NCERT Solutions for Class 11 Physics Chapter 8: Mechanical Properties of Solids

Concept. Young’s modulus Y = (F/A)/(ΔL/L) = F L/(A ΔL). If two wires are stretched by the same force F and give the same extension ΔL, then Y is proportional to L/A. So the ratio of Y for two such wires is the ratio of their (length/area) values.

Question. A steel wire of length 4.7 m and cross-sectional area 3.0×10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0×10⁻⁵ m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Given

• Steel: L_s = 4.7 m, A_s = 3.0×10⁻⁵ m²
• Copper: L_c = 3.5 m, A_c = 4.0×10⁻⁵ m²
• Same load F and same extension ΔL for both wires.

Step 1 — write the formula for each wire

Y_s = F L_s / (A_s ΔL)     Y_c = F L_c / (A_c ΔL)

Step 2 — divide (F and ΔL cancel)

Y_s/Y_c = (L_s/A_s) × (A_c/L_c) = (L_s A_c)/(L_c A_s)

Step 3 — substitute

Y_s/Y_c = (4.7 × 4.0×10⁻⁵)/(3.5 × 3.0×10⁻⁵) = 18.8/10.5 = 1.79 ≈ 1.8

So steel has Young’s modulus about 1.8 times that of copper. Check against the table: Table 8.1 gives Y(steel) = 200×10⁹ and Y(copper) = 110×10⁹ N m⁻², ratio = 1.82. Our answer agrees with the table to within the rounding of the data, which is a good sign.

Extra depth. Notice that the exponents 10⁻⁵ cancel and the final ratio is a pure number. The method is general: whenever two wires carry the same force and get the same extension, Y ∝ L/A; if they carry the same force and have the same L and A, then Y ∝ 1/ΔL (the stiffer wire stretches less). This is also how a physicist compares moduli experimentally in the Searle’s apparatus.

Conclusion: Y(steel)/Y(copper) = (L_s A_c)/(L_c A_s) = 1.79 ≈ 1.8, in agreement with the NCERT table values (200/110 = 1.82).

Concept. In the straight (Hooke) part of a stress-strain graph the slope is Young’s modulus, Y = stress/strain. The yield strength is the stress at which the curve ceases to be elastic.

Question. Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) the approximate yield strength for this material?

(a) Young’s modulus

The graph is a straight line through the origin up to a stress of about 300×10⁶ N m⁻², so the slope can be taken from any convenient point, for example the grid point at strain = 0.002 where stress = 150×10⁶ N m⁻²:

Y = stress/strain = (150×10⁶ N m⁻²)/0.002 = 75×10⁹ = 7.5×10¹⁰ N m⁻²

Check with the top end of the line: (300×10⁶)/0.004 = 7.5×10¹⁰ N m⁻² — the same, which confirms that the line is straight. A value of 7.5×10¹⁰ N m⁻² (75 GPa) is close to that of aluminium (70 GPa) in Table 8.1, so the material is probably an aluminium alloy.

(b) Approximate yield strength

The curve is linear up to the point where it reaches its maximum height of about 300×10⁶ N m⁻² near strain 0.004 and then bends over (the stress stops increasing). That point corresponds to the end of the elastic behaviour, so

σ_y ≈ 3×10⁸ N m⁻² (about 300 MPa)

Units. The vertical axis is labelled in units of 10⁶ N m⁻², so a reading of 150 means 150×10⁶ N m⁻² and not 150 N m⁻². Missing this prefix is the commonest error in this question.

Extra depth. The slope of the line is computed from stress divided by strain — the graph in the book is a “strain-stress” curve in words but is drawn with strain on the x-axis and stress on the y-axis, so the slope is still Y. If the graph had been drawn with stress on the x-axis, the slope would be 1/Y (the “compliance”). Always look at which quantity is on which axis before taking the slope.

Conclusion: Y = 7.5×10¹⁰ N m⁻² (slope of the linear part) and the yield strength ≈ 3×10⁸ N m⁻² (stress where the curve stops being linear).

Concept. The slope of the initial straight part of a stress-strain graph is Young’s modulus; the height reached by the curve before fracture is the strength of the material.

Question. The stress-strain graphs for materials A and B are shown in Fig. 8.10 (drawn to the same scale). (a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material?

(a) Greater Young’s modulus

Young’s modulus is the ratio stress/strain in the linear region, i.e. the slope of the graph near the origin. Both graphs are drawn on the same scale, so their slopes can be compared directly by looking at them. The curve for A rises almost vertically — for a small strain it reaches a large stress — while the curve for B rises less steeply.

slope(A) > slope(B)  ⇒  Y_A > Y_B

Answer (a): material A.

(b) Stronger material

The strength of a material is the stress it can bear before it yields or breaks (ultimate strength), i.e. the height reached by the curve on the stress axis. The curve of A reaches the greater stress and keeps carrying a high stress over a long range of strain, whereas the curve of B reaches only a slightly lower maximum, so more stress is needed to break A.

Answer (b): material A is the stronger (and, because it carries the stress over a larger strain range, it is also the tougher material).

Extra depth. Be careful to separate three ideas: stiffness (Young’s modulus, slope), strength (maximum stress, height) and toughness (area under the curve, energy absorbed before fracture). Material B in the figure is a stiff, brittle-looking material with less strain to failure; A is both stiffer (larger slope) and stronger here. In general a material may have a high Y but low strength (glass) or low Y but high strength.

Conclusion: A has the greater slope, hence the greater Young’s modulus, and its curve rises to a higher stress, hence A is the stronger material.

Concept. Young’s modulus is the ratio of stress to longitudinal strain; a larger Y means a smaller extension for the same stress. The extension of a coil spring involves twisting (shear) of the wire, not simple stretching of its length.

Question. State, with reasons, whether true or false: (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus.

(a) False

For the same applied stress a rubber band stretches by a much larger strain than a steel wire. Since Y = stress/strain, the larger strain means a smaller Young’s modulus. Numerically Y(steel) = 2.0×10¹¹ N m⁻² (Table 8.1) whereas Y(rubber) is only of the order of 10⁶–10⁸ N m⁻², thousands of times smaller. Rubber stretches easily, but a material that stretches more for a given load is less elastic in the physics sense (NCERT, Points to Ponder 6).

(b) True

When a helical spring (coil) is pulled, the length of the wire forming the coil does not change appreciably. Instead, every small element of the wire is twisted: the cross-section of the wire is turned about the axis of the wire, which is a shearing strain. A coil’s elongation is therefore decided by the shear modulus (modulus of rigidity) of the wire material. A coil of a material with a large G (steel, G = 84 GPa) is stiffer than one made of lead (G = 5.6 GPa), whatever the length of the wire.

Extra depth. The result for springs is derived in Class XII Physics (and engineering) as k = G d⁴/(8 N D³), where d is the wire diameter, D the coil diameter and N the number of turns: the spring constant is proportional to the shear modulus G, not to Young’s modulus. This is why a lead spring is soft even though lead is a metal.

More to know. Write the two ideas as “Y → longitudinal strain; G → shear”. Elastic moduli exist only for solids except B. Typical values (10⁹ N m⁻²): steel Y = 200, G = 84; copper Y = 110, G = 42; lead Y = 16, G = 5.6; rubber Y ≈ 0.01–0.1. For many metals G ≈ Y/3, so a coil spring that gets its stiffness from G is about three times “softer” than a wire of the same material stretched by Y. This is why springs can be extended by several centimetres while a wire of the same metal extends by a fraction of a millimetre under the same force.

Conclusion: (a) False — Y(rubber) is far less than Y(steel); (b) True — a coil is stretched mainly by twisting of the wire, so its extension is governed by the shear modulus.

Concept. A wire in series supports the weight of everything hanging below it, so the tension in each wire is different. Use ΔL = F L/(A Y) separately for each wire.

Question. Two wires of diameter 0.25 cm, one of steel and the other of brass, are loaded as in Fig. 8.11. The unloaded length of the steel wire is 1.5 m and that of the brass wire is 1.0 m. Compute the elongations of the steel and brass wires.

Given

• Diameter = 0.25 cm ⇒ radius r = 0.125 cm = 1.25×10⁻³ m
• A = π r² = 3.14 × (1.25×10⁻³)² = 4.91×10⁻⁶ m² (same for both wires)
• Y(steel) = 2.0×10¹¹ N m⁻² (Table 8.1); Y(brass) = 0.91×10¹¹ N m⁻² (standard value, not given in Table 8.1 of this book)
• g = 9.8 m s⁻²

Step 1 — tension in each wire

• Brass wire hangs from the 4.0 kg mass and supports only the 6.0 kg mass: F_b = 6.0 × 9.8 = 58.8 N.
• Steel wire is attached to the ceiling and carries both masses (4.0 + 6.0 = 10.0 kg): F_s = 10.0 × 9.8 = 98 N. (The weight of the wires is negligible.)

Step 2 — elongation of the steel wire

ΔL_s = F_s L_s/(A Y_s) = (98 × 1.5)/(4.91×10⁻⁶ × 2.0×10¹¹) = 147/(9.82×10⁵) = 1.5×10⁻⁴ m

Step 3 — elongation of the brass wire

ΔL_b = F_b L_b/(A Y_b) = (58.8 × 1.0)/(4.91×10⁻⁶ × 0.91×10¹¹) = 58.8/(4.47×10⁵) = 1.3×10⁻⁴ m

So the steel wire stretches by about 0.15 mm and the brass wire by about 0.13 mm. The strains are 1.0×10⁻⁴ and 1.3×10⁻⁴, tiny, so we are well inside the elastic limit (stress = F/A = 2.0×10⁷ and 1.2×10⁷ N m⁻², much less than the yield strengths of 250×10⁶ N m⁻² for steel).

Extra depth. The key idea is the series connection: forces in a series chain are determined by the weights below the wire, so the upper wire has the larger tension. The elongations simply add if one is asked for the total extension: ΔL_total = 1.5×10⁻⁴ + 1.3×10⁻⁴ = 2.8×10⁻⁴ m. Using Table 8.1 only, Y(brass) is not given; NCERT’s own answer uses 0.91×10¹¹ N m⁻², the standard value, and I have followed it.

Conclusion: ΔL(steel) ≈ 1.5×10⁻⁴ m and ΔL(brass) ≈ 1.3×10⁻⁴ m, because the steel wire carries 10 kg and the brass wire only 6 kg.

Concept. A cube with one face fixed to a wall and a weight hung from the opposite face is in shear: the force is parallel to the face. The shear modulus is G = (F/A)/(Δx/L), so the displacement is Δx = F L/(A G).

Question. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is attached to the opposite face. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Given

• Edge L = 10 cm = 0.10 m
• Area of the face parallel to the force A = L² = (0.10)² = 1.0×10⁻² m²
• F = mg = 100 × 9.8 = 980 N (acts vertically, i.e. parallel to the faces)
• G = 25 GPa = 25×10⁹ N m⁻²

Step 1 — shearing stress

σ_s = F/A = 980/(1.0×10⁻²) = 9.8×10⁴ N m⁻²

Step 2 — shearing strain

θ = σ_s/G = (9.8×10⁴)/(25×10⁹) = 3.92×10⁻⁶ rad

Step 3 — displacement of the free face

Δx = θ L = 3.92×10⁻⁶ × 0.10 = 3.9×10⁻⁷ m

The vertical deflection of the face is therefore about 3.9×10⁻⁷ m (0.39 µm), far too small to see: this is why solid blocks, even loaded by 100 kg, look perfectly rigid. In terms of the one-line formula Δx = F L/(A G) = (980 × 0.1)/(0.01 × 25×10⁹) = 3.9×10⁻⁷ m.

Extra depth. The angle of shear (3.9×10⁻⁶ rad ≈ 0.0002°) is tiny, so tan θ ≈ θ is an excellent approximation. The same result follows from the NCERT definition G = (F/A)/(Δx/L) = F L/(A Δx). Compare with Example 8.4 (lead slab): there the stress is 1.8×10⁶ N m⁻² and the displacement is 0.16 mm; here the stress is only 9.8×10⁴ N m⁻² and aluminium is ~4.5 times stiffer in shear, so the displacement is a thousand times smaller.

Conclusion: Δx = F L/(A G) = 3.9×10⁻⁷ m: the aluminium cube sags by only about 0.4 micrometre under 100 kg.

Concept. Each column is under compressive stress. Load per column = (total weight)/4. Compressional strain = stress/Y, with the area of a hollow cylinder = π(R² − r²).

Question. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Given

• Mass of structure M = 50,000 kg ⇒ W = Mg = 5.0×10⁴ × 9.8 = 4.9×10⁵ N
• Load on each column F = W/4 = 1.225×10⁵ N
• Inner radius r = 30 cm = 0.30 m; outer radius R = 60 cm = 0.60 m
• Y(mild steel) = 2.0×10¹¹ N m⁻² (Table 8.1)

Step 1 — area of cross-section of the hollow column

A = π(R² − r²) = 3.14 × (0.36 − 0.09) = 3.14 × 0.27 = 0.848 m²

Step 2 — compressive stress in each column

σ = F/A = (1.225×10⁵)/0.848 = 1.44×10⁵ N m⁻²

Step 3 — compressional strain

ε = σ/Y = (1.44×10⁵)/(2.0×10¹¹) = 7.2×10⁻⁷

The compressional strain of each column is about 7.2×10⁻⁷ (i.e. 7.2×10⁻⁵ %): negligible. The stress (1.4×10⁵ N m⁻²) is also far below the yield strength of steel (250×10⁶ N m⁻²), so the columns are extremely safe. A column compresses by ΔL = εL; for a 10 m column that is only about 7 micrometre.

Extra depth. Using the area of a hollow cylinder is the point of this question: the material exists only between the radii r and R, so A = π(R² − r²) = π(R − r)(R + r). A common wrong approach is to use πR² (a solid column), which gives 1.13 m² and a strain of 5.4×10⁻⁷. The hollow shape saves material at little cost to strength, which is the same idea as the I-beam of Section 8.6.

Conclusion: Compressional strain = (Mg/4)/[π(R² − r²)Y] = 7.2×10⁻⁷ for each column.

Concept. Within the elastic limit, strain = stress/Y = F/(A Y). The area is that of a rectangle, A = width × thickness, with both dimensions converted to metres.

Question. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

Given

• A = 15.2 mm × 19.1 mm = 15.2×10⁻³ m × 19.1×10⁻³ m = 2.90×10⁻⁴ m²
• F = 44,500 N
• Y(copper) = 110 GPa = 1.1×10¹¹ N m⁻² (Table 8.1)

Step 1 — tensile stress

σ = F/A = 44500/(2.90×10⁻⁴) = 1.53×10⁸ N m⁻²

Step 2 — strain from Hooke’s law

ε = σ/Y = (1.53×10⁸)/(1.1×10¹¹) = 1.4×10⁻³

The resulting strain is about 1.4×10⁻³ (0.14%). The extension of a 1 m length would be 1.4 mm. The problem says “producing only elastic deformation”: indeed the stress 1.5×10⁸ N m⁻² lies below the yield strength of copper, 2.0×10⁸ N m⁻² (Table 8.1), so Hooke’s law ε = σ/Y can be used.

Extra depth. Check units: area in m² = (15.2×10⁻³)(19.1×10⁻³) = 290.3×10⁻⁶ = 2.90×10⁻⁴ m². If you forget the milli prefix and use 15.2 × 19.1 mm² as m² the answer is wrong by 10⁶. Converting each side to metres first is safer than converting the product.

More to know. This is a one-step application of the definition Y = (F/A)/(ΔL/L): the strain ΔL/L is exactly the quantity asked for, so no length is needed. Many students think the length of the copper piece is missing; it is not needed because strain is a ratio. If the problem had also asked for the elongation, the length L would be required: ΔL = εL. Dimensions check: [F/(A Y)] = N/(m² × N m⁻²) = 1, dimensionless as strain must be. Also note the elastic energy stored per unit volume here is u = ½σε = ½ × 1.53×10⁸ × 1.4×10⁻³ ≈ 1.1×10⁵ J m⁻³ (about 0.1 J in each cm³ of the bar).

Conclusion: Strain = F/(A Y) = 44500/(2.90×10⁻⁴ × 1.1×10¹¹) = 1.4×10⁻³.

Concept. The maximum force a wire or cable can carry is limited by the maximum stress: F_max = σ_max × A. The area is that of a circular cross-section, A = π r².

Question. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N m⁻², what is the maximum load the cable can support?

Given

• r = 1.5 cm = 1.5×10⁻² m
• σ_max = 10⁸ N m⁻²
• g = 9.8 m s⁻²

Step 1 — area of cross-section

A = π r² = 3.14 × (1.5×10⁻²)² = 3.14 × 2.25×10⁻⁴ = 7.07×10⁻⁴ m²

Step 2 — maximum force (load) from stress = F/A

F_max = σ_max A = 10⁸ × 7.07×10⁻⁴ = 7.1×10⁴ N

As a mass: M = F/g = 7.07×10⁴/9.8 ≈ 7.2×10³ kg, i.e. a load of about 7 tonnes. (NCERT asks for “the maximum load”, i.e. the maximum force 7.1×10⁴ N; giving the mass as well shows that you understand the question.)

Extra depth. The limiting stress 10⁸ N m⁻² = 100 MPa is well below the yield strength of steel (250×10⁶ N m⁻² in Table 8.1); it already contains a factor of safety of 2.5. In real chairlifts and cranes a factor of safety of about ten is used (Section 8.6), because the cable also carries the dynamic loads of starting and stopping and its own weight, and because each wire is made of many thin strands.

More to know. Notice how this problem is the reverse of Examples 8.1 and 8.2: there the force is given and the stress is found, here the maximum stress is given and the force is found. Any wire problem fits the triangle F = σ × A; with σ = Yε you can also find the force for a given strain, F = Y A ε. In practice engineers use the yield strength divided by a factor of safety as the permitted stress, and the ultimate strength (400×10⁶ N m⁻² for steel) as the absolute breaking limit; a cable of this size would break at about 400×10⁶ × 7.07×10⁻⁴ ≈ 2.8×10⁵ N, about four times the safe working load.

Conclusion: F_max = σ_maxπr² = 10⁸ × 7.07×10⁻⁴ = 7.1×10⁴ N (about 7.2 tonnes).

Concept. A rigid bar stays horizontal, so all three wires of equal length stretch by the same amount. If in addition they carry the same tension, then F L/(A Y) is the same for each, so A Y must be the same for each wire.

Question. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Step 1 — tension and extension in each wire

• Tension in each wire: T = (15 × 9.8)/3 = 49 N (equal tension, as required).
• The bar is rigid and the wires are of the same length, so every wire has the same extension ΔL (otherwise the bar would tilt).

Step 2 — write Y for each wire

Y_Cu = T L/(A_Cu ΔL)    Y_Fe = T L/(A_Fe ΔL)

T, L and ΔL are the same, therefore Y_Cu A_Cu = Y_Fe A_Fe, i.e.

A_Cu/A_Fe = Y_Fe/Y_Cu

Step 3 — convert area ratio to diameter ratio (A = π d²/4)

(d_Cu/d_Fe)² = Y_Fe/Y_Cu  ⇒  d_Cu/d_Fe = √(Y_Fe/Y_Cu)

With Y(iron) = 190×10⁹ N m⁻² and Y(copper) = 110×10⁹ N m⁻² (Table 8.1):

d_Cu/d_Fe = √(190/110) = √1.727 = 1.31

The copper wires must be about 1.3 times thicker than the iron wire: d_Cu : d_Fe = 1.31 : 1 (and d_Fe/d_Cu = 0.76). The copper wire has the smaller Young’s modulus, so it needs a larger cross-section to give the same stiffness.

Extra depth. Physical reading: for a wire, F = (Y A/L) ΔL, i.e. the wire behaves like a spring of stiffness k = Y A/L. Three springs of equal length stretched by the same amount carry equal forces only if their stiffnesses are equal, so Y A is the same for all. This is the same reason why the stiffest member of a structure carries the largest share of the load: equal tension forces the design to match Y A.

Conclusion: Equal tension and equal extension give Y A = constant, hence d_Cu/d_Fe = √(Y_Fe/Y_Cu) = 1.31.

Concept. At the lowest point of a vertical circle the tension must both support the weight and supply the centripetal force: T − mg = mω² r. The extension then follows from ΔL = T L/(A Y).

Question. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.

Given

• m = 14.5 kg, r = L = 1.0 m (the wire is the radius of the circle)
• ω = 2 rev/s = 2 × 2π = 4π = 12.57 rad s⁻¹
• A = 0.065 cm² = 0.065×10⁻⁴ = 6.5×10⁻⁶ m²
• Y(steel) = 2.0×10¹¹ N m⁻², g = 9.8 m s⁻²

Step 1 — tension at the lowest point

Take upward (towards the centre) as positive. The net force towards the centre supplies mω² r:

T − mg = m ω² r  ⇒  T = m(g + ω² r)

T = 14.5 × (9.8 + 157.9 × 1.0) = 14.5 × 167.7 = 2.43×10³ N

(ω² = (12.57)² = 157.9 s⁻².)

Step 2 — elongation

ΔL = T L/(A Y) = (2.43×10³ × 1.0)/(6.5×10⁻⁶ × 2.0×10¹¹) = 2.43×10³/(1.3×10⁶) = 1.87×10⁻³ m

The wire elongates by about 1.9×10⁻³ m (1.9 mm), a strain of 1.9×10⁻³ = 0.19%.

Extra depth. At the lowest point the mass is moving fastest and the tension is greatest: T = mg + mω²r = 142 N (weight) + 2290 N (centripetal part). At the highest point the tension would be only mω²r − mg = 2148 N, still large. Strictly, the radius is the stretched length 1.0019 m, but the correction is negligible (0.19%). Honest check: the stress T/A = 3.7×10⁸ N m⁻² is larger than the yield strength of steel in Table 8.1 (250×10⁶ N m⁻²) and close to its ultimate strength (400×10⁶ N m⁻²), so a real steel wire would deform permanently or even break. NCERT assumes the wire stays elastic, as the question implies, and so does the answer above.

Conclusion: T = m(g + ω² r) = 2.43×10³ N; ΔL = T L/(A Y) = 1.87×10⁻³ m ≈ 1.9 mm.

Concept. Bulk modulus B = −Δp/(ΔV/V). For a gas at constant temperature (Boyle’s law, pV = constant), B_gas = p (the pressure itself).

Question. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, pressure increase = 100.0 atm (1 atm = 1.013×10⁵ Pa), final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Part 1 — bulk modulus of water

• Pressure increase Δp = 100 atm = 100 × 1.013×10⁵ = 1.013×10⁷ Pa
• V = 100.0 litre, |ΔV| = 100.5 − 100.0 = 0.5 litre
• Volume strain |ΔV|/V = 0.5/100.0 = 5.0×10⁻³

B = Δp/(|ΔV|/V) = (1.013×10⁷)/(5.0×10⁻³) = 2.03×10⁹ Pa

This agrees with the value of Table 8.3 for water, 2.2×10⁹ N m⁻². Note. The data say the final volume is 100.5 litre, which is larger than the initial volume; physically the water must have been compressed to 99.5 litre. NCERT has used only the magnitude of the change, since B is always positive, and so do we.

Part 2 — bulk modulus of air at constant temperature

For a given mass of gas at constant temperature pV = constant. Differentiating, pΔV + VΔp = 0, so Δp/(ΔV/V) = −p. Hence the isothermal bulk modulus of a gas equals its pressure. For air at atmospheric pressure:

B_air = p = 1.0×10⁵ Pa (Table 8.3: 1.0×10⁻⁴ × 10⁹ N m⁻²)

Part 3 — the ratio

B_water/B_air = (2.03×10⁹)/(1.0×10⁵) = 2.0×10⁴

Extra depth. Why is the ratio so large? In a gas the molecules are far apart (about ten times their own diameter) and there is mostly empty space between them; the intermolecular forces are negligible, so a small pressure rise pushes the molecules closer easily and the volume changes appreciably. In a liquid the molecules are practically in contact; to decrease the volume the molecules must be pushed closer against the strong repulsive forces that act at such small separations, so a very large pressure is needed for a small change of volume. Therefore water is about 20,000 times harder to compress than air; this is why liquids are treated as incompressible in the hydraulic machines of Chapter 9.

Conclusion: B(water) ≈ 2.0×10⁹ Pa, B(air) ≈ 1.0×10⁵ Pa, ratio ≈ 2×10⁴ because molecules in a liquid are almost touching but in a gas they are far apart.

Concept. Under a pressure increase Δp the volume of water decreases by ΔV/V = −Δp/B. Since mass is unchanged, the density rises as ρ = ρ₀/(1 − |ΔV|/V).

Question. What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03×10³ kg m⁻³?

Given

• Surface density ρ₀ = 1.03×10³ kg m⁻³ (sea water)
• Pressure increase Δp = 80.0 atm = 80 × 1.013×10⁵ = 8.10×10⁶ Pa
• Bulk modulus of water B = 2.2×10⁹ N m⁻² (Table 8.3)

Step 1 — fractional change in volume

|ΔV|/V = Δp/B = (8.10×10⁶)/(2.2×10⁹) = 3.68×10⁻³

Step 2 — density from conservation of mass

The mass m = ρ₀V₀ = ρV stays the same, with V = V₀(1 − 3.68×10⁻³). Hence

ρ = ρ₀/(1 − 3.68×10⁻³) = 1.03×10³ × 1.00370 = 1.034×10³ kg m⁻³

The density at that depth is 1.034×10³ kg m⁻³: it has increased by only 0.37% (about 3.8 kg m⁻³) even at 80 atm. (80 atm corresponds to a depth of about 800 m of sea water.)

Extra depth. The result shows why water can be treated as incompressible at everyday depths: the density change is less than half a percent even 800 m below the surface. The same computation with ΔV/V = 1.36% at 3000 m depth is worked in Example 8.5 of the text. If the question had been about air, the answer would be completely different because B_air is 10⁴ times smaller.

More to know. Sea water at the surface has density 1.03×10³ kg m⁻³ (fresh water 1.00×10³). Even at the bottom of the deepest ocean trench (about 11 km, pressure ≈ 1100 atm) the volume of water is reduced by only about 5% (the bulk modulus itself rises with pressure), so the density there is about 1.08×10³ kg m⁻³. This near-constancy of density is what lets us use the hydrostatic formula p = p₀ + ρgh with constant ρ in Chapter 9. It also explains why a submarine hull must withstand the pressure but need not allow for the water around it changing its buoyancy appreciably.

Conclusion: ρ = ρ₀/(1 − Δp/B) = 1.034×10³ kg m⁻³ at 80 atm.

Concept. The volume strain under hydraulic pressure is |ΔV|/V = Δp/B, with B the bulk modulus of glass from Table 8.3.

Question. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Given

• Δp = 10 atm = 10 × 1.013×10⁵ = 1.013×10⁶ Pa
• B(glass) = 37×10⁹ N m⁻² (Table 8.3, NCERT)

Solution

|ΔV|/V = Δp/B = (1.013×10⁶)/(37×10⁹) = 2.7×10⁻⁵

So the volume of the glass slab decreases by a fraction of 2.7×10⁻⁵ (about 0.0027 %); with the negative sign that shows a decrease, ΔV/V = −2.7×10⁻⁵. A 1 litre piece of glass would shrink by only 0.027 cm³. Notice that this is only about 1/17 of the fractional change of water at 10 atm (4.6×10⁻⁴), because B(glass) is about 17 times larger.

Extra depth. The table above shows the effect of B on volume strain, using the same pressure of 10 atm. The last row does not obey the small-strain formula, because for a gas Δp/B is not small: the actual change is obtained from Boyle’s law. For solids and liquids, ΔV/V is always very small and the linear relation is accurate. Because a slab is also compressed equally from all sides, there is no change in shape, only in volume.

More to know. Glass is brittle, which means its fracture strength in tension is small (50×10⁶ N m⁻², Table 8.1) but it is strong in compression. A hydraulic pressure compresses the slab from every side, so no tension arises anywhere inside and the slab does not crack, even at pressures where a bending load would shatter it. This is why glass portholes of deep-sea submersibles can survive hundreds of atmospheres. The result of this problem also shows an important number to remember: for typical solids the volume strain under 10 atm is of the order 10⁻⁵, which is the reason why we normally ignore the compressibility of solids.

Conclusion: ΔV/V = Δp/B = 1.013×10⁶/(37×10⁹) = 2.7×10⁻⁵ (a decrease in volume).

Concept. From B = −p/(ΔV/V) the volume contraction is |ΔV| = p V/B. First find the original volume V of the cube.

Question. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0×10⁶ Pa.

Given

• Edge = 10 cm = 0.10 m ⇒ V = (0.10)³ = 1.0×10⁻³ m³ (= 1000 cm³)
• p = 7.0×10⁶ Pa
• B(copper) = 140×10⁹ N m⁻² (Table 8.3)

Step 1 — volume strain

|ΔV|/V = p/B = (7.0×10⁶)/(140×10⁹) = 5.0×10⁻⁵

Step 2 — volume contraction

|ΔV| = (5.0×10⁻⁵) × (1.0×10⁻³ m³) = 5.0×10⁻⁸ m³ = 5.0×10⁻² cm³

The cube contracts by only 5×10⁻⁸ m³ (0.05 cm³ or 50 mm³), a fraction 5×10⁻⁵ of its volume — invisible to the eye. Each edge shortens by about one third of the fractional volume change, 1.7×10⁻⁵ of 10 cm = 1.7×10⁻³ mm (about 1.7 micrometre). 7.0×10⁶ Pa is roughly 70 atm, i.e. the pressure at a depth of about 700 m of water.

Extra depth. This is an application of “unit conversion” more than of physics: the volume must be in m³ (or consistent cm³ together with B in N cm⁻²). A safe route is to keep SI throughout: V = 1.0×10⁻³ m³, B = 1.4×10¹¹ Pa. For a cube, linear strain = (1/3)(volume strain), because V ∝ a³ gives ΔV/V = 3Δa/a for small changes.

More to know. Compare copper with other metals at the same pressure: with the Table 8.3 values the volume contraction of a 10 cm cube at 7.0×10⁶ Pa is 5.0×10⁻⁸ m³ for copper, 9.7×10⁻⁸ m³ for aluminium (B = 72 GPa), 7.0×10⁻⁸ m³ for iron (100 GPa), 2.7×10⁻⁸ m³ for nickel (260 GPa) and 3.2×10⁻⁶ m³ for water (2.2 GPa). The relative sizes are inversely proportional to B, so the materials with the largest bulk modulus resist compression best.

Conclusion: |ΔV| = p V/B = 7.0×10⁶ × 1.0×10⁻³/(1.4×10¹¹) = 5.0×10⁻⁸ m³ (0.05 cm³).

Concept. Bulk modulus B = Δp/(|ΔV|/V), so the pressure change needed for a given volume strain is Δp = B × (|ΔV|/V).

Question. How much should the pressure on a litre of water be changed to compress it by 0.10%?

Given

• Volume strain |ΔV|/V = 0.10% = 0.10/100 = 1.0×10⁻³
• B(water) = 2.2×10⁹ N m⁻² (Table 8.3)
• (The original volume, 1 litre = 10⁻³ m³, is not needed because the strain is given as a fraction; it lets us find |ΔV| = 1.0×10⁻⁶ m³ = 1 cm³.)

Solution

Δp = B × (|ΔV|/V) = (2.2×10⁹) × (1.0×10⁻³) = 2.2×10⁶ Pa

So the pressure on the water must be increased by 2.2×10⁶ Pa, i.e. by about 22 atmospheres (2.2×10⁶/1.013×10⁵ = 21.7 atm). The water is compressed by 1 cm³ out of every litre.

Extra depth. The linear relation Δp ∝ ΔV/V is valid while the strain is small; the table shows that to compress water by only 1% you need about 217 atm, which is why hydraulic fluids are treated as incompressible in machines working at a few atmospheres. In the reverse direction, relieving a pressure of 22 atm lets one litre of water expand by 1 cm³, a fact used in the design of pressure vessels filled with liquid (even a small leak of volume produces a big pressure drop).

More to know. This question completes the set of the three elastic moduli: Y for a change in length (Ex 8.1–8.3, 8.5, 8.7–8.9, 8.11), G for a change in shape (Ex 8.4, 8.6) and B for a change in volume (Ex 8.12–8.16). For every bulk-modulus problem the recipe is the same: (i) convert the pressure to pascal, (ii) write the volume strain as a pure fraction, (iii) use B = −Δp/(ΔV/V) and (iv) keep the sign (an increase in pressure gives a decrease in volume). Reciprocal of B is the compressibility k = 1/B = 4.5×10⁻¹⁰ Pa⁻¹ for water, so ΔV/V = −kΔp = 4.5×10⁻¹⁰ × 2.2×10⁶ = 1.0×10⁻³ as required.

Conclusion: Δp = B(|ΔV|/V) = 2.2×10⁹ × 10⁻³ = 2.2×10⁶ Pa (about 22 atm) is needed to compress 1 litre of water by 0.10%.