Mechanical Properties of Solids

1

The property of a body by which it tends to regain its original shape after deformation is called:Easy

A. Plasticity

B. Elasticity

C. Ductility

D. Brittleness

✅ Solution

Elasticity is the property by which a body tends to regain its original shape and size when the deforming force is removed. Plasticity is the opposite — the body does not regain its shape (e.g., putty, clay).

Answer: B — Elasticity

2

Stress is defined as:Easy

A. Force × Area

B. Restoring force per unit area

C. Change in length / Original length

D. Force / Volume

✅ Solution

Stress = Restoring Force / Area = F/ASI unit: N/m² or Pascal (Pa). Dimensions: [ML⁻¹T⁻²]. Note: stress ≠ pressure (though both have same units), as pressure is always compressive.

Answer: B

3

Young's modulus Y is defined as:Easy

A. Y = (F·A)/(L·ΔL)

B. Y = (F·L)/(A·ΔL)

C. Y = (A·ΔL)/(F·L)

D. Y = (F·ΔL)/(A·L)

✅ Solution

Y = Tensile stress / Longitudinal strain = (F/A) / (ΔL/L) = FL/(AΔL)Y is a property of the material (not shape). Unit: N/m² or Pa.

Answer: B — Y = FL/(AΔL)

4

Hooke's law is valid in the region of the stress-strain curve between:Moderate

A. O to A (proportional limit)

B. A to B (yield point)

C. B to D (plastic zone)

D. Beyond D (fracture zone)

✅ Solution

Hooke's Law (stress ∝ strain) is valid only in the linear elastic region O to A (proportional limit). In this region the graph is perfectly straight. Beyond A, stress-strain relationship is no longer linear.

Answer: A — O to A (proportional limit)

5

The yield point on a stress-strain curve represents:Moderate

A. Proportional limit

B. Point of fracture

C. Elastic limit — permanent deformation begins

D. Maximum tensile strength

✅ Solution

Point B on the stress-strain curve is the yield point (elastic limit). Beyond this, even after the load is removed, the material does NOT regain its original shape — permanent (plastic) deformation begins. The corresponding stress is called yield strength (σ_y).

Answer: C

6

The SI unit of stress is:Easy

A. Newton

B. Pascal

C. Joule

D. Dimensionless

✅ Solution

Stress = F/A. Units = N/m² = Pascal (Pa). Dimensions = [ML⁻¹T⁻²]. Note: Strain is dimensionless (ratio of two lengths). So Young's modulus also has units of Pa.

Answer: B — Pascal (Pa)

7

Shear modulus (modulus of rigidity) G is defined as:Moderate

A. G = Tensile stress / Longitudinal strain

B. G = Shearing stress / Shearing strain

C. G = Hydraulic pressure / Volume strain

D. G = Volume strain / Hydraulic pressure

✅ Solution

G = Shearing stress / Shearing strain = (F/A) / θ = F/(Aθ)Also G = (F×L)/(A×Δx). Shear modulus exists only for solids (not liquids/gases). For most materials G ≈ Y/3.

Answer: B

8

Bulk modulus B is given by:Moderate

A. B = p(ΔV/V)

B. B = −p/(ΔV/V)

C. B = (ΔV/V)/p

D. B = p + ΔV/V

✅ Solution

B = −p/(ΔV/V)The negative sign ensures B is always positive (when pressure p increases, volume decreases so ΔV is negative). Bulk modulus applies to solids, liquids AND gases.

Answer: B — B = −p/(ΔV/V)

9

The reciprocal of bulk modulus is called:Easy

A. Young's modulus

B. Compressibility

C. Shear modulus

D. Poisson's ratio

✅ Solution

Compressibility k = 1/B = −(1/p)(ΔV/V)Compressibility measures how easy it is to compress the material. Gases have very high compressibility; solids have very low. Gases are ~10⁶ times more compressible than solids.

Answer: B — Compressibility

10

Among the following, which material has the highest Young's modulus? (From NCERT Table 8.1)Moderate

A. Aluminium (70 GPa)

B. Copper (110 GPa)

C. Steel (200 GPa)

D. Glass (65 GPa)

✅ Solution

From NCERT Table 8.1: Steel Y = 200 × 10⁹ N/m² (200 GPa) which is the highest among common materials. Higher Y means the material needs a larger force to deform — it is more rigid/elastic (not in the everyday sense).

Answer: C — Steel (200 GPa)

11

Longitudinal strain is defined as:Easy

A. ΔL × L

B. L / ΔL

C. ΔL / L

D. F / A

✅ Solution

Longitudinal strain = ΔL/LIt is dimensionless (ratio of change in length to original length). Similarly, volume strain = ΔV/V, and shearing strain = Δx/L = tanθ ≈ θ.

Answer: C — ΔL/L

12

Shearing strain is equal to:Moderate

A. ΔV/V

B. ΔL/L

C. tanθ ≈ θ

D. F/A

✅ Solution

Shearing strain = Δx/L = tanθ ≈ θ (for small angles)θ is the angular displacement from vertical. Example: when you push the top face of a book while holding the bottom face fixed.

Answer: C — tanθ ≈ θ

13

Which type of stress does NOT change the shape of a body, only its volume?Moderate

A. Tensile stress

B. Shearing stress

C. Hydraulic stress

D. Compressive stress

✅ Solution

Hydraulic (or volume) stress — when a body is uniformly compressed from all sides by a fluid. Forces act perpendicular to every surface. Volume decreases but shape is preserved (no change in geometrical shape).

Answer: C — Hydraulic stress

14

Steel is preferred over copper for heavy-duty machines because:Moderate

A. Steel is lighter

B. Steel has a higher Young's modulus — more elastic

C. Steel is cheaper

D. Steel has a lower melting point

✅ Solution

Steel has Y = 200 GPa, much higher than copper (110 GPa), aluminium (70 GPa) and brass. A higher Y means steel requires a larger force to produce the same strain — it deforms less under load, making it more suitable for structural and machine applications.

Answer: B

15

A material which can be drawn into thin wires without breaking is called:Easy

A. Brittle

B. Elastic

C. Ductile

D. Plastic

✅ Solution

Ductile materials (like copper, steel) have a large distance between the ultimate tensile strength point D and fracture point E on the stress-strain curve. They can sustain large plastic deformation before fracture. Brittle materials (glass, cast iron) fracture immediately after the elastic limit.

Answer: C — Ductile

16

Elastic potential energy stored per unit volume in a strained wire is:Moderate

A. u = σε

B. u = ½σε

C. u = σ²/Y

D. u = Yε

✅ Solution

u = ½σε = ½ × stress × strain (per unit volume)Also: u = ½ × Y × ε² = ½ × σ²/Y. This is analogous to elastic PE in a spring: U = ½kx².

Answer: B — u = ½σε

17

Which of the following is most compressible?Easy

A. Steel

B. Water

C. Air

D. Copper

✅ Solution

From NCERT Table 8.3: B_air = 1.0 × 10⁻⁴ GPa (very low), B_water = 2.2 GPa, B_steel = 160 GPa. Compressibility = 1/B. Gases are ~10⁶ times more compressible than solids. Air is most compressible.

Answer: C — Air (most compressible)

18

Poisson's ratio is defined as:Moderate

A. Lateral strain / Longitudinal strain

B. Longitudinal strain / Lateral strain

C. Shearing stress / Shearing strain

D. Volume strain / Longitudinal strain

✅ Solution

Poisson's ratio σ = (Δd/d) / (ΔL/L) = Lateral strain / Longitudinal strainIt is dimensionless. For steel σ ≈ 0.28–0.30; for aluminium σ ≈ 0.33.

Answer: A

19

Substances like rubber and aorta tissue that can be stretched to large strains and still return to original shape are called:Easy

A. Plastics

B. Elastomers

C. Metals

D. Ceramics

✅ Solution

Elastomers (like rubber, elastic tissue of aorta) can be stretched to very large strains and still return to original shape. However, they do NOT obey Hooke's law over most of the elastic region — the stress-strain curve is nonlinear.

Answer: B — Elastomers

20

For a beam loaded at the centre and supported at ends, the sag δ is given by δ = Wl³/(4bd³Y). To reduce sag most effectively, you should:Hard

A. Increase breadth b

B. Increase depth d

C. Decrease Young's modulus Y

D. Increase length l

✅ Solution

δ ∝ 1/d³ and δ ∝ 1/b. Since δ is proportional to d⁻³, increasing d is FAR more effective than increasing b (which gives only d⁻¹ effect). This is why I-shaped beams have large depth — to minimise sag with minimum weight.

Answer: B — Increase depth d

21

The point D on the stress-strain curve (ultimate tensile strength) represents:Moderate

A. Elastic limit

B. Proportional limit

C. Maximum stress the material can withstand

D. Point of zero stress

✅ Solution

Point D on the stress-strain curve = Ultimate Tensile Strength (σ_u). Beyond D, additional strain is produced even by a reduced applied force, and fracture occurs at E. If D and E are close → brittle; if far → ductile.

Answer: C — Maximum stress the material can withstand

22

Shear modulus (G) exists for:Moderate

A. Solids only

B. Liquids only

C. Gases only

D. All states of matter

✅ Solution

Shear modulus exists only for SOLIDS because only solids have a definite shape. Liquids and gases cannot sustain shear stress — they flow. Young's modulus is also only for solids. Only Bulk modulus applies to all states (solid, liquid, gas).

Answer: A — Solids only

23

If a wire is stretched to double its length, the longitudinal strain is:Easy

A. 0.5

B. 1

C. 2

D. 0

✅ Solution

If original length = L, new length = 2L, then ΔL = 2L − L = L. Longitudinal strain = ΔL/L = L/L = 1

Answer: B — 1 (100% strain)

24

Which cross-sectional shape is preferred for load-bearing beams in bridges to prevent both bending and buckling?Moderate

A. Circular cross-section

B. Square cross-section

C. I-shaped cross-section

D. Triangular cross-section

✅ Solution

The I-beam (I-shaped cross-section) is used in construction because: it provides large load-bearing surface, sufficient depth to prevent bending (δ ∝ d⁻³), and reduces buckling. It also reduces weight without sacrificing strength — hence more economical.

Answer: C — I-shaped cross-section

25

For a material under tensile stress, if stress is doubled (within elastic limit), the strain:Moderate

A. Remains unchanged

B. Is halved

C. Is doubled

D. Becomes four times

✅ Solution

Within the elastic (proportional) limit, Hooke's Law holds: stress = Y × strain, i.e., stress ∝ strain. If stress doubles → strain doubles (Y is constant)

Answer: C — Strain is doubled

1

A structural steel rod (radius 10 mm, length 1.0 m) is stretched by a 100 kN force. Find (a) stress, (b) elongation, and (c) strain. Y_steel = 2.0 × 10¹¹ N m⁻². [NCERT Ex 8.1] Moderate3 Marks

✅ Solution

r = 10 mm = 10⁻² m; A = πr² = 3.14 × (10⁻²)² = 3.14 × 10⁻⁴ m² (a) Stress = F/A = 10⁵ / (3.14×10⁻⁴) = 3.18 × 10⁸ N m⁻² (b) ΔL = (F/A)×L/Y = (3.18×10⁸ × 1) / (2×10¹¹) = 1.59 × 10⁻³ m = 1.59 mm (c) Strain = ΔL/L = 1.59×10⁻³ / 1 = 1.59 × 10⁻³ = 0.16%

(a) 3.18 × 10⁸ Pa; (b) 1.59 mm; (c) 1.59 × 10⁻³

2

A copper wire (length 2.2 m) and steel wire (length 1.6 m), both diameter 3.0 mm, connected end to end. Net elongation = 0.70 mm under load W. Find W. Y_Cu = 1.1×10¹¹, Y_s = 2.0×10¹¹ N/m². [NCERT Ex 8.2] Moderate3 Marks

✅ Solution

Both wires have same tension W and same A = π(1.5×10⁻³)² = 7.07×10⁻⁶ m² ΔLc/ΔLs = (Ys/Yc) × (Lc/Ls) = (2.0/1.1) × (2.2/1.6) = 2.5 ΔLc + ΔLs = 7×10⁻⁴ m → ΔLc = 5×10⁻⁴ m, ΔLs = 2×10⁻⁴ m W = A×Yc×ΔLc/Lc = 7.07×10⁻⁶ × 1.1×10¹¹ × 5×10⁻⁴ / 2.2 ≈ 180 N

Load W ≈ 180 N

3

In a human pyramid, combined mass of all = 280 kg, performer at bottom = 60 kg. Each femur: length 50 cm, radius 2.0 cm. Y_bone = 9.4 × 10⁹ N/m². Find compression in each thighbone. [NCERT Ex 8.3] Moderate3 Marks

✅ Solution

Extra mass supported = 280 − 60 = 220 kg; Weight = 220 × 9.8 = 2156 N Force per femur = 2156/2 = 1078 N A = π(2×10⁻²)² = 1.257 × 10⁻³ m²; L = 0.5 m ΔL = FL/(YA) = (1078 × 0.5)/(9.4×10⁹ × 1.257×10⁻³) = 539 / (1.182×10⁷) = 4.56 × 10⁻⁵ m ≈ 4.56 × 10⁻³ cm

Compression ≈ 4.55 × 10⁻⁵ m (very tiny — 0.0091%)

4

A square lead slab (side 50 cm, thickness 10 cm) is subjected to a shearing force of 9.0 × 10⁴ N on its narrow face. Lower edge is fixed. Find the upper edge displacement. G_lead = 5.6 × 10⁹ N/m². [NCERT Ex 8.4] Moderate3 Marks

✅ Solution

Area of shear face = 50 cm × 10 cm = 0.5 × 0.1 = 0.05 m² Shearing stress = F/A = 9×10⁴ / 0.05 = 1.8 × 10⁶ N/m² Shearing strain = Stress/G = 1.8×10⁶ / 5.6×10⁹ = 3.21 × 10⁻⁴ Displacement Δx = strain × L = 3.21×10⁻⁴ × 0.5 = 1.6 × 10⁻⁴ m = 0.16 mm

Displacement of upper edge = 0.16 mm

5

The average depth of the Indian Ocean is 3000 m. Calculate the fractional compression ΔV/V of water. Bulk modulus of water B = 2.2 × 10⁹ N/m², g = 10 m/s². [NCERT Ex 8.5] Moderate3 Marks

✅ Solution

Hydraulic pressure at depth: p = hρg = 3000 × 1000 × 10 = 3 × 10⁷ N/m² ΔV/V = p/B = (3×10⁷) / (2.2×10⁹) = 1.364 × 10⁻² = 1.36%

Fractional compression ΔV/V = 1.36 × 10⁻² = 1.36%

6

A steel wire of length 4.7 m and cross-section 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-section 4.0 × 10⁻⁵ m² under the same load. Find the ratio Y_steel/Y_copper. [NCERT Ex 8.1 text] Moderate3 Marks

✅ Solution

Same load F and same elongation ΔL. Y = FL/(AΔL), so Y ∝ L/A (since F and ΔL are same) Y_s/Y_c = (L_s/A_s) / (L_c/A_c) = (L_s × A_c) / (L_c × A_s) = (4.7 × 4.0×10⁻⁵) / (3.5 × 3.0×10⁻⁵) = 18.8 / 10.5 ≈ 1.79

Y_steel/Y_copper ≈ 1.79

7

Two wires (diameter 0.25 cm each), steel (length 1.5 m) and brass (length 1.0 m), are loaded with 6 kg (brass) and 4 + 6 = 10 kg total (steel carries full load above it). Y_steel = 2.0×10¹¹, Y_brass = 0.91×10¹¹ N/m². Compute elongations of each wire. [NCERT Ex 8.5] Moderate3 Marks

✅ Solution

A = π(1.25×10⁻³)² = 4.91×10⁻⁶ m² Steel carries load = (4+6)kg = 10 kg; F_steel = 98 N ΔL_steel = F_s×L_s/(Y_s×A) = 98×1.5/(2×10¹¹×4.91×10⁻⁶) = 147/9.82×10⁵ ≈ 1.49×10⁻⁴ m = 0.149 mm Brass carries 6 kg; F_brass = 58.8 N ΔL_brass = 58.8×1.0/(0.91×10¹¹×4.91×10⁻⁶) = 58.8/4.47×10⁵ ≈ 1.31×10⁻⁴ m = 0.131 mm

ΔL_steel ≈ 0.149 mm; ΔL_brass ≈ 0.131 mm

8

An aluminium cube of edge 10 cm has one face fixed to a vertical wall. A 100 kg mass is attached to the opposite face. G_Al = 25 × 10⁹ N/m². What is the vertical deflection of this face? [NCERT Ex 8.6] Moderate3 Marks

✅ Solution

Shearing force F = mg = 100 × 9.8 = 980 N Area A = 0.1 × 0.1 = 0.01 m² (face area); L = 0.1 m (edge length) Shearing stress = F/A = 980/0.01 = 9.8 × 10⁴ N/m² Deflection Δx = stress × L / G = (9.8×10⁴ × 0.1) / (25×10⁹) = 3.92 × 10⁻⁷ m

Vertical deflection ≈ 3.9 × 10⁻⁷ m ≈ 0.39 μm

9

Four hollow steel cylinders (inner radius 30 cm, outer radius 60 cm) support a structure of mass 50,000 kg. Y_steel = 2×10¹¹ N/m². Calculate the compressional strain in each column. [NCERT Ex 8.7] Moderate3 Marks

✅ Solution

Weight = 50000 × 9.8 = 4.9 × 10⁵ N; Force per column = 4.9×10⁵/4 = 1.225×10⁵ N A = π(R² − r²) = π(0.6² − 0.3²) = π × 0.27 = 0.848 m² Compressional strain = F/(A×Y) = 1.225×10⁵ / (0.848 × 2×10¹¹) = 1.225×10⁵ / 1.696×10¹¹ ≈ 7.22 × 10⁻⁷

Compressional strain ≈ 7.22 × 10⁻⁷

10

A steel cable of radius 1.5 cm supports a chair lift. Maximum stress must not exceed 10⁸ N/m². What is the maximum load the cable can support? [NCERT Ex 8.9] Easy3 Marks

✅ Solution

A = πr² = 3.14 × (1.5×10⁻²)² = 3.14 × 2.25×10⁻⁴ = 7.065×10⁻⁴ m² Maximum load = Maximum stress × A = 10⁸ × 7.065×10⁻⁴ = 7.065 × 10⁴ N ≈ 7.06 × 10⁴ N

Maximum load ≈ 7.06 × 10⁴ N ≈ 7200 kg weight

11

A 14.5 kg mass at end of a 1.0 m steel wire (area 0.065 cm²) is whirled in vertical circle at ω = 2 rev/s at bottom. Y_steel = 2×10¹¹ N/m². Find elongation at the lowest point. [NCERT Ex 8.11] Hard3 Marks

✅ Solution

ω = 2 rev/s = 4π rad/s; v = ωL = 4π × 1 = 4π m/s At bottom: T − mg = mv²/L → T = mg + mv²/L = m(g + ω²L) T = 14.5 × (9.8 + 16π²×1) = 14.5 × (9.8 + 157.9) ≈ 14.5 × 167.7 = 2432 N A = 0.065×10⁻⁴ m²; ΔL = TL/(YA) ΔL = 2432×1/(2×10¹¹×6.5×10⁻⁶) = 2432/1.3×10⁶ ≈ 1.87×10⁻³ m

Elongation ≈ 1.87 × 10⁻³ m ≈ 1.87 mm

12

Initial volume of water = 100 litres. Pressure increased by 100 atm (1 atm = 1.013×10⁵ Pa). Final volume = 100.5 litres. Calculate bulk modulus of water. [NCERT Ex 8.12] Moderate3 Marks

✅ Solution

Wait — final volume is 100.5 L means it INCREASED? That's an error in reading. The volume should DECREASE under pressure. ΔV = 100.5 − 100 = 0.5 L (if given as expansion, then ΔV/V negative for compression, but here |ΔV| = 0.5 L) ΔV/V = 0.5/100 = 5×10⁻³; Δp = 100 × 1.013×10⁵ = 1.013×10⁷ Pa B = −Δp/(ΔV/V) = 1.013×10⁷ / 5×10⁻³ = 2.026×10⁹ N/m² ≈ 2.03 GPa Compare with air: B_air ≈ 10⁻⁴ GPa → B_water/B_air ≈ 2×10⁴ times larger

Bulk modulus of water ≈ 2.03 × 10⁹ Pa ≈ 2.03 GPa

13

Compute the fractional change in volume of a glass slab subjected to hydraulic pressure of 10 atm. B_glass = 37 GPa, 1 atm = 1.013×10⁵ Pa. [NCERT Ex 8.14] Easy3 Marks

✅ Solution

Pressure p = 10 × 1.013×10⁵ = 1.013×10⁶ Pa ΔV/V = p/B = 1.013×10⁶ / (37×10⁹) = 2.74 × 10⁻⁵ This is an extremely small compression (0.00274%) — showing glass has very high bulk modulus.

ΔV/V = 2.74 × 10⁻⁵ (0.00274%)

14

Determine the volume contraction of a solid copper cube (10 cm edge) subjected to hydraulic pressure of 7.0 × 10⁶ Pa. B_copper = 140 GPa. [NCERT Ex 8.15] Easy3 Marks

✅ Solution

V = (0.1)³ = 10⁻³ m³; p = 7×10⁶ Pa; B = 140×10⁹ Pa ΔV/V = p/B = 7×10⁶ / 140×10⁹ = 5 × 10⁻⁵ ΔV = (ΔV/V) × V = 5×10⁻⁵ × 10⁻³ = 5 × 10⁻⁸ m³ = 5 × 10⁻⁸ m³ = 5 × 10⁻² cm³ = 0.05 cm³

Volume contraction ΔV = 5 × 10⁻⁸ m³ = 0.05 cm³

15

How much should the pressure on 1 litre of water be changed to compress it by 0.10%? B_water = 2.2 × 10⁹ N/m². [NCERT Ex 8.16] Easy3 Marks

✅ Solution

ΔV/V = 0.10% = 0.10/100 = 10⁻³ Δp = B × (ΔV/V) = 2.2×10⁹ × 10⁻³ = 2.2 × 10⁶ Pa = 2.2 MPa ≈ 21.7 atm

Pressure change required = 2.2 × 10⁶ Pa ≈ 21.7 atm

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