NCERT Solutions · Class 11 Physics Units and Measurement

Each part is a unit-conversion, not a new calculation — get the conversion factor right and the number falls out.(a) Volume of a cube of side $\displaystyle 1$ cm, in \(\displaystyle m^3\)Volume of a cube of side \(\displaystyle a\) is \(\displaystyle V = a^3\).Convert the side to metres first, since the answer is wanted in \(\displaystyle m^3\): \[a = 1\ \text{cm} = 1\times10^{-2}\ \text{m} \]A common slip here is converting the length factor of \(\displaystyle 10^{-2}\) but forgetting to cube it. Since volume scales as the cube of length, \[V = a^3 = \left(10^{-2}\ \text{m}\right)^3 = 10^{-6}\ \text{m}^3 \](b) Surface area of a solid cylinder of radius $\displaystyle 2.0$ cm and height $\displaystyle 10.0$ cm, in \(\displaystyle (mm)^2\)A solid cylinder's total surface area includes the curved side plus both flat circular ends: \[S = \underbrace{2\pi r h}_{\text{curved surface}} + \underbrace{2\pi r^2}_{\text{two ends}} = 2\pi r(r+h) \] where \(\displaystyle r\) is the radius and \(\displaystyle h\) is the height.Substituting \(\displaystyle r = 2.0\ \text{cm}\), \(\displaystyle h = 10.0\ \text{cm}\): \[S = 2\pi (2.0\ \text{cm})(2.0\ \text{cm}+10.0\ \text{cm}) = 2\pi(2.0\ \text{cm})(12.0\ \text{cm}) = 48\pi\ \text{cm}^2 \] \[S = 48 \times 3.142\ \text{cm}^2 \approx 150.8\ \text{cm}^2 \]The radius is given to only $\displaystyle 2$ significant figures ($\displaystyle 2.0$ cm), so the product cannot be trusted beyond $\displaystyle 2$ significant figures: \(\displaystyle S \approx 1.5\times10^{2}\ \text{cm}^2\).Now convert to \(\displaystyle mm^2\). This is the step people get wrong: since \(\displaystyle 1\ \text{cm} = 10\ \text{mm}\), the area factor is \(\displaystyle 10^2 = 100\), not $\displaystyle 10$: \[1\ \text{cm}^2 = (10\ \text{mm})^2 = 100\ \text{mm}^2 \] \[S = 1.5\times10^{2}\ \text{cm}^2 \times 100\ \frac{\text{mm}^2}{\text{cm}^2} = 1.5\times10^{4}\ \text{mm}^2 \](c) Distance covered in $\displaystyle 1$ s by a vehicle moving at $\displaystyle 18$ km \(\displaystyle h^{-1}\)First convert the speed to SI units (\(\displaystyle m\,s^{-1}\)), using \(\displaystyle 1\ \text{km} = 1000\ \text{m}\) and \(\displaystyle 1\ \text{h} = 3600\ \text{s}\): \[1\ \text{km}\,h^{-1} = \frac{1000\ \text{m}}{3600\ \text{s}} = \frac{5}{18}\ \text{m}\,s^{-1} \] \[v = 18\ \text{km}\,h^{-1} = 18 \times \frac{5}{18}\ \text{m}\,s^{-1} = 5\ \text{m}\,s^{-1} \]Distance in time \(\displaystyle t\) at constant speed is \(\displaystyle d = v t\): \[d = \left(5\ \text{m}\,s^{-1}\right)(1\ \text{s}) = 5\ \text{m} \](d) Density of lead, given its relative density is $\displaystyle 11.3$Relative density (specific gravity) is defined as \[\text{Relative density} = \frac{\text{density of substance}}{\text{density of water}} \] and, being a ratio of two densities, it is a pure number — it carries no unit. That is why "$\displaystyle 11.3$" alone cannot be the answer to either blank; it must be multiplied by the density of water to get an actual density.Taking the density of water as \(\displaystyle 1\ \text{g}\,cm^{-3}\) (its value in CGS units): \[\rho_{\text{lead}} = 11.3 \times 1\ \text{g}\,cm^{-3} = 11.3\ \text{g}\,cm^{-3} \]To convert to \(\displaystyle kg\,m^{-3}\), use \(\displaystyle 1\ \text{g} = 10^{-3}\ \text{kg}\) and \(\displaystyle 1\ \text{cm}^3 = 10^{-6}\ \text{m}^3\): \[1\ \text{g}\,cm^{-3} = \frac{10^{-3}\ \text{kg}}{10^{-6}\ \text{m}^3} = 10^{3}\ \text{kg}\,m^{-3} \] \[\rho_{\text{lead}} = 11.3 \times 10^{3}\ \text{kg}\,m^{-3} = 1.13\times10^{4}\ \text{kg}\,m^{-3} \]Answer: (a) \(\displaystyle 10^{-6}\ m^3\) (b) \(\displaystyle 1.5\times10^{4}\ (mm)^2\) (c) $\displaystyle 5$ m (d) $\displaystyle 11.3$ g \(\displaystyle cm^{-3}\), i.e. \(\displaystyle 1.13\times10^{4}\ kg\,m^{-3}\)

Converting a measured quantity to another set of units means multiplying its number by the correct powers of the ratios between the units — the physical quantity itself never changes, only how many "boxes" of a given size are used to describe it.(a) $\displaystyle 1$ kg m² s⁻² = .... g cm² s⁻²Convert the mass unit and the length unit separately, then multiply.Mass: \(\displaystyle 1\,\text{kg} = 10^{3}\,\text{g} \)Length, squared because the unit is \(\displaystyle \text{m}^2 \): \(\displaystyle 1\,\text{m} = 10^{2}\,\text{cm} \), so \(\displaystyle 1\,\text{m}^2 = (10^{2})^{2}\,\text{cm}^2 = 10^{4}\,\text{cm}^2 \)Time is already seconds on both sides, so it carries through unchanged.\[1\,\text{kg m}^2\text{s}^{-2} = (10^{3}\,\text{g}) \times (10^{4}\,\text{cm}^2) \times \text{s}^{-2} = 10^{7}\,\text{g cm}^2\text{s}^{-2} \]This is the joule-to-erg conversion in disguise: \(\displaystyle 1\,\text{kg m}^2\text{s}^{-2} \) is one joule and \(\displaystyle 1\,\text{g cm}^2\text{s}^{-2} \) is one erg, so this line also shows \(\displaystyle 1\,\text{J} = 10^{7}\,\text{erg} \).(b) $\displaystyle 1$ m = ..... lyA light year is a unit of distance, not of time — it is how far light travels in one year. Use \(\displaystyle d = ct \), with \(\displaystyle c = 3.00\times10^{8}\,\text{m s}^{-1} \) and one year taken as $\displaystyle 365.25$ days.\[1\,\text{year} = 365.25 \times 24 \times 3600\,\text{s} = 3.156\times10^{7}\,\text{s} \]\[1\,\text{ly} = c \times (1\,\text{year}) = (3.00\times10^{8}\,\text{m s}^{-1})(3.156\times10^{7}\,\text{s}) = 9.46\times10^{15}\,\text{m} \]Inverting the ratio gives $\displaystyle 1$ metre in light years:\[1\,\text{m} = \frac{1}{9.46\times10^{15}}\,\text{ly} = 1.06\times10^{-16}\,\text{ly} \](c) $\displaystyle 3.0$ m s⁻² = .... km h⁻²Convert length and time separately. Because the unit is \(\displaystyle \text{s}^{-2} \), the time conversion factor must be squared, not applied once — this is the step most people miss.Length: \(\displaystyle 1\,\text{m} = 10^{-3}\,\text{km} \)Time: \(\displaystyle 1\,\text{s} = \dfrac{1}{3600}\,\text{h} \), so \(\displaystyle 1\,\text{s}^{-1} = 3600\,\text{h}^{-1} \), and squaring, \(\displaystyle 1\,\text{s}^{-2} = (3600)^{2}\,\text{h}^{-2} = 1.296\times10^{7}\,\text{h}^{-2} \)\[3.0\,\text{m s}^{-2} = 3.0 \times (10^{-3}\,\text{km}) \times (1.296\times10^{7}\,\text{h}^{-2}) = 3.888\times10^{4}\,\text{km h}^{-2} \]The given value, $\displaystyle 3.0$, carries only two significant figures, so the final answer is rounded to two:\[3.0\,\text{m s}^{-2} = 3.9\times10^{4}\,\text{km h}^{-2} \](d) G = $\displaystyle 6.67$ × $\displaystyle 10$⁻¹¹ N m² (kg)⁻² = .... (cm)³ s⁻² g⁻¹First rewrite the newton in base units, using Newton's second law \(\displaystyle F = ma \), so \(\displaystyle 1\,\text{N} = 1\,\text{kg m s}^{-2} \):\[1\,\text{N m}^2\text{kg}^{-2} = (1\,\text{kg m s}^{-2}) \times \text{m}^2 \times \text{kg}^{-2} = 1\,\text{m}^3\text{s}^{-2}\text{kg}^{-1} \]Now convert metres to centimetres and kilograms to grams. Since \(\displaystyle 1\,\text{m} = 10^{2}\,\text{cm} \), \(\displaystyle 1\,\text{m}^3 = 10^{6}\,\text{cm}^3 \). Since \(\displaystyle 1\,\text{kg} = 10^{3}\,\text{g} \), inverting gives \(\displaystyle 1\,\text{kg}^{-1} = 10^{-3}\,\text{g}^{-1} \).\[1\,\text{N m}^2\text{kg}^{-2} = (10^{6}\,\text{cm}^3) \times \text{s}^{-2} \times (10^{-3}\,\text{g}^{-1}) = 10^{3}\,\text{cm}^3\text{s}^{-2}\text{g}^{-1} \]Apply this factor to \(\displaystyle G \):\[G = 6.67\times10^{-11} \times 10^{3}\,\text{cm}^3\text{s}^{-2}\text{g}^{-1} = 6.67\times10^{-8}\,\text{cm}^3\text{s}^{-2}\text{g}^{-1} \]The $\displaystyle 6.67$ already carries three significant figures and every conversion factor used is an exact power of ten, so the answer keeps three significant figures.Answer: (a) \(\displaystyle 10^{7}\) g cm² s⁻²; (b) \(\displaystyle 1.06\times10^{-16}\) ly; (c) \(\displaystyle 3.9\times10^{4}\) km h⁻²; (d) \(\displaystyle 6.67\times10^{-8}\) cm³ s⁻² g⁻¹

A physical quantity's size doesn't change when you change units — only the number and the unit trade off, so \(\displaystyle n_1u_1=n_2u_2\) must hold between any two unit systems. That single invariance is what lets you convert a quantity without knowing anything else about it, provided you know its dimensional formula.Step $\displaystyle 1$: Find the dimensional formula of energy.A calorie is a unit of energy, and energy equals work. By Newton's second law, force is mass times acceleration: \[[\text{Force}] = [M][LT^{-2}] = [MLT^{-2}] \] Work is force times displacement: \[[\text{Energy}] = [\text{Work}] = [MLT^{-2}][L] = [M^{1}L^{2}T^{-2}] \] So energy has dimensions \(\displaystyle a=1\) in mass, \(\displaystyle b=2\) in length, \(\displaystyle c=-2\) in time.Step $\displaystyle 2$: Write the general unit-conversion rule.If a quantity has dimensional formula \(\displaystyle M^{a}L^{b}T^{c}\), and it is measured as \(\displaystyle n_1\) in a unit system built from base units \(\displaystyle M_1, L_1, T_1\), and as \(\displaystyle n_2\) in another system built from \(\displaystyle M_2, L_2, T_2\), then because the quantity itself is unchanged, \[n_1\big[M_1^{a}L_1^{b}T_1^{c}\big]=n_2\big[M_2^{a}L_2^{b}T_2^{c}\big] \] so that \[n_2=n_1\left(\frac{M_1}{M_2}\right)^{a}\left(\frac{L_1}{L_2}\right)^{b}\left(\frac{T_1}{T_2}\right)^{c} \]Step $\displaystyle 3$: Identify the two systems.System $\displaystyle 1$ is the ordinary SI system, in which \(\displaystyle 1\text{ J}=1\text{ kg}\,\text{m}^2\text{s}^{-2}\), so \[M_1=1\text{ kg},\quad L_1=1\text{ m},\quad T_1=1\text{ s},\quad n_1=4.2 \] (since $\displaystyle 1$ calorie \(\displaystyle \approx 4.2\) J, i.e. \(\displaystyle 4.2\) times the SI energy unit).System $\displaystyle 2$ is the new system described in the problem: \[M_2=\alpha\text{ kg},\quad L_2=\beta\text{ m},\quad T_2=\gamma\text{ s} \]Step $\displaystyle 4$: Substitute into the conversion formula, using \(\displaystyle a=1,\ b=2,\ c=-2\). \[n_2=4.2\left(\frac{1\text{ kg}}{\alpha\text{ kg}}\right)^{1}\left(\frac{1\text{ m}}{\beta\text{ m}}\right)^{2}\left(\frac{1\text{ s}}{\gamma\text{ s}}\right)^{-2} \] The kg, m, s cancel in each ratio, leaving pure numbers: \[n_2=4.2\times\alpha^{-1}\times\beta^{-2}\times\gamma^{-2\times(-1)} \]This last step is the one place a sign slip creeps in: the time exponent in the dimensional formula is \(\displaystyle c=-2\), and the conversion factor is \(\displaystyle (T_1/T_2)^{c}=(1/\gamma)^{-2}\). Flipping a ratio to the power \(\displaystyle -2\) flips it back over, so \(\displaystyle (1/\gamma)^{-2}=\gamma^{2}\), not \(\displaystyle \gamma^{-2}\). Keeping that sign straight gives \[n_2=4.2\,\alpha^{-1}\beta^{-2}\gamma^{2} \]Step $\displaystyle 5$: State the result.So in the new system, where the units of mass, length and time are \(\displaystyle \alpha\) kg, \(\displaystyle \beta\) m and \(\displaystyle \gamma\) s respectively, \[1\text{ calorie}=4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}\ \text{(new units of energy)} \] which is exactly the magnitude the problem asks you to show. The coefficient \(\displaystyle 4.2\) is carried unchanged throughout — it is the given approximate conversion factor (two significant figures) and no arithmetic rounding is introduced by the unit change, since \(\displaystyle \alpha,\beta,\gamma\) are left as symbols.Answer: \(\displaystyle 1\text{ calorie}=4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}\) in the new system of units, obtained from energy's dimensional formula \(\displaystyle [M^1L^2T^{-2}]\) via \(\displaystyle n_2=n_1(M_1/M_2)^1(L_1/L_2)^2(T_1/T_2)^{-2}\).

A number is never "large" or "small" by itself — a dimensioned quantity only has a size relative to another quantity of the same kind.Length, mass, speed, and so on are not pure numbers; each carries a unit, and a unit is itself the chosen "standard" quantity of that kind. When you say a length is \(\displaystyle 5 \) m, you are really saying it is $\displaystyle 5$ times the standard metre. So the phrase "large" or "small" is shorthand for a ratio: \[\text{"quantity } X \text{ is large"} \;\equiv\; \frac{X}{X_{\text{standard}}} \gg 1 \] and "small" means the same ratio is \(\displaystyle \ll 1 \). Until you say what \(\displaystyle X_{\text{standard}} \) is, the ratio has no value, so calling \(\displaystyle X \) "large" or "small" carries no information — the same object can be enormous next to one reference and tiny next to another. A person is "tall" next to a child and "short" next to a giraffe; nothing about the person's own height changed, only the object being compared against. This is why the statement in the question is exact: a dimensional quantity's size is meaningful only in comparison with another quantity of the same dimensions, and calling it large or small without naming that comparison quantity is meaningless.With that in mind, look at each statement. Some name their standard of comparison already — those are fine as they stand. Others say "large"/"small"/"great" with no comparison object stated at all — those need a standard supplied.(a) "Atoms are very small objects." No comparison object is named — small compared to what? An atom (radius \(\displaystyle \sim 10^{-10} \) m) is enormous next to an atomic nucleus (\(\displaystyle \sim 10^{-15} \) m) but minuscule next to a football. The statement must name the reference: Reframed: Atoms are small in size compared to the everyday objects we handle, such as a football or a grain of sand.(b) "A jet plane moves with great speed." Again no reference speed is stated. A jet is fast next to a bicycle, but its speed is a small fraction of the speed of sound at supersonic threshold, and utterly negligible next to the speed of light. Reframed: A jet plane moves with great speed compared to a bicycle or a car (though its speed is small compared to the speed of light).(c) "The mass of Jupiter is very large." Large compared to what mass? Jupiter's mass is large next to the Earth's, but it is small next to the Sun's mass. Reframed: The mass of Jupiter is very large compared to the mass of the Earth (though it is small compared to the mass of the Sun).(d) "The air inside this room contains a large number of molecules." "Large" here is again relative — large compared to what count? The number is enormous next to, say, the number of people in the room, but it is a small fraction of the number of molecules in a room-sized volume of a denser gas or a larger enclosure. Reframed: The air inside this room contains a large number of molecules compared to the number of molecules in a small test tube of air (or compared to the number of people in the room).(e) "A proton is much more massive than an electron." This statement already names its comparison object — the electron — explicitly. It is not saying "a proton is very massive" in isolation; it is directly comparing two quantities of the same dimension (mass), so it is already meaningful and needs no reframing. (Numerically, \(\displaystyle m_p \approx 1836\, m_e \), confirming the comparison is valid as stated.)(f) "The speed of sound is much smaller than the speed of light." This too already states its standard of comparison — the speed of light — so it is already a valid, meaningful comparison and needs no reframing. (Indeed \(\displaystyle v_{\text{sound}} \approx 340 \) m/s while \(\displaystyle c \approx 3\times10^{8} \) m/s, so the ratio is about \(\displaystyle 10^{-6} \), genuinely "much smaller.")The pattern across all six: whenever a comparison object is missing, "large/small/great" is meaningless and must be reframed by naming a reference of the same dimension; whenever the comparison object is already named (as in e, f), the statement is already meaningful and stands unchanged.**Answer: A dimensional quantity's magnitude is only meaningful as a ratio to another quantity of the same kind, so "large"/"small" without naming that reference carries no information. (a) Atoms are small compared to everyday objects like a football. (b) A jet plane is fast compared to a bicycle or car. (c) Jupiter's mass is large compared to the Earth's mass. (d) The room's air has a large number of molecules compared to a small container of air. (e) and (f) already name their comparison object (electron; speed of light) and need no reframing.

A "new unit of length" defined by \(\displaystyle c=1\) simply means: $\displaystyle 1$ new unit is the distance light travels in $\displaystyle 1$ second — so with the speed of light set equal to $\displaystyle 1$ in this system, a distance and a time (in seconds) become numerically the same.Setting up the definitionSpeed is defined by the relation\[v = \frac{\text{distance}}{\text{time}} \]Here the new unit of length, call it \(\displaystyle L_{\text{new}}\), is chosen so that the speed of light in vacuum has the value\[c = 1\ \frac{L_{\text{new}}}{\text{s}} \]That is, light covers exactly \(\displaystyle 1\) new unit of length in \(\displaystyle 1\) second. This is the same idea as a "light-year," except the base time unit here is the second rather than the year — so \(\displaystyle 1\ L_{\text{new}} = 1\) light-second.Converting the given time to secondsThe time taken by light to travel from the Sun to the Earth is given as $\displaystyle 8$ min $\displaystyle 20$ s. Converting fully to seconds ($\displaystyle 1$ min = $\displaystyle 60$ s):\[t = 8 \times 60\ \text{s} + 20\ \text{s} = 480\ \text{s} + 20\ \text{s} = 500\ \text{s} \]A step people trip on: don't leave the time as "$\displaystyle 8$ min $\displaystyle 20$ s" and try to multiply — mixed units have no single numerical value to plug into \(\displaystyle v=d/t\). Convert everything to one unit (seconds) first.Finding the distanceUsing \(\displaystyle v=d/t\) with \(\displaystyle v=c=1\ L_{\text{new}}/\text{s}\) and \(\displaystyle t = 500\ \text{s}\):\[d = c \times t = \left(1\ \frac{L_{\text{new}}}{\text{s}}\right) \times (500\ \text{s}) \]The seconds cancel, leaving\[d = 500\ L_{\text{new}} \]So the Sun–Earth distance, expressed in this new unit, is numerically identical to the light travel time in seconds — because by construction \(\displaystyle 1\) new unit of length \(\displaystyle \equiv\) the distance light covers in \(\displaystyle 1\) s.Significant figures: the given time, $\displaystyle 8$ min $\displaystyle 20$ s, converts exactly to $\displaystyle 500$ s (no measurement uncertainty is introduced by the conversion), so the result is kept as $\displaystyle 500$, with all three digits significant.\[d = 500\ (\text{new unit of length}) \]Answer: The Sun–Earth distance is $\displaystyle 500$ new units of length (i.e., $\displaystyle 500$ "light-seconds," since $\displaystyle 1$ new unit is defined as the distance light travels in $\displaystyle 1$ s).

The most precise instrument is the one with the smallest least count — precision means how fine a length it can resolve, not how large an object it can hold.The least count (LC) of a length-measuring instrument is the smallest change in length it can detect. Comparing the three instruments means comparing their least counts.(a) Vernier callipersFor a vernier calliper, \[\text{LC} = \frac{\text{value of one main scale division (MSD)}}{\text{number of divisions on the vernier scale, } N} \] A standard vernier calliper has \(\displaystyle \text{MSD} = 1\ \text{mm} \), and here \(\displaystyle N = 20 \). So \[\text{LC}_{\text{vernier}} = \frac{1\ \text{mm}}{20} = 0.05\ \text{mm} = 0.005\ \text{cm} = 5\times10^{-3}\ \text{cm} \](b) Screw gaugeFor a screw gauge, \[\text{LC} = \frac{\text{pitch}}{\text{number of divisions on the circular scale}} \] Here the pitch is \(\displaystyle 1\ \text{mm} \) and the circular scale has \(\displaystyle 100 \) divisions, so \[\text{LC}_{\text{screw gauge}} = \frac{1\ \text{mm}}{100} = 0.01\ \text{mm} = 0.001\ \text{cm} = 1\times10^{-3}\ \text{cm} \] The trap here is stopping at "$\displaystyle 100$ divisions" and assuming more divisions automatically means finer measurement — the pitch has to be divided by that number; a coarse pitch with many divisions is not automatically finer than a fine pitch with fewer.(c) Optical instrumentAn optical instrument that uses interference of light can resolve a length down to about the wavelength of the light it uses. Visible light has a wavelength of roughly \(\displaystyle \lambda \approx 6000\ \text{Å} \) (a representative value in the middle of the visible range), and \[1\ \text{Å} = 10^{-8}\ \text{cm} \] so \[\text{LC}_{\text{optical}} \approx 6000\times10^{-8}\ \text{cm} = 6\times10^{-5}\ \text{cm} \]Comparing the three least counts\[\text{LC}_{\text{vernier}} = 5\times10^{-3}\ \text{cm}, \qquad \text{LC}_{\text{screw gauge}} = 1\times10^{-3}\ \text{cm}, \qquad \text{LC}_{\text{optical}} \approx 6\times10^{-5}\ \text{cm} \]The optical instrument's least count is smaller than the screw gauge's by a factor of about $\displaystyle 17$, and smaller than the vernier calliper's by a factor of about 80. The smallest least count belongs to the instrument that can resolve the finest difference in length, so it is the most precise of the three, followed by the screw gauge, with the vernier calliper the least precise.Answer: The optical instrument is the most precise — its least count (\(\displaystyle \approx 6\times10^{-5}\ \text{cm} \), set by the wavelength of light) is far smaller than that of the screw gauge (\(\displaystyle 1\times10^{-3}\ \text{cm} \)) or the vernier callipers (\(\displaystyle 5\times10^{-3}\ \text{cm} \)).

The magnification tells you how many times bigger the image looks than the real object — divide it out to get back to the true size.A microscope of magnification \(\displaystyle m \) makes an object look \(\displaystyle m \) times wider than it actually is. If \(\displaystyle w_{\text{obs}} \) is the width you see in the field of view and \(\displaystyle w_{\text{real}} \) is the true width of the object, the two are related by\[m = \frac{w_{\text{obs}}}{w_{\text{real}}} \]so\[w_{\text{real}} = \frac{w_{\text{obs}}}{m} \]Here the student's $\displaystyle 20$ observations give an average observed width (this is what appears in the microscope's field of view, already magnified) of\[w_{\text{obs}} = 3.5 \text{ mm} \]and the magnification is\[m = 100 \]Averaging the $\displaystyle 20$ readings is what keeps this a genuine estimate — it cancels out the small random errors of judging the edge of the hair under the microscope, which is why the problem bothers to tell you $\displaystyle 20$ observations were taken rather than just one.Substituting,\[w_{\text{real}} = \frac{3.5 \text{ mm}}{100} = 0.035 \text{ mm} \]Writing this in scientific notation, keeping it to the two significant figures carried by the measured $\displaystyle 3.5$ mm (the magnification $\displaystyle 100$ is a fixed instrument specification, not a measured quantity, so it does not limit the precision):\[w_{\text{real}} = 3.5 \times 10^{-2} \text{ mm} \]It is worth converting this into a unit that matches everyday intuition about hair thickness. Since \(\displaystyle 1 \text{ mm} = 10^{3} \, \mu\text{m} \),\[w_{\text{real}} = 3.5 \times 10^{-2} \text{ mm} \times 10^{3} \, \frac{\mu\text{m}}{\text{mm}} = 35 \, \mu\text{m} \]which is the same thickness written as \(\displaystyle 3.5 \times 10^{-5} \text{ m} \).Answer: The thickness of the hair is \(\displaystyle 3.5 \times 10^{-2} \text{ mm} = 35 \, \mu\text{m} \) (to two significant figures).

(a) Wind the thread around the scale many times and divide — a single pass against the millimetre marks buries the whole diameter inside the scale's own reading error.A metre scale can be read to about $\displaystyle 1$ mm at best (half the smallest division). A thread's diameter is far smaller than that, so holding a single strand against the scale tells you nothing — the diameter is smaller than the uncertainty of the scale itself.Instead, wrap the thread closely around a uniform rod (or the scale), coil touching coil, with no gaps and no overlaps, for a counted number of turns N — say $\displaystyle 40$ or 50. Measure the total length L that these N turns occupy along the scale. Then\[d = \frac{L}{N} \]This is the method of measuring a small length by multiplying it up. The scale's reading uncertainty, about \(\displaystyle \pm 1 \) mm, attaches to L, not to a single turn, so it gets divided by N when you compute d — a $\displaystyle 40$-fold reduction in error for $\displaystyle 40$ turns. Winding evenly (no gaps to inflate L, no overlaps to shrink it) and counting the turns correctly while wrapping are what make this trick work.(b) No — the least count keeps shrinking on paper, but the screw gauge's real accuracy is capped by mechanical imperfections that don't shrink with it.The least count of a screw gauge is\[\text{L.C.} = \frac{\text{pitch}}{\text{number of divisions on the circular scale}} \]Here the pitch is $\displaystyle 1.0$ mm and the circular scale carries $\displaystyle 200$ divisions, so\[\text{L.C.} = \frac{1.0\ \text{mm}}{200} = 5.0\times10^{-3}\ \text{mm} = 5.0\ \mu\text{m} \]Cutting the circular scale into more divisions — $\displaystyle 1000$ instead of $\displaystyle 200$, say — shrinks this number on paper (to \(\displaystyle 1.0\ \text{mm}/1000 = 1.0\times10^{-3}\) mm), which looks like unlimited precision is available just for the marking.It is not. A screw gauge's actual accuracy is set by defects that do not get smaller just because the scale does: backlash (a bit of dead play when the screw's direction of turning reverses), non-uniformity of the screw thread itself along its length, and friction between screw and frame. These contribute an error to every reading that stays roughly fixed in size — so once the least count is pushed below it, refining the scale further buys nothing. Past that point the divisions are also so close together that the eye cannot reliably tell which one lines up with the reference line, so reading error grows even as the nominal least count shrinks. So: no, the accuracy cannot be increased without limit merely by adding divisions.(c) Random errors partly cancel when readings are averaged, and that cancellation gets more complete the more readings go into the average.Each single measurement of the rod's diameter with vernier callipers carries a small random error — from exactly where the jaws come to rest, from parallax while reading the vernier scale, from the rod's cross-section not being a perfect circle at every point along its length. These errors scatter above and below the true diameter more or less unpredictably from one trial to the next.When n independent readings \(\displaystyle d_1, d_2, \ldots, d_n\) are averaged,\[\bar d = \frac{d_1+d_2+\cdots+d_n}{n} \]and the statistical spread of this mean about the true value falls as\[\sigma_{\bar d} = \frac{\sigma}{\sqrt n} \]where \(\displaystyle \sigma \) is the spread of a single reading. This is the key point people miss: it is the number of readings, not any change in the instrument, that buys the improvement. Going from n = $\displaystyle 5$ to n = $\displaystyle 100$ shrinks the uncertainty in the mean by a factor of \(\displaystyle \sqrt{100/5} = \sqrt{20} \approx 4.5 \), because in a set of $\displaystyle 100$ the positive and negative random deviations cancel far more thoroughly than they can in a set of only $\displaystyle 5$, where a single unusually high or low reading can still pull the mean noticeably away from the true diameter. That is why $\displaystyle 100$ measurements are expected to yield a mean diameter closer to the rod's actual diameter than $\displaystyle 5$ measurements do.Answer: (a) Coil the thread in N close, non-overlapping turns around the scale, measure the coil's total length L, and take the diameter as d = L/N, dividing the scale's reading error by N. (b) No — the least count is $\displaystyle 1.0$ mm/$\displaystyle 200$ = 5.0x10^-$\displaystyle 3$ mm, but past a point backlash, thread-manufacturing imperfections, and friction fix the real error, so cutting the circular scale into more divisions stops improving accuracy. (c) Random errors partly cancel on averaging, with the uncertainty in the mean falling as $\displaystyle 1$/sqrt(n); $\displaystyle 100$ readings therefore give a mean diameter much closer to the true value than $\displaystyle 5$ readings do.

Area scales as the square of length, so to get a linear magnification from two areas you must take a square root — you cannot compare the areas directly and call that "the" magnification.Call the linear magnification \(\displaystyle m \). If every length in the photograph is stretched by a factor \(\displaystyle m \) when it is projected, then a two‑dimensional area — being length \(\displaystyle \times\) length — is stretched by \(\displaystyle m^2 \): \[\frac{A_{\text{screen}}}{A_{\text{slide}}} = m^{2} \]Get both areas into the same unit first. The slide's picture area is given in \(\displaystyle cm^2\) and the screen's in \(\displaystyle m^2\); they must match before you divide. Using \(\displaystyle 1\ cm = 10^{-2}\ m\), so \(\displaystyle 1\ cm^{2} = 10^{-4}\ m^{2}\): \[A_{\text{slide}} = 1.75\ cm^{2} = 1.75\times10^{-4}\ m^{2} \] \[A_{\text{screen}} = 1.55\ m^{2} \](The "$\displaystyle 35$ mm" describing the slide is just the film format — it plays no role in this calculation since the picture area on the slide is already given directly.)Areal magnification. Substitute into the ratio: \[m^{2} = \frac{A_{\text{screen}}}{A_{\text{slide}}} = \frac{1.55\ m^{2}}{1.75\times10^{-4}\ m^{2}} = 8.86\times10^{3} \]Carrying more digits before rounding: \(\displaystyle m^2 = \dfrac{1.55}{1.75}\times10^{4} = 8857.14\ldots \)Linear magnification. Take the square root: \[m = \sqrt{8857.14\ldots} = 94.11\ldots \]Both given quantities ($\displaystyle 1.75$ and $\displaystyle 1.55$) carry $\displaystyle 3$ significant figures, so the final answer is rounded to $\displaystyle 3$ significant figures — not left with the extra digits the square root produced: \[m \approx 94.1 \]This says each linear dimension of the house's image on the screen is about $\displaystyle 94$ times the corresponding dimension on the slide — a sensible number for a photograph blown up from a small negative to fill a projection screen.Answer: The linear magnification of the projector–screen arrangement is \(\displaystyle m \approx 94.1 \) (i.e., about \(\displaystyle 94\times\)).

Significant figures are the digits that actually carry precision — leading zeros never do, but a zero sitting between or after other digits usually does.Three rules settle every part here: 1. Every non-zero digit is significant. 2. A zero sandwiched between non-zero digits (a "captive" zero) is significant — it cannot be dropped without changing the number. 3. A zero that only serves to place the decimal point (a leading zero, or one used purely as a power-of-ten placeholder) is not significant. Once a number is written with an explicit decimal point, though, a trailing zero after the decimal point is a deliberate statement of precision and counts as significant.Going through each value:(a) \(\displaystyle 0.007 \ \mathrm{m^2} \) Rewrite it in scientific notation to see the placeholders clearly: \(\displaystyle 0.007 = 7 \times 10^{-3} \). The zeros before the $\displaystyle 7$ only fix where the decimal point sits — they carry no information about precision, so they are not counted. Significant figures: just the $\displaystyle 7$. → $\displaystyle 1$ significant figure.(b) \(\displaystyle 2.64 \times 10^{24} \ \mathrm{kg} \) Numbers already written in the form \(\displaystyle a \times 10^{n} \) are the easiest case: every digit in \(\displaystyle a \) is significant, and the power of ten is not a digit at all — it only scales the value. Digits: $\displaystyle 2$, $\displaystyle 6$, 4. → $\displaystyle 3$ significant figures.(c) \(\displaystyle 0.2370 \ \mathrm{g\ cm^{-3}} \) The leading zero (before the $\displaystyle 2$) is a placeholder — not significant. But the trailing zero after the $\displaystyle 7$ is written deliberately, after a decimal point that already has non-zero digits before it; if it weren't meant to be trusted, the number would simply have been written as 0.237. That final zero states "we know this to the fourth digit," so it counts. Digits: $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 7$, 0. → $\displaystyle 4$ significant figures.(d) \(\displaystyle 6.320 \ \mathrm{J} \) Here $\displaystyle 6$, $\displaystyle 3$, $\displaystyle 2$ are non-zero digits, all significant. The trailing $\displaystyle 0$ comes after the decimal point in a number that already has non-zero digits ahead of it, so by rule $\displaystyle 3$ it is a genuine statement of precision, not a placeholder. Digits: $\displaystyle 6$, $\displaystyle 3$, $\displaystyle 2$, 0. → $\displaystyle 4$ significant figures.(e) \(\displaystyle 6.032 \ \mathrm{N\ m^{-2}} \) $\displaystyle 6$ and $\displaystyle 3$ and $\displaystyle 2$ are non-zero and significant. The $\displaystyle 0$ sits between the $\displaystyle 6$ and the $\displaystyle 3$ — a captive zero — so by rule $\displaystyle 2$ it is significant too. Digits: $\displaystyle 6$, $\displaystyle 0$, $\displaystyle 3$, 2. → $\displaystyle 4$ significant figures.(f) \(\displaystyle 0.0006032 \ \mathrm{m^2} \) Write it as \(\displaystyle 6.032 \times 10^{-4} \) to separate the placeholders from the real digits: all four leading zeros (0.0006...) exist only to position the decimal point and are not significant. What remains, $\displaystyle 6032$, is exactly the digit string from part (e): the $\displaystyle 0$ here is again a captive zero (between $\displaystyle 6$ and $\displaystyle 3$), hence significant. Digits: $\displaystyle 6$, $\displaystyle 0$, $\displaystyle 3$, 2. → $\displaystyle 4$ significant figures.The pattern to keep in memory: leading zeros are never significant (they're just place-holders for the decimal point), trailing zeros after an explicit decimal point are significant, and any zero caught between two non-zero digits is always significant.Answer: (a) $\displaystyle 1$, (b) $\displaystyle 3$, (c) $\displaystyle 4$, (d) $\displaystyle 4$, (e) $\displaystyle 4$, (f) $\displaystyle 4$ significant figures.