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The Vault
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The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Until now every body you met was rigid, but real steel cables stretch and real bridges sag. This chapter treats deformation quantitatively through stress and strain, and Hooke's law links the two by a material constant called the modulus of elasticity. Learning to read a stress-strain graph tells you when a material bends back and when it stays bent for good.
The SI unit of strain is:
Answer
It has no unit — strain is a ratio of two lengths (or two volumes), so the units cancel and it is a pure number. N/m² is the unit of stress and of the moduli of elasticity, which is why the two are often confused.
Two wires of the same material and the same length have radii in the ratio 1 : 2. When the same force is applied to each, the ratio of their elongations is:
Answer
4 : 1 — from Y = FL/(AΔL), ΔL = FL/(AY), so for the same F, L and Y the elongation is inversely proportional to the area, and hence to r². With radii in the ratio 1 : 2 the areas are in the ratio 1 : 4, so the elongations are in the ratio 4 : 1. The thinner wire stretches four times as much.
Which of the following materials is most elastic in the physics sense?
Answer
Steel, because it has a large Young's modulus — in physics, elasticity means the ability to resist deformation and to recover the original shape, measured by the modulus. Steel needs a very large stress for a given strain and springs back precisely, so it is more elastic than rubber, which stretches enormously for a small stress and does not follow Hooke's law. Everyday language uses 'elastic' in the opposite sense, which is the trap in this question.
A wire is stretched so that its strain is 0.002 and the stress in it is 4×10⁸ N/m². Young's modulus of the material is:
Answer
2×10¹¹ N/m² — using Y = stress/strain = 4×10⁸/0.002 = 4×10⁸/2×10⁻³ = 2×10¹¹ N/m². This is close to the value for steel, which is a useful order-of-magnitude check.
Assertion (A): Steel is preferred over copper for the cables of a suspension bridge. Reason (R): The Young's modulus of steel is greater than that of copper.
Answer
Both A and R are true and R is the correct explanation of A — Young's modulus of steel is roughly twice that of copper, so for the same load and the same cross-section a steel cable stretches only about half as much. A bridge deck must not sag appreciably as traffic loads change, so the stiffer material is the right engineering choice, and the higher modulus is precisely the reason. Steel's higher tensile strength is an additional benefit, but the stiffness argument alone justifies the assertion.
A metal wire of length 2 m and cross-sectional area 1×10⁻⁶ m² stretches by 1 mm under a load of 100 N. Calculate Young's modulus of the metal.
Answer
Y = FL/(AΔL) Y = (100 × 2)/(1×10⁻⁶ × 1×10⁻³) Y = 200/(1×10⁻⁹) Y = 2×10¹¹ N/m². Remember to convert the 1 mm elongation to 1×10⁻³ m before substituting.
A solid metal cube of side 0.1 m is subjected to a uniform pressure increase of 5×10⁶ Pa. If the bulk modulus of the metal is 1.25×10¹¹ Pa, find the change in its volume.
Answer
Original volume V = (0.1)³ = 1×10⁻³ m³. Bulk modulus B = −P/(ΔV/V), so the magnitude of the fractional change is: ΔV/V = P/B = 5×10⁶/1.25×10¹¹ = 4×10⁻⁵. ΔV = 4×10⁻⁵ × 1×10⁻³ = 4×10⁻⁸ m³. The volume decreases by 4×10⁻⁸ m³, that is by 0.04 cm³, a tiny fraction as expected for a metal with a very large bulk modulus.
Derive an expression for the elastic potential energy stored per unit volume in a stretched wire. A steel wire of length 3 m and cross-sectional area 2×10⁻⁶ m² is stretched by 1.5 mm; taking Y = 2×10¹¹ N/m², find the energy stored in it.
Answer
Derivation: consider a wire of natural length L and cross-sectional area A, being stretched slowly by a gradually increasing force. Suppose at some stage the extension is x and the corresponding force is F. From the definition of Young's modulus, Y = (F/A)/(x/L), so F = YAx/L. The force is not constant during the stretching — it grows from zero to its final value — so the work must be integrated. The work done in producing a further infinitesimal extension dx is dW = F dx = (YA/L) x dx. Integrating from x = 0 to the final extension ΔL: W = (YA/L) ∫x dx from 0 to ΔL = (YA/L)(ΔL²/2) = ½ (YAΔL/L) ΔL. Since YAΔL/L is just the final stretching force F, this is W = ½FΔL — the familiar 'half the force times the extension', the factor ½ arising because the average force during stretching is half the final force. Energy per unit volume: divide by the volume AL, u = W/(AL) = (½FΔL)/(AL) = ½ × (F/A) × (ΔL/L) = ½ × stress × strain. Using stress = Y × strain, this can also be written u = ½Y(strain)² = (stress)²/2Y. Calculation: strain = ΔL/L = 1.5×10⁻³/3 = 5×10⁻⁴. Stress = Y × strain = 2×10¹¹ × 5×10⁻⁴ = 1×10⁸ N/m². u = ½ × 1×10⁸ × 5×10⁻⁴ = 2.5×10⁴ J/m³. Volume = AL = 2×10⁻⁶ × 3 = 6×10⁻⁶ m³. Energy stored = 2.5×10⁴ × 6×10⁻⁶ = 0.15 J.
Draw and describe the stress-strain curve for a ductile metal, marking the proportional limit, elastic limit, yield point, ultimate tensile strength and fracture point. Explain what happens to the specimen in each region and why a crane's lifting cable must always be operated well below the yield point.
Answer
Description of the curve: plot stress on the vertical axis against strain on the horizontal axis for a metal wire loaded steadily to breaking. 1. From the origin to point P, the proportional limit, the graph is a straight line. Here stress ∝ strain, Hooke's law holds exactly, and the slope of this line is Young's modulus. Removing the load returns the wire to its exact original length. 2. From P to E, the elastic limit, the curve bends slightly but the deformation is still fully recoverable. Hooke's law no longer holds, yet the wire still returns to its original length when unloaded. 3. Just beyond E lies the yield point Y. Past this point the wire acquires a permanent set: on unloading it does not come back to its original length, and a residual strain remains. 4. From Y onward is the plastic region, where a large increase in strain is produced by a modest increase in stress. The metal flows and the wire visibly thins and lengthens. A ductile metal such as copper or mild steel has a long plastic region, which is exactly why it can be drawn into wires. 5. The highest point on the curve, U, is the ultimate tensile strength — the maximum stress the material can withstand. Beyond U a narrow neck forms at some point along the wire, the local area shrinks rapidly, and the curve falls until fracture occurs at point F. Why a crane cable is kept well below the yield point: once the yield point is crossed, the cable lengthens permanently and never recovers, and each subsequent lift adds more permanent strain and thins the cable further. That thinning raises the actual stress for the same load, so the process runs away towards fracture with no obvious warning to the operator. Engineers therefore fix a working stress several times smaller than the yield stress; the ratio of the two is called the factor of safety, and it also absorbs shock loads, sudden jerks, corrosion and gradual fatigue that a static calculation would miss.
Read the following and answer the questions that follow: A workshop tests two vertical wires of equal length 4 m hanging from a rigid support. Wire P is steel with Y = 2×10¹¹ Pa and area 1×10⁻⁶ m², and wire Q is copper with Y = 1.1×10¹¹ Pa and the same area. Each is loaded with a 20 kg mass (g = 10 m/s²). (a) Calculate the stress in each wire. (b) Calculate the elongation of each wire. (c) Explain why both wires have the same stress but different elongations. (d) The load on the steel wire is slowly increased until the wire snaps. Sketch in words how the graph of load against extension would look up to that moment.
Answer
(a) The load is the same for both: F = mg = 20 × 10 = 200 N. The areas are equal, so stress = F/A = 200/1×10⁻⁶ = 2×10⁸ Pa in each wire. (b) For the steel wire: ΔL = FL/(AY) = (200 × 4)/(1×10⁻⁶ × 2×10¹¹) = 800/(2×10⁵) = 4×10⁻³ m = 4 mm. For the copper wire: ΔL = (200 × 4)/(1×10⁻⁶ × 1.1×10¹¹) = 800/(1.1×10⁵) ≈ 7.27×10⁻³ m ≈ 7.3 mm. (c) Stress depends only on the applied force and the cross-sectional area, and both are identical here, so the stress must be the same. Elongation, however, is governed by strain = stress/Y, and Young's modulus is a property of the material. Copper's smaller modulus means it yields more strain for the same stress, so the copper wire stretches roughly 1.8 times as far — the ratio 2×10¹¹/1.1×10¹¹ ≈ 1.8, exactly matching 7.3/4. (d) The graph would begin as a straight line through the origin, since load ∝ extension while Hooke's law holds; the slope of this line is YA/L. Near the elastic limit the line would start to curve gently to the right as extension grows faster than load. At the yield point the curve would flatten sharply, with the wire extending a great deal for very little extra load. It would then rise slowly to a maximum at the ultimate tensile strength, after which a neck forms; from there the curve falls as the load the wire can carry drops, ending abruptly at fracture. The area under this graph at any stage gives the work done in stretching the wire.
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