JEE Physics Mechanical Properties Of Solids MCQs Set 12

Answer: (c) infinity

Explanation:
1. A perfectly rigid body does not deform at all.
2. So the strain is zero for any stress and Y = stress/strain is infinite.

Teacher's Note:
1. Rigid means infinite Y.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The definite volume of solids (strong inter-atomic forces) makes them nearly incompressible, whereas gases have no fixed volume.
2. So the reason explains the assertion.

Teacher's Note:
1. Bulk modulus of solids is large.

Answer: (b) Hydraulic stress

Explanation:
1. Hydraulic stress acts equally on all surfaces.
2. It changes only the volume, not the shape.

Teacher's Note:
1. Uniform pressure keeps the shape.

Answer: (d) \( YX^2/2 \)

Explanation:
1. Energy per unit volume = ½ × stress × strain.
2. = ½ (YX)(X) = YX²/2.

Teacher's Note:
1. u = ½ × stress × strain.

Answer: (b) \( 1.2\times10^{-2} \)

Explanation:
1. P = ρgh = 10³ × 10 × 2700 = 2.7 × 10⁷ Pa.
2. ΔV/V = P × compressibility = 2.7 × 10⁷ × 45.4 × 10⁻¹¹ = 1.2 × 10⁻².

Teacher's Note:
1. Strain = pressure × compressibility.

Answer: (c) bulk modulus

Explanation:
1. Fluids cannot sustain shear and do not have a fixed length.
2. They resist only volume change, which is described by the bulk modulus.

Teacher's Note:
1. Fluids have only B.

Answer: (d) one-fourth for wire A as for wire B

Explanation:
1. Δl ∝ 1/A ∝ 1/d².
2. Wire A's diameter is twice, so its extension is one-fourth of B's.

Teacher's Note:
1. Extension varies as 1/d².

Answer: (b) 0.004

Explanation:
1. Shear strain at the surface φ = rθ/L.
2. = 0.01 × 0.8/2 = 0.004.

Teacher's Note:
1. φ = rθ/L.

Answer: (b) \( \frac12\frac{\rho gL^2}{Y} \)

Explanation:
1. The stress at a distance x from the free end is ρgx.
2. ΔL = ∫(ρgx/Y)dx from 0 to L = ρgL²/2Y.

Teacher's Note:
1. Integrate the varying stress.

Answer: (a) 2 : 1

Explanation:
1. Equal extension requires F ∝ Y for the same length and area.
2. The weights are in the ratio 2 : 1.

Teacher's Note:
1. Δl = Fl/AY.

Answer: (c) III only

Explanation:
1. I is true, since a hollow shaft has a greater torsional rigidity for the same mass.
2. II is true.
3. The torque needed for a given twist is inversely proportional to the length, so a long rod is easier to twist; III is wrong.

Teacher's Note:
1. Twisting torque is inversely proportional to length.

Answer: (d) I, II and III

Explanation:
1. Liquids cannot be sheared elastically, so G = 0.
2. Heating weakens the inter-atomic bonds, lowering Y.
3. Poisson's ratio is a ratio of two strains, so it has no unit.

Teacher's Note:
1. All three statements are correct.

Answer: (c) stress

Explanation:
1. Stress = restoring force per unit area, equal in magnitude to the applied force per unit area at equilibrium.
2. Strain is the fractional deformation.

Teacher's Note:
1. Stress is force per area.

Answer: (b) the same load

Explanation:
1. The breaking load depends on the cross-sectional area and the material, not on the length.
2. So the wire holds the same load.

Teacher's Note:
1. Breaking force depends on area.

Answer: (c) 1.0

Explanation:
1. Copper: Δl = FL/AY = 500 × 8/(0.5 × 10⁻⁴ × 10¹¹) = 0.8 mm.
2. Steel: 500 × 4/(0.5 × 10⁻⁴ × 2 × 10¹¹) = 0.2 mm.
3. Total = 1.0 mm.

Teacher's Note:
1. Elongations of series wires add.

Answer: (c) \( 250\ \text{m s}^{-1} \)

Explanation:
1. k = YA/L = 5 × 10⁸ × 25 × 10⁻⁶/0.1 = 1.25 × 10⁵ N/m.
2. Stored energy ½kx² = ½ × 1.25 × 10⁵ × 0.05² = 156.25 J.
3. ½mv² = 156.25 gives v = √(2 × 156.25/0.005) = 250 m/s.

Teacher's Note:
1. Elastic energy becomes kinetic energy.

Answer: (d) 0.03

Explanation:
1. V ∝ a³, so ΔV/V = 3Δa/a = 3 × 1% = 3%.
2. The bulk strain is 0.03.

Teacher's Note:
1. Volume changes by three times the fractional length change.

Answer: (b) remains constant

Explanation:
1. The ratio stress/strain is the modulus, a constant of the material in the elastic region.
2. So it stays constant.

Teacher's Note:
1. Modulus does not change.

Answer: (a) Steel

Explanation:
1. Elasticity is judged by the Young's modulus, which is the largest for steel among these (about 2 × 10¹¹ N/m²).
2. A larger modulus means a smaller strain for the same stress.

Teacher's Note:
1. Steel has the largest Young's modulus here.

Answer: (b) 2.450 J

Explanation:
1. k = mg/x = 4 × 9.8/0.02 = 1960 N/m.
2. W = ½kx² = ½ × 1960 × 0.05² = 2.45 J.

Teacher's Note:
1. Find k first.

Answer: (b) 0.0098%

Explanation:
1. Stress = 9.8 N/10⁻⁶ m² = 9.8 × 10⁶ N/m².
2. Strain = 9.8 × 10⁶/10¹¹ = 9.8 × 10⁻⁵ = 0.0098%.

Teacher's Note:
1. Strain = stress/Y.

Answer: (c) shape

Explanation:
1. A tangential force causes layers to slide over one another.
2. The shape changes while the volume stays the same.

Teacher's Note:
1. Shear changes shape, not volume.