NCERT Exemplar Solutions Class 11 Physics Mechanical Properties of Solids — Very Short, Short and Long Answer Questions Q14–29

NCERT Exemplar Problems, Class 11 Physics — Mechanical Properties of Solids, Very Short Answer Questions 14 to 18, Short Answer Questions 19 to 23 and Long Answer Questions 24 to 29. All 16 questions solved. The book numbers them 9.14 to 9.29.

Three ideas carry this page. Hooke’s law in the form $\Delta l = Fl/(AY)$, applied piece by piece when the tension varies along a wire, as it does in a heavy hanging wire or a spinning rod. A modulus is stress over strain, so a rigid body has infinite moduli and the pressure needed to squeeze a ball follows from $p = B\,\Delta V/V$. Prevented expansion becomes stress: a rod that cannot grow when heated carries a stress $Y\alpha\,\Delta T$.

Key insight. When the tension changes along a body, do not use one value of $F$. Write the tension at a general section, find the stretch of a thin slice, $d(\Delta l) = T\,dx/(AY)$, and add up the slices. Questions 25 and 26 are both done this way.

Very Short Answer Questions

Question 14

The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?

Solution. Stress $= Y \times$ strain. For the same strain, the material with the larger Young’s modulus carries the larger stress, so steel does.

Question 15

Is stress a vector quantity?

Solution. No. Stress is force per unit area, but both the force and the area are directed: the area has its own orientation, given by its normal. One stress value needs two directions (the direction of the force and the orientation of the surface it acts on), so stress is a tensor (of rank two), not a vector. It does not add by the vector parallelogram law.

Question 16

Identical springs of steel and copper are equally stretched. On which, more work will have to be done?

Solution. The springs are identical in shape and size; only the material differs. The stiffness of a coil spring is proportional to the shear modulus of its wire, and steel’s is larger than copper’s, so $k_{\text{steel}} > k_{\text{copper}}$. “Equally stretched” means the same extension $x$, and the work done is

$$W = \tfrac{1}{2}kx^2,$$

so more work is done on the steel spring.

Printed key disagreement. The key prints copper. That answer holds only if “equally stretched” is read as “stretched by equal forces”: then $W = F^2/(2k)$ and the softer copper spring needs more work. The natural reading of the question is equal stretch, and for that the answer is steel.

Question 17

What is the Young’s modulus for a perfect rigid body?

Solution. A perfectly rigid body has zero strain under any stress. $Y = \text{stress}/\text{strain}$ with zero strain is infinite.

Question 18

What is the Bulk modulus for a perfect rigid body?

Solution. Its volume does not change under any pressure, so $\Delta V/V = 0$ and $B = -\Delta p/(\Delta V/V)$ is infinite.

Short Answer Questions

Question 19

A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of the same material of length $2L$ and radius $2r$, is pulled by a force $2f$. Find the increase in length of this wire.

Solution. The same material means the same $Y$. For the first wire

$$l = \frac{fL}{\pi r^2 Y}.$$

For the second wire

$$l’ = \frac{(2f)(2L)}{\pi(2r)^2Y} = \frac{4fL}{4\pi r^2Y} = l.$$

Doubling the force and the length each doubles the extension, but doubling the radius quadruples the area, and the factors cancel.

Question 20

A steel rod ($Y = 2.0\times10^{11}\ \text{N m}^{-2}$; and $\alpha = 10^{-5}\ ^\circ\text{C}^{-1}$) of length $1$ m and area of cross-section $1\ \text{cm}^2$ is heated from $0^\circ\text{C}$ to $200^\circ\text{C}$, without being allowed to extend or bend. What is the tension produced in the rod?

(The book prints $\alpha = 10^{-50}$, an obvious misprint for $10^{-5}$, the value for steel and the one the answer key uses.)

Solution. If free, the rod would expand by a strain $\alpha\,\Delta T = 10^{-5}\times200 = 2\times10^{-3}$. The supports prevent this, which is the same as compressing it back by that strain. The force needed is

$$F = YA\,\alpha\,\Delta T = 2\times10^{11}\times10^{-4}\times2\times10^{-3} = 4\times10^4\ \text{N}.$$

The rod pushes outwards on its supports, so this is a compressive force in the rod. The length of $1$ m does not enter: thermal stress depends only on $Y\alpha\,\Delta T$.

Question 21

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1\%$. (The bulk modulus of rubber is $9.8\times10^8\ \text{N m}^{-2}$; and the density of sea water is $10^3\ \text{kg m}^{-3}$.)

Solution. The extra pressure needed for a volume strain of $10^{-3}$ is

$$p = B\,\frac{\Delta V}{V} = 9.8\times10^8\times10^{-3} = 9.8\times10^5\ \text{Pa}.$$

The pressure at depth $h$ exceeds that at the surface by $\rho gh$, so

$$h = \frac{p}{\rho g} = \frac{9.8\times10^5}{10^3\times9.8} = 100\ \text{m}.$$

Question 22

A truck is pulling a car out of a ditch by means of a steel cable that is $9.1$ m long and has a radius of $5$ mm. When the car just begins to move, the tension in the cable is $800$ N. How much has the cable stretched? (Young’s modulus for steel is $2\times10^{11}\ \text{N m}^{-2}$.)

Solution.

$$\Delta l = \frac{FL}{\pi r^2 Y} = \frac{800\times9.1}{\pi\times(5\times10^{-3})^2\times2\times10^{11}} = \frac{7280}{1.571\times10^7} = 4.6\times10^{-4}\ \text{m}.$$

The cable stretches by about half a millimetre. (The key rounds this to $0.5\times10^{-3}$ m.)

Question 23

Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?

Solution. The ivory ball. Both arrive with the same kinetic energy. During impact the ivory ball deforms elastically and recovers its shape almost completely, so most of the energy stored in the deformation is returned as kinetic energy for the rebound. Wet clay deforms plastically: it stays flattened, and the energy goes into permanent deformation and heat, so it barely rises at all.

Long Answer Questions

Question 24

Consider a long steel bar under a tensile stress due to forces $\mathbf{F}$ acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle $\theta$ with the length. What are the tensile and shearing stresses on this plane?

(a) For what angle is the tensile stress a maximum?

(b) For what angle is the shearing stress a maximum?

Solution. Let the normal cross-section of the bar have area $A$. The plane $aa’$ makes an angle $\theta$ with the length, so it cuts the bar more obliquely and its area is larger:

$$A’ = \frac{A}{\sin\theta}.$$

The part of the bar to one side of $aa’$ is held in equilibrium by the force $F$ across that plane. Resolve $F$ along the normal to the plane and along the plane. Because the plane makes $\theta$ with $F$, the normal to it makes $\tfrac{\pi}{2} – \theta$ with $F$:

$$F_N = F\sin\theta, \qquad F_P = F\cos\theta.$$

Dividing each by the area $A’$:

$$\text{tensile stress} = \frac{F\sin\theta}{A/\sin\theta} = \frac{F}{A}\sin^2\theta, \qquad \text{shearing stress} = \frac{F\cos\theta}{A/\sin\theta} = \frac{F}{A}\sin\theta\cos\theta = \frac{F\sin2\theta}{2A}.$$

(a) $\sin^2\theta$ is largest at $\theta = \pi/2$: the ordinary cross-section, where the tensile stress is $F/A$.

(b) $\sin2\theta$ is largest at $2\theta = \pi/2$, that is $\theta = \pi/4$, where the shearing stress is $F/(2A)$. This is why ductile materials pulled in tension often fail along planes at $45^\circ$.

Question 25

(a) A steel wire of mass $\mu$ per unit length with a circular cross section has a radius of $0.1$ cm. The wire is of length $10$ m when measured lying horizontal, and hangs from a hook on the wall. A mass of $25$ kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains $\ll$ longitudinal strains, find the extension in the length of the wire. The density of steel is $7860\ \text{kg m}^{-3}$ (Young’s modulus $Y = 2\times10^{11}\ \text{N m}^{-2}$).

(b) If the yield strength of steel is $2.5\times10^8\ \text{N m}^{-2}$, what is the maximum weight that can be hung at the lower end of the wire?

Solution. (a) The tension is not the same all along the wire: a section higher up also holds the wire below it. Measure $x$ upwards from the load. The section at $x$ supports the load $Mg$ and a length $x$ of wire:

$$T(x) = Mg + \mu g x.$$

A slice $dx$ at that height stretches by $d(\Delta l) = \dfrac{T(x)\,dx}{AY}$. Adding the slices over the whole length $L$:

$$\Delta l = \frac{1}{AY}\int_0^L(Mg + \mu gx)\,dx = \frac{1}{AY}\left(MgL + \frac{\mu gL^2}{2}\right) = \frac{L}{AY}\left(Mg + \frac{mg}{2}\right),$$

where $m = \mu L$ is the mass of the wire. The wire’s own weight counts as if half of it hung at the end. Now the numbers: $A = \pi(10^{-3})^2 = 3.14\times10^{-6}\ \text{m}^2$, and $m = 7860\times3.14\times10^{-6}\times10 = 0.247$ kg. With $g = 9.8\ \text{m s}^{-2}$,

$$\Delta l = \frac{10}{3.14\times10^{-6}\times2\times10^{11}}\,(245 + 1.21) = 3.9\times10^{-3}\ \text{m}.$$

The wire’s own weight adds only about $0.5\%$; almost all of the $4$ mm comes from the $25$ kg load.

(b) The tension is greatest at the top, where it is $(M + m)g$. Yield begins when this equals the yield stress times the area:

$$(M+m)g = 2.5\times10^8\times3.14\times10^{-6} = 785\ \text{N}.$$

The wire weighs only $2.4$ N, so the largest weight that can hang at the end is about $783$ N, or roughly $7.8\times10^2$ N, a mass of about $80$ kg. (The key takes $g = 10$, writes $M = 25\pi$ kg and then calls it about $75$ kg; $25\pi$ is $78.5$ kg.)

Question 26

A steel rod of length $2l$, cross sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

Solution. Let the rod spin with angular speed $\omega$ and have mass per length $\mu = M/(2l)$. Consider one half, and a section at distance $x$ from the axis. The part of the rod beyond $x$ moves in a circle, and the tension at $x$ supplies its centripetal force:

$$T(x) = \int_x^l \mu\omega^2 s\,ds = \frac{\mu\omega^2}{2}\,(l^2 – x^2).$$

The tension is largest at the centre and zero at the free end. A slice $dx$ at $x$ stretches by $T(x)\,dx/(AY)$, so one half stretches by

$$\delta = \frac{\mu\omega^2}{2AY}\int_0^l (l^2 – x^2)\,dx = \frac{\mu\omega^2}{2AY}\cdot\frac{2l^3}{3} = \frac{\mu\omega^2l^3}{3AY}.$$

Both halves stretch, so the total extension is

$$2\delta = \frac{2\mu\omega^2l^3}{3AY} = \frac{M\omega^2l^2}{3AY}.$$

(The key’s working reaches $\mu\omega^2l^3/(3YA)$ for one half but its last line drops a power of $l$; with $\mu = M/2l$ the result above is the dimensionally correct one.)

Question 27

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by $\Delta T$. Find change in the angle ABC. [Coeff. of linear expansion for Cu is $\alpha_1$, Coeff. of linear expansion for Al is $\alpha_2$.]

Solution. The angle $B = \theta$ is opposite the aluminium side AC. With $AB = BC = l_1$ and $AC = l_2$, the isosceles triangle gives

$$\sin\frac{\theta}{2} = \frac{l_2}{2l_1}.$$

Differentiate (take logarithms first):

$$\frac{1}{2}\cot\frac{\theta}{2}\,d\theta = \frac{dl_2}{l_2} – \frac{dl_1}{l_1} = \alpha_2\Delta T – \alpha_1\Delta T.$$

At $\theta = 60^\circ$, $\cot 30^\circ = \sqrt3$, so

$$d\theta = \frac{2(\alpha_2 – \alpha_1)\,\Delta T}{\sqrt3}.$$

This has the sign it should. If the aluminium side expands more than the copper sides ($\alpha_2 > \alpha_1$, as it does in practice), the side opposite B grows relative to the sides enclosing B, and the angle at B opens.

Printed key error. The key reaches $d\theta = 2(\alpha_1 – \alpha_2)\Delta T/\sqrt3$, the same size with the opposite sign. Its working differentiates $2l_1l_3\cos\theta = l_1^2 + l_3^2 – l_2^2$ correctly to a term $-2l_1l_3\sin\theta\,d\theta$, then writes it as $+l^2\sin\theta\,d\theta$ in the next line. With the sign kept, the result above follows (a direct sympy check of $\theta = 2\sin^{-1}\!\big[l_2/(2l_1)\big]$ agrees).

Question 28

In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\dfrac{Y\pi r^4}{4R}$. $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Solution. Let the tree have height $h$, so its weight is $W = \rho g\,\pi r^2h$, acting at the centre of gravity at height $h/2$. When the trunk bends into an arc of radius $R$, the centre of gravity moves sideways by a distance $d$ from the vertical through the base. In the right triangle formed by the centre of curvature, the base and the centre of gravity,

$$R^2 = (R – d)^2 + \left(\frac{h}{2}\right)^2 \quad\Rightarrow\quad d \approx \frac{h^2}{8R}\quad (d \ll R).$$

The weight then exerts a toppling torque $Wd$ about the base. The tree is about to buckle when this equals the elastic restoring torque:

$$\rho g\,\pi r^2h\cdot\frac{h^2}{8R} = \frac{Y\pi r^4}{4R}.$$

$R$ cancels, which is what makes a critical height meaningful:

$$h^3 = \frac{2Yr^2}{\rho g} \quad\Rightarrow\quad h = \left(\frac{2Y}{\rho g}\right)^{1/3} r^{2/3}.$$

Taller trees need disproportionately thicker trunks: $r \propto h^{3/2}$.

Question 29

A stone of mass $m$ is tied to an elastic string of negligible mass and spring constant $k$. The unstretched length of the string is $L$ and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.

(a) Find the distance $y$ from the top when the mass comes to rest for an instant, for the first time.

(b) What is the maximum velocity attained by the stone in this drop?

(c) What shall be the nature of the motion after the stone has reached its lowest point?

Solution. (a) For the first $L$ the string is slack and the stone falls freely. After that the string stretches by $y – L$. At the lowest point the stone is momentarily at rest, so all the gravitational energy lost has gone into the string:

$$mgy = \tfrac{1}{2}k(y – L)^2 \quad\Rightarrow\quad \tfrac12ky^2 – (kL + mg)y + \tfrac12kL^2 = 0.$$

Taking the root with $y > L$:

$$y = \frac{(kL + mg) + \sqrt{m^2g^2 + 2mgkL}}{k} = L + \frac{mg}{k} + \frac{\sqrt{m^2g^2 + 2mgkL}}{k}.$$

(b) The stone speeds up as long as the downward pull of gravity exceeds the string’s upward pull, so it is fastest where they balance: extension $x_0 = mg/k$. Energy conservation at that point gives

$$\tfrac12mv^2 = mg\left(L + \frac{mg}{k}\right) – \tfrac12k\left(\frac{mg}{k}\right)^2 = mgL + \frac{m^2g^2}{2k},$$

$$v_{\max} = \sqrt{2gL + \frac{mg^2}{k}}.$$

(c) While the string is stretched, the equation of motion is

$$m\frac{d^2y}{dt^2} = mg – k(y – L) \quad\Rightarrow\quad \frac{d^2y}{dt^2} = -\frac{k}{m}\left(y – y_0\right),\qquad y_0 = L + \frac{mg}{k},$$

which is simple harmonic motion about $y_0$ with angular frequency $\omega = \sqrt{k/m}$. So from the lowest point the stone rises in SHM about $y_0$. This is the key’s answer.

It is worth finishing the thought. The amplitude of this SHM is $\sqrt{m^2g^2 + 2mgkL}/k$, which is larger than $mg/k$, so the stone reaches $y = L$ while still moving upwards. There the string goes slack, the stone moves freely under gravity, and (with no energy lost) it rises back to P. The motion repeats, so it is periodic, but it is simple harmonic only during the part with the string taut.

Common mistakes

• Answering copper in Question 16 without saying what is equal. At equal extension the stiffer steel spring needs more work.
• Letting the length enter thermal stress in Question 20. The stress is $Y\alpha\,\Delta T$ whatever the length.
• Using one value of tension in Question 25. The tension grows towards the top; the wire’s own weight counts as half of it at the end.
• Forgetting the second half of the rod in Question 26. Each half stretches by $\mu\omega^2l^3/(3AY)$.
• Losing the sign in Question 27. Check it physically: a faster-expanding opposite side must open the angle.
• Treating the whole of Question 29 as SHM. The string can only pull, so above $y = L$ the stone is in free fall.

Practise next

• Problem. A brass rod of cross-sectional area $2\ \text{cm}^2$ is clamped between two rigid walls at $20^\circ$C and heated to $70^\circ$C. Taking $\alpha = 2\times10^{-5}\ ^\circ\text{C}^{-1}$ and $Y = 1\times10^{11}\ \text{N m}^{-2}$, find the force the rod exerts on the walls.

Show answer

$2\times10^4$ N. The thermal stress is $Y\alpha\,\Delta T = 10^{11}\times2\times10^{-5}\times50 = 10^8\ \text{N m}^{-2}$, and the force is this times $2\times10^{-4}\ \text{m}^2$, as in Question 20.

• Problem. A solid steel ball (bulk modulus $1.6\times10^{11}\ \text{N m}^{-2}$) is lowered to a depth of $1000$ m in sea water of density $10^3\ \text{kg m}^{-3}$. Taking $g = 9.8\ \text{m s}^{-2}$, find the fractional decrease in its volume.

Show answer

$6.1\times10^{-5}$. The extra pressure is $\rho gh = 9.8\times10^6$ Pa, so $\Delta V/V = p/B = 9.8\times10^6/1.6\times10^{11} \approx 6.1\times10^{-5}$. Steel is about $160$ times stiffer than the rubber of Question 21.

• Exemplar Mechanical Properties of Solids: Multiple Choice Questions — the same ideas tested option by option.
• NCERT textbook Chapter 8 — Mechanical Properties of Solids for the theory of stress, strain and the elastic moduli.