Q.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.

Showing the 12 most recent of 66 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A rubber hose 50 cm long and internal diameter 1 cm is stretched to 60 cm. The internal diameter of stretched hose is (Poisson's ratio of the rubber $\sigma=0.5$) (A) 8 mm (B) 9 mm (C) 10 mm (D) 7 mm

   ›Reveal solutionSolution

   Stretching a rubber hose lengthwise causes it to contract sideways — the amount is governed by Poisson's ratio, which links the lateral (diameter) strain to the longitudinal (length) strain. Answer: (B) 9 mm.

   Concept and Intuition

   When a material is stretched along its length, it typically becomes thinner in the perpendicular directions — this is a nearly universal elastic effect, quantified by Poisson's ratio $\sigma=-\dfrac{\text{lateral strain}}{\text{longitudinal strain}}$ (the negative sign reflects that the two strains have opposite signs: length increases, diameter decreases). A larger $\sigma$ (up to a maximum of 0.5 for an incompressible material like rubber) means more sideways contraction per unit of stretch.

   Step-by-Step Solution

   1. Longitudinal strain: the hose stretches from 50 cm to 60 cm, so

   $\varepsilon_{long}=\frac{\Delta L}{L}=\frac{60-50}{50}=\frac{10}{50}=0.2.$

   2. Lateral strain from Poisson's ratio:

   $\varepsilon_{lat}=-\sigma\,\varepsilon_{long}=-0.5\times0.2=-0.1.$

   (The diameter shrinks by 10%.)

   3. New diameter:

   $d'=d(1+\varepsilon_{lat})=1\ cm\times(1-0.1)=0.9\ cm=9\ mm.$ …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A book of dimension 4 cm × 1.5 cm × 10 cm is kept in a way that the 10 cm edge is vertical. A horizontal force of 3 N is applied at the top face. If the shear modulus of the book is $2\times10^5\ Nm^{-2}$, then the horizontal displacement of the top face will be (A) 0.5 mm (B) 2.5 mm (C) 5 mm (D) 10 mm

   ›Reveal solutionSolution

   This is a straightforward shear-modulus computation: identify the shear area (the horizontal cross-section, 4 cm × 1.5 cm) and the shear length (the vertical height, 10 cm), then solve $\Delta x = FL/(AG)$ to get 2.5 mm.

   Concept and Intuition

   Shear modulus $G$ relates a tangential (shearing) stress to the resulting shear strain: $G = \dfrac{\text{shear stress}}{\text{shear strain}} = \dfrac{F/A}{\Delta x/L}$, where $A$ is the area of the face parallel to the applied force (the face across which layers slide relative to each other), and $L$ is the distance perpendicular to that sliding, over which the shear angle builds up. Picture the book as a stack of horizontal layers, glued at the bottom; pushing the top layer sideways shears every layer relative to the one below it.

   Step-by-Step Solution

   1. The book's three edges are 4 cm, 1.5 cm, and 10 cm. It's held with the 10 cm edge vertical, so the base (bottom, fixed) and top (where the force is applied) faces are each $4\ \text{cm} \times 1.5\ \text{cm}$.
   2. Shear area: $A = 4 \times 1.5 = 6\ \text{cm}^2 = 6\times10^{-4}\ \text{m}^2$.
   3. The height over which the shear deformation develops (bottom fixed, top displaced) is the vertical edge: $L = 10\ \text{cm} = 0.1\ \text{m}$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A copper wire of negligible mass, length 1 m and area of cross-section $10^{-6}m^2$ is kept on a smooth horizontal table. One end of the wire is fixed and other end of the wire attached with a ball of mass 1 kg. If the ball and wire are rotating with $20\ rad\,s^{-1}$, an elongation of $10^{-3}m$ is observed in the wire, find its Young's modulus (A) $4 \times 10^{11}Nm^{-2}$ (B) $8 \times 10^{11}Nm^{-2}$ (C) $4 \times 10^{8}Nm^{-2}$ (D) $400\ Nm^{-2}$

   ›Reveal solutionSolution

   Tests combining circular-motion (centripetal force) with the definition of Young's modulus. Answer: $4\times10^{11}\ Nm^{-2}$.

   Concept and Intuition

   The wire, being massless, only needs to supply the centripetal force required to keep the ball moving in a circle — this force IS the tension in the wire, and it is this tension that stretches the wire elastically. Once we know the tension, the elongation observed lets us back out the material's Young's modulus via $Y=\dfrac{\text{stress}}{\text{strain}}=\dfrac{T/A}{\Delta L/L}$.

   Step-by-Step Solution

   1. Centripetal force needed for the ball: $F_c=m\omega^2 r$, where $r$ is the radius of the circular path, essentially equal to the wire's length $L=1$ m (elongation of $10^{-3}$ m is negligible compared to 1 m).
   2. $F_c=1\times(20)^2\times1=400$ N. This is the tension $T$ in the (massless) wire.
   3. Young's modulus: $Y=\dfrac{T/A}{\Delta L/L}=\dfrac{TL}{A\,\Delta L}$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The elastic behavior of a material for linear stress to linear strain is given in the figure (graph: Strain on the y-axis vs Stress in $N m^{-2}$ on the x-axis, a straight line through the origin passing through the points $(20, 10\times10^{-11})$, $(40, 20\times10^{-11})$ and $(60, 30\times10^{-11})$). The energy density for a linear strain of $4\times10^{-4}$ is (material is elastic upto linear strain of $4\times10^{-4}$) (A) $20000\ J\,m^{-3}$ (B) $16000\ J\,m^{-3}$ (C) $12000\ J\,m^{-3}$ (D) $8000\ J\,m^{-3}$

   ›Reveal solutionSolution

   This tests reading Young's modulus off a strain-vs-stress graph and using the elastic energy density formula. The answer is $16000\ J\,m^{-3}$.

   Concept and Intuition

   Elastic potential energy stored per unit volume in a strained material is $u = \tfrac12 \times \text{stress} \times \text{strain} = \tfrac12 Y (\text{strain})^2$, where $Y$ is Young's modulus (stress/strain). Since the given graph plots strain on the y-axis against stress on the x-axis, its slope directly gives $1/Y$ (the reciprocal of Young's modulus), not $Y$ itself — a detail worth watching for.

   Step-by-Step Solution

   1. From the graph, slope $= \dfrac{\Delta(\text{strain})}{\Delta(\text{stress})} = \dfrac{10\times10^{-11} - 0}{20 - 0} = 5\times10^{-12}\ m^2/N$.
   2. Since strain $= \dfrac{1}{Y}\times\text{stress}$, the slope equals $\dfrac{1}{Y}$: so $Y = \dfrac{1}{5\times10^{-12}} = 2\times10^{11}\ N\,m^{-2}$.
   3. Energy density for strain $\varepsilon = 4\times10^{-4}$ (material still elastic here, as given): $u = \tfrac12 Y \varepsilon^2$. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A spring is stretched by applying a load to its free end. The strain produced in the spring is (A) Volumetric (B) Shear (C) Longitudinal and shear (D) Volumetric and longitudinal

   ›Reveal solutionSolution

   When a spring is stretched, the wire experiences both longitudinal strain (due to axial tension) and shear strain (due to twisting of the coils), so the correct answer is (C).

   The key to this question is understanding that a spring is not a simple straight rod — it’s a coiled wire. When you pull on a spring, the wire itself undergoes a combination of deformations. Let’s break down why both longitudinal and shear strain appear.

   1. First, recall what strain types mean.

      • Longitudinal strain is the change in length per unit length along the direction of the applied force.
      • Shear strain is the angular distortion caused by forces parallel to a face.
      • Volumetric strain is the change in volume per unit volume, which occurs under uniform pressure (hydrostatic stress), not uniaxial tension.

   2. Consider the geometry of a spring.

      A typical helical spring is made of wire wound into a helix. When you apply a load along the axis of the spring, each small segment of the wire is not simply pulled straight — it is also twisted. This is because the coils are at an angle to the axis.

   3. Identify the dominant stress in the wire.

      The applied axial load creates a torque on each cross-section of the wire. This torque produces shear stress (and thus shear strain) in the wire. At the same time, the axial component of the load also stretches the wire along its own length, producing longitudinal strain. So the wire experiences both.

   4. Why not volumetric strain? …

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6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.When a wire of length 'L' and radius 'r', fixed at one end is stretched by a force F, the increase in its length is 'x'. If another wire made of same material of length '2L' and radius '2r' is stretched by force '2F', the increase in its length will be (A) $x$ (B) $2x$ (C) $\dfrac{x}{2}$ (D) $\dfrac{x}{4}$

   ›Reveal solutionSolution

   Tests the dependence of elastic extension on length, area and force via Young's modulus. All three scale factors (length ×2, area ×4, force ×2) cancel exactly, so the extension is unchanged: $x' = x$.

   Concept and Intuition

   Young's modulus relates stress and strain: $Y = \dfrac{F/A}{x/L}$, so the extension is

   $x = \dfrac{FL}{AY}$

   This tells us extension grows with force and original length, but shrinks with cross-sectional area (a thicker wire stretches less for the same force). Since area depends on $r^2$, doubling the radius has a much stronger (quadrupling) effect on reducing the extension than doubling the force or length has on increasing it — this is the intuition for why the numbers might cancel exactly here.

   Step-by-Step Solution

   1. Original wire: $x = \dfrac{FL}{\pi r^2 Y}$ (using $A = \pi r^2$).
   2. New wire has $F' = 2F$, $L' = 2L$, $r' = 2r$, same material so same $Y$.
   3. New area: $A' = \pi (2r)^2 = 4\pi r^2$.
   4. New extension: …

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7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A wire of weight W and area of cross-section A elongates under its own weight. If Y is the Young's modulus and $\sigma$ is the Poisson's ratio of the material of the wire, then the fractional change in the radius of the wire is (A) $\dfrac{2\sigma W}{AY}$ (B) $\dfrac{\sigma W}{2AY}$ (C) $\dfrac{\sigma W}{3AY}$ (D) $\dfrac{\sigma W}{AY}$

   ›Reveal solutionSolution

   A wire hanging under its own weight has tension varying linearly from $W$ at the top to zero at the bottom, so its average longitudinal stress (and hence strain) is effectively half of what a uniformly-loaded wire of the same weight would show. The average fractional radial change carries the same characteristic factor of $\tfrac12$, giving $\dfrac{\sigma W}{2AY}$.

   Concept and Intuition

   This is the "wire hangs under its own weight" variant of Hooke's law problems. Unlike a wire loaded only at its free end (where tension is uniform $=W$ throughout), a wire supporting its own weight has internal tension that depends on position: any cross-section only has to support the weight of the wire below it, so tension is maximum ($W$) at the point of suspension and zero at the free end. This linear variation is exactly why the elongation formula for a self-weighted wire, $\delta L = \dfrac{WL}{2AY}$, carries a factor of $\tfrac12$ compared to the uniformly-loaded case $\delta L = \dfrac{WL}{AY}$. Poisson's ratio then converts this longitudinal effect into a lateral (radial) contraction, and the same averaging carries the factor of $\tfrac12$ through.

   Step-by-Step Solution

   1. Let the wire have length $L$, cross-section $A$, weight $W$, and let $y$ be measured from the free (bottom) end.
   2. The tension at position $y$ supports only the weight of the wire below it: $T(y) = W\cdot\dfrac{y}{L}$.
   3. Local longitudinal stress: $\sigma_{long}(y) = \dfrac{T(y)}{A} = \dfrac{Wy}{LA}$; local longitudinal strain: $\epsilon(y) = \dfrac{\sigma_{long}(y)}{Y} = \dfrac{Wy}{LAY}$. …

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8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.When a uniform bar of length '$l$' breadth '$b$' and thickness '$d$' was supported by rigid supports near the ends and loaded at the center by a vehicle of mass $M$, it sags the bar by an amount $\delta$ is (Young's modulus of material is $Y$) (A) $\dfrac{Ml^3}{4bd^3Y}$ (B) $\dfrac{Yl^3}{4Mgbd^3}$ (C) $\dfrac{Yd^3}{4Mgl^3b}$ (D) $\dfrac{Mgl^3}{4bd^3Y}$

   ›Reveal solutionSolution

   This tests the standard beam-bending (elevation/depression) formula for a bar supported at both ends and loaded centrally. Answer: $\delta = \dfrac{Mgl^3}{4bd^3Y}$.

   Concept and Intuition

   When a rectangular bar rests on supports near its two ends and a load is placed at its center, the bar bends (sags). This depression $\delta$ depends on the applied load, the bar's geometry (length, breadth, thickness), and the elastic property (Young's modulus) of the material — this is the physics behind the classic 'uniform bending' / Searle's bar experiment.

   Step-by-Step Solution

   1. The load applied at the center is the weight of the vehicle: $W = Mg$.
   2. The standard formula (derived from beam theory, a standard NCERT-level result) for depression at the center of a bar of length $l$, breadth $b$, thickness $d$ supported at both ends is:

   $\delta = \dfrac{Wl^3}{4bd^3Y}$

   3. Substituting $W = Mg$: $\delta = \dfrac{Mgl^3}{4bd^3Y}$ …

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9. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The work done in stretching a wire by 1 mm is 2 J. The work necessary for stretching another wire of the same material but triple the radius and one third of length by 1 mm is (A) 8 J (B) 27 J (C) 54 J (D) 18 J

   ›Reveal solutionSolution

   Tests how elastic work-done depends on a wire's geometry (radius and length) through Young's modulus, for the same fixed extension.

   Concept and Intuition

   Stretching a wire stores elastic potential energy. For a given material and a given absolute extension $\Delta L$, a thicker wire needs a much larger force (force $\propto$ area) while a shorter wire is stiffer (force $\propto 1/L$) — both make the work done larger. Since work is $\tfrac12\times\text{force}\times\text{extension}$, it inherits both dependences.

   Step-by-Step Solution

   1. From Young's modulus, $Y = \dfrac{F L}{A\,\Delta L} \Rightarrow F = \dfrac{YA\,\Delta L}{L}$.
   2. Work done in stretching by $\Delta L$: $W = \tfrac12 F \Delta L = \dfrac{YA(\Delta L)^2}{2L}$.
   3. With $A=\pi r^2$: $W = \dfrac{Y\pi r^2(\Delta L)^2}{2L} \propto \dfrac{r^2}{L}$ (same material $Y$, same $\Delta L=1$ mm for both wires).
   4. Wire 1: radius $r$, length $L$, $W_1 = 2\,\text{J}$. Wire 2: radius $3r$, length $L/3$. …

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10. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
    Q.The areas of cross-section of two wires A and B of same length made of different materials are $2 \times 10^{-6}\ \mathrm{m^2}$ and $4 \times 10^{-6}\ \mathrm{m^2}$ respectively. If the ratio of Young's moduli of materials of the wires A and B is 2 : 3, the elongations in the wires A and B are 1.2 mm and 1.8 mm respectively, then the ratio of energies stored in the wires A and B is (A) 8 : 27 (B) 2 : 3 (C) 4 : 27 (D) 4 : 9

    ›Reveal solutionSolution

    Combining the given Young's modulus ratio, area ratio, and elongation ratio in the elastic energy formula gives an energy ratio of 4:27.

    Concept and Intuition

    The elastic potential energy stored in a stretched wire can be written using Hooke's-law-based expressions. Starting from $F = \dfrac{YA}{L}\Delta L$ (Young's modulus relation) and energy $U=\tfrac12F\Delta L$, we get

    $U = \frac12\cdot\frac{YA}{L}(\Delta L)^2$

    Since both wires have the same length $L$, that factor cancels in the ratio, leaving only $Y$, $A$, and $(\Delta L)^2$ to compare.

    Step-by-Step Solution

    1. Energy formula: $U = \dfrac{YA(\Delta L)^2}{2L}$.
    2. Ratio (with common $L$ cancelling): $\frac{U_A}{U_B} = \frac{Y_A}{Y_B}\cdot\frac{A_A}{A_B}\cdot\left(\frac{\Delta L_A}{\Delta L_B}\right)^2$ …

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11. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.A uniform metal wire is suspended from a rigid ceiling and a solid sphere is attached to the second end of the wire. If the radius of the sphere is doubled and then immersed in a liquid whose density is 60% of the density of the material of the sphere, then the percentage increase in the elongation of the wire is (A) $420$ (B) $120$ (C) $220$ (D) $320$

    ›Reveal solutionSolution

    This tests Hooke's law elongation combined with buoyancy — doubling the sphere's radius increases its volume 8-fold, while immersion in a liquid reduces the effective weight by the buoyant fraction, netting a $220\%$ increase in elongation.

    Concept and Intuition

    Elongation of a wire under Young's modulus is $\Delta L = \dfrac{FL}{AY}$, directly proportional to the tension $F$ in the wire (since $L$, $A$, $Y$ of the wire are unchanged). Initially the tension is just the sphere's weight in air. After the radius is doubled, the volume (and hence mass, and hence weight) scales as $r^3$, i.e., by a factor of $2^3=8$. But now the sphere is immersed in a liquid, so buoyancy reduces the effective weight — the net downward force becomes weight minus buoyant force, which scales with $(\rho_{sphere}-\rho_{liquid})$ instead of just $\rho_{sphere}$. Combining the volume increase (8×) with the reduced effective density fraction gives the new tension as a multiple of the old, from which the percentage increase follows.

    Step-by-Step Solution

    1. Let the sphere's material density be $\rho$ and original radius $r$. Initial tension (weight in air): $T_1 = \rho \left(\tfrac43\pi r^3\right) g$.
    2. New radius $=2r$; new volume $= \tfrac43\pi(2r)^3 = 8\times\tfrac43\pi r^3$ (8× the original volume).
    3. Liquid density $=0.6\rho$. Effective (net) density supporting the wire $= \rho - 0.6\rho = 0.4\rho$. …

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12. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
    Q.A wire can sustain a weight of 100 kg before it breaks. The wire is cut into two equal parts. Without breaking each part can hold weight up to (A) 50 kg (B) 200 kg (C) 100 kg (D) 40 kg

    ›Reveal solutionSolution

    Breaking strength depends on cross-sectional area, not length, so cutting the wire in half does not change how much weight each half can hold: still 100 kg.

    Concept and Intuition

    A material has a characteristic breaking (ultimate tensile) stress $\sigma_{max}$. The maximum load a wire can bear before breaking is $F_{max}=\sigma_{max}\times A$, where $A$ is the cross-sectional area. This is a property of the cross-section, independent of how long the wire is.

    Step-by-Step Solution

    1. The original wire (full length) breaks at 100 kg, so $\sigma_{max}A = 100\,\mathrm{kg}\times g$ (in force units).
    2. Cutting the wire into two equal parts changes only the length of each piece; the cross-sectional area $A$ and the material (hence $\sigma_{max}$) remain unchanged. …

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