NCERT Solutions · Class 11 Physics Units and Measurement

Convert every length to the same unit first — mixing cm and m is the trap here.The sheet's dimensions are given as \(\displaystyle l = 4.234\ \text{m} \), \(\displaystyle b = 1.005\ \text{m} \), \(\displaystyle t = 2.01\ \text{cm} \).Since \(\displaystyle 1\ \text{cm} = 10^{-2}\ \text{m}\), \[t = 2.01\ \text{cm} = 2.01\times10^{-2}\ \text{m} = 0.0201\ \text{m}. \]Surface area of a rectangular (cuboidal) sheet.A rectangular sheet of metal, thought of as a thin box of length \(\displaystyle l\), breadth \(\displaystyle b\) and thickness \(\displaystyle t\), has total surface area \[A = 2(lb + bt + tl). \]Substituting the values in metres: \[lb = 4.234 \times 1.005 = 4.25517\ \text{m}^2 \] \[bt = 1.005 \times 0.0201 = 0.0202005\ \text{m}^2 \] \[tl = 0.0201 \times 4.234 = 0.0851034\ \text{m}^2 \]Adding these, \[lb+bt+tl = 4.25517 + 0.0202005 + 0.0851034 = 4.3604739\ \text{m}^2 \]so \[A = 2 \times 4.3604739 = 8.7209478\ \text{m}^2. \]Volume of the sheet.Volume of a rectangular block is \[V = l \times b \times t = (lb)\times t = 4.25517 \times 0.0201 = 0.085528917\ \text{m}^3. \]Now apply significant figures — do this only at the very end, not after each line.Look at how precisely each measurement is known: In a product (or a mix of sums and products, as here), the answer cannot be trusted to more significant figures than the least precisely known quantity that entered it. Thickness is known to only $\displaystyle 3$ significant figures, so both the area and the volume — since both are built from products involving \(\displaystyle t\) — must be rounded to $\displaystyle 3$ significant figures. Carrying the extra digits (the ...9478, ...528917) through the calculation and only cutting them off now avoids compounding rounding error at each step.\[A = 8.7209478\ \text{m}^2 \;\longrightarrow\; 8.72\ \text{m}^2 \] \[V = 0.085528917\ \text{m}^3 = 8.5528917\times10^{-2}\ \text{m}^3 \;\longrightarrow\; 8.55\times10^{-2}\ \text{m}^3 \]**Answer: Area \(\displaystyle \approx 8.72\ \text{m}^2 \), Volume \(\displaystyle \approx 8.55\times10^{-2}\ \text{m}^3 \) (each rounded to $\displaystyle 3$ significant figures, matching the $\displaystyle 3$-significant-figure precision of the thickness measurement).

Adding measurements of very different precision does not make the result more precise than the least precise one — it can only be as good as the sourest number in the sum.Step $\displaystyle 1$ — Put every mass in the same unit.The grocer's balance reads to the nearest \(\displaystyle 0.01\text{ kg}\) (that is what "$\displaystyle 2.30$ kg" tells you — two decimal places, three significant figures). The gold pieces are weighed on a much finer balance, to the nearest \(\displaystyle 0.01\text{ g}\).\[20.15\text{ g} = 0.02015\text{ kg}, \qquad 20.17\text{ g} = 0.02017\text{ kg} \]Step $\displaystyle 2$ — Add the raw numbers first, then apply the rule for addition.The rule: when you add measured quantities, the sum can be trusted only to the number of decimal places of the least precisely known term — you don't gain precision by adding a fuzzy number to two sharp ones.\[2.30\text{ kg} + 0.02015\text{ kg} + 0.02017\text{ kg} = 2.34032\text{ kg} \]Here \(\displaystyle 2.30\text{ kg}\) is uncertain already in the second decimal place (it could hide anything from \(\displaystyle 2.295\) to \(\displaystyle 2.305\text{ kg}\)), so nothing past two decimal places in the sum means anything.\[\boxed{2.34032\text{ kg} \rightarrow 2.34\text{ kg}} \](a) Total mass = \(\displaystyle 2.34\text{ kg}\).Step $\displaystyle 3$ — The difference of the two gold pieces.Both gold-piece masses were read to the same precision, \(\displaystyle 0.01\text{ g}\) (four significant figures each), so subtracting them is safe to that same precision — no rounding surprises here since the two terms are equally precise.\[20.17\text{ g} - 20.15\text{ g} = 0.02\text{ g} \]This aside is the one place people trip up: \(\displaystyle 0.02\text{ g}\) looks like it has only one significant figure, and it does — subtraction can shrink the number of significant figures even when both inputs had four. That is not a mistake; it is exactly what significant-figure bookkeeping is supposed to show. The mass difference is genuinely known only to \(\displaystyle \pm 0.01\text{ g}\) or so, because that is the resolution of the instrument that produced the two starting numbers.(b) Difference in masses = \(\displaystyle 0.02\text{ g}\).Answer: (a) Total mass \(\displaystyle = 2.34\text{ kg}\). (b) Difference in masses of the two pieces \(\displaystyle = 0.02\text{ g}\).

A physical formula cannot mix quantities that don't have matching dimensions — the "$\displaystyle 1$" here is a pure number, so whatever is subtracted from it must be a pure number too.The relation the boy is trying to recall is Einstein's expression for how mass appears to increase with speed in special relativity:\[m = \dfrac{m_0}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}} \]where He wrote it as\[m = \dfrac{m_0}{\sqrt{1-v^{2}}} \]which has forgotten where \(\displaystyle c\) belongs. To find the right spot, check dimensions rather than guessing.Step $\displaystyle 1$ — Look at what is being subtracted from 1. The number \(\displaystyle 1\) is dimensionless (it is a pure number, no units attached). In any valid equation you can only add or subtract quantities that have the same dimensions. So whatever sits next to the \(\displaystyle 1\) inside the bracket must also be dimensionless.Step $\displaystyle 2$ — Check the dimensions of \(\displaystyle v^2\) alone. Speed \(\displaystyle v\) has dimensions \(\displaystyle [\mathrm{LT}^{-1}]\), so \(\displaystyle v^2\) has dimensions \(\displaystyle [\mathrm{L^2T^{-2}}]\). That is not dimensionless — so \(\displaystyle 1 - v^2\) is dimensionally meaningless, exactly the mistake the boy's version contains.Step $\displaystyle 3$ — Find a quantity with the same dimensions as \(\displaystyle v\) to divide it by. The speed of light \(\displaystyle c\) also has dimensions \(\displaystyle [\mathrm{LT}^{-1}]\), the same as \(\displaystyle v\). Forming the ratio\[\frac{v}{c} \]cancels the dimensions completely: \(\displaystyle \left[\dfrac{\mathrm{LT^{-1}}}{\mathrm{LT^{-1}}}\right] = [\,\text{dimensionless}\,]\).Step $\displaystyle 4$ — Square it and place it where \(\displaystyle v^2\) was. Since \(\displaystyle \dfrac{v}{c}\) is dimensionless, so is \(\displaystyle \left(\dfrac{v}{c}\right)^2 = \dfrac{v^2}{c^2}\). This is the term that belongs under the square root, replacing the bare \(\displaystyle v^2\):\[m = \dfrac{m_0}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}} \]A quick check on the physics confirms this is sensible: when \(\displaystyle v \ll c\) (everyday speeds), \(\displaystyle v^2/c^2 \approx 0\), the bracket becomes \(\displaystyle \approx 1\), and \(\displaystyle m \approx m_0\) — ordinary Newtonian mechanics, where mass doesn't depend on speed, is recovered. As \(\displaystyle v \to c\), the denominator \(\displaystyle \to 0\) and \(\displaystyle m \to \infty\), which is the relativistic statement that no massive particle can be accelerated to the speed of light. Neither of these sensible limits would hold if \(\displaystyle c\) were placed anywhere else (say, multiplying \(\displaystyle m_0\), or added outside the root) — only inserting it as \(\displaystyle v^2/c^2\) keeps the equation both dimensionally consistent and physically correct.Answer: The missing \(\displaystyle c\) goes in the denominator of the \(\displaystyle v^2\) term, as \(\displaystyle v^2/c^2\): \[m=\dfrac{m_{0}}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}} \]

Model the hydrogen atom as a tiny sphere, find the volume of one atom, then scale up by Avogadro's number for a mole.The size given, \(\displaystyle 0.5\ \text{Å} \), is the atomic radius \(\displaystyle r\). Converting to metres using \(\displaystyle 1\ \text{Å} = 10^{-10}\ \text{m}\):\[r = 0.5\ \text{Å} = 0.5 \times 10^{-10}\ \text{m} = 5 \times 10^{-11}\ \text{m} \]Volume of one atom. Treating the atom as a sphere, its volume is \(\displaystyle V_{\text{atom}} = \dfrac{4}{3}\pi r^{3}\), where \(\displaystyle r\) is the radius found above.\[r^{3} = \left(5\times10^{-11}\ \text{m}\right)^{3} = 125\times10^{-33}\ \text{m}^{3} = 1.25\times10^{-31}\ \text{m}^{3} \]\[V_{\text{atom}} = \frac{4}{3}\pi \left(1.25\times10^{-31}\ \text{m}^{3}\right) \approx 5.24\times10^{-31}\ \text{m}^{3} \]Scale up to one mole. A mole of hydrogen atoms contains Avogadro's number of atoms, \(\displaystyle N_A = 6.02\times10^{23}\ \text{mol}^{-1}\). The total atomic volume of the mole is the volume of one atom multiplied by the number of atoms in a mole — this is a straight multiplication, not an addition, because every atom contributes its own separate volume:\[V_{\text{mole}} = N_A \, V_{\text{atom}} = \left(6.02\times10^{23}\ \text{mol}^{-1}\right)\left(5.24\times10^{-31}\ \text{m}^{3}\right) \]\[V_{\text{mole}} \approx 3.15\times10^{-7}\ \text{m}^{3} \]Rounding. The only measured quantity here is the atomic size, \(\displaystyle 0.5\) Å — a value carrying just one significant figure. Avogadro's number and \(\displaystyle \pi\) are known to far more figures than that and cannot manufacture precision the data does not have, so the answer must be quoted to one significant figure, not the three-figure number the raw arithmetic happens to produce.\[V_{\text{mole}} \approx 3\times10^{-7}\ \text{m}^{3} \]Answer: The total atomic volume of one mole of hydrogen atoms is about \(\displaystyle 3\times10^{-7}\ \text{m}^{3}\).

A mole's molar volume (how much space the gas fills) and its molecular volume (how much space the molecules themselves would take up if packed together) are two completely different numbers — this ratio measures the size of that gap.Step $\displaystyle 1$ — Molar volume in SI unitsAt STP, one mole of an ideal gas occupies the molar volume \[V_m = 22.4\ \text{L} \] Since \(\displaystyle 1\ \text{L} = 10^{-3}\ \text{m}^3 \), \[V_m = 22.4 \times 10^{-3}\ \text{m}^3 = 2.24 \times 10^{-2}\ \text{m}^3 \]Step $\displaystyle 2$ — Volume actually occupied by one hydrogen moleculeThe problem gives the size of a hydrogen molecule as about $\displaystyle 1$ Å — that is a diameter, not a radius, so it must be halved before it goes into the sphere formula: \[d = 1\ \text{Å} = 1\times10^{-10}\ \text{m} \quad\Rightarrow\quad r = \frac{d}{2} = 5\times10^{-11}\ \text{m} \]Treating the molecule as a sphere, its volume is \[v = \frac{4}{3}\pi r^3 \] \[r^3 = (5\times10^{-11}\ \text{m})^3 = 1.25\times10^{-31}\ \text{m}^3 \] \[v = \frac{4}{3}\pi (1.25\times10^{-31}\ \text{m}^3) = 5.24\times10^{-31}\ \text{m}^3 \]Step $\displaystyle 3$ — Volume occupied by a mole of these moleculesAvogadro's number, \(\displaystyle N_A = 6.022\times10^{23}\ \text{mol}^{-1} \), is the number of molecules in one mole. Multiplying the volume of one molecule by \(\displaystyle N_A \) gives the total space the molecules themselves would fill if packed with no gaps between them — call this the molecular volume \(\displaystyle V_{\text{mol}} \): \[V_{\text{mol}} = N_A \times v = (6.022\times10^{23}\ \text{mol}^{-1})(5.24\times10^{-31}\ \text{m}^3) = 3.15\times10^{-7}\ \text{m}^3 \]Step $\displaystyle 4$ — The ratio\[\frac{V_m}{V_{\text{mol}}} = \frac{2.24\times10^{-2}\ \text{m}^3}{3.15\times10^{-7}\ \text{m}^3} = 7.1\times10^{4} \]The size given for the molecule, "about $\displaystyle 1$ Å," carries only one significant figure, and that figure gets cubed on the way to a volume, so quoting more than one significant figure in the final ratio would be false precision. Rounding once, at the end: \[\frac{V_m}{V_{\text{mol}}} \approx 7\times10^{4} \]Why the ratio is so largeA ratio of \(\displaystyle 7\times10^{4}\) means that if every hydrogen molecule in a mole of gas were packed together with no space between them, they would take up only about \(\displaystyle \dfrac{1}{70000} \) of the volume the gas actually occupies at STP. Essentially all the rest is empty space between molecules that sit, on average, far apart compared to their own size. This is exactly the assumption behind the ideal gas law: molecules are treated as point particles with negligible volume compared to the container, and that approximation works well only because a real gas truly is this empty. It is also why gases compress so much more easily than liquids or solids — there is a huge amount of empty space left to squeeze out before the molecules themselves get in each other's way.Answer: The ratio of molar volume to the molecular volume of a mole of hydrogen is about \(\displaystyle 7\times10^{4}\) (precisely \(\displaystyle 7.1\times10^{4}\) from the given numbers); it is so large because gas molecules occupy only a tiny fraction of the space between them, so almost all of a gas's volume is empty space.

What you perceive as "speed" here is really an angular speed, and angular speed falls off with distance even when the true relative speed does not.Take the ground to be the fixed frame, and let the train move with speed \(\displaystyle v \) relative to the ground. Every object fixed to the ground — a nearby tree, a distant hilltop, the Moon, a star — has zero velocity in the ground frame. By the law of relative velocity, \[\vec{v}_{\text{object, train}} \;=\; \vec{v}_{\text{object, ground}} - \vec{v}_{\text{train, ground}} \;=\; -\vec{v} \] so every one of these objects, near or far, appears to move backward relative to you with exactly the same speed \(\displaystyle v\). The linear (relative) speed is not what distinguishes a tree from the Moon. What distinguishes them is how fast the direction to that object — the line of sight — swings around you, and that depends on distance.Angular speed of the line of sight. Let \(\displaystyle d\) be the perpendicular distance of the object from the track (this stays constant as you travel). Measure your position \(\displaystyle x\) along the track from the point directly opposite the object, so \(\displaystyle x=0\) at the instant the object is abeam. The line of sight makes an angle \(\displaystyle \theta\) with the perpendicular, where \(\displaystyle \tan\theta = x/d\). Differentiating with respect to time, \[\sec^2\theta \,\frac{d\theta}{dt} = \frac{1}{d}\frac{dx}{dt} = \frac{v}{d} \quad\Longrightarrow\quad \omega \equiv \frac{d\theta}{dt} = \frac{v}{d}\cos^2\theta . \] This is largest, \(\displaystyle \omega_{\max} = v/d \), exactly when the object is abeam — which is also when you notice it most. So the angular speed at which an object sweeps across your view is inversely proportional to its distance \(\displaystyle d\), even though every object shares the same linear speed \(\displaystyle v\). This is the everyday version of parallax, the same idea this chapter uses to measure the distance to the Moon and to stars.Putting numbers to it (take a fast train, \(\displaystyle v = 25\ \text{m/s} \approx 90\ \text{km/h}\), just to see the scale of the effect — the argument does not depend on this exact value): \[\omega = \frac{v}{d} = \frac{25\ \text{m/s}}{10\ \text{m}} = 2.5\ \text{rad/s} \approx 140^\circ/\text{s}. \] It sweeps through your field of view in a fraction of a second — this is why nearby objects seem to whip past. \[\omega = \frac{25}{5\times10^{3}} = 5\times10^{-3}\ \text{rad/s} \approx 0.3^\circ/\text{s}, \] already $\displaystyle 500$ times slower — barely perceptible over a glance. \[\omega = \frac{25}{3.84\times10^{8}} \approx 6.5\times10^{-8}\ \text{rad/s}. \] Even watched for a full minute, it shifts by only \(\displaystyle \omega\,\Delta t \approx 4\times10^{-6}\ \text{rad}\), under one arc-second — far below the unaided eye's resolution of about one arc-minute (\(\displaystyle \sim 3\times10^{-4}\ \text{rad}\)). It cannot be seen to move at all. \[\omega \sim \frac{25}{9.5\times10^{15}} \approx 3\times10^{-15}\ \text{rad/s}, \] utterly undetectable — real stellar parallax needs a baseline as large as the Earth's orbit (\(\displaystyle \sim 3\times10^{11}\ \text{m}\), taken six months apart), not the width of a train window.Where the common mix-up happens: it is tempting to think the far objects are "really" moving with you or are somehow exempt from the relative-velocity rule. They are not — they have the identical relative velocity \(\displaystyle v\) as the trees do. The difference is entirely in the angular rate \(\displaystyle \omega = v/d\) at which your line of sight to them turns, and that rate collapses as \(\displaystyle d\) grows, so distant objects fall behind your view too slowly to notice and so seem to hang motionless, or "follow" you, exactly as the Moon appears to on any night walk.Answer: Every object outside the train, near or far, has the same relative velocity \(\displaystyle v\) (opposite to the train's motion), but what you see is the angular speed \(\displaystyle \omega = v/d\) at which your line of sight to it turns. For nearby trees and houses \(\displaystyle d\) is small, so \(\displaystyle \omega\) is large and they appear to rush backward; for hilltops, the Moon, and stars \(\displaystyle d\) is enormous, so \(\displaystyle \omega\) is negligibly small (well below the eye's resolving power) and they appear stationary. This is the same parallax effect used elsewhere in this chapter to measure large astronomical distances.

A hot plasma has no fixed shape, so intuition says "gas-like" and low density — but density depends on how tightly packed the matter is, not on its phase, and gravity can pack a gas as tightly as a solid.Before calculating, the natural guess is that since the Sun is entirely ionized gas (a plasma) at every point — never solid or liquid — its mass density should fall in the range typical of gases, roughly \(\displaystyle 1\) kg/m³ or less (compare ordinary air at about \(\displaystyle 1.2\) kg/m³). Solids and liquids, by contrast, typically have densities of order \(\displaystyle 10^{3}\) to \(\displaystyle 10^{4}\) kg/m³, because their atoms are packed close together by strong interatomic bonds. A plasma has no such bonds holding its particles together, so the guess is that it should be far more dilute.Now check this guess with the data. Density is mass per unit volume, \[\rho = \frac{M}{V} \] where \(\displaystyle M\) is the mass enclosed and \(\displaystyle V\) is the volume it occupies.Treating the Sun as a sphere of radius \(\displaystyle R\), its volume is \[V = \frac{4}{3}\pi R^{3} \]Substitute \(\displaystyle R = 7.0 \times 10^{8}\) m: \[R^{3} = (7.0\times10^{8}\,\text{m})^{3} = 343 \times 10^{24}\,\text{m}^3 = 3.43\times10^{26}\,\text{m}^3 \]\[V = \frac{4}{3}\pi \times 3.43\times10^{26}\,\text{m}^3 = 4.189 \times 3.43\times10^{26}\,\text{m}^3 \approx 1.44\times10^{27}\,\text{m}^3 \]Now divide the given mass by this volume. Using \(\displaystyle M = 2.0\times10^{30}\) kg: \[\rho = \frac{M}{V} = \frac{2.0\times10^{30}\,\text{kg}}{1.44\times10^{27}\,\text{m}^3} \]\[\rho = \frac{2.0}{1.44}\times10^{30-27}\,\text{kg/m}^3 = 1.39\times10^{3}\,\text{kg/m}^3 \]Both pieces of given data (\(\displaystyle 2.0\times10^{30}\) kg and \(\displaystyle 7.0\times10^{8}\) m) carry only two significant figures, so the final density must also be rounded to two significant figures — only at this last step, not in the intermediate volume or ratio: \[\rho \approx 1.4\times10^{3}\,\text{kg/m}^3 \]This is the surprise: \(\displaystyle 1.4\times10^{3}\) kg/m³ is not in the range of ordinary gases at all — it is comparable to (in fact somewhat higher than) the density of water (\(\displaystyle 1.0\times10^{3}\) kg/m³) and squarely in the range of solids and liquids. So the initial guess based on "it's all plasma, so it must be gas-like" turns out to be wrong.The physical reason is that phase (solid, liquid, gas, plasma) is a statement about bonding and structure, not about how densely matter is packed. In an ordinary gas, nothing holds the molecules close together, so they spread out. Inside the Sun, however, the enormous self-gravitation of a mass of \(\displaystyle 2.0\times10^{30}\) kg compresses the plasma inward with tremendous force, especially near the core, squeezing the ionized matter to a density comparable to solids and liquids even though no solid or liquid bonding exists anywhere in it. Gravity, not molecular bonding, is doing the packing.Answer: The Sun's average mass density works out to \(\displaystyle \rho \approx 1.4\times10^{3}\) kg/m³ — comparable to the density range of solids and liquids (denser than water), not to ordinary gases, even though the Sun is entirely plasma. This is because gravitational self-compression, not interatomic bonding, packs the matter this tightly.