NCERT Solutions · Class 11 Physics Mechanical Properties of Solids

At the lowest point the wire has two jobs at once: it holds the weight up and it supplies the centripetal force. Both loads pull on the same wire, so the tension there is bigger than the weight alone — and it is that total tension, not \(\displaystyle mg\), that stretches the wire.Step $\displaystyle 1$ — Convert everything to SI, including the angular velocity.The rate of turning is given in revolutions per second. One revolution is \(\displaystyle 2\pi\) radians, and every formula below (\(\displaystyle a_c=\omega^2 r\)) needs \(\displaystyle \omega\) in radians per second:\[\omega = 2\pi \nu = 2\pi \times 2.0\ \text{rev/s} = 4\pi\ \text{rad/s} = 12.566\ \text{rad s}^{-1} \]This is the step people get wrong. Writing \(\displaystyle \omega = 2\ \text{rad s}^{-1}\) straight off the page loses a factor of \(\displaystyle (2\pi)^2 \approx 39.5\) in the centripetal term.The area is given in \(\displaystyle \text{cm}^2\), and \(\displaystyle 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2\) (the conversion factor is squared too):\[A = 0.065\ \text{cm}^2 = 0.065 \times 10^{-4}\ \text{m}^2 = 6.5\times10^{-6}\ \text{m}^2 \]Also: \(\displaystyle m = 14.5\ \text{kg}\), \(\displaystyle L = 1.0\ \text{m}\) (the unstretched length, which is also the radius of the circle), \(\displaystyle g = 9.8\ \text{m s}^{-2}\), and for steel Young's modulus \(\displaystyle Y = 2.0\times10^{11}\ \text{N m}^{-2}\).Step $\displaystyle 2$ — Newton's second law along the wire, at the lowest point.At the bottom of the circle the centre of the circle is directly above the mass, so the net upward force must equal the centripetal force \(\displaystyle m\omega^2 L\). Taking upward as positive, with \(\displaystyle T\) the tension (up) and \(\displaystyle mg\) the weight (down):\[T - mg = m\omega^{2}L \qquad\Longrightarrow\qquad T = m\left(g + \omega^{2}L\right) \]Here \(\displaystyle T\) is the tension in newtons, \(\displaystyle m\) the mass in kg, \(\displaystyle g\) the acceleration due to gravity, \(\displaystyle \omega\) the angular velocity in rad s\(\displaystyle ^{-1}\), and \(\displaystyle L\) the radius of the circular path.Substituting:\[\omega^{2}L = (12.566\ \text{s}^{-1})^{2}\times 1.0\ \text{m} = 157.91\ \text{m s}^{-2} \] \[T = 14.5\ \text{kg}\times\left(9.8 + 157.91\right)\ \text{m s}^{-2} = 14.5 \times 167.71 = 2431.8\ \text{N} \]Worth pausing on the two pieces: the weight contributes only \(\displaystyle mg = 14.5\times 9.8 = 142.1\ \text{N}\), while the whirling contributes \(\displaystyle m\omega^2 L = 14.5 \times 157.91 = 2289.7\ \text{N}\). The spinning dominates by a factor of about $\displaystyle 16$ — the wire is nowhere near just "holding a hanging mass".Step $\displaystyle 3$ — Apply the definition of Young's modulus.Young's modulus is longitudinal stress divided by longitudinal strain:\[Y = \frac{\text{stress}}{\text{strain}} = \frac{T/A}{\Delta L/L} \qquad\Longrightarrow\qquad \Delta L = \frac{T L}{A Y} \]where \(\displaystyle \Delta L\) is the elongation, \(\displaystyle L\) the original length, \(\displaystyle A\) the cross-sectional area and \(\displaystyle Y\) Young's modulus.\[\Delta L = \frac{2431.8\ \text{N}\times 1.0\ \text{m}}{\left(6.5\times10^{-6}\ \text{m}^{2}\right)\left(2.0\times10^{11}\ \text{N m}^{-2}\right)} = \frac{2431.8\ \text{N}\cdot\text{m}}{1.3\times10^{6}\ \text{N}} \]The newtons cancel and metres survive, as an elongation must:\[\Delta L = 1.871\times10^{-3}\ \text{m} \]Step $\displaystyle 4$ — Round once, at the end.The length \(\displaystyle 1.0\ \text{m}\) and the area \(\displaystyle 0.065\ \text{cm}^2\) each carry two significant figures, so the answer is limited to two:\[\Delta L \approx 1.9\times10^{-3}\ \text{m} = 1.9\ \text{mm} \]The wire stretches along its own length, i.e. radially outward — downward — at the lowest point.A note on the printed key. NCERT's answer for this exercise, \(\displaystyle 1.539\times10^{-4}\ \text{m}\), is not what the question as printed gives. That number is exactly what you get by putting \(\displaystyle \omega = 2\ \text{rad s}^{-1}\) into the same working: then \(\displaystyle \omega^2 L = 4.0\ \text{m s}^{-2}\), \(\displaystyle T = 14.5\times(9.8+4.0) = 200.1\ \text{N}\), and \(\displaystyle \Delta L = 200.1/(1.3\times10^{6}) = 1.539\times10^{-4}\ \text{m}\) — the key's digits reproduced exactly. So the key used the number "$\displaystyle 2$" as though it were already in rad s\(\displaystyle ^{-1}\), while the question says $\displaystyle 2$ rev/s. Converting revolutions to radians multiplies the centripetal term by \(\displaystyle (2\pi)^2 \approx 39.5\), which is why the two results differ by roughly a factor of $\displaystyle 12$ once the weight term is included. Take the question at its word and the elongation is \(\displaystyle 1.9\ \text{mm}\).Answer: \(\displaystyle \Delta L = 1.9\times10^{-3}\ \text{m}\) (about \(\displaystyle 1.9\ \text{mm}\)), the wire stretching along its length — downward — at the lowest point; the textbook's printed \(\displaystyle 1.539\times10^{-4}\ \text{m}\) follows from misreading $\displaystyle 2$ rev/s as $\displaystyle 2$ rad/s.

Bulk modulus tells you how much pressure it takes to squeeze a fixed amount of substance into a slightly smaller volume — the larger the number, the harder the substance resists being compressed.The bulk modulus \(\displaystyle B \) is defined by \[B = -\frac{\Delta P}{\Delta V / V} \] where \(\displaystyle \Delta P\) is the change in pressure, \(\displaystyle V\) is the original volume, and \(\displaystyle \Delta V\) is the resulting change in volume. The minus sign is there only because volume shrinks when pressure rises; since the question asks for the size of \(\displaystyle B\), it is enough to work with the magnitudes \(\displaystyle |\Delta P|\) and \(\displaystyle |\Delta V|\).Bulk modulus of waterConvert the pressure increase to SI units, using the conversion factor the question gives: \[\Delta P = 100.0\ \text{atm} \times 1.013\times10^{5}\ \text{Pa/atm} = 1.013\times10^{7}\ \text{Pa} \]The volume change is the difference between the two stated volumes: \[|\Delta V| = 100.5\ \text{L} - 100.0\ \text{L} = 0.5\ \text{L}, \qquad V = 100.0\ \text{L} \](As printed, the "final" volume is larger than the initial one, which looks backwards for a compression — squeezing water harder should shrink it, not grow it. What the bulk-modulus formula needs is only the size of the volume change relative to the original volume, so the arithmetic is unaffected by this; physically, water under $\displaystyle 100$ atm of extra pressure would occupy $\displaystyle 99.5$ L, not $\displaystyle 100.5$ L, but the magnitude of the change is the same $\displaystyle 0.5$ L either way.)The fractional volume change is \[\frac{|\Delta V|}{V} = \frac{0.5\ \text{L}}{100.0\ \text{L}} = 5.0\times10^{-3} \]So \[B_{\text{water}} = \frac{\Delta P}{|\Delta V|/V} = \frac{1.013\times10^{7}\ \text{Pa}}{5.0\times10^{-3}} = 2.026\times10^{9}\ \text{Pa} \]A short aside on precision: $\displaystyle 0.5$ L is a difference of two volumes each known only to the nearest $\displaystyle 0.1$ L, so on its own it carries just one reliable digit. The rest of the data in this problem are given to four figures, so the answer is quoted to that same precision, \(\displaystyle B_{\text{water}} = 2.026\times10^{9}\ \text{Pa}\) — but the honest uncertainty in a real measurement like this is set by how precisely that small volume change was actually read off, not by how many digits are printed.Bulk modulus of air, for comparisonAt constant temperature a gas obeys Boyle's law, \(\displaystyle PV = \text{constant}\). Differentiating gives \(\displaystyle P\,dV + V\,dP = 0\), so \(\displaystyle -\dfrac{dP}{dV/V} = P\). The isothermal bulk modulus of a gas is therefore just its own pressure: \[B_{\text{air}} = P_{\text{atm}} = 1.013\times10^{5}\ \text{Pa} \]The ratio\[\frac{B_{\text{water}}}{B_{\text{air}}} = \frac{2.026\times10^{9}\ \text{Pa}}{1.013\times10^{5}\ \text{Pa}} = 2.0\times10^{4} \]Water is roughly $\displaystyle 20,000$ times less compressible than air.Why the ratio is so large: In a gas, the molecules sit far apart with mostly empty space between them, so even a modest rise in pressure crowds them noticeably closer together — little resistance, small bulk modulus. In a liquid such as water, the molecules are already touching, sitting almost at the spacing where the short-range repulsive force between them takes over; squeezing water further means fighting that steep repulsion for almost no gain in volume, so an enormous pressure produces only a tiny fractional squeeze. That difference — nearly empty space in a gas versus already-packed molecules in a liquid — is exactly what makes liquids behave as nearly incompressible compared with gases.Answer: Bulk modulus of water ≈ \(\displaystyle 2.026\times10^{9}\ \text{Pa}\); bulk modulus of air (isothermal) ≈ \(\displaystyle 1.013\times10^{5}\ \text{Pa}\); ratio \(\displaystyle B_{\text{water}}/B_{\text{air}} \approx 2.0\times10^{4}\) — water is about $\displaystyle 20,000$ times harder to compress than air, because its molecules are already packed against each other's repulsive limit while air's molecules are far apart and easily pushed together.

A pressure increase squeezes any fluid's volume very slightly, and that shrinkage is exactly what the bulk modulus measures — so working from water's bulk modulus tells you how much denser it gets at depth.By definition, if a fixed mass of fluid at volume \(\displaystyle V_1 \) has its pressure raised by \(\displaystyle \Delta P \), its fractional decrease in volume is\[\frac{\Delta V}{V_1} = \frac{\Delta P}{B} \]where \(\displaystyle B \) is the bulk modulus of the fluid. For water this chapter's table of bulk moduli gives \(\displaystyle B = 2.2\times10^{9}\ \text{N m}^{-2} \) — the same standard value used throughout this set of problems.Because the mass \(\displaystyle m \) of the water sample does not change as it is compressed, volume and density are tied together through \(\displaystyle V=m/\rho \). Write \(\displaystyle \rho_1 \) for the surface density and \(\displaystyle \rho_2 \) for the density at depth, so \(\displaystyle V_1=m/\rho_1 \) and \(\displaystyle V_2=m/\rho_2 \):\[\frac{\Delta V}{V_1}=\frac{V_1-V_2}{V_1}=1-\frac{V_2}{V_1}=1-\frac{\rho_1}{\rho_2} \]Combining the two expressions,\[1-\frac{\rho_1}{\rho_2}=\frac{\Delta P}{B}\qquad\Rightarrow\qquad \rho_2=\frac{\rho_1}{1-\Delta P/B} \]Since \(\displaystyle \Delta P/B \) will come out much smaller than $\displaystyle 1$, use the binomial approximation \(\displaystyle (1-x)^{-1}\approx 1+x \):\[\rho_2\approx \rho_1\left(1+\frac{\Delta P}{B}\right) \]Now substitute. The pressure at depth is $\displaystyle 80.0$ atm, which is so much larger than the roughly $\displaystyle 1$ atm at the surface that the surface pressure can be neglected; take \(\displaystyle \Delta P = 80.0\ \text{atm} \):\[\Delta P = 80.0 \times 1.013\times10^{5}\ \text{Pa} = 8.104\times10^{6}\ \text{Pa} \]\[\frac{\Delta P}{B}=\frac{8.104\times10^{6}\ \text{Pa}}{2.2\times10^{9}\ \text{Pa}}=3.68\times10^{-3} \]This ratio is both the fractional drop in volume and (to this approximation) the fractional rise in density — under $\displaystyle 0.4$%, which is exactly why liquids are described as "nearly incompressible."\[\rho_2\approx 1.03\times10^{3}\ \text{kg m}^{-3}\left(1+3.68\times10^{-3}\right) \]\[\rho_2\approx 1.03\times10^{3}\ \text{kg m}^{-3}+3.8\ \text{kg m}^{-3} \]\[\rho_2\approx 1.034\times10^{3}\ \text{kg m}^{-3} \]A note on precision: \(\displaystyle B \) itself is known to only $\displaystyle 2$ significant figures, so the correction term \(\displaystyle 3.8\ \text{kg m}^{-3}\) really carries an uncertainty of a few tenths. Rounding it away and reporting just \(\displaystyle 1.0\times10^{3}\ \text{kg m}^{-3}\) would erase the entire point of the calculation — that pressure does raise water's density, just by a very small amount — so the extra figure is kept, exactly as this book's own answer does for this problem.Answer: The density of water at that depth is about \(\displaystyle 1.034\times10^{3}\ \text{kg m}^{-3}\), roughly \(\displaystyle 4\ \text{kg m}^{-3}\) (about $\displaystyle 0.4$%) higher than the surface value of \(\displaystyle 1.03\times10^{3}\ \text{kg m}^{-3}\), confirming that water is nearly incompressible even at $\displaystyle 80$ atm.

A hydraulic pressure squeezes a body equally from every side, so it is the bulk modulus — not Young's modulus — that connects it to the change in volume.When a solid is subjected to a uniform pressure \(\displaystyle p\) acting inward from all directions (a hydraulic pressure), its volume shrinks by a fractional amount \(\displaystyle \dfrac{\Delta V}{V}\), and the two are related by the bulk modulus \(\displaystyle B\):\[B = \frac{p}{\Delta V / V} \]Here \(\displaystyle p\) is the applied pressure, \(\displaystyle \Delta V\) is the change in volume, \(\displaystyle V\) is the original volume, and \(\displaystyle B\) is a property of the material — for glass, \(\displaystyle B = 37 \times 10^{9}\ \text{N m}^{-2}\), the tabulated bulk modulus for glass. Rearranging for the quantity asked,\[\frac{\Delta V}{V} = \frac{p}{B} \]Convert the pressure to SI units. The pressure is given in atmospheres, but \(\displaystyle B\) is in pascals, so the two must be brought to the same unit before dividing — this is the step it is easy to skip. Using \(\displaystyle 1\ \text{atm} = 1.013 \times 10^{5}\ \text{Pa}\),\[p = 10\ \text{atm} = 10 \times 1.013 \times 10^{5}\ \text{Pa} = 1.013 \times 10^{6}\ \text{Pa} \]Substitute.\[\frac{\Delta V}{V} = \frac{1.013 \times 10^{6}\ \text{Pa}}{37 \times 10^{9}\ \text{Pa}} = 2.7378\ldots \times 10^{-5} \]Both the pressure ($\displaystyle 10$ atm) and the bulk modulus ($\displaystyle 37$ × $\displaystyle 10$⁹) are given to two significant figures, so the result is rounded to two significant figures. Note that \(\displaystyle \Delta V/V\) is a ratio of two volumes, so it carries no unit — it is a pure number, the fraction of the slab's volume that disappears under the squeeze.\[\frac{\Delta V}{V} \approx 2.7 \times 10^{-5} \]This is an extremely small fraction — about $\displaystyle 27$ parts in a million — which is exactly why glass is called nearly incompressible: even ten atmospheres of squeezing pressure barely dents its volume.Answer: The fractional change in volume is \(\displaystyle \dfrac{\Delta V}{V} \approx 2.7 \times 10^{-5}\) (a decrease of about $\displaystyle 27$ millionths of the original volume).

Under all-round (hydraulic) pressure a solid changes its volume, not its shape — so the elastic constant you need here is the bulk modulus \(\displaystyle B\), not Young's modulus \(\displaystyle Y\).The bulk modulus is defined as hydraulic stress divided by volume strain:\[B = -\,\frac{p}{\Delta V / V} \]where \(\displaystyle p\) is the hydraulic pressure applied to the body, \(\displaystyle V\) is its original volume, and \(\displaystyle \Delta V\) is the change in volume. The minus sign is part of the definition: pressing inward (\(\displaystyle p\) positive) always shrinks the body (\(\displaystyle \Delta V\) negative), so \(\displaystyle B\) itself comes out positive.What we are given, and what we look up: An aside on the step people get wrong: copper's Young's modulus is about \(\displaystyle 1.1 \times 10^{11}\ \text{Pa}\) and its shear modulus is \(\displaystyle 42 \times 10^{9}\ \text{Pa}\). Neither belongs here. Young's modulus governs stretching along one direction, the shear modulus governs a change of shape at constant volume; only \(\displaystyle B\) governs squeezing from every side at once.Original volume of the cube:\[V = a^{3} = (0.10\ \text{m})^{3} = 1.0 \times 10^{-3}\ \text{m}^{3} = 1000\ \text{cm}^{3} \]Rearranging the definition for the size of the volume change,\[|\Delta V| = \frac{p\,V}{B} \]Substituting:\[|\Delta V| = \frac{(7.0 \times 10^{6}\ \text{Pa}) \times (1.0 \times 10^{-3}\ \text{m}^{3})}{1.4 \times 10^{11}\ \text{Pa}} = \frac{7.0 \times 10^{3}\ \text{Pa}\cdot\text{m}^{3}}{1.4 \times 10^{11}\ \text{Pa}} = 5.0 \times 10^{-8}\ \text{m}^{3} \]The units behave: \(\displaystyle \text{Pa} \cdot \text{m}^{3} / \text{Pa} = \text{m}^{3}\), a volume, as it must be.Converting to the units the question is phrased in, with \(\displaystyle 1\ \text{m}^{3} = 10^{6}\ \text{cm}^{3}\):\[|\Delta V| = 5.0 \times 10^{-8}\ \text{m}^{3} \times \frac{10^{6}\ \text{cm}^{3}}{1\ \text{m}^{3}} = 5.0 \times 10^{-2}\ \text{cm}^{3} \]The sign matters as much as the size: \(\displaystyle \Delta V\) is negative, so this is a contraction of the cube, not an expansion. The volume strain is\[\frac{|\Delta V|}{V} = \frac{5.0 \times 10^{-8}\ \text{m}^{3}}{1.0 \times 10^{-3}\ \text{m}^{3}} = 5.0 \times 10^{-5} \]that is, about \(\displaystyle 0.005\%\) — a reassuringly tiny squeeze for a metal under $\displaystyle 70$ atmospheres.On significant figures: the pressure \(\displaystyle 7.0 \times 10^{6}\ \text{Pa}\) carries two, and the tabulated \(\displaystyle B = 1.4 \times 10^{11}\ \text{Pa}\) carries two, so the result is quoted to two significant figures: \(\displaystyle 5.0 \times 10^{-2}\ \text{cm}^{3}\), i.e. \(\displaystyle 0.050\ \text{cm}^{3}\). No rounding was done before this final step.A note on the answer printed at the back of the book. The key gives \(\displaystyle 0.058\ \text{cm}^{3}\). That figure does not follow from the data in this chapter. Work backwards from it and the bulk modulus it implies is\[B = \frac{p\,V}{|\Delta V|} = \frac{(7.0 \times 10^{6}\ \text{Pa})(1.0 \times 10^{-3}\ \text{m}^{3})}{5.8 \times 10^{-8}\ \text{m}^{3}} \approx 1.2 \times 10^{11}\ \text{Pa} \]but the book's own table of bulk moduli lists copper at \(\displaystyle 140\ \text{GPa} = 1.4 \times 10^{11}\ \text{Pa}\), which is also the accepted handbook value for copper. The printed \(\displaystyle 0.058\ \text{cm}^{3}\) is therefore inconsistent with the table the question expects you to use, and \(\displaystyle 0.050\ \text{cm}^{3}\) is the value that follows from it. If your teacher wants the printed key reproduced, show this line of reasoning alongside it.Answer: The copper cube contracts by \(\displaystyle |\Delta V| = 5.0 \times 10^{-2}\ \text{cm}^{3} = 0.050\ \text{cm}^{3}\) (a decrease in volume; volume strain \(\displaystyle 5.0 \times 10^{-5}\)), using \(\displaystyle B_{\text{copper}} = 1.4 \times 10^{11}\ \text{Pa}\), quoted to two significant figures.

Water resists a squeeze so strongly that even a tiny percentage compression needs a pressure change measured in millions of pascals — this is the bulk modulus at work.When a fluid is squeezed uniformly, the bulk modulus \(\displaystyle B \) tells you how much pressure change \(\displaystyle \Delta P \) is needed to produce a given fractional change in volume \(\displaystyle \dfrac{\Delta V}{V} \):\[B = \frac{\Delta P}{\left(\dfrac{\Delta V}{V}\right)} \]Here \(\displaystyle \Delta P \) is the increase in pressure applied, \(\displaystyle V \) is the original volume, and \(\displaystyle \Delta V \) is the resulting decrease in volume — the ratio \(\displaystyle \Delta V/V \) is a pure fraction (no units), so the size of the sample (a litre, a bathtub, an ocean) never enters the answer. This is the detail that trips people up: there is no need to convert the litre into \(\displaystyle \text{m}^3 \) at all, because \(\displaystyle V \) cancels out of the calculation.The fractional compression. "Compress it by $\displaystyle 0.10$%" means\[\frac{\Delta V}{V} = \frac{0.10}{100} = 1.0 \times 10^{-3} \]The bulk modulus of water. Water is one of the standard materials tabulated in this chapter, with\[B_{\text{water}} = 2.2 \times 10^{9}\ \text{Pa} \]This huge value (compared to, say, a gas) is exactly why liquids are called nearly incompressible: it takes an enormous pressure to shrink water's volume by even a fraction of a percent.Solving for \(\displaystyle \Delta P \). Rearranging the bulk-modulus relation,\[\Delta P = B_{\text{water}} \times \frac{\Delta V}{V} \]Substituting the values:\[\Delta P = \left(2.2 \times 10^{9}\ \text{Pa}\right) \times \left(1.0 \times 10^{-3}\right) \]\[\Delta P = 2.2 \times 10^{6}\ \text{Pa} \]Significant figures. Both inputs — the $\displaystyle 0.10$% and the bulk modulus $\displaystyle 2.2$ × $\displaystyle 10$⁹ Pa — carry two significant figures, so the answer is rounded once, at the end, to two significant figures: \(\displaystyle 2.2 \times 10^{6}\ \text{Pa} \).As a sanity check, since atmospheric pressure is about \(\displaystyle 1.0 \times 10^{5}\ \text{Pa} \), this pressure increase is roughly $\displaystyle 22$ times atmospheric pressure — a large but physically reasonable pressure to compress water by just one part in a thousand.Answer: The pressure must be increased by about \(\displaystyle 2.2 \times 10^{6}\ \text{Pa} \) (≈ $\displaystyle 22$ atmospheres).