Q.A brass wire $1.8\ \text{m}$ long at $27\ ^\circ\text{C}$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39\ ^\circ\text{C}$, what is the tension developed in the wire, if its diameter is $2.0\ \text{mm}$? Coefficient of linear expansion of brass $= 2.0 \times 10^{-5}\ \text{K}^{-1}$; Young's modulus of brass $= 0.91 \times 10^{11}\ \text{Pa}$.

Showing the 12 most recent of 19 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A wire, made of a certain material of length-l and area of cross section-a can withstand a maximum load $=\mathrm{W}$ without breaking. If, another wire of the same material and crosssectional area is used with double the original length, what will be the maximum load that the wire can withstand, without breaking? (A) Will be halved to 0.5 W (B) Will be doubled to 2 W (C) Remains the same $=\mathrm{W}$ (D) Would be four times $=4 \mathrm{~W}$

   ›Reveal solutionSolution

   The maximum load a wire can withstand without breaking depends only on its material and cross-sectional area, not on its length. Therefore, doubling the length leaves the maximum load unchanged. The correct option is (C).

   The key concept here is tensile strength — the maximum stress a material can endure before breaking. Stress is defined as force per unit area:

   $\sigma = \frac{F}{A}$

   For a given material, the breaking stress is a fixed property (assuming no defects). The wire’s length does not appear in this formula. So, if the cross-sectional area stays the same, the maximum force (load) that causes breaking remains the same, regardless of length.

   Let’s walk through it step by step:

   1. Recall the definition of breaking stress The wire breaks when the tensile stress reaches the material’s ultimate tensile strength, $\sigma_{\text{max}}$.

   $\sigma_{\text{max}} = \frac{W}{a}$

   Here, $W$ is the maximum load (force) and $a$ is the cross-sectional area. This is the condition for the original wire.

   2. Consider the new wire

      The new wire is made of the same material (so $\sigma_{\text{max}}$ is identical) and has the same cross-sectional area $a$. Its length is doubled, but length does not appear in the stress equation.

   3. Set up the breaking condition for the new wire

      Let the new maximum load be $W'$. Then:

   $\sigma_{\text{max}} = \frac{W'}{a}$

   Since $\sigma_{\text{max}}$ is the same as before, we have:

   $\frac{W}{a} = \frac{W'}{a}$

   which simplifies to $W' = W$.

   4. Why length doesn’t matter here …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A light rod of length 1 m is suspended from ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of material $X$ and is of cross-section $0.1 \mathrm{~cm}^2$. and the other of material Y of cross-section $0.3 \mathrm{~cm}^2 . \mathrm{A}$ weight is hung from the wire at a point to produce equal strain in the wires. The ratio of Young's moduli of wires A to B is $3: 1$. The location of the point from one end of the wire is (A) 0.2 m (B) 0.75 m (C) 0.5 m (D) 0.25 m

   ›Reveal solutionSolution

   Equal strain forces equal tension in both wires, so the load hangs at the rod's midpoint, $0.5\,\text{m}$ from either end.

   Setting up the equal-strain condition

   Strain in a wire is $\varepsilon = \dfrac{\text{stress}}{Y} = \dfrac{F}{A\,Y}$, where $F$ is the tension, $A$ the cross-section and $Y$ Young's modulus.

   For equal strain in the two wires:

   $\frac{F_X}{A_X\,Y_X} = \frac{F_Y}{A_Y\,Y_Y}$

   Given $A_X = 0.1\,\text{cm}^2$, $A_Y = 0.3\,\text{cm}^2$ and $Y_X : Y_Y = 3 : 1$:

   $A_X\,Y_X = 0.1 \times 3 = 0.3, \qquad A_Y\,Y_Y = 0.3 \times 1 = 0.3$

   Since $A_X Y_X = A_Y Y_Y$, the equal-strain condition reduces to $F_X = F_Y$.

   Locating the load …

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3. KCET 2026Set C21 markMCQ
   Q.There are two wires of same material and same length while the diameter of second wire is two times the diameter of the first wire. Then the ratio of extensions produced in the wires by applying same load will be (A) 1:1 (B) 1:2 (C) 2:1 (D) 4:1

   ›Reveal solutionSolution

   Extension from Young's modulus, $e = FL/(AY)$, depends inversely on cross-sectional area for fixed force, length, and material.

   Step 1 — Set up the two wires

   Both wires share the same material (same Young's modulus $Y$), same length $L$, and the same applied load $F$. Let wire 1 have diameter $d$ and wire 2 have diameter $2d$.

   $A_1 = \frac{\pi d^2}{4}, \qquad A_2 = \frac{\pi (2d)^2}{4} = 4\left(\frac{\pi d^2}{4}\right) = 4A_1$

   Step 2 — Compare the extensions …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.One end of a nylon rope of length 1 and diameter 10 mm is fixed to free limb. A monkey weighing 100 N jumps to catch the free end and stays there. The change in diameter of the rope is (Young's modulus of the wire is Y and Poisson Ratio is $\sigma$ ) (A) $\frac{4000 \sigma}{\pi Y}$ (B) $\frac{400 \sigma}{\pi Y}$ (C) $\frac{40000 \sigma}{\pi Y}$ (D) $\frac{40000 \pi}{\sigma Y}$

   ›Reveal solutionSolution

   The change in diameter is found from the lateral strain (Poisson effect) caused by the axial tensile stress. The answer is $\frac{40000 \sigma}{\pi Y}$, so option (C) is correct.

   ---

   Concept & Intuition

   When the monkey hangs on the rope, the rope stretches lengthwise (axial strain). Because of the Poisson effect, a material that stretches in one direction contracts in the perpendicular directions. Here, the diameter decreases. The change in diameter is proportional to the original diameter, the axial strain, and the Poisson ratio. The axial strain itself comes from the tensile stress (weight divided by cross-sectional area) divided by Young's modulus. So we just need to compute the stress carefully, then apply the Poisson relation.

   ---

   Step-by-step solution

   1. Find the cross-sectional area of the rope

      Diameter $d = 10\ \text{mm} = 10^{-2}\ \text{m}$.

      Area $A = \frac{\pi d^2}{4} = \frac{\pi (10^{-2})^2}{4} = \frac{\pi \times 10^{-4}}{4} = \frac{\pi}{4} \times 10^{-4}\ \text{m}^2$.

   2. Compute the axial tensile stress

      The monkey’s weight $W = 100\ \text{N}$ is the tensile force.

      Stress $\sigma_{\text{axial}} = \frac{W}{A} = \frac{100}{\frac{\pi}{4} \times 10^{-4}} = \frac{100 \times 4}{\pi \times 10^{-4}} = \frac{400}{\pi \times 10^{-4}} = \frac{400 \times 10^4}{\pi} = \frac{4 \times 10^6}{\pi}\ \text{Pa}$.

   3. Find the axial strain

      By Hooke’s law, axial strain $\varepsilon_{\text{axial}} = \frac{\sigma_{\text{axial}}}{Y} = \frac{4 \times 10^6}{\pi Y}$.

   4. Relate lateral strain to axial strain via Poisson ratio

      Poisson ratio $\sigma$ (here the symbol given in the problem) is defined as

      $\sigma = -\frac{\text{lateral strain}}{\text{axial strain}}$. …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A wire of negligible mass having uniform area of cross section ' A ' and young modulus ' Y ' is used to suspend a point mass ' m '. The point mass executes simple harmonic motion in a vertical plane with a period ' T ', then the length of the wire is : (A) $L=\frac{T Y^2 A}{4 \pi^2 m}$ (B) $L=\frac{T^2 Y A}{4 \pi m^2}$ (C) $L=\frac{T Y^2 A}{4 \pi m^2}$ (D) $L=\frac{T^2 Y A}{4 \pi^2 m}$

   ›Reveal solutionSolution

   The period of a mass on a vertical wire is determined by the wire's stiffness, which comes from its Young's modulus and geometry. Using the formula for the spring constant of a stretched wire and the period of a simple harmonic oscillator, we find $L = \frac{T^2 Y A}{4 \pi^2 m}$, which corresponds to option (D).

   The key concept here is that a wire under tension behaves like a spring for small oscillations. When the mass is displaced vertically, the wire stretches and exerts a restoring force proportional to the extension, exactly like Hooke's law. The "spring constant" $k$ for a wire of length $L$, cross-sectional area $A$, and Young's modulus $Y$ is $k = \frac{Y A}{L}$. Then the period of simple harmonic motion for a mass $m$ on such a spring is $T = 2\pi \sqrt{\frac{m}{k}}$. Combining these gives the length directly.

   1. Recall the spring constant for a wire under tension. Young's modulus is defined as $Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}$. Rearranging:

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

   This is exactly Hooke's law $F = k \Delta L$, so the effective spring constant is

   $k = \frac{Y A}{L}.$

   2. Write the period of a mass-spring system. For a simple harmonic oscillator, the period is

   $T = 2\pi \sqrt{\frac{m}{k}}.$

   3. Substitute the expression for $k$.

   $T = 2\pi \sqrt{\frac{m}{Y A / L}} = 2\pi \sqrt{\frac{m L}{Y A}}.$

   4. Solve for $L$. Square both sides:

   $T^2 = 4\pi^2 \frac{m L}{Y A}.$

   Multiply both sides by $Y A$ and divide by $4\pi^2 m$:

   $$$$

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.Young's modulus of the material of wires X and Y are in the ratio 4:1 and the areas of cross sections of the wires X and Y are in the ratio $2: 1$. If the same amount of load is applied to both the wires, the ratio of elongation produced in the wires X and Y will be: (Assume length of the wires X and Y initially are the same) (A) $1: 8$ (B) $1: 1$ (C) $8: 1$ (D) $1: 2$

   ›Reveal solutionSolution

   Using Hooke’s law for elastic deformation, elongation is inversely proportional to Young’s modulus and cross‑sectional area. Given the ratios, the elongation ratio of X to Y is 1:8, so option (A) is correct.

   Concept & Intuition

   When the same load (force) is applied to two wires of equal initial length, the amount each stretches depends on how stiff the material is (Young’s modulus $Y$) and how thick the wire is (cross‑sectional area $A$). A larger Young’s modulus means the material resists stretching more; a larger area also resists stretching more. So elongation is inversely proportional to both $Y$ and $A$. The problem gives ratios for $Y$ and $A$, so we combine them to find the ratio of elongations.

   Step‑by‑step reasoning

   1. Recall the elongation formula For a wire of length $L$, cross‑sectional area $A$, Young’s modulus $Y$, under a tensile force $F$, the elongation $\Delta L$ is given by Hooke’s law:

   $\Delta L = \frac{F L}{A Y}.$

   Here $F$ and $L$ are the same for both wires (same load, same initial length), so $\Delta L \propto \frac{1}{A Y}$.

   2. Write the given ratios

      • Young’s modulus: $Y_X : Y_Y = 4 : 1$ → $\frac{Y_X}{Y_Y} = 4$.
      • Cross‑sectional area: $A_X : A_Y = 2 : 1$ → $\frac{A_X}{A_Y} = 2$.

   3. Set up the ratio of elongations

   $\frac{\Delta L_X}{\Delta L_Y} = \frac{ \frac{1}{A_X Y_X} }{ \frac{1}{A_Y Y_Y} } = \frac{A_Y Y_Y}{A_X Y_X}.$

   4. Substitute the known ratios $$$$

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.The temperature of a wire is doubled. The Young's modulus of elasticity (A) Will decrease (B) Will also double (C) Will become four times (D) Will remain the same

   ›Reveal solutionSolution

   Young's modulus measures a material's stiffness, which depends on interatomic forces. When temperature rises, atoms vibrate more, weakening these bonds and reducing the modulus. Doubling the temperature (in Kelvin) decreases Young's modulus, so the correct choice is (A).

   Concept & Intuition

   Young's modulus $Y$ is defined as the ratio of stress to strain in the elastic region:

   $Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}.$

   It reflects how strongly atoms resist being pulled apart. At the atomic level, the modulus is proportional to the curvature of the interatomic potential energy curve near equilibrium. When temperature increases, atoms gain thermal energy and vibrate with larger amplitudes. This effectively reduces the "stiffness" of the bonds — the potential well becomes shallower on average — so the material becomes more compliant. Hence, Young's modulus decreases with increasing temperature for most solids.

   Step-by-step reasoning

   1. Recall the temperature dependence of elastic moduli For most crystalline solids, Young's modulus decreases approximately linearly with temperature over a wide range. The empirical relation is

   $Y(T) = Y_0 \left[1 - \alpha (T - T_0)\right],$

   where $\alpha$ is a positive constant (the temperature coefficient of the modulus). Doubling the temperature (say from $T$ to $2T$) means a large increase, so $Y$ will drop.

   2. Why it cannot double or stay the same

      • If $Y$ doubled, the material would become twice as stiff — opposite to the known effect of heating (bonds weaken).
      • If $Y$ remained the same, temperature would have no effect on bond strength, which contradicts the observed thermal expansion and softening.
      • Becoming four times is even more unrealistic.

   3. Consider the atomic picture …

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8. COMEDK 2024Set 2024-M1 markMCQ
   Q.If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ respectively, then the corresponding ratio of increase in their lengths would be (A) $\frac{a}{3 b^2 c}$ (B) $\frac{3 a}{2 b^2 c}$ (C) $\frac{2 a}{3 b^2 c}$ (D) $\frac{2 \mathrm{ab}^2}{c}$

   ›Reveal solutionSolution

   Using $\Delta L=\frac{FL}{\pi r^2 Y}$ for each wire, with the brass wire carrying $3Mg$ and the steel wire $2Mg$, the ratio of extensions is $\frac{2a}{3b^2c}$. The correct option is (C).

   Concept

   A loaded wire extends by $\Delta L=\dfrac{FL}{AY}=\dfrac{FL}{\pi r^2 Y}$. The two wires carry different tensions — the upper (brass) wire supports both hanging masses while the lower (steel) wire supports only the bottom one — so their tensions differ before the given length/radius/modulus ratios are applied.

   Solution

   1. Tensions. $F_{\text{brass}}=(M+2M)g=3Mg$, $F_{\text{steel}}=2Mg$.
   2. Extension formula. $\Delta L=\dfrac{FL}{\pi r^2 Y}$ for each wire.
   3. Given ratios (steel : brass). $\dfrac{L_s}{L_b}=a,\ \dfrac{r_s}{r_b}=b,\ \dfrac{Y_s}{Y_b}=c.$
   4. Ratio of extensions. …

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9. COMEDK 2023Set 2023-E1 markMCQ
   Q.A man grows into a giant such that his height increases to 8 times his original height. Assuming that his density remains same, the stress in the leg will change by a factor of (A) $2 \sqrt{2}$ (B) 4 (C) 16 (D) 8

   ›Reveal solutionSolution

   (This is exactly why giants are structurally impossible at the same material strength - stress grows in proportion to height.)

   Concept: scaling of stress with linear size for a geometrically similar body of constant density.

   Stress in the leg = (weight supported) / (cross-sectional area of the leg)

   = (density * volume * g) / area

   Under a uniform scale-up by a factor n = 8 (all lengths multiplied by 8, density unchanged):

   • volume scales as L^3 -> 8^3 = 512 times, so weight scales 512 times …

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10. COMEDK 2023Set 2023-M1 markMCQ
    Q.Two wire of same material having radius in ratio 2 : 1 and lengths in ratio 1: 2. If same force is applied on them, then ratio of their change in length will be (A) $1: 1$ (B) $1: 2$ (C) $1: 4$ (D) $1: 8$

    ›Reveal solutionSolution

    Using $\Delta l\propto L/r^2$ with lengths $1{:}2$ and radii $2{:}1$ gives a change-in-length ratio of $1{:}8$.

    Elongation under a force $F$:

    $\Delta l=\frac{FL}{AY}=\frac{FL}{\pi r^2 Y}.$

    Same material ($Y$ common) and same force $F$, so $\Delta l\propto \dfrac{L}{r^2}$.

    Given $r_1:r_2=2:1$ and $L_1:L_2=1:2$: …

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11. COMEDK 2023Set 2023-M1 markMCQ
    Q.Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3 A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount? (A) 4F (B) 6F (C) 9F (D) F

    ›Reveal solutionSolution

    Equal volume forces $L_2=L_1/3$; with $F\propto A/L$ for the same $\Delta l$, the second wire needs $3\times3=9$ times the force.

    Required force for a given extension $\Delta l$:

    $F=\frac{YA\,\Delta l}{L}.$

    Equal volume $V=AL$ means $L=V/A$, so:

    $F=\frac{YA\,\Delta l}{V/A}=\frac{YA^2\,\Delta l}{V}\propto A^2.$ …

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12. COMEDK 2022Set 20221 markMCQ
    Q.A copper and a steel wire of same diameter are connected end to end. A deforming force F$_1$ is applied to the wire which causes an elongation of 1 cm. The two wires will have (A) the same stress (B) different stress (C) the same strain (D) different strain

    ›Reveal solutionSolution

    [!TLDR]

    In a series (end-to-end) arrangement both wires carry the same force over the same area, so they experience the same stress.

    Concept

    Stress is defined as force per unit area, $\sigma = F/A$ (CBSE Class 11 elasticity). For wires joined end to end, the applied force is transmitted equally through each section.

    Solution

    The two wires are connected end to end, so the same force $F_1$ acts on both. They have the same diameter, hence the same cross-sectional area $A$. Therefore

    $\sigma_{Cu} = \frac{F_1}{A} = \sigma_{steel},$ …

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