $L_1 = 1.0\text{ m}$ and circular radius $r_1 = 1.0\text{ mm}$ ($Y_1 = 2.0 \times 10^{11}\text{ Pa}$) joined end-to-end to a brass segment of length $L_2 = 2.0\text{ m}$ and circular radius $r_2 = 2.0\text{ mm}$ ($Y_2 = 1.0 \times 10^{11}\text{ Pa}$). A tensile load $F = 314.16\text{ N} = 100\pi\text{ N}$ is applied to the composite wire.
(a) Determine the tensile stress in the steel and brass segments.
(b) Calculate the individual elongations $\Delta L_1, \Delta L_2$ and the total elongation $\Delta L_{\text{total}}$.
(c) Determine the effective equivalent stiffness $k_{\text{eq}}$ of the composite wire.
Solution:
Cross-sectional areas:
$$A_1 = \pi r_1^2 = \pi (1.0 \times 10^{-3})^2 = \pi \times 10^{-6}\text{ m}^2$$
$$A_2 = \pi r_2^2 = \pi (2.0 \times 10^{-3})^2 = 4\pi \times 10^{-6}\text{ m}^2$$
(a) Tensile Stress in Each Segment:
Because the segments are connected in series, the tensile force $F = 100\pi\text{ N}$ is identical in both:
$$\sigma_1 = \frac{F}{A_1} = \frac{100\pi\text{ N}}{\pi \times 10^{-6}\text{ m}^2} = \mathbf{1.0 \times 10^8\text{ N/m}^2 = 100\text{ MPa}}$$
$$\sigma_2 = \frac{F}{A_2} = \frac{100\pi\text{ N}}{4\pi \times 10^{-6}\text{ m}^2} = \mathbf{2.5 \times 10^7\text{ N/m}^2 = 25\text{ MPa}}$$
(b) Individual and Total Elongations:
$$\Delta L_1 = \frac{\sigma_1 L_1}{Y_1} = \frac{(1.0 \times 10^8\text{ Pa})(1.0\text{ m})}{2.0 \times 10^{11}\text{ Pa}} = 0.50 \times 10^{-3}\text{ m} = \mathbf{0.50\text{ mm}}$$
$$\Delta L_2 = \frac{\sigma_2 L_2}{Y_2} = \frac{(2.5 \times 10^7\text{ Pa})(2.0\text{ m})}{1.0 \times 10^{11}\text{ Pa}} = 0.50 \times 10^{-3}\text{ m} = \mathbf{0.50\text{ mm}}$$
$$\mathbf{\Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 = 0.50\text{ mm} + 0.50\text{ mm} = 1.00\text{ mm}}$$
(c) Equivalent Spring Stiffness:
$$k_1 = \frac{Y_1 A_1}{L_1} = \frac{(2.0 \times 10^{11})(\pi \times 10^{-6})}{1.0} = 2.0\pi \times 10^5\text{ N/m}$$
$$k_2 = \frac{Y_2 A_2}{L_2} = \frac{(1.0 \times 10^{11})(4\pi \times 10^{-6})}{2.0} = 2.0\pi \times 10^5\text{ N/m}$$
Since $k_1 = k_2 = 2.0\pi \times 10^5\text{ N/m}$:
$$k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2} = \frac{k_1}{2} = 1.0\pi \times 10^5\text{ N/m} \approx \mathbf{3.14 \times 10^5\text{ N/m}}$$
Verification: $\Delta L_{\text{total}} = \frac{F}{k_{\text{eq}}} = \frac{100\pi}{\pi \times 10^5} = 10^{-3}\text{ m} = 1.0\text{ mm}$.
Example 2 (Calculus Integration of a Linearly Tapered Wire):
A circular wire of length $L$ and Young’s modulus $Y$ has a radius that tapers linearly from $r_1$ at end $x = 0$ to $r_2$ at end $x = L$. A tensile stretching force $F$ is applied at both ends.
(a) Determine the radius $r(x)$ of the cross-section at a distance $x$ from end $x = 0$.
(b) Derive the total elongation $\Delta L$ of the wire by setting up and evaluating a differential calculus integral.
(c) Verify that for uniform cylindrical cross-section ($r_1 = r_2 = R$), the formula reduces to standard Hooke’s law.
Solution:
(a) Radius Function $r(x)$:
Since the radius tapers linearly from $r_1$ at $x = 0$ to $r_2$ at $x = L$:
$$\mathbf{r(x) = r_1 + \left(\frac{r_2 – r_1}{L}\right) x}$$
(b) Calculus Integration for Elongation:
Consider an infinitesimal slice of thickness $dx$ at distance $x$. The cross-sectional area of this slice is:
$$A(x) = \pi [r(x)]^2$$
The elongation of this thin element under tensile force $F$ is:
$$d(\Delta L) = \frac{F dx}{A(x) Y} = \frac{F dx}{\pi Y [r(x)]^2}$$
Let $u = r(x) = r_1 + \left(\frac{r_2 – r_1}{L}\right) x \implies du = \left(\frac{r_2 – r_1}{L}\right) dx \implies dx = \left(\frac{L}{r_2 – r_1}\right) du$.
When $x = 0 \implies u = r_1$; when $x = L \implies u = r_2$.
Integrating over the entire length:
$$\Delta L = \int_{r_1}^{r_2} \frac{F}{\pi Y u^2} \left(\frac{L}{r_2 – r_1}\right) du = \frac{F L}{\pi Y (r_2 – r_1)} \int_{r_1}^{r_2} u^{-2} du$$
$$\Delta L = \frac{F L}{\pi Y (r_2 – r_1)} \left[-\frac{1}{u}\right]_{r_1}^{r_2} = \frac{F L}{\pi Y (r_2 – r_1)} \left[\frac{1}{r_1} – \frac{1}{r_2}\right] = \frac{F L}{\pi Y (r_2 – r_1)} \left[\frac{r_2 – r_1}{r_1 r_2}\right]$$
Canceling $(r_2 – r_1)$:
$$\mathbf{\Delta L = \frac{F L}{\pi r_1 r_2 Y}}$$
(c) Limiting Case Verification:
If $r_1 = r_2 = R$, the cross-sectional area is $A = \pi R^2$:
$$\Delta L = \frac{F L}{\pi R R Y} = \frac{F L}{\pi R^2 Y} = \frac{F L}{A Y}$$
This matches the standard cylindrical wire formula exactly.
Example 3 (Dynamics of Sudden Release & Energy Balance – JEE Advanced):
A weightless elastic wire of length $L$, cross-sectional area $A$, and Young’s modulus $Y$ hangs vertically from a rigid support.
(a) A block of mass $M$ is attached to the lower end and lowered gradually until static equilibrium is reached. Find the equilibrium elongation $\Delta L_0$ and stored strain energy $U_0$.
(b) If the block of mass $M$ is attached and released suddenly from rest when the wire is completely unstretched, determine the maximum elongation $\Delta L_{\max}$ during the motion.
(c) Compare the maximum elongation to the equilibrium elongation and describe the resulting motion.
Solution:
Effective spring constant of the wire: $k = \frac{Y A}{L}$.
(a) Gradual Loading (Static Equilibrium):
At equilibrium, upward restoring force equals downward weight:
$$k \Delta L_0 = M g \implies \mathbf{\Delta L_0 = \frac{M g}{k} = \frac{M g L}{Y A}}$$
Stored elastic strain energy at equilibrium:
$$\mathbf{U_0 = \frac{1}{2} k (\Delta L_0)^2 = \frac{1}{2} M g \Delta L_0 = \frac{M^2 g^2 L}{2 Y A}}$$
(b) Sudden Release from Rest:
When released from rest at $x = 0$ (velocity $v = 0$):
At maximum elongation $\Delta L_{\max}$, the mass momentarily comes to rest again ($v = 0$).
By conservation of mechanical energy between the release point ($x = 0$) and maximum extension ($x = \Delta L_{\max}$):
$$\Delta K + \Delta U_g + \Delta U_{\text{elastic}} = 0$$
$$0 – M g \Delta L_{\max} + \frac{1}{2} k (\Delta L_{\max})^2 = 0$$
Dividing by $\Delta L_{\max} \ne 0$:
$$\frac{1}{2} k \Delta L_{\max} = M g \implies \mathbf{\Delta L_{\max} = \frac{2 M g}{k} = 2 \Delta L_0 = \frac{2 M g L}{Y A}}$$
(c) Nature of Motion:
The maximum elongation is precisely **twice** the equilibrium elongation ($\Delta L_{\max} = 2\Delta L_0$).
The mass executes simple harmonic motion (SHM) about the equilibrium position $x = \Delta L_0$ with amplitude:
$$A_{\text{shm}} = \Delta L_0 = \frac{M g L}{Y A}$$
The angular frequency of oscillation is:
$$\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{Y A}{M L}}$$
Example 4 (Elongation of a Rod Rotating in a Horizontal Plane):
A uniform thin metal rod of mass $M$, length $L$, cross-sectional area $A$, and Young’s modulus $Y$ is pivoted at one end on a smooth horizontal table and rotates with constant angular velocity $\omega$.
(a) Determine the internal tension $T(x)$ at distance $x$ from the pivot axis.
(b) Derive the total elongation $\Delta L$ produced in the rotating rod.
Solution:
Linear mass density of the rod: $\lambda = \frac{M}{L}$.
(a) Tension Distribution $T(x)$:
Consider a section of the rod from distance $x$ to the free outer end $L$. The tension $T(x)$ at position $x$ provides the necessary centripetal force for the outer portion of length $(L – x)$:
For an infinitesimal element of mass $dm = \lambda dr$ at distance $r$ from the pivot:
$$dF_c = (dm) r \omega^2 = (\lambda dr) r \omega^2$$
Integrating from $r = x$ to $r = L$:
$$T(x) = \int_x^L \lambda \omega^2 r dr = \lambda \omega^2 \left[\frac{r^2}{2}\right]_x^L = \frac{M \omega^2}{2 L} (L^2 – x^2)$$
$$\mathbf{T(x) = \frac{M \omega^2}{2 L} (L^2 – x^2)}$$
At the pivot ($x = 0$): $T(0) = \frac{1}{2} M \omega^2 L$.
At the free tip ($x = L$): $T(L) = 0$.
(b) Total Elongation $\Delta L$:
An element $dx$ at distance $x$ experiences an extension:
$$d(\Delta L) = \frac{T(x) dx}{A Y} = \frac{M \omega^2}{2 A Y L} (L^2 – x^2) dx$$
Integrating from $x = 0$ to $x = L$:
$$\Delta L = \frac{M \omega^2}{2 A Y L} \int_0^L (L^2 – x^2) dx = \frac{M \omega^2}{2 A Y L} \left[ L^2 x – \frac{x^3}{3} \right]_0^L$$
$$\Delta L = \frac{M \omega^2}{2 A Y L} \left[ L^3 – \frac{L^3}{3} \right] = \frac{M \omega^2}{2 A Y L} \left(\frac{2}{3} L^3\right) = \mathbf{\frac{M \omega^2 L^2}{3 A Y}}$$
Practice Worksheet: Hooke’s Law & Young’s Modulus
Section A: Single Correct Questions (Q1 to Q5)
Q1. A structural metal wire of Young’s modulus $Y$ is stretched such that the tensile stress developed in it is $\sigma$. The elastic potential energy stored per unit volume of the wire is:
(A) $\frac{\sigma^2}{2 Y}$
(B) $\frac{\sigma^2}{Y}$
(C) $\frac{2 \sigma^2}{Y}$
(D) $\frac{\sigma^2}{4 Y}$
Q2. A circular metal wire of length $L$ and Young’s modulus $Y$ has a radius that tapers uniformly from $r_1$ at one end to $r_2$ at the opposite end. If equal stretching forces $F$ are applied at both ends, the total elongation of the wire is:
(A) $\frac{F L}{\pi r_1 r_2 Y}$
(B) $\frac{2 F L}{\pi (r_1 + r_2) Y}$
(C) $\frac{F L}{\pi (r_1^2 + r_2^2) Y}$
(D) $\frac{4 F L}{\pi (r_1 + r_2)^2 Y}$
Q3. A steel wire of original length $L$ and cross-sectional area $A$ has an effective spring constant $k$ and Young’s modulus $Y$. If the wire is cut into two equal halves of length $L/2$, the effective spring constant and Young’s modulus of each half are respectively:
(A) $2k, Y$
(B) $\frac{k}{2}, Y$
(C) $2k, \frac{Y}{2}$
(D) $k, Y$
Q4. A light elastic wire of length $L$, cross-sectional area $A$, and Young’s modulus $Y$ hangs vertically from a ceiling. If a block of mass $M$ is attached to the lower end and released suddenly from rest when the wire is unstretched, the maximum elongation produced in the wire is:
(A) $\frac{2 M g L}{A Y}$
(B) $\frac{M g L}{A Y}$
(C) $\frac{M g L}{2 A Y}$
(D) $\frac{4 M g L}{A Y}$
Q5. A uniform thin metal rod of mass $M$, length $L$, cross-sectional area $A$, and Young’s modulus $Y$ is rotated horizontally about a vertical axis passing through one of its ends with constant angular velocity $\omega$. The total elongation of the rod is:
(A) $\frac{M \omega^2 L^2}{3 A Y}$
(B) $\frac{M \omega^2 L^2}{2 A Y}$
(C) $\frac{M \omega^2 L^2}{6 A Y}$
(D) $\frac{2 M \omega^2 L^2}{3 A Y}$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Two elastic wires of lengths $L_1, L_2$, cross-sectional areas $A_1, A_2$, and Young’s moduli $Y_1, Y_2$ are connected together. Which of the following statements is/are correct?
(A) When connected in series end-to-end and stretched by force $F$, the same tensile force $F$ acts on both segments.
(B) In series connection, the effective stiffness satisfies $\frac{1}{k_{\text{eq}}} = \frac{L_1}{Y_1 A_1} + \frac{L_2}{Y_2 A_2}$.
(C) When connected in parallel side-by-side to support a rigid horizontal bar, both wires undergo identical elongation if the bar remains horizontal.
(D) In parallel connection, the effective stiffness is $k_{\text{eq}} = \frac{Y_1 A_1}{L_1} + \frac{Y_2 A_2}{L_2}$.
Q7. At the microscopic atomic scale, the elastic behavior of crystalline solids is modeled as atoms connected by interatomic springs. Which of the following statements is/are correct?
(A) If $k_0$ is the interatomic spring constant and $r_0$ is the equilibrium interatomic separation, then Young’s modulus is given by $Y = \frac{k_0}{r_0}$.
(B) For typical metals where $Y \approx 2 \times 10^{11}\text{ N/m}^2$ and $r_0 \approx 2.5 \times 10^{-10}\text{ m}$, the interatomic force constant is of the order of $50\text{ N/m}$.
(C) Interatomic potential energy curves are asymmetric about the equilibrium separation $r_0$.
(D) The asymmetry of the interatomic potential energy curve is the microscopic physical origin of thermal expansion in solids.
Q8. A weightless elastic wire hangs vertically. A body of mass $M$ is attached to the lower end and gently released so that it stretches the wire slowly to its final static equilibrium extension $\Delta L_0$. Which of the following statements is/are correct?
(A) The loss in gravitational potential energy of the mass is $M g \Delta L_0$.
(B) The elastic potential energy stored in the stretched wire at equilibrium is $\frac{1}{2} M g \Delta L_0$.
(C) Exactly half of the gravitational potential energy lost is dissipated as thermal energy and internal damping.
(D) If the mass had been dropped from the unstretched position, it would execute simple harmonic motion about the equilibrium position with amplitude $\Delta L_0$.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A mass $M$ attached to the lower end of a vertical wire causes an elongation $\Delta L_0$ when lowered gradually to equilibrium. When released suddenly from the unextended position, it reaches a maximum elongation $\Delta L_{\max} = n \Delta L_0$. Find the integer value of $n$.
Q10. A uniform thin rod of mass $M$, length $L$, cross-sectional area $A$, and Young’s modulus $Y$ is rotated horizontally about a fixed pivot at one end with constant angular velocity $\omega$. The total elongation produced in the rod can be expressed as $\Delta L = \frac{M \omega^2 L^2}{k A Y}$. Determine the integer value of $k$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Elastic Strain Energy Density: $u = \frac{\sigma^2}{2Y}$ |
| Q2 | (A) | Tapered Wire Calculus Integration: $\Delta L = \frac{FL}{\pi r_1 r_2 Y}$ |
| Q3 | (A) | Stiffness Scaling vs. Material Invariance: $k \propto 1/L \implies 2k$, $Y$ constant |
| Q4 | (A) | Sudden Loading Dynamics: $\Delta L_{\max} = 2 \Delta L_0 = \frac{2MgL}{AY}$ |
| Q5 | (A) | Rotating Rod Centripetal Elongation: $\Delta L = \frac{M\omega^2 L^2}{3AY}$ |
| Q6 | (A, B, C, D) | Series and Parallel Combinations of Elastic Wires |
| Q7 | (A, B, C, D) | Interatomic Spring Model ($Y = k_0/r_0$) & Potential Asymmetry |
| Q8 | (A, B, C, D) | Work-Energy Dissipation Theorem in Gradual vs Sudden Wire Loading |
| Q9 | 2 | Sudden vs. Gradual Elongation Ratio: $\Delta L_{\max} / \Delta L_0 = 2 \implies n = 2$ |
| Q10 | 3 | Rotating Rod Integral Factor: $\Delta L = \frac{M\omega^2 L^2}{3AY} \implies k = 3$ |
Detailed Explanations
Solution 1:
The work done per unit volume in stretching an elastic wire is:
$$u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} \sigma \varepsilon$$
From Hooke’s law, $\varepsilon = \frac{\sigma}{Y}$. Substituting $\varepsilon$:
$$u = \frac{1}{2} \sigma \left(\frac{\sigma}{Y}\right) = \mathbf{\frac{\sigma^2}{2 Y}}$$
Correct Answer: (A)
Solution 2:
Radius of the cross-section at distance $x$ from the smaller end:
$$r(x) = r_1 + \left(\frac{r_2 – r_1}{L}\right) x$$
Elongation of element of thickness $dx$:
$$d(\Delta L) = \frac{F dx}{\pi [r(x)]^2 Y}$$
Integrating from $x = 0$ to $x = L$:
$$\Delta L = \frac{F}{\pi Y} \int_0^L \frac{dx}{\left[r_1 + \left(\frac{r_2 – r_1}{L}\right) x\right]^2} = \frac{F}{\pi Y} \left[ \frac{-L}{(r_2 – r_1) \left[r_1 + \left(\frac{r_2 – r_1}{L}\right) x\right]} \right]_0^L$$
$$\Delta L = \frac{F L}{\pi Y (r_2 – r_1)} \left[ \frac{1}{r_1} – \frac{1}{r_2} \right] = \frac{F L}{\pi Y (r_2 – r_1)} \left[\frac{r_2 – r_1}{r_1 r_2}\right] = \mathbf{\frac{F L}{\pi r_1 r_2 Y}}$$
Correct Answer: (A)
Solution 3:
Effective spring stiffness of a wire is:
$$k = \frac{Y A}{L} \propto \frac{1}{L}$$
When the wire is cut in half ($L’ = L/2$):
$$k’ = \frac{Y A}{L / 2} = 2 \left(\frac{Y A}{L}\right) = \mathbf{2k}$$
Young’s modulus $Y$ is an intensive material property that depends purely on interatomic forces and temperature; it does not change when dimensions change. Hence, $Y$ remains unchanged.
Correct Answer: (A)
Solution 4:
When the block is released suddenly from rest at the unstretched position ($x = 0$):
Let maximum extension be $\Delta L_{\max}$. At this maximum elongation point, velocity is momentarily zero ($v = 0$).
By conservation of mechanical energy:
$$\Delta K + \Delta U_g + \Delta U_{\text{elastic}} = 0$$
$$0 – M g \Delta L_{\max} + \frac{1}{2} k (\Delta L_{\max})^2 = 0 \implies \frac{1}{2} k \Delta L_{\max} = M g$$
$$\Delta L_{\max} = \frac{2 M g}{k} = \frac{2 M g}{\frac{Y A}{L}} = \mathbf{\frac{2 M g L}{A Y}}$$
Correct Answer: (A)
Solution 5:
At distance $x$ from the rotation pivot, the tension $T(x)$ provides the centripetal force for the outer segment of length $(L – x)$:
$$T(x) = \int_x^L \left(\frac{M}{L} dr\right) r \omega^2 = \frac{M \omega^2}{2 L} (L^2 – x^2)$$
Elongation of element $dx$:
$$d(\Delta L) = \frac{T(x) dx}{A Y} = \frac{M \omega^2}{2 A Y L} (L^2 – x^2) dx$$
Integrating from $x = 0$ to $x = L$:
$$\Delta L = \frac{M \omega^2}{2 A Y L} \int_0^L (L^2 – x^2) dx = \frac{M \omega^2}{2 A Y L} \left[L^2 (L) – \frac{L^3}{3}\right] = \frac{M \omega^2}{2 A Y L} \left(\frac{2 L^3}{3}\right) = \mathbf{\frac{M \omega^2 L^2}{3 A Y}}$$
Correct Answer: (A)
Solution 6:
– (A) True: In a series chain, force balance requires equal tension $F$ across every cross-section.
– (B) True: Total elongation $\Delta L = \Delta L_1 + \Delta L_2 \implies \frac{F}{k_{\text{eq}}} = \frac{F}{k_1} + \frac{F}{k_2} \implies \frac{1}{k_{\text{eq}}} = \frac{L_1}{Y_1 A_1} + \frac{L_2}{Y_2 A_2}$.
– (C) True: In parallel, keeping the support bar horizontal enforces identical elongation $\Delta L_1 = \Delta L_2$.
– (D) True: Total force $F = F_1 + F_2 \implies k_{\text{eq}} \Delta L = k_1 \Delta L + k_2 \Delta L \implies k_{\text{eq}} = k_1 + k_2 = \frac{Y_1 A_1}{L_1} + \frac{Y_2 A_2}{L_2}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: Longitudinal strain $\varepsilon = \Delta r / r_0$; stress $\sigma = F / r_0^2 = (k_0 \Delta r) / r_0^2$. Young’s modulus $Y = \sigma / \varepsilon = \frac{k_0 \Delta r / r_0^2}{\Delta r / r_0} = \frac{k_0}{r_0}$.
– (B) True: $k_0 = Y r_0 \approx (2 \times 10^{11}\text{ Pa})(2.5 \times 10^{-10}\text{ m}) = 50\text{ N/m}$.
– (C) True: The interatomic potential curve rises much more steeply for compression ($r < r_0$) than for expansion ($r > r_0$).
– (D) True: As thermal energy increases, the average atomic separation shifts toward larger values due to this asymmetry, generating macroscopic thermal expansion.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) True: Loss in gravitational potential energy is $M g \Delta L_0$.
– (B) True: Stored elastic strain energy is $\frac{1}{2} k (\Delta L_0)^2 = \frac{1}{2} M g \Delta L_0$.
– (C) True: $M g \Delta L_0 – \frac{1}{2} M g \Delta L_0 = \frac{1}{2} M g \Delta L_0$ is dissipated as heat/damping.
– (D) True: Without damping, sudden release oscillates between $x = 0$ and $x = 2 \Delta L_0$ with amplitude $\Delta L_0$ and equilibrium position $\Delta L_0$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
Static equilibrium elongation under gradual load:
$$\Delta L_0 = \frac{M g}{k}$$
Maximum elongation under sudden release from rest:
$$\Delta L_{\max} = \frac{2 M g}{k} = 2 \Delta L_0$$
Taking the ratio:
$$\frac{\Delta L_{\max}}{\Delta L_0} = 2$$
$$\mathbf{n = 2}$$
Correct Answer: 2
Solution 10:
As derived in Example 4 and Question 5:
$$\Delta L = \frac{M \omega^2 L^2}{3 A Y}$$
Comparing with $\frac{M \omega^2 L^2}{k A Y}$:
$$\mathbf{k = 3}$$
Correct Answer: 3