NCERT Solutions Class 11 Physics Chapter 8 — Mechanical Properties of Solids

NCERT Class 11 Physics — Mechanical Properties of Solids, Chapter 8 Exercises. All 16 questions solved.

Every exercise at the end of Chapter 8 asks how much a solid deforms under a load. Questions 8.1 to 8.4 are about reading and comparing moduli; 8.5 and 8.7 to 8.11 are about stretching or compressing wires, columns and cables; 8.6 is about shear; and 8.12 to 8.16 are about the change of volume under pressure. Each calculation uses one of three definitions:

• Young’s modulus — for a change of length, $Y = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}$.
• Shear modulus — for a change of shape, $G = \dfrac{F/A}{\Delta x/L}$, where the force is parallel to the face of area $A$ and $\Delta x$ is the sideways shift of a face a distance $L$ from the fixed one.
• Bulk modulus — for a change of volume, $B = -\dfrac{p}{\Delta V/V}$; the minus sign says that an increase in pressure reduces the volume.

The moduli come from the book’s Tables 8.1 to 8.3: $Y$ for steel $2.0\times10^{11}$, copper $1.1\times10^{11}$ and iron $1.9\times10^{11}\ \text{Pa}$; $B$ for water $2.2\times10^9$, glass $3.7\times10^{10}$ and copper $1.4\times10^{11}\ \text{Pa}$. We take $g = 9.8\ \text{m s}^{-2}$.

Key insight. Every question here is one modulus rearranged, so the real work is in the two inputs: the force the piece actually carries, and the area it acts on. Get the force right — the upper wire in 8.5 carries both masses, each column in 8.7 carries a quarter of the load, and the wire in 8.11 carries the weight plus the centripetal force — and pick the modulus that matches the deformation: length (Young’s), shape (shear, 8.6) or volume (bulk, 8.12 to 8.16). What is left is one line of arithmetic.

Question 8.1

A steel wire of length $4.7\ \text{m}$ and cross-sectional area $3.0\times10^{-5}\ \text{m}^2$ stretches by the same amount as a copper wire of length $3.5\ \text{m}$ and cross-sectional area of $4.0\times10^{-5}\ \text{m}^2$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Solution. From $Y = \dfrac{FL}{A\,\Delta L}$, both wires carry the same load $F$ and have the same extension $\Delta L$, so those cancel in the ratio and only $L/A$ is left:

$$\frac{Y_s}{Y_c} = \frac{L_s/A_s}{L_c/A_c} = \frac{4.7/(3.0\times10^{-5})}{3.5/(4.0\times10^{-5})} = \frac{1.567\times10^5}{0.875\times10^5} = 1.79$$

The steel wire is longer and thinner, yet stretches no more than the copper one, so steel must be the stiffer material.

Question 8.2

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?

Question 8.2 (a)

Solution. Young’s modulus is the slope of the straight, proportional part of the graph, stress divided by strain. The line passes through a convenient grid point, a stress of $150\times10^6\ \text{N m}^{-2}$ at a strain of $0.002$:

$$Y = \frac{150\times10^6}{0.002} = 7.5\times10^{10}\ \text{N m}^{-2}$$

Question 8.2 (b)

Solution. The yield strength is the stress at which the material stops following Hooke’s law and begins to give way. Here the straight line runs almost to the top and then bends over at a stress of about $300\times10^6\ \text{N m}^{-2}$.

Question 8.3

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?

Question 8.3 (a)

Solution. Young’s modulus is the slope of the initial straight part of a stress-strain graph: the stress needed per unit strain. The graphs share a scale, so the slopes can be compared directly. A’s graph rises almost vertically from the origin, far more steeply than B’s, so a given stress produces less strain in A.

Question 8.3 (b)

Solution. The strength of a material is the greatest stress it can bear before it breaks, its ultimate strength, which is the highest point of its stress-strain curve (this is the “ultimate strength” of Table 8.1). A’s curve climbs to a slightly higher stress than B’s before it turns down, so A can withstand the greater stress and is the stronger material. The two peaks are close in the figure, so this is a comparison to read carefully rather than at a glance; A is also the stiffer of the two, from (a).

Question 8.4

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.

Question 8.4 (a)

Solution. Young’s modulus is the stress needed per unit strain. Rubber stretches enormously under a small stress, while steel hardly stretches under a large one, so rubber’s modulus is far smaller. Steel’s $2.0\times10^{11}\ \text{Pa}$ is thousands of times that of rubber. Everyday speech calls rubber “more elastic” because it stretches more, but in physics a material that deforms less for a given load has the larger modulus.

Question 8.4 (b)

Solution. When a coil spring is pulled, the wire it is made of hardly changes length at all. Each turn is twisted, so the wire changes shape rather than length. A change of shape at constant volume is a shear, so the spring’s stiffness is set by the shear modulus of its wire.

Question 8.5

Two wires of diameter $0.25\ \text{cm}$, 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 $1.5\ \text{m}$ and that of brass wire is $1.0\ \text{m}$. Compute the elongations of the steel and the brass wires.

Solution. The tension in each wire is the weight of everything hanging below it. The steel wire holds up both masses; the brass wire holds only the lower one:

$$T_{\text{steel}} = (4.0 + 6.0)\times9.8 = 98\ \text{N}, \qquad T_{\text{brass}} = 6.0\times9.8 = 58.8\ \text{N}$$

Both wires have radius $0.125\ \text{cm}$, so $A = \pi(1.25\times10^{-3})^2 = 4.91\times10^{-6}\ \text{m}^2$. Rearranging $Y = FL/(A\,\Delta L)$ gives $\Delta L = FL/(AY)$. For steel $Y = 2.0\times10^{11}\ \text{Pa}$ (Table 8.1); brass is not in Table 8.1, and the standard value $Y = 0.91\times10^{11}\ \text{Pa}$ is used.

$$\Delta L_{\text{steel}} = \frac{98\times1.5}{4.91\times10^{-6}\times2.0\times10^{11}} = 1.50\times10^{-4}\ \text{m}$$

$$\Delta L_{\text{brass}} = \frac{58.8\times1.0}{4.91\times10^{-6}\times0.91\times10^{11}} = 1.32\times10^{-4}\ \text{m}$$

Question 8.6

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

Solution. The weight of the mass acts downwards along the free face, parallel to it, so it shears the cube rather than stretching it. The force is spread over the face, of area $A = (0.10)^2 = 0.01\ \text{m}^2$:

$$\text{shear stress} = \frac{F}{A} = \frac{100\times9.8}{0.01} = 9.8\times10^4\ \text{N m}^{-2}$$

$$\text{shear strain} = \frac{\text{stress}}{G} = \frac{9.8\times10^4}{25\times10^9} = 3.92\times10^{-6}$$

The shear strain is $\Delta x/L$, the deflection divided by the distance of the loaded face from the fixed one, which is the edge, $L = 0.10\ \text{m}$:

$$\Delta x = 3.92\times10^{-6}\times0.10 = 3.92\times10^{-7}\ \text{m}$$

Note on the book’s answer. The key prints $4\times10^{-6}\ \text{m}$. That number is the shear strain, $3.92\times10^{-6}$, which has no unit, not the deflection. A deflection of $4\times10^{-6}\ \text{m}$ on a $0.10\ \text{m}$ cube would be a strain of $4\times10^{-5}$, needing a shear stress of $25\times10^9\times4\times10^{-5} = 1.0\times10^6\ \text{N m}^{-2}$ — ten times what the $100\ \text{kg}$ mass provides.

Question 8.7

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

Solution. The load is shared equally, so each column carries a quarter of the weight:

$$F = \frac{50{,}000\times9.8}{4} = 1.225\times10^5\ \text{N}$$

Only the steel carries load, and in a hollow column that is the ring between the two radii:

$$A = \pi(r_2^2 – r_1^2) = \pi(0.60^2 – 0.30^2) = 0.848\ \text{m}^2$$

The stress is $F/A = 1.44\times10^5\ \text{N m}^{-2}$, and with $Y = 2.0\times10^{11}\ \text{Pa}$ for steel,

$$\text{strain} = \frac{F}{AY} = \frac{1.225\times10^5}{0.848\times2.0\times10^{11}} = 7.22\times10^{-7}$$

Note on the book’s answer. The key prints $2.8\times10^{-6}$. That is what you get by putting the whole weight, $4.9\times10^5\ \text{N}$, on the cross-section of a single column ($2.89\times10^{-6}$). The question says the load is shared uniformly by four columns, so each carries a quarter and the strain is a quarter of that, $7.22\times10^{-7}$.

Question 8.8

A piece of copper having a rectangular cross-section of $15.2\ \text{mm}\times19.1\ \text{mm}$ is pulled in tension with $44{,}500\ \text{N}$ force, producing only elastic deformation. Calculate the resulting strain?

Solution. The piece is pulled along its length, so this is tensile stress and Young’s modulus applies (not the shear modulus). The stress is

$$\sigma = \frac{F}{A} = \frac{44{,}500}{(15.2\times10^{-3})(19.1\times10^{-3})} = \frac{44{,}500}{2.90\times10^{-4}} = 1.53\times10^8\ \text{N m}^{-2}$$

This is below copper’s yield strength of $200\times10^6\ \text{N m}^{-2}$ (Table 8.1), which is consistent with the deformation being elastic. With $Y = 1.1\times10^{11}\ \text{Pa}$ for copper,

$$\text{strain} = \frac{\sigma}{Y} = \frac{1.53\times10^8}{1.1\times10^{11}} = 1.39\times10^{-3}$$

Note on the book’s answer. The key prints $0.127$. As a strain that is a hundred times too large: it would mean the copper stretched by $12.7\%$, needing a stress of $0.127\times1.1\times10^{11} = 1.4\times10^{10}\ \text{N m}^{-2}$, some 35 times copper’s ultimate strength. The key’s digits match a strain of $0.128\%$, which is what an older value of copper’s modulus, $1.2\times10^{11}\ \text{Pa}$, gives; with the book’s own Table 8.1 the strain is $1.39\times10^{-3}$, or $0.14\%$.

Question 8.9

A steel cable with a radius of $1.5\ \text{cm}$ supports a chairlift at a ski area. If the maximum stress is not to exceed $10^8\ \text{N m}^{-2}$, what is the maximum load the cable can support?

Solution. Stress is load per unit area, so the largest load is the largest allowed stress times the area of the cable’s cross-section:

$$F_{\max} = \sigma_{\max}\,\pi r^2 = 10^8\times\pi(1.5\times10^{-2})^2 = 10^8\times7.07\times10^{-4} = 7.07\times10^4\ \text{N}$$

Question 8.10

A rigid bar of mass $15\ \text{kg}$ is supported symmetrically by three wires each $2.0\ \text{m}$ 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.

Solution. Each wire carries a third of the weight, $T = 15\times9.8/3 = 49\ \text{N}$. The bar is rigid and hangs level, so all three wires stretch by the same amount; and they have the same length, so they all have the same strain $\Delta L/L$. From $Y = \dfrac{T/A}{\Delta L/L}$, the tension is $T = YA\,(\Delta L/L)$. Equal tension and equal strain therefore require the same value of $YA$ for every wire, that is $Y d^2$ must be the same:

$$Y_{\text{Cu}}\,d_{\text{Cu}}^2 = Y_{\text{Fe}}\,d_{\text{Fe}}^2 \quad\Rightarrow\quad \frac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\frac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} = \sqrt{\frac{1.9\times10^{11}}{1.1\times10^{11}}} = 1.31$$

Copper is the less stiff metal, so its wires must be thicker to take the same load with the same stretch.

Note on the book’s answer. The key prints $1.25$. That follows from taking copper’s Young’s modulus as $1.2\times10^{11}\ \text{Pa}$ (which gives $1.26$); with the $1.1\times10^{11}\ \text{Pa}$ of the book’s Table 8.1 the ratio is $1.31$. The difference is only the value of the constant.

Question 8.11

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

Solution. At the lowest point the tension pulls up, towards the centre, and the weight pulls down. Their difference supplies the centripetal force $m\omega^2r$ that keeps the mass on its circle:

$$T – mg = m\omega^2 r \quad\Rightarrow\quad T = m(g + \omega^2 r)$$

The angular velocity must be in radians per second: one revolution is $2\pi$ radians, so $\omega = 2\times2\pi = 4\pi = 12.57\ \text{rad s}^{-1}$ and $\omega^2 = 157.9\ \text{s}^{-2}$. The radius is the length of the wire, $1.0\ \text{m}$ (its stretch is far too small to matter):

$$T = 14.5\,(9.8 + 157.9\times1.0) = 14.5\times167.7 = 2432\ \text{N}$$

With $A = 0.065\ \text{cm}^2 = 6.5\times10^{-6}\ \text{m}^2$ and $Y = 2.0\times10^{11}\ \text{Pa}$ for steel:

$$\Delta L = \frac{TL}{AY} = \frac{2432\times1.0}{6.5\times10^{-6}\times2.0\times10^{11}} = 1.87\times10^{-3}\ \text{m}$$

Note on the book’s answer. The key prints $1.539\times10^{-4}\ \text{m}$. That is exactly the result of taking $\omega = 2\ \text{rad s}^{-1}$: then $T = 14.5(9.8 + 4) = 200.1\ \text{N}$ and $\Delta L = 200.1/(1.3\times10^6) = 1.539\times10^{-4}\ \text{m}$. But the question gives $2$ revolutions per second, which is $4\pi\ \text{rad s}^{-1}$, and the elongation is $1.87\times10^{-3}\ \text{m}$, about $1.9\ \text{mm}$.

Question 8.12

Compute the bulk modulus of water from the following data: Initial volume $= 100.0$ litre, Pressure increase $= 100.0\ \text{atm}$ ($1\ \text{atm} = 1.013\times10^5\ \text{Pa}$), Final volume $= 99.5$ litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

The book prints the final volume as $100.5$ litre. That is a misprint: an increase in pressure compresses the water, so its volume must fall, to $99.5$ litre. Only the size of the change, $0.5$ litre, enters the answer, and the book’s own answer uses it.

Solution. The pressure increase is $p = 100.0\times1.013\times10^5 = 1.013\times10^7\ \text{Pa}$, and the volume strain is $\Delta V/V = -0.5/100.0 = -5.0\times10^{-3}$:

$$B = -\frac{p}{\Delta V/V} = \frac{1.013\times10^7}{5.0\times10^{-3}} = 2.026\times10^9\ \text{Pa}$$

Air. For a gas kept at constant temperature, $pV$ is constant, so a small pressure change $\Delta p$ gives $V\,\Delta p + p\,\Delta V = 0$ and $B = -\Delta p/(\Delta V/V) = p$. At atmospheric pressure, $B_{\text{air}} \approx 1.0\times10^5\ \text{Pa}$, which is the value in Table 8.3. The ratio is

$$\frac{B_{\text{water}}}{B_{\text{air}}} = \frac{2.026\times10^9}{1.0\times10^5} \approx 2\times10^4$$

Why so large. In water the molecules are already packed against one another, and squeezing them closer is resisted by strong short-range repulsive forces, so even a large pressure changes the volume very little. In air the molecules are far apart, with mostly empty space between them and only weak forces, so the gas is easily compressed.

Question 8.13

What is the density of water at a depth where pressure is $80.0\ \text{atm}$, given that its density at the surface is $1.03\times10^3\ \text{kg m}^{-3}$?

The book prints the surface density as “$1.03\times103\ \text{kg m}^{-3}$”. The exponent has slipped onto the line; it is $1.03\times10^3\ \text{kg m}^{-3}$, the density of sea water.

Solution. The pressure compresses the water, so a fixed mass occupies less volume and its density goes up. Using the bulk modulus of water, $B = 2.2\times10^9\ \text{Pa}$ (Table 8.3), and $p = 80.0\times1.013\times10^5 = 8.10\times10^6\ \text{Pa}$, the fractional change in volume is

$$\frac{|\Delta V|}{V} = \frac{p}{B} = \frac{8.10\times10^6}{2.2\times10^9} = 3.68\times10^{-3}$$

The mass is unchanged, so the density scales inversely with the volume:

$$\rho’ = \frac{\rho}{1 – |\Delta V|/V} = \frac{1.03\times10^3}{1 – 3.68\times10^{-3}} = 1.034\times10^3\ \text{kg m}^{-3}$$

Strictly, the increase over the surface pressure is $79.0\ \text{atm}$, not $80.0$, but that changes the answer only in the fifth figure.

Question 8.14

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of $10\ \text{atm}$.

Solution. A hydraulic pressure acts on every face, so it changes the volume, and the bulk modulus applies. For glass $B = 37\times10^9\ \text{Pa}$ (Table 8.3), and $p = 10\times1.013\times10^5 = 1.013\times10^6\ \text{Pa}$:

$$\frac{\Delta V}{V} = -\frac{p}{B} = -\frac{1.013\times10^6}{3.7\times10^{10}} = -2.74\times10^{-5}$$

The volume decreases by $2.74\times10^{-5}$ of itself, that is $0.0027\%$.

Note on the book’s answer. The key prints $0.0027$. The digits are right but the per cent sign has been dropped: the fractional change is $0.0027\%$, or $2.7\times10^{-5}$, not $0.0027$.

Question 8.15

Determine the volume contraction of a solid copper cube, $10\ \text{cm}$ on an edge, when subjected to a hydraulic pressure of $7.0\times10^6\ \text{Pa}$.

Solution. The cube’s volume is $V = (0.10)^3 = 1.0\times10^{-3}\ \text{m}^3$. A hydraulic pressure changes volume, so the relevant modulus is the bulk modulus of copper, $B = 140\times10^9\ \text{Pa}$ (Table 8.3). From $B = p/(|\Delta V|/V)$:

$$|\Delta V| = \frac{pV}{B} = \frac{7.0\times10^6\times1.0\times10^{-3}}{1.4\times10^{11}} = 5.0\times10^{-8}\ \text{m}^3 = 0.050\ \text{cm}^3$$

Note on the book’s answer. The key prints $0.058\ \text{cm}^3$. That follows from a modulus of $1.2\times10^{11}\ \text{Pa}$, which is close to copper’s Young’s modulus, not the $1.4\times10^{11}\ \text{Pa}$ bulk modulus the book’s Table 8.3 gives. A change of volume under hydraulic pressure is governed by the bulk modulus, and with it the contraction is $0.050\ \text{cm}^3$.

Question 8.16

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

In the book this question is followed by a stray line, “carry one quarter of the load.”, left over from some other text; it has no bearing on the question.

Solution. Compressing by $0.10\%$ means $|\Delta V|/V = 1.0\times10^{-3}$. From the definition of the bulk modulus, with $B = 2.2\times10^9\ \text{Pa}$ for water:

$$p = B\,\frac{|\Delta V|}{V} = 2.2\times10^9\times1.0\times10^{-3} = 2.2\times10^6\ \text{Pa}$$

The answer is the same for any quantity of water, since only the fractional change matters; the “litre” is not needed. This is about 22 atmospheres, a large pressure for a tiny squeeze, which is why water is usually treated as incompressible.

Common mistakes

• Question 8.4(a): thinking that the material which stretches more is “more elastic”. In physics the opposite holds: Young’s modulus is the stress needed per unit strain, so the material that stretches less under a given load, steel, has the larger modulus.
• Question 8.5: giving the steel wire only the $4.0\ \text{kg}$ mass. A wire’s tension is the weight of everything below it. The steel wire holds the brass wire and both masses, $98\ \text{N}$. A second slip is using the diameter, $0.25\ \text{cm}$, in place of the radius in $\pi r^2$, which makes the area four times too large.
• Question 8.6: stopping at the shear strain. The strain, $3.9\times10^{-6}$, has no unit. The deflection is the strain times the edge of the cube, $3.9\times10^{-7}\ \text{m}$; the book’s key makes exactly this slip.
• Question 8.7: forgetting to share the load, or using the full disc. Each of the four columns carries a quarter of the weight, and only the ring of steel, $\pi(r_2^2 – r_1^2)$, carries it, not $\pi r_2^2$.
• Question 8.8: reaching for the shear modulus. A piece “pulled in tension” is stretched along its length, so its strain is set by Young’s modulus. Using copper’s $42\ \text{GPa}$ shear modulus gives a strain nearly three times too large.
• Question 8.11: using $2\ \text{rad s}^{-1}$ for $2$ revolutions per second. One revolution is $2\pi$ radians, so $\omega = 4\pi\ \text{rad s}^{-1}$. The difference is a factor of $4\pi^2 \approx 39$ in the centripetal force. Students also sometimes forget to add the weight to the tension at the lowest point.
• Question 8.13: making the water less dense. Pressure squeezes a fixed mass into a smaller volume, so the density rises, $\rho’ = \rho/(1 – |\Delta V|/V)$, rather than falls.
• Question 8.15: using Young’s modulus for a change of volume. A hydraulic pressure acts on all faces and changes the volume, so the bulk modulus, $140\ \text{GPa}$ for copper, is the right constant.

Practise next

• Chapter 9, Mechanical Properties of Fluids — pressure, and the other half of the bulk modulus story: how fluids transmit and respond to it.
• Chapter 4, Laws of Motion — tension, equilibrium and circular motion, which fix the forces in 8.5, 8.10 and 8.11 before any modulus is used.