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NCERT Solutions · Class 11 Physics Mechanical Properties of Solids

16 questions · 10 still being checked

Exercises 8.1–8.10 (part 1 of 2)

  1. Exercise 8.1

    A steel wire of length 4.7\displaystyle 4.7 m and cross-sectional area 3.0\displaystyle 3.0 × 10−5\displaystyle 10^{-5} m2\displaystyle m^{2} stretches by the same amount as a copper wire of length 3.5\displaystyle 3.5 m and cross-sectional area of 4.0\displaystyle 4.0 × 10−5\displaystyle 10^{-5} m2\displaystyle m^{2} under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
    NCERT’s answer
    1.$\displaystyle 8$
    Young's modulus compares stress to strain, and here both wires carry the same load and the same extension — so the ratio collapses to just their lengths and areas.Young's modulus is defined as\[Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{F L}{A\,\Delta L} \]where \(\displaystyle F\) is the stretching force (the load), \(\displaystyle A\) is the cross-sectional area, \(\displaystyle L\) is the original length, and \(\displaystyle \Delta L\) is the extension produced by that load.Write this for each wire. Let the subscript \(\displaystyle s\) denote steel and \(\displaystyle c\) denote copper.\[Y_s = \frac{F L_s}{A_s\,\Delta L_s}, \qquad Y_c = \frac{F L_c}{A_c\,\Delta L_c} \]The problem states the two wires are stretched by the same load (so \(\displaystyle F\) is identical in both) and stretch by the same amount (so \(\displaystyle \Delta L_s = \Delta L_c\)). This is the detail that is easy to miss — it is not that the two wires have the same strain, only the same absolute stretch \(\displaystyle \Delta L\), and it is exactly this common \(\displaystyle F\) and common \(\displaystyle \Delta L\) that cancel when the two expressions are divided.\[\frac{Y_s}{Y_c} = \frac{F L_s /(A_s\,\Delta L_s)}{F L_c /(A_c\,\Delta L_c)} = \frac{L_s}{A_s}\times\frac{A_c}{L_c} = \frac{L_s A_c}{L_c A_s} \]Now substitute the given values:\[L_s = 4.7\ \text{m}, \quad A_s = 3.0\times10^{-5}\ \text{m}^2 \] \[L_c = 3.5\ \text{m}, \quad A_c = 4.0\times10^{-5}\ \text{m}^2 \]\[\frac{Y_s}{Y_c} = \frac{(4.7\ \text{m})(4.0\times10^{-5}\ \text{m}^2)}{(3.5\ \text{m})(3.0\times10^{-5}\ \text{m}^2)} \]\[= \frac{18.8\times10^{-5}}{10.5\times10^{-5}} = \frac{18.8}{10.5} \]\[= 1.79047\ldots \]All four input quantities are given to two significant figures, so the quotient is rounded to two significant figures at the end (not mid-calculation):\[\frac{Y_s}{Y_c} \approx 1.8 \]Since lengths and areas cancel units, this ratio is a pure number — it carries no units of its own, unlike \(\displaystyle Y\) itself (which would be in pascals, Pa).**Answer: \(\displaystyle Y_{\text{steel}} : Y_{\text{copper}} \approx 1.8 : 1 \) (steel's Young's modulus is about $\displaystyle 1.8$ times that of copper).

  2. Exercise 8.2

    NCERT_Question_Class11_Physics_Ch8_Q8-2
    Figure 8.9\displaystyle 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? Fig. 8.9\displaystyle 8.9
    NCERT’s answer
    (a)
    From the given graph for a stress of $\displaystyle 150$ × \(\displaystyle 10^{6}\) N \(\displaystyle m^{-2}\) the strain is $\displaystyle 0.002$ (b) Approximate yield strength of the material is $\displaystyle 3$ × \(\displaystyle 10^{8}\) N \(\displaystyle m^{-2}\)
    The straight part of a stress–strain graph is Hooke's-law territory: its slope is Young's modulus, and the stress at which that straight part ends is the yield strength. So this whole question is "find the slope of the line, then find where the line stops being a line."First, read the axes. Stress \(\displaystyle \sigma \) is on the vertical axis in units of \(\displaystyle 10^{6}\ \text{N m}^{-2} \), marked \(\displaystyle 0, 50, 100, \ldots, 300\); strain \(\displaystyle \varepsilon \) is on the horizontal axis, marked \(\displaystyle 0, 0.001, 0.002, 0.003, 0.004\). Strain is a ratio of two lengths, so it has no units. The curve starts at the origin, rises as a straight line, bends over near the top right, reaches a maximum and then droops slightly before the plot ends.(a) Young's modulusYoung's modulus is defined by Hooke's law for a stretched wire:\[Y = \frac{\text{tensile stress}}{\text{longitudinal strain}} = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L/L} \]where \(\displaystyle F \) is the stretching force, \(\displaystyle A \) the cross-sectional area, \(\displaystyle L \) the original length and \(\displaystyle \Delta L \) the extension. On this graph \(\displaystyle \sigma \) and \(\displaystyle \varepsilon \) are plotted directly, so \(\displaystyle Y \) is simply the slope of the straight portion — and only of the straight portion, because \(\displaystyle Y \) is defined where stress and strain are proportional.Two points read off that straight portion:
    it starts at the origin, \(\displaystyle (\varepsilon_1, \sigma_1) = (0,\ 0) \);
    it passes through the crossing of the vertical rule at strain \(\displaystyle 0.002\) and the horizontal rule at \(\displaystyle 150 \times 10^{6}\ \text{N m}^{-2}\), so \(\displaystyle (\varepsilon_2, \sigma_2) = (0.002,\ 150 \times 10^{6}\ \text{N m}^{-2}) \).
    Substituting into the slope:\[Y = \frac{\sigma_2 - \sigma_1}{\varepsilon_2 - \varepsilon_1} = \frac{150 \times 10^{6}\ \text{N m}^{-2} - 0}{0.002 - 0} \]\[Y = \frac{150 \times 10^{6}}{2 \times 10^{-3}}\ \text{N m}^{-2} = 75 \times 10^{9}\ \text{N m}^{-2} \]\[Y = 7.5 \times 10^{10}\ \text{N m}^{-2} \]Check it with a third point on the same straight stretch: where the vertical rule at strain \(\displaystyle 0.003\) meets the line, the stress is about \(\displaystyle 225 \times 10^{6}\ \text{N m}^{-2}\) (a shade below halfway between the \(\displaystyle 200\) and \(\displaystyle 250\) marks), and\[\frac{225 \times 10^{6}\ \text{N m}^{-2}}{0.003} = 7.5 \times 10^{10}\ \text{N m}^{-2} \]— the same value, so the line really is straight over that whole range. Note \(\displaystyle Y \) carries the units of stress alone, \(\displaystyle \text{N m}^{-2} \) (= Pa), because strain is dimensionless. It is a scalar, so there is no direction to quote. Reading a graph to better than two figures is not honest, so \(\displaystyle Y = 7.5 \times 10^{10}\ \text{N m}^{-2} \), i.e. \(\displaystyle 75\ \text{GPa}\), to two significant figures.(b) Approximate yield strengthThe yield strength \(\displaystyle \sigma_y \) is the stress at the point where the material stops obeying Hooke's law — where the graph first curves away from the straight line and the deformation stops being fully elastic.Tracking along the line: it is still straight where it crosses strain \(\displaystyle 0.003\) at \(\displaystyle 225 \times 10^{6}\ \text{N m}^{-2}\), and still straight at about \(\displaystyle (0.0035,\ 262 \times 10^{6}\ \text{N m}^{-2})\). It first visibly peels away from that straight line at about \(\displaystyle (0.004,\ 295 \times 10^{6}\ \text{N m}^{-2})\), just under the top rule at \(\displaystyle 300 \times 10^{6}\ \text{N m}^{-2}\). Past there the curve flattens, reaches its highest point at roughly \(\displaystyle 295 \times 10^{6}\ \text{N m}^{-2}\), and then bends downwards, ending near \(\displaystyle 285 \times 10^{6}\ \text{N m}^{-2}\) at a strain of about \(\displaystyle 0.0044\) — that turnover is the material giving way, which confirms the bend just before it is the yield region and not a slip of the pen.So the stress at the bend is, to the accuracy the graph allows (the question asks only for an approximate value, so one significant figure is all that is justified):\[\sigma_y \approx 300 \times 10^{6}\ \text{N m}^{-2} = 3 \times 10^{8}\ \text{N m}^{-2} \]This too is a scalar with units \(\displaystyle \text{N m}^{-2} \).Answer: (a) \(\displaystyle Y = 7.5 \times 10^{10}\ \text{N m}^{-2} \) ($\displaystyle 75$ GPa), the slope of the straight portion, read from the origin to the point \(\displaystyle (0.002,\ 150 \times 10^{6}\ \text{N m}^{-2})\); (b) yield strength \(\displaystyle \approx 3 \times 10^{8}\ \text{N m}^{-2} \), the stress at which the curve leaves the straight line, at about \(\displaystyle (0.004,\ 295 \times 10^{6}\ \text{N m}^{-2})\).

  3. Exercise 8.3

    NCERT_Question_Class11_Physics_Ch8_Q8-3
    The stress-strain graphs for materials A and B are shown in Fig. 8.10. Fig. 8.10\displaystyle 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?

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    NCERT’s answer
    (a)
    Material A (b) Strength of a material is determined by the amount of stress required to cause fracture: material A is stronger than material B.
    Everything here comes off two features of Fig. $\displaystyle 8.10$: the slope of the first straight stretch of each graph (that slope is Young's modulus) and the height of the highest point each graph reaches (that height is the strength).Before comparing anything, use the licence the question gives you — "the graphs are drawn to the same scale". The two boxes are the same size (measuring the printed figure, the two stress axes are the same length and the two strain axes are the same length), so equal heights mean equal stress and equal widths mean equal strain in the two panels. Neither axis carries numbers, so every reading below is quoted as a fraction of the axis length, e.g. "$\displaystyle 0.47$ of the stress axis" means the point sits $\displaystyle 47$% of the way up from the strain axis to the arrowhead. That is enough for a comparison, which is all that is asked.(a) Which has the greater Young's modulus?In the elastic region a solid obeys Hooke's law,\[\sigma = Y\,\varepsilon , \qquad Y=\frac{\sigma}{\varepsilon} \]where \(\displaystyle \sigma \) is the stress \(\displaystyle =F/A \) in \(\displaystyle \mathrm{N\,m^{-2}} \), \(\displaystyle \varepsilon \) is the strain \(\displaystyle =\Delta L/L \) (a pure number, no units), and \(\displaystyle Y \) is Young's modulus, also in \(\displaystyle \mathrm{N\,m^{-2}} \). So \(\displaystyle Y \) is the slope of the initial straight portion of a stress–strain graph: the steeper that first straight line, the larger \(\displaystyle Y \).What the figure shows:
    Graph A leaves the origin steeply and is still straight as it passes the height $\displaystyle 0.47$ of the stress axis, which it reaches at only about $\displaystyle 0.09$ of the strain axis. It has clearly bent over by about $\displaystyle (0.12, 0.60)$.
    Graph B leaves the origin along a much longer, gentler straight line. It passes that same height, $\displaystyle 0.47$ of the stress axis, at about $\displaystyle 0.15$ of the strain axis, and stays straight until roughly $\displaystyle (0.25, 0.76)$.
    Compare them at that one common stress, \(\displaystyle \sigma = 0.47 \) of the stress axis. Since \(\displaystyle Y = \sigma/\varepsilon \) with \(\displaystyle \sigma \) the same for both,\[\frac{Y_A}{Y_B}=\frac{\sigma/\varepsilon_A}{\sigma/\varepsilon_B}=\frac{\varepsilon_B}{\varepsilon_A}\approx\frac{0.15}{0.09}\approx 1.7 \]For the same stress, A strains only about three-fifths as much as B, so A is the stiffer material: \(\displaystyle Y_A > Y_B \), by a factor of roughly $\displaystyle 1.7$ as this graph is drawn ($\displaystyle 2$ significant figures is as much as an unnumbered axis can support).(b) Which is the stronger material?Strength is a different idea from stiffness. The strength of a material is fixed by the stress needed to make it fracture — the greatest stress it can carry before it gives way — so on a stress–strain graph you read the height of the topmost point, not the slope.Reading the two peaks against the same stress axis:
    A rises to its maximum, about $\displaystyle 0.85$ of the stress axis, at a strain of about $\displaystyle 0.58$, then turns downwards (the necking stage) and the curve stops at about $\displaystyle (0.70, 0.79)$ — that end of the line is fracture.
    B reaches its maximum, about $\displaystyle 0.84$ of the stress axis, much earlier, at a strain of about $\displaystyle 0.30$; it then dips and wobbles and the curve stops at about $\displaystyle (0.50, 0.82)$.
    A's summit is the higher one, so A withstands the greater stress before fracturing: A is the stronger material. Be honest about the margin — on this schematic the two peaks are close ($\displaystyle 0.85$ against $\displaystyle 0.84$ of the same axis). What is unmistakable is that A carries a stress near that peak over a huge stretch of strain, from about $\displaystyle 0.35$ all the way to fracture at $\displaystyle 0.70$, whereas B's graph has ended by a strain of 0.50. A is both the stronger and by far the more ductile; B is the more brittle of the two.Two traps worth naming, because the figure sets both:
    Steeper does not mean stronger. Part (a) is about stiffness (resisting stretch), part (b) about breaking. A happens to win both here, but a very stiff material can still snap at a low stress.
    B's straight portion climbs higher than A's (B is still elastic at $\displaystyle 0.74$ of the stress axis, while A has already yielded by about $\displaystyle 0.60$). That makes B's elastic limit — its yield strength — the higher one. It is not the breaking stress, which is what "stronger" means here.
    Answer: (a) Material A — the straight part of its graph is the steeper, giving \(\displaystyle Y_A \approx 1.7\,Y_B \) as read off the figure (same stress $\displaystyle 0.47$ of the stress axis reached at strain $\displaystyle 0.09$ for A but $\displaystyle 0.15$ for B). (b) Material A — its curve reaches the greater maximum stress before fracture (about $\displaystyle 0.85$ of the stress axis against about $\displaystyle 0.84$ for B) and sustains it out to a far larger strain, fracturing only at strain ≈ $\displaystyle 0.70$ where B's curve has ended by ≈ 0.50.

  4. Exercise 8.4

    Read the following two statements below carefully and state, with reasons, if it is true or false. The Young’s modulus of rubber is greater than that of steel; The stretching of a coil is determined by its shear modulus.

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    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    (a)
    False (b) True
    "More elastic" in physics means "harder to deform," and that flips the everyday sense of the word — steel, not rubber, is the more elastic material.Statement (a): False.Young's modulus is defined for longitudinal stretching as \[Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} \] where \(\displaystyle F\) is the stretching force, \(\displaystyle A\) the cross-sectional area, \(\displaystyle L\) the original length, and \(\displaystyle \Delta L\) the extension. A larger \(\displaystyle Y\) means a given stress produces a smaller strain — the material springs back more readily and resists being deformed. That is what "elastic" means to a physicist, even though in everyday speech we call rubber "elastic" because it stretches a lot.Rubber does exactly the opposite of resisting deformation: hang even a small load on a rubber band and it stretches enormously, so \(\displaystyle \Delta L/L\) is large for a small \(\displaystyle F/A\). Steel under the same stress stretches by only a tiny fraction of its length — a steel wire loaded well within its elastic limit shows a strain thousands of times smaller than rubber under a comparable stress. Since \[Y = \frac{\text{stress}}{\text{strain}}, \] a much smaller strain for the same stress means a much larger \(\displaystyle Y\). Numerically, \(\displaystyle Y_{\text{steel}} \approx 2\times10^{11}\ \text{Pa}\) while \(\displaystyle Y_{\text{rubber}} \approx 10^{5}\text{–}10^{6}\ \text{Pa}\) — steel's Young's modulus is larger by roughly five orders of magnitude.So the statement has it backwards: steel is the stiffer, more elastic material, and its Young's modulus is far greater than that of rubber, not the other way round.Statement (b): True.Picture what actually happens inside the wire when a helical spring (a coil) is stretched by a force \(\displaystyle F\) applied along its axis. The wire itself is not being pulled end-to-end the way a straight rod is. Because the wire is wound in a helix of coil radius \(\displaystyle R\), the axial force exerts a torque \(\displaystyle F R\) about the axis of the wire at every cross-section. This torque twists the wire — it produces a shearing (torsional) strain in the material, not a simple tensile strain.The amount by which the spring stretches for a given force therefore depends on how resistant the wire material is to this twisting, which is exactly what the shear modulus (also called the modulus of rigidity), \(\displaystyle G\), measures: \[G = \frac{\text{shearing stress}}{\text{shearing strain}}. \] A larger \(\displaystyle G\) means the wire twists less for a given torque, so the coil extends less for a given load. The extension of the spring is thus governed by \(\displaystyle G\), not by the Young's modulus of the wire material — Young's modulus describes resistance to straight-line stretching, which is not the deformation occurring here.This is the step that trips people up: a spring visibly "stretches," so it is tempting to reach for Young's modulus. But the deformation happening inside the wire is a twist, so the relevant elastic constant is the shear modulus.Answer: (a) is false — Young's modulus of steel is far greater than that of rubber, since steel strains far less than rubber under the same stress; (b) is true — stretching a coil twists its wire, so the extension is governed by the shear modulus, not the Young's modulus.

  5. Exercise 8.5

    Two wires of diameter 0.25\displaystyle 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\displaystyle 1.5 m and that of brass wire is 1.0\displaystyle 1.0 m. Compute the elongations of the steel and the brass wires. Fig. 8.11\displaystyle 8.11
    NCERT’s answer
    1.$\displaystyle 5$ × \(\displaystyle 10^{-4}\) m (steel); $\displaystyle 1.3$ × \(\displaystyle 10^{-4}\) m (brass)
    Two wires joined end to end don't share the load equally — each one carries the weight of everything hanging below its own lower end.In Fig. $\displaystyle 8.11$ the steel wire is fixed to a rigid support at the top; a mass of $\displaystyle 4.0$ kg is tied at the joint where the brass wire is attached below it, and a mass of $\displaystyle 6.0$ kg hangs from the free lower end of the brass wire. So the steel wire has to hold up both masses, while the brass wire only has to hold up the one below it.Step $\displaystyle 1$: Cross-sectional area (same for both wires)Diameter \(\displaystyle d = 0.25 \text{ cm} = 2.5\times10^{-3}\text{ m} \), so the radius is \(\displaystyle r = 1.25\times10^{-3}\text{ m} \).\[A = \pi r^2 = \pi\,(1.25\times10^{-3}\text{ m})^2 = 4.91\times10^{-6}\text{ m}^2 \]Step $\displaystyle 2$: Tension in each wireA hanging mass pulls on the wire with its weight, \(\displaystyle W = Mg\), not with its mass — the wire feels newtons, not kilograms. Take \(\displaystyle g = 9.8\text{ m/s}^2\).Steel wire carries both masses: \[F_{\text{steel}} = (4.0+6.0)\text{ kg} \times 9.8\text{ m/s}^2 = 98\text{ N} \]Brass wire carries only the lower mass: \[F_{\text{brass}} = 6.0\text{ kg} \times 9.8\text{ m/s}^2 = 58.8\text{ N} \]Step $\displaystyle 3$: Young's modulus gives the elongationYoung's modulus is \(\displaystyle Y = \dfrac{F/A}{\Delta L/L} \), where \(\displaystyle F\) is the stretching force, \(\displaystyle A\) the cross-sectional area, \(\displaystyle L\) the unstretched length, and \(\displaystyle \Delta L\) the elongation. Solving for \(\displaystyle \Delta L\):\[\Delta L = \frac{FL}{AY} \]Using the standard values \(\displaystyle Y_{\text{steel}} = 2.0\times10^{11}\text{ Pa} \) and \(\displaystyle Y_{\text{brass}} = 0.91\times10^{11}\text{ Pa} \):Steel wire (\(\displaystyle L = 1.5\text{ m}\)): \[\Delta L_{\text{steel}} = \frac{98\text{ N}\times 1.5\text{ m}}{(4.91\times10^{-6}\text{ m}^2)(2.0\times10^{11}\text{ Pa})} = \frac{147}{9.82\times10^{5}}\text{ m} = 1.497\times10^{-4}\text{ m} \]Brass wire (\(\displaystyle L = 1.0\text{ m}\)): \[\Delta L_{\text{brass}} = \frac{58.8\text{ N}\times 1.0\text{ m}}{(4.91\times10^{-6}\text{ m}^2)(0.91\times10^{11}\text{ Pa})} = \frac{58.8}{4.47\times10^{5}}\text{ m} = 1.316\times10^{-4}\text{ m} \]The given data (diameter, lengths, masses) all carry two significant figures, so both elongations round to two significant figures at the end — not before.\[\Delta L_{\text{steel}} \approx 1.5\times10^{-4}\text{ m}, \qquad \Delta L_{\text{brass}} \approx 1.3\times10^{-4}\text{ m} \]Answer: The steel wire elongates by about \(\displaystyle 1.5\times10^{-4}\text{ m}\) ($\displaystyle 0.15$ mm), and the brass wire elongates by about \(\displaystyle 1.3\times10^{-4}\text{ m}\) ($\displaystyle 0.13$ mm).

  6. Exercise 8.6

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

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    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    Deflection = $\displaystyle 4$ × \(\displaystyle 10^{-6}\) m
    The load hangs alongside the fixed face, so the force is tangential to it — this is shear, and the modulus to use is \(\displaystyle G\), not Young's modulus \(\displaystyle Y\).The cube's back face is glued to a vertical wall. The weight of the $\displaystyle 100$ kg mass pulls straight down on the opposite (free) face. That downward force lies in the plane of the face rather than perpendicular to it, so it does not stretch or squash the cube — it slides the free face down relative to the fixed one, tilting the top and bottom faces into parallelograms. That is a shear deformation.Step $\displaystyle 1$ — the tangential force.The force is the weight of the hanging mass: \[F = mg = (100\ \text{kg})(9.8\ \text{m s}^{-2}) = 980\ \text{N} \]Here \(\displaystyle m\) is the hanging mass and \(\displaystyle g = 9.8\ \text{m s}^{-2}\) is the acceleration due to gravity. A quick aside on the step people skip: the data gives a mass in kilograms, and a modulus needs a force in newtons. Feeding $\displaystyle 100$ into the stress formula instead of $\displaystyle 980$ is the most common way this question goes wrong.Step $\displaystyle 2$ — the area the force acts over.Shear stress is force divided by the area of the face being sheared — the free face itself, of edge \(\displaystyle L = 10\ \text{cm} = 0.10\ \text{m}\): \[A = L^2 = (0.10\ \text{m})^2 = 1.0 \times 10^{-2}\ \text{m}^2 \]Step $\displaystyle 3$ — the shear stress.\[\text{shear stress} = \frac{F}{A} = \frac{980\ \text{N}}{1.0\times 10^{-2}\ \text{m}^2} = 9.8\times 10^{4}\ \text{N m}^{-2} = 9.8\times 10^{4}\ \text{Pa} \]Step $\displaystyle 4$ — the shear strain, from the definition of the shear modulus.The shear modulus \(\displaystyle G\) is defined as \[G = \frac{\text{shear stress}}{\text{shear strain}} = \frac{F/A}{\theta} \] where \(\displaystyle \theta\) is the shear strain — the angle (in radians) through which a vertical edge of the cube tips. Note the unit: \(\displaystyle 25\ \text{GPa} = 25 \times 10^{9}\ \text{Pa}\), since one gigapascal is \(\displaystyle 10^{9}\ \text{Pa}\).Rearranging, \[\theta = \frac{F/A}{G} = \frac{9.8\times 10^{4}\ \text{Pa}}{25\times 10^{9}\ \text{Pa}} = 3.92\times 10^{-6} \]Shear strain is a ratio of two stresses' worth of the same unit, so it is a pure number — and because it is an angle it is in radians, not degrees. It is far too small for the degree/radian distinction to be visible here, but the small-angle relation used next is only valid in radians.Step $\displaystyle 5$ — turn the strain into a deflection.For a small angle, \(\displaystyle \theta \approx \Delta x / L\), where \(\displaystyle \Delta x\) is how far the free face slides and \(\displaystyle L\) is the distance between the fixed face and the free face — i.e. the depth of the cube, \(\displaystyle 0.10\ \text{m}\). So \[\Delta x = \theta L = (3.92\times 10^{-6})(0.10\ \text{m}) = 3.92\times 10^{-7}\ \text{m} \]The data carry two significant figures (\(\displaystyle g = 9.8\ \text{m s}^{-2}\), \(\displaystyle G = 25\ \text{GPa}\), \(\displaystyle L = 10\ \text{cm}\)), so rounding once, at the end, to two significant figures: \[\Delta x = 3.9\times 10^{-7}\ \text{m} \approx 4\times 10^{-7}\ \text{m} = 0.39\ \mu\text{m} \]The displacement is a vector, so it needs its direction: the free face moves vertically downward, in the direction of the hanging weight, relative to the face fixed to the wall.A note on the printed key. The answer key gives \(\displaystyle 4\times 10^{-6}\ \text{m}\). That is ten times too large; the data given in the question support \(\displaystyle 4\times 10^{-7}\ \text{m}\). You can see the factor of ten directly: the strain is \(\displaystyle 9.8\times 10^{4}/25\times 10^{9} = 3.92\times 10^{-6}\), and multiplying that dimensionless strain by \(\displaystyle L = 0.10\ \text{m}\) — not by \(\displaystyle 1\ \text{m}\) — moves the exponent down one, to \(\displaystyle 10^{-7}\). Getting \(\displaystyle 4\times 10^{-6}\ \text{m}\) would require an edge of \(\displaystyle 1\ \text{m}\), not \(\displaystyle 10\ \text{cm}\). Treat \(\displaystyle 4\times 10^{-7}\ \text{m}\) as the correct value.Answer: \(\displaystyle \Delta x \approx 3.9\times 10^{-7}\ \text{m}\) (about \(\displaystyle 4\times 10^{-7}\ \text{m}\), i.e. \(\displaystyle 0.39\ \mu\text{m}\)), the loaded face sliding vertically downward relative to the fixed face.

  7. Exercise 8.7

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

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    2.$\displaystyle 8$ × \(\displaystyle 10^{-6}\)
    "Assuming the load distribution to be uniform" means each column carries only a quarter of the weight — that quarter, not the whole $\displaystyle 50,000$ kg, is the force that goes into the stress.The law in play is the definition of Young's modulus \(\displaystyle Y\): for a rod or column loaded along its length,\[Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{\Delta L/L} \qquad\Longrightarrow\qquad \text{strain}=\frac{\Delta L}{L}=\frac{F}{AY} \]where \(\displaystyle F\) is the force pressing along the column's axis, \(\displaystyle A\) is the cross-sectional area it acts over, \(\displaystyle \Delta L/L\) is the fractional change in length, and \(\displaystyle Y\) is Young's modulus of the material. For steel, Table $\displaystyle 8.1$ of the book gives \(\displaystyle Y = 200\times10^{9}\ \mathrm{N\,m^{-2}} = 2.0\times10^{11}\ \mathrm{N\,m^{-2}}\). Strain is a ratio of two lengths, so it has no unit.Step $\displaystyle 1$ — the force on one column. The structure's weight is its mass times \(\displaystyle g\):\[W = Mg = (50{,}000\ \mathrm{kg})(9.8\ \mathrm{m\,s^{-2}}) = 4.9\times10^{5}\ \mathrm{N} \]The aside worth pausing on: $\displaystyle 50,000$ kg is a mass, not a force. It must be multiplied by \(\displaystyle g\) before it can appear in a stress, and the newton is the unit that belongs in \(\displaystyle F/A\).Four identical columns share this weight equally, so each one pushes back with\[F=\frac{W}{4}=\frac{4.9\times10^{5}\ \mathrm{N}}{4}=1.225\times10^{5}\ \mathrm{N} \]Step $\displaystyle 2$ — the cross-sectional area of one column. Each column is hollow: the steel occupies the annulus between the inner radius \(\displaystyle r\) and the outer radius \(\displaystyle R\). Converting to metres, \(\displaystyle r = 30\ \mathrm{cm} = 0.30\ \mathrm{m}\) and \(\displaystyle R = 60\ \mathrm{cm} = 0.60\ \mathrm{m}\), and\[A=\pi\left(R^{2}-r^{2}\right)=\pi\left[(0.60\ \mathrm{m})^{2}-(0.30\ \mathrm{m})^{2}\right] =\pi\left(0.36-0.09\right)\ \mathrm{m^{2}}=\pi(0.27\ \mathrm{m^{2}}) \]\[A = 0.8482\ \mathrm{m^{2}} \]A second trap: the load-bearing area is \(\displaystyle \pi(R^{2}-r^{2})\), not \(\displaystyle \pi(R-r)^{2}\). Squaring the wall thickness would give \(\displaystyle \pi(0.30)^2 = 0.283\ \mathrm{m^2}\), about a third of the true area, and would inflate the strain threefold.Step $\displaystyle 3$ — the compressive stress in one column.\[\text{stress}=\frac{F}{A}=\frac{1.225\times10^{5}\ \mathrm{N}}{0.8482\ \mathrm{m^{2}}}=1.4442\times10^{5}\ \mathrm{N\,m^{-2}} \]Step $\displaystyle 4$ — the strain.\[\text{strain}=\frac{\text{stress}}{Y}=\frac{1.4442\times10^{5}\ \mathrm{N\,m^{-2}}}{2.0\times10^{11}\ \mathrm{N\,m^{-2}}}=7.221\times10^{-7} \]The units cancel completely, as they must for a strain. The radii are given to two significant figures, so rounding once, at the end: \(\displaystyle 7.2\times10^{-7}\). It is a compressional strain, meaning each column is shortened, not stretched, by this fraction of its length — a $\displaystyle 4$ m column would settle by about \(\displaystyle 3\ \mu\mathrm{m}\).On the number printed in the book. The answer key gives \(\displaystyle 2.8\times10^{-6}\), which is almost exactly four times the value above. That factor of four is the four columns: putting the entire \(\displaystyle 4.9\times10^{5}\ \mathrm{N}\) on a single column's area gives a stress of \(\displaystyle 5.78\times10^{5}\ \mathrm{N\,m^{-2}}\) and a strain of \(\displaystyle 2.9\times10^{-6}\) with \(\displaystyle Y = 2.0\times10^{11}\ \mathrm{N\,m^{-2}}\) (and \(\displaystyle 2.8\times10^{-6}\) if the mild-steel value \(\displaystyle Y \approx 2.06\times10^{11}\ \mathrm{N\,m^{-2}}\) is used instead of the table's steel entry). The printed answer therefore drops the division of the load among the four columns, which is the one thing the phrase "assuming the load distribution to be uniform" was put in the question to tell you to do. A factor of $\displaystyle 4$ is not a rounding disagreement, so the key is in error here; the strain per column is \(\displaystyle 7.2\times10^{-7}\).Answer: The compressional strain of each column is \(\displaystyle 7.2\times10^{-7}\) — dimensionless, and a shortening (compression) of each column along its length. (The book's printed \(\displaystyle 2.8\times10^{-6}\) is four times too large: it loads all $\displaystyle 50,000$ kg onto one column instead of sharing it among the four.)

  8. Exercise 8.8

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

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    0.$\displaystyle 127$
    Strain is stress divided by Young's modulus, and it is a pure number — no unit at all. Nothing in this question asks for an elongation, so no length is needed; the cross-section, the pull and the modulus of copper are the whole of it.Step $\displaystyle 1$ — the area of cross-section, in square metres.The two sides are given in millimetres. Convert before multiplying, because \(\displaystyle 1\ \text{mm}^2 = 10^{-6}\ \text{m}^2\) — squaring the prefix is where this calculation usually goes wrong by a factor of a million.\[A = (15.2 \times 10^{-3}\ \text{m}) \times (19.1 \times 10^{-3}\ \text{m}) = 290.32 \times 10^{-6}\ \text{m}^2 = 2.9032 \times 10^{-4}\ \text{m}^2 \]Step $\displaystyle 2$ — the tensile stress.Tensile stress is the force applied normal to the cross-section divided by that area,\[\sigma = \frac{F}{A} \]where \(\displaystyle F = 44{,}500\ \text{N}\) is the pulling force and \(\displaystyle A\) the area found above.\[\sigma = \frac{44{,}500\ \text{N}}{2.9032 \times 10^{-4}\ \text{m}^2} = 1.5328 \times 10^{8}\ \text{N m}^{-2} \]Step $\displaystyle 3$ — Young's modulus of copper.Young's modulus \(\displaystyle Y\) is defined as the ratio of tensile stress \(\displaystyle \sigma\) to longitudinal strain \(\displaystyle \varepsilon\):\[Y = \frac{\sigma}{\varepsilon} \]Table $\displaystyle 8.1$ of this chapter lists copper at \(\displaystyle Y = 110 \times 10^{9}\ \text{N m}^{-2} = 1.1 \times 10^{11}\ \text{N m}^{-2}\), and the worked Example $\displaystyle 8.2$ in the same chapter uses exactly that figure for copper. This is the number the question expects you to look up — it is not derivable from the data given.Step $\displaystyle 4$ — the strain.Rearranging the definition, \(\displaystyle \varepsilon = \sigma / Y\):\[\varepsilon = \frac{1.5328 \times 10^{8}\ \text{N m}^{-2}}{1.1 \times 10^{11}\ \text{N m}^{-2}} = 1.393 \times 10^{-3} \]The newtons per square metre cancel top and bottom, which is why strain carries no unit — it is \(\displaystyle \Delta L / L\), a length over a length.Significant figures. The force and both edge lengths are given to three significant figures, but the tabulated modulus \(\displaystyle 1.1 \times 10^{11}\ \text{N m}^{-2}\) carries only two, and a quotient can be no more precise than its least precise input. Rounding once, at the end:\[\varepsilon \approx 1.4 \times 10^{-3} \](If you keep the unrounded \(\displaystyle 1.39 \times 10^{-3}\), that is the same result carried to one extra digit; \(\displaystyle 1.4 \times 10^{-3}\) is the honest statement of it.) As a percentage this is a stretch of about \(\displaystyle 0.14\%\) of the bar's length.A check that the deformation really is elastic, as the question stipulates. Table $\displaystyle 8.1$ gives copper a yield strength of \(\displaystyle \sigma_y = 200 \times 10^{6}\ \text{N m}^{-2} = 2.0 \times 10^{8}\ \text{N m}^{-2}\). Our stress, \(\displaystyle 1.53 \times 10^{8}\ \text{N m}^{-2}\), sits below that, so the bar is still on the linear part of its stress–strain curve and Hooke's law applies. Consistent.A note on the printed answer. The answers appendix gives \(\displaystyle 0.127\) for this question, and that number cannot be right. A strain of \(\displaystyle 0.127\) means the copper bar stretches by \(\displaystyle 12.7\%\) of its length. Copper yields at a strain of only \(\displaystyle \sigma_y / Y = (2.0 \times 10^{8})/(1.1 \times 10^{11}) \approx 1.8 \times 10^{-3}\), i.e. under \(\displaystyle 0.2\%\), by this book's own table — so \(\displaystyle 0.127\) is some seventy times past the point where the bar stops behaving elastically, contradicting the question's own words "producing only elastic deformation." The printed figure looks like a lost power of ten: read as \(\displaystyle 1.27 \times 10^{-3}\) it is at least physically sensible, but even that would require \(\displaystyle Y \approx 1.2 \times 10^{11}\ \text{N m}^{-2}\) for copper, whereas Table $\displaystyle 8.1$ of this very chapter says \(\displaystyle 1.1 \times 10^{11}\ \text{N m}^{-2}\), which gives \(\displaystyle 1.4 \times 10^{-3}\). Work it from the book's own table and \(\displaystyle 1.4 \times 10^{-3}\) is what comes out.Answer: strain \(\displaystyle \approx 1.4 \times 10^{-3}\) (dimensionless, about \(\displaystyle 0.14\%\)); unrounded, \(\displaystyle 1.39 \times 10^{-3}\), from a tensile stress of \(\displaystyle 1.53 \times 10^{8}\ \text{N m}^{-2}\) divided by \(\displaystyle Y_{\text{copper}} = 1.1 \times 10^{11}\ \text{N m}^{-2}\). The appendix value \(\displaystyle 0.127\) is an error.

  9. Exercise 8.9

    A steel cable with a radius of 1.5\displaystyle 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108\displaystyle 10^{8} N m−2\displaystyle m^{-2}, what is the maximum load the cable can support ?
    NCERT’s answer
    7.$\displaystyle 07$ × $\displaystyle 10$ $\displaystyle 4$ N
    Stress is force divided by area, so the maximum force the cable can carry is the maximum stress times its cross-sectional area — you need the area of the cable's circular cross-section, not its length or anything else.Stress is defined as \[\sigma = \frac{F}{A} \] where \(\displaystyle F\) is the force (the load) carried by the cable and \(\displaystyle A\) is the area of its cross-section. Rearranging for the maximum load the cable can bear before the stress limit is reached, \[F_{\text{max}} = \sigma_{\text{max}} \, A \]Finding the cross-sectional areaThe cable's cross-section is a circle of radius \(\displaystyle r = 1.5\ \text{cm} = 1.5 \times 10^{-2}\ \text{m}\). A cable this size is a fat rope, not a wire — the radius is centimetres, so it must be converted to metres before it goes anywhere near an SI formula.\[A = \pi r^2 = \pi \left(1.5 \times 10^{-2}\ \text{m}\right)^2 = \pi \times 2.25 \times 10^{-4}\ \text{m}^2 = 7.069 \times 10^{-4}\ \text{m}^2 \]Applying the stress limitThe maximum stress allowed is \(\displaystyle \sigma_{\text{max}} = 10^{8}\ \text{N m}^{-2}\). Substituting,\[F_{\text{max}} = \sigma_{\text{max}} \, A = \left(10^{8}\ \text{N m}^{-2}\right)\left(7.069 \times 10^{-4}\ \text{m}^2\right) \]\[F_{\text{max}} = 7.069 \times 10^{4}\ \text{N} \]The radius was given to two significant figures ($\displaystyle 1.5$ cm), so the final result is stated to that precision: \[F_{\text{max}} \approx 7.1 \times 10^{4}\ \text{N} \]This is a force — the weight the cable can carry — not a mass. "Load" here means the pull the cable resists, so the answer belongs in newtons; do not divide by \(\displaystyle g\) unless a mass is explicitly asked for.Answer: The cable can support a maximum load of about \(\displaystyle 7.1 \times 10^{4}\ \text{N}\) (\(\displaystyle \approx 7.07 \times 10^{4}\ \text{N}\) before rounding).

  10. Exercise 8.10

    A rigid bar of mass 15\displaystyle 15 kg is supported symmetrically by three wires each 2.0\displaystyle 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.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    \(\displaystyle D_{copper}\)/\(\displaystyle D_{iron}\) = $\displaystyle 1.25$
    A rigid bar that stays level must stretch by the same amount at every wire — and that geometric fact, not the load, is what fixes the diameter ratio here.Picture the bar hanging from the three vertical wires, each originally the same length \(\displaystyle L = 2.0 \) m: a copper wire at each end, the iron wire exactly in the middle. The bar is rigid, so it cannot bend between the supports, and by symmetry there is nothing to make it tilt toward one end — it stays horizontal. A rigid bar that stays horizontal and doesn't rotate must move down by the same distance at every point along it, so the extension of the middle (iron) wire has to equal the extension of the end (copper) wires: \[\Delta L_{\text{Cu}} = \Delta L_{\text{Fe}} = \Delta L \] Since all three wires start at the same length \(\displaystyle L\), this also means their strains are equal: \[\frac{\Delta L_{\text{Cu}}}{L} = \frac{\Delta L_{\text{Fe}}}{L} \]Now bring in Young's modulus, \(\displaystyle Y = \dfrac{F/A}{\Delta L/L} \), where \(\displaystyle F\) is the tension in a wire, \(\displaystyle A\) is its cross-sectional area, and \(\displaystyle \Delta L/L\) is its strain. Solving for the tension, \[F = Y\,A\left(\frac{\Delta L}{L}\right) \]Write this once for a copper wire and once for the iron wire. The condition we're asked to satisfy is that every wire carries the same tension \(\displaystyle T\): \[T = Y_{\text{Cu}}\,A_{\text{Cu}}\left(\frac{\Delta L}{L}\right) = Y_{\text{Fe}}\,A_{\text{Fe}}\left(\frac{\Delta L}{L}\right) \]The strain \(\displaystyle \Delta L/L\) is the same on both sides — that was the whole point of the rigid-bar argument — so it cancels straight out: \[Y_{\text{Cu}}\,A_{\text{Cu}} = Y_{\text{Fe}}\,A_{\text{Fe}} \]This is the step it's easy to miss: the $\displaystyle 15$ kg mass and the $\displaystyle 2.0$ m length never entered the calculation. They set how large the actual tension and elongation turn out to be, but the ratio of areas depends only on the ratio of the two Young's moduli — that is why the problem gives numbers you don't end up using.Writing the area of a wire as \(\displaystyle A = \dfrac{\pi}{4}d^2 \), the \(\displaystyle \pi/4\) cancels from both sides, leaving \[\frac{A_{\text{Cu}}}{A_{\text{Fe}}} = \frac{d_{\text{Cu}}^{\,2}}{d_{\text{Fe}}^{\,2}} = \frac{Y_{\text{Fe}}}{Y_{\text{Cu}}} \]Using the standard Young's-modulus values for these two metals, \(\displaystyle Y_{\text{Cu}} = 1.1\times10^{11}\ \text{N m}^{-2}\) for copper and \(\displaystyle Y_{\text{Fe}} = 1.9\times10^{11}\ \text{N m}^{-2}\) for (wrought) iron, \[\frac{d_{\text{Cu}}^{\,2}}{d_{\text{Fe}}^{\,2}} = \frac{1.9\times10^{11}}{1.1\times10^{11}} = \frac{19}{11} = 1.727 \] \[\frac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{1.727} = 1.314 \]The Young's-modulus data going in are only good to two significant figures ($\displaystyle 1.1$ and $\displaystyle 1.9$), so the ratio should be rounded to the same precision, giving \(\displaystyle 1.3\).This makes physical sense, not just algebraic sense: copper is the softer, more stretchy metal (lower \(\displaystyle Y\)), so for a copper wire to carry the same tension as the stiffer iron wire while stretching by the same amount, it needs more cross-sectional area — a fatter wire. (As a check, this equal-tension condition also fixes the actual tension: the total weight is shared as \(\displaystyle 2T + T = 3T\), so \(\displaystyle T = \dfrac{mg}{3} = \dfrac{15 \times 9.8}{3} = 49\ \text{N}\) in every wire — consistent, though the question only asks for the diameter ratio.)Answer: \(\displaystyle d_{\text{Cu}} : d_{\text{Fe}} = \sqrt{19/11} \approx 1.3 : 1\) — the copper wires must be about $\displaystyle 1.3$ times thicker (in diameter) than the iron wire.