JEE Physics Mechanical Properties Of Solids MCQs Set 04

Answer: (c) \( \dfrac{W_1+3W/4}{S} \)

Explanation:
1. The part of the wire below the point at 3L/4 from the lower end has weight 3W/4.
2. The section at that point carries this weight plus W1, so stress = (W1 + 3W/4)/S.

Teacher's Note:
1. Tension at any section equals the weight hanging below it.
2. Divide by the area to get stress.

Answer: (b) 0.004

Explanation:
1. Angle of shear φ satisfies L φ = r θ.
2. φ = r θ / L = 0.01 × 0.8 / 2 = 0.004.

Teacher's Note:
1. The outer surface of the rod moves through an arc r θ while the length is L.
2. Shear angle is that arc divided by L.

Answer: (b) 0.09°

Explanation:
1. L φ = r θ, so φ = r θ / L = (2 × 10⁻³ × 45°) / 1.
2. This gives φ = 0.09°.

Teacher's Note:
1. The relation holds for any unit of angle if θ and φ are in the same unit.
2. Keep r and L in metres.

Answer: (d) ℓ

Explanation:
1. The area is A = V/ℓ, so strain = F/(A Y) = F ℓ / (V Y).
2. Hence strain is proportional to ℓ.

Teacher's Note:
1. A fixed volume means area and length change together.
2. Express the area in terms of length before substituting.

Answer: (a) 0.5 e

Explanation:
1. e = F ℓ / (π r² Y).
2. For tension 2F and radius 2r, e' = 2F ℓ / (π 4r² Y) = e/2.

Teacher's Note:
1. Elongation is proportional to F and inversely proportional to r².
2. Doubling both gives a net factor 2/4.

Answer: (d) Nearly 226 N

Explanation:
1. F = Y A (strain) = 0.9 × 10¹¹ × π (0.4 × 10⁻³)² × 0.005.
2. This is about 226 N.

Teacher's Note:
1. Use the radius, not the diameter, in the area.
2. 0.5% means a strain of 0.005.

Answer: (d) \( \dfrac{r_2^2}{r_1^2} \)

Explanation:
1. Stress = F / (π r²), and F is the same for both wires.
2. So stress1 / stress2 = r2² / r1².

Teacher's Note:
1. For equal load, the thinner wire has the larger stress.
2. Stress is inversely proportional to the area.

Answer: (b) 1 : 2

Explanation:
1. For identical wires and equal loads, extension is inversely proportional to Y.
2. Ratio = 11 : 22 = 1 : 2.

Teacher's Note:
1. Larger Y means a stiffer wire and smaller extension.
2. Take the ratio in reverse order.

Answer: (a) \( \dfrac{\Delta V}{V} \propto C \)

Explanation:
1. Bulk modulus B = p / (ΔV/V) and compressibility C = 1/B.
2. So ΔV/V = p C, which is proportional to C for constant p.

Teacher's Note:
1. Compressibility is the reciprocal of bulk modulus.
2. A more compressible object changes volume more.

Answer: (a) 4.9 × 10⁸ N/m²

Explanation:
1. Extra pressure at the bottom = ρ g h = 1000 × 9.8 × 100 = 9.8 × 10⁵ Pa.
2. B = Δp / (ΔV/V) = 9.8 × 10⁵ / 0.002 = 4.9 × 10⁸ N/m².

Teacher's Note:
1. 0.2% means a fractional change of 0.002.
2. The bulk modulus of a solid should be large.

Answer: (a) 188 kPa

Explanation:
1. Δp = B × (ΔV/V) = 4700 × 10⁶ × 4 × 10⁻⁵.
2. This gives 1.88 × 10⁵ Pa = 188 kPa.

Teacher's Note:
1. 0.004% = 4 × 10⁻⁵ as a fraction.
2. The total volume of 200 litres is not needed.

Answer: (c) 0.4

Explanation:
1. Y = 2η(1 + σ), so 2.8 η = 2η(1 + σ).
2. 1 + σ = 1.4, so σ = 0.4.

Teacher's Note:
1. Poisson's ratio of ordinary materials lies between 0 and 0.5.
2. That rules out options 2.4 and 1.2.

Answer: (d) 112

Explanation:
1. Force constant k = Y × r0, where r0 is the interatomic spacing.
2. k = 2 × 10¹¹ × 5.6 × 10⁻¹⁰ = 112 N/m.

Teacher's Note:
1. Each atomic bond acts like a spring.
2. Y = (k/r0) comes from F/A = Y × (Δr/r0).

Answer: (b) 0.7

Explanation:
1. Poisson's ratio for a stable material lies between −1 and +0.5.
2. The value 0.7 is outside this range.

Teacher's Note:
1. The upper limit 0.5 corresponds to an incompressible material.
2. Remember that Poisson's ratio cannot exceed 0.5.

Answer: (a) rises

Explanation:
1. The wire stores elastic potential energy while it is stretched.
2. When it breaks suddenly, this energy is released inside the wire as heat, so the temperature rises.

Teacher's Note:
1. Elastic energy cannot disappear.
2. It is converted into thermal energy when the wire breaks.

Answer: (b) 6 × 10⁻³ J

Explanation:
1. The force rises from zero to mg = 4 N as the wire stretches.
2. Work = ½ × 4 × 3 × 10⁻³ = 6 × 10⁻³ J.

Teacher's Note:
1. Use the average force, ½ F x, for a gradually applied load.
2. F x alone would be twice as large.

Answer: (d) 314 N

Explanation:
1. F = Y A (strain) = 4 × 10¹¹ × π (0.5 × 10⁻³)² × 10⁻³.
2. This is about 314 N.

Teacher's Note:
1. The area uses the radius 0.5 mm.
2. A strain limit gives the force limit directly.

Answer: (a) \( \sqrt{\dfrac{2E}{Y}} \)

Explanation:
1. Energy per unit volume E = ½ Y (strain)².
2. So strain = √(2E/Y).

Teacher's Note:
1. Square the strain in the energy formula.
2. Check that the answer has no unit.

Answer: (d) (a), (b) and (c)

Explanation:
1. Steel has a much larger Young's modulus than rubber, so (a) is true.
2. Energy = ½ F ΔL = Y A (ΔL)²/(2L), so (b) is true, and breaking stress is a property of the material, so (c) is true.

Teacher's Note:
1. Elasticity means resistance to deformation and is measured by the modulus.
2. Steel resists deformation better than rubber.

Answer: (c) half as elastic potential energy and half as thermal energy

Explanation:
1. The elastic energy stored in the stretched wire is ½ Mg ℓ.
2. The remaining ½ Mg ℓ is dissipated as heat during the oscillations that settle down.

Teacher's Note:
1. Energy is conserved.
2. Only half goes into the wire as elastic energy, because the force grows gradually from 0 to Mg.

Answer: (d) OD

Explanation:
1. Load = (Y A / L) × elongation, so the slope is proportional to the area A for the same material and length.
2. The thickest wire has the largest slope, which is line OD.

Teacher's Note:
1. Slope of load against elongation is Y A / L.
2. Greater thickness means greater slope.

Answer: (b) Δℓ versus ℓ²

Explanation:
1. A = V/ℓ, so Δℓ = F ℓ / (A Y) = F ℓ² / (V Y).
2. Δℓ is proportional to ℓ², so the graph of Δℓ against ℓ² is a straight line through the origin.

Teacher's Note:
1. Express the area through the fixed volume.
2. A graph is a straight line when the quantities plotted are directly proportional.

Answer: (c) 2 : 1

Explanation:
1. The lower ends stay level when both wires have the same extension.
2. Extension = F L / (A Y), so F is proportional to Y and the weights are in the ratio 2 : 1.

Teacher's Note:
1. Same extension, same length and area means force proportional to Y.
2. The stiffer wire carries more weight.

Answer: (c) \( \dfrac{p}{3B} \)

Explanation:
1. ΔV/V = p / B and V is proportional to R³, so ΔV/V = 3 ΔR/R.
2. Therefore ΔR/R = p / (3B).

Teacher's Note:
1. Volume depends on the cube of the radius.
2. The fractional change in radius is one third of the fractional change in volume.

Answer: (a) 2 × 10⁹ N/m²

Explanation:
1. Strain = extension / 1 m, so the strain goes from 2 × 10⁻³ to 4 × 10⁻³ while the stress goes from 4 × 10⁶ to 8 × 10⁶ N/m².
2. Y = Δstress / Δstrain = 4 × 10⁶ / (2 × 10⁻³) = 2 × 10⁹ N/m².

Teacher's Note:
1. Y is the slope of the stress-strain line.
2. Convert kN/m² to N/m² and mm to m.

Answer: (d) all of these

Explanation:
1. Within the elastic limit, strain is proportional to stress, extension is proportional to stress (for a given wire) and increase in length is proportional to the weight hung.
2. Each pair gives a straight line through the origin.

Teacher's Note:
1. Hooke's law gives direct proportionality.
2. Any such pair of directly proportional quantities gives a straight line.

Answer: (b) brittle and ductile

Explanation:
1. A brittle material breaks soon after reaching its ultimate strength, so the two points are close.
2. A ductile material stretches a lot between the ultimate strength and the fracture point, so the points are far apart.

Teacher's Note:
1. Brittle materials fracture with little plastic deformation.
2. Ductile materials such as copper can be drawn into wires.