NCERT Exemplar Solutions Class 11 Physics Mechanical Properties of Solids — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 11 Physics — Mechanical Properties of Solids, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 9.1 to 9.13; here they are Questions 1 to 13. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Stress is force per area, strain is change per original size. A modulus is the ratio of the two, so it is a property of the material, not of the length or thickness of a particular sample.
• Hooke’s law in a wire. $\Delta l = \dfrac{Fl}{AY}$, which is all you need for wires sharing a load.
• Kinds of strain. Longitudinal strain changes length, shear strain changes shape at constant volume, volumetric strain changes volume.
• Reading a stress-strain curve. The linear stretch shows the elastic range; a short gap between the ultimate strength and the fracture point marks a brittle material.

Key insight. In almost every question here, first decide which quantity is shared. Wires in series share the force; wires holding a rigid bar level share the extension; a material property such as breaking stress does not care about length at all.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

Modulus of rigidity of ideal liquids is

Solution. The modulus of rigidity is shear stress divided by shear strain. A liquid cannot sustain a shear stress at rest: the smallest tangential force makes it flow, so the shear strain keeps growing without any resisting stress. For an ideal (non-viscous) liquid the resisting shear stress is exactly zero, so $G = 0$.

Question 2

The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

Solution. A wire breaks when the stress in it reaches the breaking stress of the material. The breaking load is therefore $F_{\max} = (\text{breaking stress}) \times A$. It depends on the material and the cross-sectional area, and not on the length. Cutting the wire to half its length leaves $A$ unchanged, so the maximum load is the same.

Question 3

The temperature of a wire is doubled. The Young’s modulus of elasticity

Solution. Young’s modulus measures how strongly the interatomic forces resist a change in spacing. Heating increases the amplitude of atomic vibration and the mean spacing, and the restoring forces at larger spacing are weaker. A given stress then produces a larger strain, so $Y$ falls as the temperature rises. (Near the melting point it falls steeply.)

Question 4

A spring is stretched by applying a load to its free end. The strain produced in the spring is

Solution. Look at the spring at two levels. The spring as a whole gets longer, so there is a change in its length: a longitudinal strain. But the wire it is wound from is not being pulled along its own length; each turn is twisted, so the material of the wire undergoes a shear (torsional) strain. That is why the stiffness of a coil spring depends on the shear modulus of its wire. Both are present, so the answer is (c).

Option (b) alone describes only the wire, and (d) alone only the overall length. Volume does not change appreciably, so (a) is wrong.

Question 5

A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

Solution. The bar is rigid and stays horizontal, so all three wires stretch by the same amount $\Delta l$. For each wire

$$T = \frac{YA\,\Delta l}{l}.$$

With equal $T$, equal $l$ and equal $\Delta l$, the product $YA$ must be the same for copper and iron:

$$Y_{\text{copper}}\,\frac{\pi d_{\text{copper}}^2}{4} = Y_{\text{iron}}\,\frac{\pi d_{\text{iron}}^2}{4} \quad\Rightarrow\quad \frac{d_{\text{copper}}}{d_{\text{iron}}} = \sqrt{\frac{Y_{\text{iron}}}{Y_{\text{copper}}}}.$$

Copper has the smaller $Y$, so the copper wires must be the thicker ones.

Question 6

A mild steel wire of length $2L$ and cross-sectional area $A$ is stretched, well within elastic limit, horizontally between two pillars (Fig. 9.1). A mass $m$ is suspended from the mid point of the wire. Strain in the wire is

The figure shows the wire fixed between the tops of two pillars a distance $2L$ apart; the mass hangs from the midpoint, which sags a vertical distance $x$ below the original line, so each half becomes a slanting segment.

Solution. Each half of the wire becomes the hypotenuse of a right triangle with sides $L$ and $x$, so the new length is $2\sqrt{L^2 + x^2}$. The strain is

$$\varepsilon = \frac{2\sqrt{L^2+x^2} – 2L}{2L} = \sqrt{1 + \frac{x^2}{L^2}} – 1 \approx \frac{x^2}{2L^2},$$

using $(1+u)^{1/2} \approx 1 + u/2$ for small $u = x^2/L^2$ (the sag is small because the wire is within its elastic limit). Option (b) is the slope of the wire, not its strain, and (c) and (d) are not even dimensionless.

Question 7

A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (Fig. 9.2). It can be done in one of the following three ways;

The tension in the strings will be

Solution. Let each string make an angle $\theta$ with the vertical. Only the vertical components hold the frame up, so

$$2T\cos\theta = Mg \quad\Rightarrow\quad T = \frac{Mg}{2\cos\theta}.$$

$T$ is smallest when $\cos\theta$ is largest, that is when $\theta = 0$ and the strings hang vertically, as in arrangement (b), where $T = Mg/2$. In (a) and (c) the strings slope, so part of each tension is wasted in horizontal components that cancel each other, and the tension must be larger.

Question 8

Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.

Solution. The load is applied at a single point, the centre of the bottom face. Steel has a very large Young’s modulus, so the strain is tiny everywhere: the rod lengthens slightly and keeps its shape. Rubber has a modulus thousands of times smaller, so it deforms visibly, and it deforms most where the stress is concentrated, right under the point of attachment. The centre of the bottom face is pulled down further than the rim, so the bottom edge is drawn out into a tapered tip at the centre.

Question 9

The stress-strain graphs for two materials are shown in Fig. 9.3 (assume same scale).

Solution. Read two things from each curve.

• Extent of the elastic (linear) region. In (i) the curve leaves the straight line after a small strain. In (ii) it stays straight almost up to the ultimate tensile strength. On the same scale, (ii) is elastic over the larger range of strain, so (c) is true.
• Gap between ultimate strength and fracture. In (i) there is a long plastic stretch after the ultimate strength before it breaks: it is ductile. In (ii) fracture follows almost at once: it is brittle. So (d) is true.

Option (a) states the conclusion of (d) but gives a wrong reason. Brittleness is about how little plastic deformation precedes fracture, not a consequence of being elastic over a large range, so the “hence” makes (a) false. Option (b) contradicts both readings.

Question 10

A wire is suspended from the ceiling and stretched under the action of a weight $F$ suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.

Solution. Cut the (massless) wire at any cross-section and look at the lower part. It is pulled down by the weight $F$ and must be pulled up by the upper part with a force $F$, so the tension at every section is $F$: (d) is true. The stress is this force spread over the area, $F/A$: (a) is true.

The ceiling’s pull and the weight act on opposite ends of the wire; they do not add to give $2F$, and they do not cancel to give zero stress. They are what keeps the wire in equilibrium while it is stretched, so (b) and (c) are false.

Question 11

A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are $1.0\ \text{mm}^2$ and $2.0\ \text{mm}^2$, respectively. $\left(Y_{Al} = 70\times10^9\ \text{N m}^{-2}\ \text{and}\ Y_{steel} = 200\times10^9\ \text{N m}^{-2}\right)$

The figure shows the rod hanging from a ceiling by the steel wire A at its left end and the aluminium wire B at its right end, with a mass $m$ hung from a point on the rod.

Solution. Hang $m$ at a distance $d$ from end A. Taking moments about A, the tensions are

$$T_B = mg\,\frac{d}{l}, \qquad T_A = mg\,\frac{l-d}{l}.$$

Equal stress. $T_A/A_A = T_B/A_B$ with $A_B = 2A_A$ gives $T_B = 2T_A$, so $d = 2l/3$. The mass must hang nearer B: (b) is true, (a) and (c) are false.

Equal strain. $\dfrac{T_A}{A_AY_{steel}} = \dfrac{T_B}{A_BY_{Al}}$ gives

$$\frac{T_A}{T_B} = \frac{A_AY_{steel}}{A_BY_{Al}} = \frac{1\times200}{2\times70} = \frac{10}{7},$$

so $T_A > T_B$ and $d = \dfrac{7}{17}\,l \approx 0.41\,l$. The mass must hang nearer A: (d) is true.

Question 12

For an ideal liquid

Solution. An ideal liquid is taken to be incompressible: no pressure, however large, changes its volume. With $\Delta V = 0$ for any $\Delta p$, the bulk modulus $B = -\Delta p/(\Delta V/V)$ is infinite, so (a) is true. And as in Question 1, an ideal liquid offers no resistance to shear, so its shear modulus is zero: (d) is true.

Question 13

A copper and a steel wire of the same diameter are connected end to end. A deforming force $F$ is applied to this composite wire which causes a total elongation of $1$ cm. The two wires will have

Solution. The wires are in series, so the same force $F$ passes through both. They have the same diameter, hence the same area, so the stress $F/A$ is the same in each: (a) is true. Strain is stress divided by Young’s modulus, and copper and steel have different $Y$ ($Y_{\text{steel}} > Y_{\text{copper}}$), so the strains differ, copper stretching more per unit length: (d) is true.

Common mistakes

• Making the breaking load depend on length in Question 2. Breaking is set by stress, and length does not enter the stress.
• Picking only “shear” in Question 4. The wire twists, but the spring as a whole also gets longer.
• Equating tensions without equal extensions in Question 5. A level rigid bar forces equal $\Delta l$, which is what links $Y$ to the area.
• Using $x/L$ for the strain in Question 6. The extension is second order in the sag, $x^2/2L^2$.
• Calling material (i) brittle in Question 9 because it bends over early. Brittleness is a short gap between the ultimate strength and fracture, which is material (ii).
• Adding the ceiling’s force and the weight in Question 10. The tension in the wire is $F$, not $2F$.
• Putting the mass at the middle for equal stress in Question 11. Wire B is twice as thick, so it must carry twice the tension.

Practise next

• Problem. A steel wire of length $2$ m and cross-sectional area $1\ \text{mm}^2$ carries a load of $50$ kg. Taking $Y = 2\times10^{11}\ \text{N m}^{-2}$ and $g = 9.8\ \text{m s}^{-2}$, find its extension.

• Problem. Two wires P and Q of the same material carry the same load. The length of P is half that of Q and its diameter is twice that of Q. Find the ratio of the extension of P to that of Q.

• Exemplar Mechanical Properties of Solids: Very Short, Short and Long Answer Questions — thermal stress, bulk modulus at depth, a hanging heavy wire, a spinning rod and the critical height of a tree.
• NCERT textbook Chapter 8 — Mechanical Properties of Solids for the theory of stress, strain and the elastic moduli.