Exercises — Mechanical Properties of Solids

16 questions✓ Free · step-by-step

8.13 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.1
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 of 4.0 × 10⁻⁵ m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Hint. Write Y = FL/(A·ΔL) for each wire separately, then divide one by the other. The load and the stretch are the same for both, so those two symbols will cancel — you never need their numerical values.
Step 1 — Write Young's modulus for each wire. From the definition Y = stress/strain = (F/A)/(ΔL/L), rearranging gives

Y = FL/(A·ΔL)

For steel: Y_s = F·L_s/(A_s·ΔL_s) For copper: Y_c = F·L_c/(A_c·ΔL_c)

Step 2 — Use what the question tells you is shared. The phrase "under a given load" means the same force F acts on both, and "stretches by the same amount" means ΔL_s = ΔL_c. So both F and ΔL are common to the two wires, which is why they will cancel in the ratio and why the question never gives you their values.

Step 3 — Take the ratio. Y_s/Y_c = [F·L_s/(A_s·ΔL)] ÷ [F·L_c/(A_c·ΔL)]

Cancelling F and ΔL:

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

Step 4 — Substitute the numbers. Y_s/Y_c = (4.7 × 4.0 × 10⁻⁵)/(3.0 × 10⁻⁵ × 3.5) = (1.88 × 10⁻⁴)/(1.05 × 10⁻⁴) = 1.79

Step 5 — Sanity check against Table 8.1. The table gives Y_steel = 200 × 10⁹ Pa and Y_copper = 110 × 10⁹ Pa, a ratio of 1.82. Our answer of 1.79 sits right on top of that, so the working is sound.

✦ Y_steel : Y_copper ≈ 1.8 : 1 — steel is about 1.8 times as stiff as copper.
Where students slip. Inverting the area ratio. Since Y = FL/(A·ΔL), the area sits in the denominator, so the copper area A_c must end up on top when you form Y_s/Y_c. Writing (L_s/L_c)×(A_s/A_c) gives 1.01 instead of 1.79 — a suspiciously round answer that should make you re-check.
8.23 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.2 — depends on Fig. 8.9, read from the PDF at 190dpi
Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
Hint. Young's modulus is the slope of the straight part of the graph, so pick two points on the straight section — the origin is one of them. Watch the axis units: stress is plotted in units of 10⁶ N m⁻².
Step 1 — Read the axes carefully. The vertical axis is stress in units of 10⁶ N m⁻², marked 0 to 300. The horizontal axis is strain, marked 0 to 0.004, and strain has no units.

Step 2 — Young's modulus is the slope of the linear region. Hooke's law holds only where the graph is straight, and there Y = stress/strain. Reading a convenient point on the straight portion: at a strain of 0.002 the stress is 150 × 10⁶ N m⁻².

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

The origin is the second point on the line, so this single reading is enough to fix the slope.

Step 3 — Find the yield strength from the graph. The yield point is where the curve stops being straight and the material begins to deform permanently. On this graph the curve rises to a maximum of about 300 × 10⁶ N m⁻² and then turns over.

Yield strength ≈ 300 × 10⁶ N m⁻² = 3 × 10⁸ N m⁻²

Step 4 — Check the answer is physically reasonable. Comparing with Table 8.1, a Young's modulus of 75 GPa is close to aluminium's 70 GPa, and the yield strength is of the right order for a structural metal. That agreement is why we can trust a value read off a graph.

✦ (a) Y ≈ 7.5 × 10¹⁰ N m⁻² (75 GPa) (b) yield strength ≈ 3 × 10⁸ N m⁻²
Where students slip. Forgetting the 10⁶ factor on the stress axis and quoting Y = 75000 N m⁻². Always read the multiplier printed in the axis label before dividing — it is part of every value on that axis.
8.32 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.3 — depends on Fig. 8.10, re-read at 560dpi because the slope comparison decides the answer
The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are 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?
Hint. Two different features answer the two parts. Young's modulus is about the steepness of the initial straight portion; strength is about how high the curve gets before the material breaks. Note the question says the graphs share a scale — without that you could compare neither.
Step 1 — Understand why "drawn to the same scale" is stated. Stress-strain graphs can only be compared if both axes mean the same thing on both plots. The question supplies that condition deliberately, so any visual comparison you make is legitimate.

Step 2 — (a) Compare the slopes of the straight portions. Young's modulus is the slope of the initial linear region, since Y = stress/strain there. Material A's straight section climbs very steeply — it gains a large stress over a very small strain. Material B's straight section is noticeably less steep, taking a larger strain to reach a comparable stress.

A steeper line means a larger stress-to-strain ratio, so A has the greater Young's modulus.

Step 3 — (b) Compare the maximum stress reached. The strength of a material is measured by the largest stress it can withstand before it fractures — its ultimate tensile strength, which is the height of the peak of the curve. Material A's curve rises to a slightly higher peak stress than B's before turning over and ending.

Because A withstands the greater stress before breaking, A is the stronger material.

Step 4 — Keep the two ideas apart. Stiffness (Young's modulus) and strength are different properties. Stiffness is how hard it is to deform a material at all; strength is how much stress it takes to break it. A material can be stiff but brittle, or flexible but very strong.

✦ (a) A has the greater Young's modulus (b) A is the stronger material
Where students slip. Answering from the width of the curve — thinking that the material which stretches further along the strain axis must be stronger. Extending further before breaking describes ductility, not strength. Strength is the height of the curve, not its length.
8.43 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.4
Read the following two statements below carefully and state, with reasons, if it is 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.
Hint. For (a), ask which material needs the bigger stress to produce the same strain. For (b), think about what actually happens to the wire of a spring when you pull the spring's ends apart — does the wire get longer, or does it twist?
Step 1 — (a) is FALSE. Young's modulus is Y = stress/strain, so a large Y means a large stress is needed to produce even a small strain. Rubber stretches enormously under a modest pull, which means its strain is huge for a small stress, so its Young's modulus is small. Steel barely stretches at all under the same stress, which is exactly what a large Y describes.

Table 8.1 gives Y_steel = 200 × 10⁹ Pa, while rubber is of the order of 10⁷ Pa — smaller by a factor of around ten thousand.

The textbook's Points to Ponder warns about precisely this confusion: in everyday speech we call the material that stretches more "more elastic", but in physics the material that stretches less for a given load is the more elastic one. Rubber stretching further makes it less stiff, not more.

Step 2 — (b) is TRUE. When you pull the two ends of a coiled spring apart, the spring gets longer — but the wire it is made from does not. Follow the wire around one turn and you find that the deformation is a twist: each element of the wire is sheared relative to its neighbour, the turns rotating slightly to open the helix out.

Since the deformation of the material is a shearing one rather than a stretching one, the resistance to it is governed by the shear modulus (also called the modulus of rigidity), not by Young's modulus.

✦ (a) False — steel's Young's modulus is far greater than rubber's. (b) True — extending a coil twists its wire, so the shear modulus governs it.
Where students slip. Marking (a) true because rubber "is more elastic" in the everyday sense. Stretching more means a larger strain for the same stress, which by Y = stress/strain makes Young's modulus smaller, not larger.
8.55 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.5 — depends on Fig. 8.11, read from the PDF. NOTE: Table 8.1 does not list brass, so Y_brass = 0.91 × 10¹¹ Pa is taken as a standard value from outside the chapter.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Hint. The two wires do not carry the same load. Work out what hangs below each wire — the steel wire is above the 4.0 kg mass, so it must support that mass and everything beneath it as well.
Step 1 — Read the loads off the arrangement. From Fig. 8.11, the steel wire hangs from the ceiling, the 4.0 kg mass is attached at its lower end, the brass wire hangs from that mass, and the 6.0 kg mass hangs at the very bottom.

The brass wire carries only what is below it: 6.0 kg. The steel wire carries everything below it: 4.0 + 6.0 = 10.0 kg.

This is the whole difficulty of the question, and it is settled by reading the figure rather than by any formula.

Step 2 — Find the common cross-sectional area. Both wires have diameter 0.25 cm = 2.5 × 10⁻³ m, so the radius is 1.25 × 10⁻³ m.

A = πr² = π(1.25 × 10⁻³)² = 4.909 × 10⁻⁶ m²

Step 3 — Elongation of the steel wire. F = 10.0 × 9.8 = 98 N, L = 1.5 m, Y_steel = 2.0 × 10¹¹ Pa (Table 8.1)

ΔL = FL/(AY) = (98 × 1.5)/(4.909 × 10⁻⁶ × 2.0 × 10¹¹) = 147/(9.818 × 10⁵) = 1.50 × 10⁻⁴ m

Step 4 — Elongation of the brass wire. F = 6.0 × 9.8 = 58.8 N, L = 1.0 m, Y_brass = 0.91 × 10¹¹ Pa

ΔL = (58.8 × 1.0)/(4.909 × 10⁻⁶ × 0.91 × 10¹¹) = 58.8/(4.467 × 10⁵) = 1.32 × 10⁻⁴ m

✦ Steel wire stretches about 1.5 × 10⁻⁴ m (0.15 mm); brass wire stretches about 1.3 × 10⁻⁴ m (0.13 mm).
Where students slip. Loading the steel wire with only 4.0 kg. The steel wire is above both masses, so it carries 10.0 kg in total — the brass wire and the 6.0 kg below simply hang from the 4.0 kg block. Using 4.0 kg gives 0.06 mm, well under half the correct value.
8.63 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.6
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 then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Hint. The weight hangs alongside the face, so it acts parallel to that face rather than perpendicular to it — that makes this a shearing problem, not a stretching one. The area to use is the area of the face the force acts along.
Step 1 — Identify the type of stress. The cube is fixed to a vertical wall and the load hangs from the opposite face. The weight therefore acts downward, parallel to that face, while the wall holds the far side. Two equal and opposite forces acting parallel to opposite surfaces is precisely shearing stress, so the shear modulus governs this problem.

Step 2 — Compute the shearing stress. The face has area A = 10 cm × 10 cm = 0.1 m × 0.1 m = 0.01 m². The applied force is the weight F = 100 × 9.8 = 980 N.

Shearing stress = F/A = 980/0.01 = 9.8 × 10⁴ N m⁻²

Step 3 — Compute the shearing strain. From the definition G = shearing stress/shearing strain,

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

This angle is in radians and is extremely small, which is why the small-angle treatment is valid.

Step 4 — Convert the strain into a deflection. Shearing strain is defined as the sideways displacement divided by the height over which it occurs, so Δx = θ × L where L is the edge of the cube:

Δx = 3.92 × 10⁻⁶ × 0.1 = 3.92 × 10⁻⁷ m

Step 5 — Comment on the size. That is under a millionth of a metre for a 100 kg load. Metals are extremely resistant to shear, which is why solid metal blocks feel completely rigid in ordinary use.

✦ Vertical deflection ≈ 3.92 × 10⁻⁷ m (about 0.4 micrometres).
Where students slip. Using Young's modulus because the deflection is a length. The direction of the force relative to the face is what decides the modulus, not the direction of the answer — a force along the face always means shear.
8.73 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.7
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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Hint. Two things need care: the load is shared between four columns, and the columns are hollow, so the load-bearing area is a ring — the outer circle minus the inner one, not the full circle.
Step 1 — Find the load on a single column. The load is uniformly distributed over four columns, so each carries one quarter of the total weight:

F = (50,000 × 9.8)/4 = 490,000/4 = 1.225 × 10⁵ N

Step 2 — Find the cross-sectional area of one hollow column. The column is a tube, so the material occupies an annulus between the inner and outer radii. With R = 0.60 m and r = 0.30 m:

A = π(R² − r²) = π(0.60² − 0.30²) = π(0.36 − 0.09) = π(0.27) = 0.8482 m²

Step 3 — Compute the compressional stress. Stress = F/A = (1.225 × 10⁵)/0.8482 = 1.444 × 10⁵ N m⁻²

Step 4 — Convert stress into strain. Young's modulus of steel from Table 8.1 is 2.0 × 10¹¹ Pa, and the same value applies in compression as in tension, since the chapter states that for a given material the strain produced is the same whether the stress is tensile or compressive.

Strain = stress/Y = (1.444 × 10⁵)/(2.0 × 10¹¹) = 7.22 × 10⁻⁷

Step 5 — Interpret the number. A strain of 7 × 10⁻⁷ means each column shortens by less than one part in a million of its length. Steel columns are chosen for exactly this reason: a 50-tonne structure barely compresses them at all.

✦ Compressional strain of each column ≈ 7.2 × 10⁻⁷ (dimensionless).
Where students slip. Taking the area as πR² and ignoring the hollow centre, which overstates the area by a third and understates the strain. A hollow column's load-bearing area is the ring of material only.
8.82 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.8
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?
Hint. Convert both millimetre dimensions to metres before multiplying them, or the area will be out by a factor of a million. The phrase "only elastic deformation" is your permission to use Hooke's law.
Step 1 — Note what "only elastic deformation" buys you. Hooke's law, and with it Y = stress/strain, holds only in the linear part of the stress-strain curve. The question states the deformation is purely elastic, which is why you are entitled to use Young's modulus here at all.

Step 2 — Find the cross-sectional area. Convert to metres first: 15.2 mm = 15.2 × 10⁻³ m and 19.1 mm = 19.1 × 10⁻³ m.

A = (15.2 × 10⁻³) × (19.1 × 10⁻³) = 2.903 × 10⁻⁴ m²

Step 3 — Compute the tensile stress. Stress = F/A = 44,500/(2.903 × 10⁻⁴) = 1.533 × 10⁸ N m⁻²

Step 4 — Convert stress to strain using copper's Young's modulus. From Table 8.1, Y_copper = 110 × 10⁹ = 1.1 × 10¹¹ Pa.

Strain = stress/Y = (1.533 × 10⁸)/(1.1 × 10¹¹) = 1.39 × 10⁻³

Step 5 — Check it really is still elastic. Table 8.1 gives copper a yield strength of 200 × 10⁶ N m⁻². Our stress of 153 × 10⁶ N m⁻² is below that, so the copper is indeed still in its elastic region and the question's statement is consistent.

✦ Strain ≈ 1.4 × 10⁻³ (that is, about 0.14%).
Where students slip. Multiplying 15.2 × 19.1 in millimetres and then treating the product as square metres. The area is 290.3 mm², which is 2.903 × 10⁻⁴ m² — a factor of 10⁶ smaller, and forgetting it makes the strain come out a million times too small.
8.92 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.9
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?
Hint. Stress is force per unit area, so the maximum force is simply the maximum stress multiplied by the cable's cross-sectional area. Decide whether the question wants an answer as a force in newtons or a mass in kilograms.
Step 1 — Find the cross-sectional area of the cable. The radius is 1.5 cm = 0.015 m, and the cable's cross-section is a circle:

A = πr² = π(0.015)² = π(2.25 × 10⁻⁴) = 7.069 × 10⁻⁴ m²

Step 2 — Convert the stress limit into a force limit. Since stress = F/A, the largest force the cable can carry before the stress limit is reached is

F_max = stress × A = (10⁸) × (7.069 × 10⁻⁴) = 7.069 × 10⁴ N

Step 3 — Express it as a load in kilograms. A load of mass m exerts a force mg, so

m = F_max/g = (7.069 × 10⁴)/9.8 = 7213 kg

Step 4 — Sanity check against the material. Table 8.1 gives steel a yield strength of 250 × 10⁶ N m⁻². The design limit of 10⁸ N m⁻² is well below that, which is what you would expect of a cable carrying people — engineers deliberately work far inside the yield point rather than at it.

✦ Maximum force ≈ 7.07 × 10⁴ N, corresponding to a load of about 7.2 × 10³ kg.
Where students slip. Leaving the radius in centimetres, so the area comes out as 7.07 cm² treated as m². Convert to metres before squaring, since the squaring turns a factor of 100 into a factor of 10,000.
8.103 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.10
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.
Hint. The bar is rigid and stays horizontal, which forces all three wires to stretch by the same amount. Combine that with equal tensions and equal lengths, and see what has to be true of A and Y.
Step 1 — Extract the three conditions hidden in the wording. • The wires are each 2.0 m long, so L is the same for all three. • The question requires each to have the same tension, so F is the same. • The bar is rigid and supported symmetrically, so it stays horizontal, which forces every wire to stretch by the same ΔL.

The mass of 15 kg never enters the calculation — it would only matter if you were asked for the actual tension.

Step 2 — Apply the elongation formula. ΔL = FL/(AY)

With F, L and ΔL all common to the three wires, the product AY must be the same for each, since everything else in the equation is fixed:

A_copper × Y_copper = A_iron × Y_iron

Step 3 — Convert to a ratio of areas. A_copper/A_iron = Y_iron/Y_copper

From Table 8.1: Y_iron (wrought) = 190 × 10⁹ Pa and Y_copper = 110 × 10⁹ Pa.

A_copper/A_iron = 190/110 = 1.727

Step 4 — Convert areas to diameters. Area goes as the square of the diameter, A = πd²/4, so the diameters are in the ratio of the square roots:

d_copper/d_iron = √(1.727) = 1.31

Step 5 — Check the logic makes sense. Copper has the smaller Young's modulus, so it is the less stiff material. To stretch by the same amount under the same pull it must be the thicker wire, which is exactly what a ratio greater than 1 says.

✦ d_copper : d_iron ≈ 1.31 : 1 — the copper wires must be about 1.3 times the diameter of the iron one.
Where students slip. Stopping at the ratio of areas and quoting 1.73 as the answer. The question asks for the ratio of diameters, so you must take the square root, because area depends on the square of the diameter.
8.115 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.11
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.
Hint. At the lowest point the tension has two jobs: holding the weight up and supplying the centripetal force. Both point the same way relative to the mass, so they add. Also convert 2 revolutions per second into radians per second.
Step 1 — Find the tension at the lowest point. At the bottom of the circle the tension acts upward and the weight acts downward, and the net upward force must provide the centripetal acceleration toward the centre, which is directly above:

T − mg = mω²r T = m(g + ω²r)

This is the step that makes the problem a mechanics question as much as an elasticity one.

Step 2 — Convert the angular velocity. ω = 2 rev/s = 2 × 2π rad/s = 4π = 12.57 rad/s

Step 3 — Evaluate the tension. With m = 14.5 kg, r = 1.0 m and g = 9.8 m s⁻²:

ω²r = (12.57)² × 1.0 = 157.9 m s⁻² T = 14.5 × (9.8 + 157.9) = 14.5 × 167.7 = 2432 N

Notice the centripetal term is sixteen times larger than gravity here, so the whirling — not the weight — is what stresses the wire.

Step 4 — Convert the area and find the elongation. A = 0.065 cm² = 0.065 × 10⁻⁴ m² = 6.5 × 10⁻⁶ m² Y_steel = 2.0 × 10¹¹ Pa (Table 8.1)

ΔL = TL/(AY) = (2432 × 1.0)/(6.5 × 10⁻⁶ × 2.0 × 10¹¹) = 2432/(1.3 × 10⁶) = 1.87 × 10⁻³ m

✦ Elongation ≈ 1.87 × 10⁻³ m, that is about 1.9 mm.
Where students slip. Using T = mg and ignoring the circular motion entirely, which gives only 0.11 mm. At the lowest point the tension must both support the weight and supply the centripetal force, so the two contributions add.
8.125 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.12
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.
Hint. Use B = Δp/(|ΔV|/V), taking the magnitude of the volume change so the modulus comes out positive. For the comparison, the isothermal bulk modulus of a gas is equal to its pressure — about one atmosphere.
Step 1 — Assemble the quantities. Δp = 100.0 atm = 100.0 × 1.013 × 10⁵ = 1.013 × 10⁷ Pa V = 100.0 litre, and the volume change has magnitude |ΔV| = 0.5 litre.

The volume strain is a ratio, so the litres cancel and there is no need to convert to cubic metres.

Volume strain = |ΔV|/V = 0.5/100.0 = 5.0 × 10⁻³

Step 2 — Compute the bulk modulus. B = Δp/(|ΔV|/V) = (1.013 × 10⁷)/(5.0 × 10⁻³) = 2.026 × 10⁹ N m⁻²

This sits right beside the 2.2 × 10⁹ Pa listed for water in Table 8.3, which confirms the arithmetic.

Step 3 — Compare with air. For a gas compressed at constant temperature, the bulk modulus equals the pressure itself, so at atmospheric pressure B_air ≈ 1.0 × 10⁵ Pa. Table 8.3 gives the same figure, 1.0 × 10⁻⁴ GPa.

B_water/B_air = (2.026 × 10⁹)/(1.0 × 10⁵) ≈ 2 × 10⁴

Water is about twenty thousand times harder to compress than air.

Step 4 — Explain the ratio in simple terms. In a gas the molecules are far apart, with a great deal of empty space between them, so squeezing simply pushes them closer together and a modest pressure produces a large fractional change in volume.

In a liquid the molecules are already almost touching, and pushing them any closer means working against strong short-range repulsion between them. There is very little free space left to remove, which is why the same pressure produces only a tiny volume change and the bulk modulus is enormous by comparison.

✦ B_water ≈ 2.03 × 10⁹ N m⁻²; B_water/B_air ≈ 2 × 10⁴, because a liquid's molecules are already close-packed while a gas is mostly empty space.
Where students slip. Carrying the minus sign from B = −p/(ΔV/V) into the final answer and reporting a negative bulk modulus. The sign in the definition exists only to make B positive when a pressure increase shrinks the volume — the modulus itself is always positive.
8.133 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.13
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m⁻³?
Hint. The mass of a sample of water does not change as it is compressed — only its volume does. Work out the fractional change in volume from the bulk modulus, then use density = mass/volume.
Step 1 — Find the fractional change in volume. Δp = 80.0 atm = 80.0 × 1.013 × 10⁵ = 8.104 × 10⁶ Pa From Table 8.3, B_water = 2.2 × 10⁹ Pa.

From B = Δp/(|ΔV|/V):

|ΔV|/V = Δp/B = (8.104 × 10⁶)/(2.2 × 10⁹) = 3.684 × 10⁻³

So the water is compressed to about 99.63% of its surface volume.

Step 2 — Relate density to volume. Compressing water does not create or destroy any of it, so the mass m of a given sample stays fixed while the volume shrinks. Since density is ρ = m/V, a smaller volume means a larger density:

ρ' = m/V' where V' = V(1 − |ΔV|/V)

Step 3 — Substitute. ρ' = ρ/(1 − |ΔV|/V) = (1.03 × 10³)/(1 − 3.684 × 10⁻³) = (1.03 × 10³)/0.996316 = 1.034 × 10³ kg m⁻³

Step 4 — Comment on the size of the effect. At a depth corresponding to 80 atmospheres — roughly 800 metres of sea water — the density has risen by only about 0.4%. This is why water is treated as incompressible in nearly every practical calculation, and it is the quantitative justification for that assumption rather than a hand wave.

✦ Density at that depth ≈ 1.034 × 10³ kg m⁻³.
Where students slip. Multiplying the surface density by (1 − ΔV/V) instead of dividing. Compression makes water denser, not less dense, so the answer must come out larger than 1.03 × 10³ — a quick check that catches the slip immediately.
8.142 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.14
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Hint. "Fractional change in volume" is exactly the volume strain ΔV/V, so you can read the answer straight off the definition of bulk modulus without ever needing the slab's actual size.
Step 1 — Recognise that no dimensions are needed. The question asks for a fraction, and the definition B = Δp/(ΔV/V) rearranges directly to

ΔV/V = Δp/B

So the size and shape of the slab are irrelevant, which is why none are given.

Step 2 — Convert the pressure. Δp = 10 atm = 10 × 1.013 × 10⁵ = 1.013 × 10⁶ Pa

Step 3 — Take the bulk modulus of glass from Table 8.3. B_glass = 37 × 10⁹ Pa = 3.7 × 10¹⁰ Pa

Step 4 — Compute the fractional change. ΔV/V = (1.013 × 10⁶)/(3.7 × 10¹⁰) = 2.74 × 10⁻⁵

Step 5 — Interpret it. That is a change of under three thousandths of one per cent. Note the answer is dimensionless — a fractional change has no units, since it is a volume divided by a volume, and quoting it in m³ would be wrong.

✦ Fractional change in volume ≈ 2.7 × 10⁻⁵ (a decrease).
Where students slip. Using Young's modulus for glass (65 GPa from Table 8.1) rather than the bulk modulus (37 GPa from Table 8.3). Hydraulic pressure acts perpendicular to every surface at once and changes volume, so the bulk modulus is the one that applies.
8.153 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.15
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.
Hint. Here the actual volume is needed, because the question asks for the contraction itself and not just the fraction. Work out the fractional change first, then multiply by the cube's volume.
Step 1 — Find the volume of the cube. The edge is 10 cm = 0.1 m, so

V = (0.1)³ = 1.0 × 10⁻³ m³

(That is one litre, which is a useful thing to recognise.)

Step 2 — Find the fractional change in volume. From Table 8.3, B_copper = 140 × 10⁹ = 1.4 × 10¹¹ Pa. The pressure is already in pascals, so no conversion is needed:

ΔV/V = Δp/B = (7.0 × 10⁶)/(1.4 × 10¹¹) = 5.0 × 10⁻⁵

Step 3 — Convert the fraction into an actual volume change. ΔV = V × (ΔV/V) = (1.0 × 10⁻³) × (5.0 × 10⁻⁵) = 5.0 × 10⁻⁸ m³

Step 4 — Express it in a unit you can picture. Since 1 m³ = 10⁶ cm³,

ΔV = 5.0 × 10⁻⁸ × 10⁶ = 0.05 cm³

Step 5 — Sense-check. A pressure of 7 × 10⁶ Pa is about 70 atmospheres, and it squeezes a litre of copper by only one twentieth of a cubic centimetre — roughly a drop. That is what a bulk modulus of 140 GPa means in practice.

✦ Volume contraction ≈ 5.0 × 10⁻⁸ m³, which is 0.05 cm³.
Where students slip. Stopping at the fractional change of 5 × 10⁻⁵ and offering it as the contraction. The question asks for a volume, so the fraction must still be multiplied by the cube's 10⁻³ m³.
8.162 marksNCERT Cl-11 Physics Part II, Ch8 Exercises, Q8.16
How much should the pressure on a litre of water be changed to compress it by 0.10%?
Hint. Turn the percentage into a plain fraction first — 0.10% is 0.10/100. Then rearrange the bulk modulus definition to give the pressure rather than the strain.
Step 1 — Convert the percentage into a volume strain. A compression of 0.10% means

ΔV/V = 0.10/100 = 1.0 × 10⁻³

Getting this conversion right is the whole question; the rest is one substitution.

Step 2 — Rearrange the definition of bulk modulus. From B = Δp/(ΔV/V),

Δp = B × (ΔV/V)

Step 3 — Substitute, using water's bulk modulus. From Table 8.3, B_water = 2.2 × 10⁹ Pa.

Δp = (2.2 × 10⁹) × (1.0 × 10⁻³) = 2.2 × 10⁶ Pa

Step 4 — Express the answer in atmospheres for a sense of scale. Since 1 atm = 1.013 × 10⁵ Pa,

Δp = (2.2 × 10⁶)/(1.013 × 10⁵) ≈ 21.7 atm

Step 5 — Notice what this tells you. It takes more than twenty atmospheres to squeeze water by a mere one tenth of one per cent. The stated volume of one litre never enters the calculation, because the compression was given as a fraction rather than as an absolute amount.

✦ The pressure must be increased by about 2.2 × 10⁶ Pa (roughly 22 atm).
Where students slip. Reading 0.10% as 0.10 rather than 0.001, which inflates the answer by a factor of a hundred to 2.2 × 10⁸ Pa. Always divide a percentage by 100 before it enters a formula.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part II textbook, Reprint 2026-27 (keph201.pdf, 13 pages), cross-checked against the official CBSE curriculum 2026-27, Subject Code 042, Unit VII — one end-of-chapter Exercises set (16 questions, 8.1-8.16). Table 8.1 (Young's moduli), Table 8.3 (bulk moduli) and the three exercise figures did not linearise under text extraction and were rendered as images and read visually; Fig. 8.10 was re-read at 560dpi because the comparison of the two initial slopes reverses the answer to Q8.3. Note Table 8.1 does not list brass, which Q8.5 requires — see that solution's sourceNote.. Questions are referenced from the NCERT textbook for identification.

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