Class 11 Physics Chapter 8: Mechanical Properties of Solids worksheet with answer key

Answer: (b) [ML⁻¹T⁻²]

Stress = F/A, so [MLT⁻²]/[L²] = [ML⁻¹T⁻²]. [MLT⁻²] is force itself; [M⁰L⁰T⁰] belongs to strain.

Answer: (b) solid < liquid < gas

Bulk modulus is largest for solids and smallest for gases, and compressibility is 1/B, so solid < liquid < gas: atoms in solids are most tightly coupled, gas molecules least.

Answer: (d) Its elastic region is very large, but it does not obey Hooke's law over most of it.

NCERT notes that for aortic tissue the elastic region is very large, Hooke's law fails over most of it, and there is no well defined plastic region; such materials are elastomers.

Answer: (a) elastic limit

The yield point B is also called the elastic limit; the stress there is the yield strength σ_y. The proportional limit A comes earlier, and D and E come later.

Answer: (c) brittle

When the ultimate strength point D and the fracture point E are close, the material is brittle. A ductile material shows a long stretch between D and E.

Shearing strain = Δx/L = 1.0/5.0 = 0.20. Shearing stress = F/A = 0.50/(0.050 × 0.050) = 2.0 × 10² N m⁻². G = stress/strain = 200/0.20 = 1.0 × 10³ N m⁻².

• strain = Δx/L = 0.20 (1)
• G = (F/A)/strain = 1.0 × 10³ N m⁻² (1)

Empirical means it is found from experiment and observation, not derived from a deeper principle; it holds for most materials for small deformations. Elastomers, such as rubber or the tissue of the aorta, can be stretched to large strains but do not show stress proportional to strain.

• empirical: found from experiment, valid for most materials at small deformation (1)
• elastomers, e.g. rubber or aortic tissue (1)

No. Stress cannot be assigned one specific direction: it describes the restoring force per unit area across a section, and its direction depends on which section is considered. Only the force acting on the portion of the body on a specified side of a section has a definite direction, so stress is not a vector quantity.

• no, stress is not a vector (1)
• it cannot be assigned a single direction; only the force on one side of a chosen section has one (1)

At extension x the force is F = YAx/L (from Y = FL/(AΔL)). For a further extension dx the work done is dW = F dx = (YA/L) x dx. So W = ∫₀^l (YA/L) x dx = ½ (YA/L) l² = ½ × Y × (l/L)² × AL. Since Y(l/L) = stress and l/L = strain, W = ½ × stress × strain × volume. This work is stored as elastic potential energy, so u = W/(AL) = ½ σε.

• F = YAx/L and dW = F dx (1)
• integrates to ½(YA/L)l² (1)
• rewrites as ½ × stress × strain × volume and u = ½σε (1)

Horizontal bar means equal extensions; with equal L and A, T = YAΔL/L ∝ Y, so T_s/T_a = 200/70. With T_s + T_a = 300 N: T_s = 300 × 200/270 = 222 N and T_a = 77.8 N. Moments about the steel end: T_a × 1.0 = 300 × x, so x = 77.8/300 = 0.26 m from the steel end. Extension = T_sL/(AY_s) = 222 × 1.0/(1.0 × 10⁻⁶ × 2.0 × 10¹¹) = 1.11 × 10⁻³ m.

• equal extension gives T ∝ Y, T_s = 222 N, T_a = 77.8 N (1)
• moments give x = 0.26 m from the steel wire (1)
• extension = 1.11 × 10⁻³ m (1)

Flat: b = 0.30 m, d = 0.050 m. On edge: b = 0.050 m, d = 0.30 m. δ ∝ 1/(bd³), so δ_flat/δ_edge = (0.050 × 0.30³)/(0.30 × 0.050³) = (0.30/0.050)² = 36. On edge the plank is deep and thin; unless the load is exactly in the right place it can buckle sideways, so a deep thin section is not safe for a moving load.

• δ ∝ 1/(bd³) with b and d for each position (1)
• ratio = 36 (1)
• deep thin plank buckles unless the load is exactly placed (1)