NCERT Exemplar Solutions Class 11 Physics Units and Measurement — Long Answer Questions Q36–44

NCERT Exemplar Problems, Class 11 Physics — Units and Measurement, Long Answer Questions 36 to 44. All 9 questions solved. The book numbers them 2.36 to 2.44.

Three ideas carry this page. Dimensional analysis checks an equation, converts between systems of units, and even predicts the form of a law when the quantities it depends on are known. Errors propagate through powers: the percentage error of $a^pb^q\cdots$ is $p$ times that of $a$ plus $q$ times that of $b$, and so on, and the result is rounded to the digit the error allows. Angular size is size over distance, which is why a telescope can enlarge a planet but not a star.

Key insight. Questions 36, 39, 40 and 41 are all the same move: write each quantity as $\text{M}^a\text{L}^b\text{T}^c$ and match the powers of M, L and T. Once the three equations are written down, the rest is algebra.

A note on the syllabus: Questions 42 (the oleic acid film) and 43 (the parsec and angular sizes) draw on the sections on measuring large and small distances that the rationalised NCERT Chapter 1 no longer covers. They are solved here for completeness.

Long Answer Questions

Question 36

A new system of units is proposed in which unit of mass is $\alpha$ kg, unit of length $\beta$ m and unit of time $\gamma$ s. How much will $5$ J measure in this new system?

Solution. Energy has dimensions $\text{ML}^2\text{T}^{-2}$, so $1\ \text{J} = 1\ \text{kg m}^2\,\text{s}^{-2}$. The new units are $\alpha$ kg, $\beta$ m and $\gamma$ s, so $1\ \text{kg} = \tfrac{1}{\alpha}$ new unit of mass, $1\ \text{m} = \tfrac1\beta$ new unit of length and $1\ \text{s} = \tfrac1\gamma$ new unit of time. Then

$$1\ \text{J} = \frac{1}{\alpha}\left(\frac1\beta\right)^2\left(\frac1\gamma\right)^{-2} = \frac{\gamma^2}{\alpha\beta^2}\ \text{new units},$$

and $5$ J becomes $\dfrac{5\gamma^2}{\alpha\beta^2}$. The check: if the new unit of time is longer ($\gamma > 1$), the same energy is a bigger number, because the new unit of energy, $\alpha\beta^2/\gamma^2$ J, is smaller.

Question 37

The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as

$$v = \frac{\pi}{8}\,\frac{Pr^4}{\eta l}$$

where $P$ is the pressure difference between the two ends of the pipe and $\eta$ is coefficient of viscosity of the liquid having dimensional formula $\text{ML}^{-1}\text{T}^{-1}$. Check whether the equation is dimensionally correct.

Solution. The left side is a volume per second, $\text{L}^3\text{T}^{-1}$. On the right $\pi/8$ is a pure number, and

$$\left[\frac{Pr^4}{\eta l}\right] = \frac{[\text{ML}^{-1}\text{T}^{-2}][\text{L}^4]}{[\text{ML}^{-1}\text{T}^{-1}][\text{L}]} = \frac{\text{ML}^3\text{T}^{-2}}{\text{MT}^{-1}} = \text{L}^3\text{T}^{-1}.$$

Both sides match, so the equation is dimensionally correct. (This is Poiseuille’s formula; dimensional analysis cannot confirm the factor $\pi/8$, only the powers.)

Question 38

A physical quantity $X$ is related to four measurable quantities $a$, $b$, $c$ and $d$ as follows: $X = a^2b^3c^{5/2}d^{-2}$. The percentage error in the measurement of $a$, $b$, $c$ and $d$ are $1\%$, $2\%$, $3\%$ and $4\%$, respectively. What is the percentage error in quantity $X$? If the value of $X$ calculated on the basis of the above relation is $2.763$, to what value should you round off the result.

Solution. For a product of powers the relative errors add, each multiplied by the magnitude of its power (a negative power still adds, since errors do not cancel):

$$\frac{\Delta X}{X} = 2\frac{\Delta a}{a} + 3\frac{\Delta b}{b} + \frac52\frac{\Delta c}{c} + 2\frac{\Delta d}{d} = 2(1) + 3(2) + 2.5(3) + 2(4) = 23.5\%.$$

The absolute error is $0.235\times2.763 = 0.65$. An error of that size already affects the first decimal place, so the digits after it mean nothing, and $X$ is rounded to $2.8$.

Question 39

In the expression $P = El^2m^{-5}G^{-2}$, $E$, $m$, $l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.

Solution. The dimensions are $[E] = \text{ML}^2\text{T}^{-2}$, $[l] = \text{ML}^2\text{T}^{-1}$, $[m] = \text{M}$ and $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$. So

$$[P] = (\text{ML}^2\text{T}^{-2})(\text{M}^2\text{L}^4\text{T}^{-2})(\text{M}^{-5})(\text{M}^2\text{L}^{-6}\text{T}^4).$$

Adding the powers: M: $1 + 2 – 5 + 2 = 0$; L: $2 + 4 – 6 = 0$; T: $-2 – 2 + 4 = 0$. Hence $[P] = \text{M}^0\text{L}^0\text{T}^0$, and $P$ is dimensionless.

Question 40

If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Solution. Write each of M, L and T as $c^p h^q G^r$ with $[c] = \text{LT}^{-1}$, $[h] = \text{ML}^2\text{T}^{-1}$, $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$. Matching powers gives three equations each time:

$$\text{M}: q – r,\qquad \text{L}: p + 2q + 3r,\qquad \text{T}: -p – q – 2r.$$

• Mass (right sides $1, 0, 0$): $p = q = \tfrac12$, $r = -\tfrac12$, so $M \propto \sqrt{\dfrac{ch}{G}}$.
• Length (right sides $0, 1, 0$): $p = -\tfrac32$, $q = r = \tfrac12$, so $L \propto \sqrt{\dfrac{hG}{c^3}}$.
• Time (right sides $0, 0, 1$): $p = -\tfrac52$, $q = r = \tfrac12$, so $T \propto \sqrt{\dfrac{hG}{c^5}}$.

These are the Planck mass, length and time. A check: $L/T = c$, as it must be.

Question 41

An artificial satellite is revolving around a planet of mass $M$ and radius $R$, in a circular orbit of radius $r$. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis, that

$$T = \frac{k}{R}\sqrt{\frac{r^3}{g}},$$

where $k$ is a dimensionless constant and $g$ is acceleration due to gravity.

Solution. Kepler’s law fixes the dependence on $r$: $T^2 \propto r^3$, so $T \propto r^{3/2}$. The period also depends on the planet, through $g$ and $R$. Take $T = k\,r^{3/2}g^xR^y$ and match dimensions:

$$[\text{T}] = [\text{L}^{3/2}]\,[\text{LT}^{-2}]^x\,[\text{L}]^y.$$

T: $1 = -2x$, so $x = -\tfrac12$. L: $0 = \tfrac32 + x + y = \tfrac32 – \tfrac12 + y$, so $y = -1$. Therefore

$$T = k\,r^{3/2}g^{-1/2}R^{-1} = \frac{k}{R}\sqrt{\frac{r^3}{g}}.$$

This agrees with the full result $T = 2\pi\sqrt{r^3/(GM)}$ once $GM = gR^2$ is used, which gives $k = 2\pi$.

Question 42

In an experiment to estimate the size of a molecule of oleic acid $1$ mL of oleic acid is dissolved in $19$ mL of alcohol. Then $1$ mL of this solution is diluted to $20$ mL by adding alcohol. Now $1$ drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions:

(a) Why do we dissolve oleic acid in alcohol?

(b) What is the role of lycopodium powder?

(c) What would be the volume of oleic acid in each mL of solution prepared?

(d) How will you calculate the volume of $n$ drops of this solution of oleic acid?

(e) What will be the volume of oleic acid in one drop of this solution?

Solution.

(a) Oleic acid does not dissolve in water, but it does dissolve in alcohol. Diluting it in alcohol lets a very small, known amount of oleic acid be delivered in a drop; the alcohol then dissolves in the water and evaporates, leaving the oleic acid to spread as a single layer.

(b) Lycopodium powder floats as a fine, even layer on the water. When the drop spreads, the oleic acid film pushes the powder aside, leaving a clear circular patch whose edge is easy to see, so the diameter, and hence the area, of the film can be measured.

(c) The first solution is $1$ mL of acid in $20$ mL, and $1$ mL of that is diluted again to $20$ mL:

$$\frac{1}{20}\times\frac{1}{20} = \frac{1}{400}\ \text{mL of oleic acid per mL of solution}.$$

(d) Let the solution fall drop by drop from a burette into a measuring cylinder and count the number of drops $n$ needed to make a known volume, say $1$ mL. The volume of $n$ drops is then read directly, and one drop is $1/n$ mL.

(e) If $n$ drops make $1$ mL, one drop is $1/n$ mL of solution, which contains

$$\frac{1}{400}\times\frac1n = \frac{1}{400n}\ \text{mL of oleic acid}.$$

The thickness of the film, and so the size of a molecule, is this volume divided by the area of the film.

Question 43

(a) How many astronomical units (A.U.) make $1$ parsec?

(b) Consider a sunlike star at a distance of $2$ parsecs. When it is seen through a telescope with $100$ magnification, what should be the angular size of the star? Sun appears to be $(1/2)^\circ$ from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than $1$ arc minute.

(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about $1/2$ A.U. from the earth. Calculate what size it will appear when seen through the same telescope.

(Comment: This is to illustrate why a telescope can magnify planets but not stars.)

Solution.

(a) One parsec is the distance at which $1$ A.U. subtends $1$ second of arc. Since $1” = \dfrac{\pi}{180\times3600}$ rad,

$$1\ \text{pc} = \frac{1\ \text{A.U.}}{\pi/(180\times3600)} = \frac{180\times3600}{\pi}\ \text{A.U.} = 206\,265\ \text{A.U.} \approx 2\times10^5\ \text{A.U.}$$

(b) Angular size falls inversely with distance. A sunlike star at $2$ pc $= 4\times10^5$ A.U. appears

$$\frac{(1/2)^\circ}{4\times10^5} = 1.25\times10^{-6}\ \text{degree} = 7.5\times10^{-5}\ \text{arcmin}.$$

Magnified $100$ times it is $7.5\times10^{-3}$ arcmin, still far below the $1$ arcmin the eye can resolve through the atmosphere. The star remains a point: the telescope cannot show its disc.

(c) From Question 25(c) the sun is about $100$ times the earth’s diameter, so Mars, at half the earth’s diameter, is $\tfrac{1}{200}$ of the sun’s diameter. At $1$ A.U. it would subtend $\tfrac{1}{200}\times\tfrac12^\circ = \tfrac{1}{400}$ degree; at $\tfrac12$ A.U. it subtends twice that, $\tfrac{1}{200}$ degree $= 0.3$ arcmin. Through the telescope:

$$100\times0.3′ = 30′ = \tfrac12^\circ.$$

That is well above the $1$ arcmin limit, so Mars shows a clear disc. The telescope magnifies a planet because the planet is close enough for its magnified size to beat the atmospheric blur; a star is so far away that even $100$ times magnification leaves it a point. (For comparison, the real Mars, $6780$ km across at $0.5$ A.U., gives $31’$ at $100\times$.)

Printed key errors. In (b) the key works the star at $1$ parsec, giving $15\times10^{-3}$ arcmin instead of $7.5\times10^{-3}$ arcmin for the $2$ parsecs in the question. In (c) it takes $D_{\text{earth}}/D_{\text{sun}} = \tfrac{1}{400}$ “from Answer 2.25(c)”, but 2.25(c) gives $\tfrac{1}{100}$; with $\tfrac{1}{400}$ it gets $7.5$ arcmin instead of $30$ arcmin. Neither slip changes the conclusion.

Question 44

Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass ($m$) to energy ($E$) as $E = mc^2$, where $c$ is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where $1\ \text{MeV} = 1.6\times10^{-13}$ J; the masses are measured in unified atomic mass unit (u) where $1\ \text{u} = 1.67\times10^{-27}$ kg.

(a) Show that the energy equivalent of $1$ u is $931.5$ MeV.

(b) A student writes the relation as $1\ \text{u} = 931.5$ MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Solution.

(a) $E = mc^2$ for $m = 1$ u. With the precise values $1\ \text{u} = 1.6605\times10^{-27}$ kg, $c = 2.998\times10^8$ m/s and $1\ \text{MeV} = 1.602\times10^{-13}$ J,

$$E = \frac{1.6605\times10^{-27}\times(2.998\times10^8)^2}{1.602\times10^{-13}}\ \text{MeV} = 931.5\ \text{MeV}.$$

With the rounded values printed in the question, $1.67\times10^{-27}\times(3\times10^8)^2 = 1.503\times10^{-10}$ J, which is $939$ MeV. The $1\%$ gap comes entirely from rounding $1.6605$ to $1.67$; $931.5$ MeV needs the constants to four figures.

(b) A mass cannot equal an energy: the left side has dimensions $\text{M}$ and the right side $\text{ML}^2\text{T}^{-2}$. The relation must carry the factor $c^2$ that converts one into the other:

$$1\ \text{u}\times c^2 = 931.5\ \text{MeV}.$$

Common mistakes

• Inverting the conversion in Question 36. A larger unit of time makes the number bigger, so $\gamma$ goes in the numerator: $5\gamma^2/(\alpha\beta^2)$.
• Subtracting the error for a negative power in Question 38. Errors always add; $d^{-2}$ contributes $+2\times4\%$.
• Keeping too many digits in Question 38. A $23.5\%$ error makes the second decimal place meaningless: $2.8$, not $2.76$.
• Forgetting the factor $2$ in distance in Question 43(b). The star is at $2$ parsecs, which halves its angular size.
• Writing $1\ \text{u} = 931.5$ MeV in Question 44. A mass and an energy cannot be equal; the $c^2$ is essential.

Practise next

• A quantity is given by $P = \dfrac{a^3b^2}{c\sqrt d}$. The percentage errors in $a$, $b$, $c$ and $d$ are $1\%$, $3\%$, $4\%$ and $2\%$. Find the percentage error in $P$.

• In a new system the unit of mass is $2$ kg, the unit of length is $3$ m and the unit of time is $4$ s. How much does $10$ J measure in this system?

• Exemplar Units and Measurement: Very Short and Short Answer Questions — the preceding questions on orders of magnitude, the vernier, systems of units and precision.
• Exemplar Units and Measurement: Multiple Choice Questions — significant figures, errors and dimensions in objective form.
• NCERT textbook Chapter 1 — Units and Measurement for the theory of SI units, significant figures, errors and dimensions.