NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurement
2.1. Fill in the Blanks
(a) The volume of a cube of side 1 cm is equal to ..... m3
Ans. Volume of cube,
V = (1 cm)3 = (10-2 m)3 = 10-6 m3.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2
Ans. Surface area = curved area + area on top /base
= 2πrh + 2πr2 = 2πr(h + r)
r = 2 cm = 20 mm
h = 10 cm = 100 mm
Surface area = 2πr (h + r)
= 2 × 3.14 × 20 (100 + 20)
= 15072 mm2
Hence, answer is 15072 mm2
(c) A vehicle moving with a speed of 18 km h–1 covers .... m in 1 s
Ans. Speed of vehicle = 18 km/h
1 km = 1000 m
1 hr = 60 x 60 = 3600 s
$$ \text{1 km/hr} = \frac{1000m}{3600s} = \frac{5}{18} \text{m/s} $$
18 km/h = (18 × 1000)/3600
= 5 m/s
Distance traveled by the vehicle in 1 s = 5 m
(d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3.
The Relative density of lead is 11.3 g cm-3
⇒ 11.3 x 103 kg m– 3
[1 kilogram = 103g, 1 meter = 102 cm]
⇒ 11.3 x 103 kg m– 3
2.2. Fill in the blanks by suitable conversion of units
(a) 1 kg m2 s–2 = ….g cm2 s–2
Ans. 1 kg m2 s–2 = 1 kg × 1m2 × 1 s – 2
We know that,
1 kg = 103 g
1 m = 100 cm = 102 cm
When the values are put together, we get:
1 kg × 1m2 × 1 s-2 = 103g × (102 cm)2 × 1 s-2
= 103g × 104 cm2 × 1 s-2
= 107 g cm2 s-2
⇒ 1 kg m2 s-2 = 107 g cm2 s-2
(b) 1 m = ….. ly
Ans. Using the formula,
Distance = speed × time
Speed of light = 3 × 108 m/s
Time = 1 year
= 365 days
= 365 × 24 hr
= 365 × 24 × 60 × 60 sec
Put these values in the formula mentioned above, we get:
One light year distance
= (3 × 108 m/s) × (365 × 24 × 60 × 60)
= 9.46 × 1015 m
9.46 × 1015 m = 1 ly
So that, 1 m = 1/9.46 × 1015 ly
= 1.06 × 10–16 ly
⇒ 1 meter = 1.06 × 10–16 ly
(c) 3.0 m s–2 = …. km h–2
$$ \textbf{Ans. } \text{1 km = 1000 m so that 1 m }= \frac{1}{1000} \text{km} \\ \text{3.0 m} s^{-2} = 3.0 \left( \frac{1}{1000}\text{km}\right) \left(\frac{1}{3600}\text{hour} \right)^{-2} \\ = 3.0 × 10^{-3}\text{km} × \left(\frac{1}{3600}\right)^{-2}h^{-2}$$
= 3 × 10-3 km × (3600)2 h-2
= 3.88 x 104 km h-2
⇒ 3.0 m s-2 = 3.88 x 104 km h-2
(d) G = 6.67 × 10–11 N m2 (kg)–2 = …. (cm)3s–2 g–1
Ans. G = 6.67 × 10-11 N m2 (kg)-2
We know that,
1N = 1 kg m s-2
1 kg = 103 g
1 m = 100 cm = 102 cm
Put the values together, we get:
⇒ 6.67 × 10-11 Nm2 kg-2
= 6.67 × 10-11 × (1 kg m s -2) (1 m2) (1 kg-2)
Solve the following and cancel out the units, we get :
⇒ 6.67 × 10-11 × (1 kg-1 × 1 m3 × 1 s-2)
Put the above values together to convert kg to g and m to cm
⇒ 6.67 x 10-11 × (103 g)-1 × (102 cm)3 × (1s-2)
⇒ 6.67 x 10-8 cm3 s-2 g -1
⇒ G = 6.67 × 10-11 Nm2(kg)-2
= 6.67 × 10-8 (cm)3 s-2 g -1
2.3. A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γ 2 in terms of the new units.
Ans. 1 cal = 4.2 kg m2 s–2
| SI | New System |
| n1 = 4.2 | n2 = ? |
| M1 = 1 kg | M2 = α kg |
| L1 = 1 m | L2 = β m |
| T1 = 1 s | T2 = γ second |
Dimensional formula of energy is [M1L2T–2]
Comparing with [MaLbTc], we get
a = 1, b = 2, c = – 2
Now,
$$n_2 = n_1\begin{bmatrix} \frac{M_1}{M_2} \end{bmatrix}^a\begin{bmatrix} \frac{L_1}{L_2} \end{bmatrix}^b \begin{bmatrix} \frac{T_1}{T_2} \end{bmatrix}^c \\ =4.2\begin{bmatrix} \frac{1\space kg}{α\space kg} \end{bmatrix}^{1}\begin{bmatrix} \frac{1\space m}{β\space m} \end{bmatrix}^{2} \begin{bmatrix} \frac{1\space s}{γ \space s} \end{bmatrix}^{-2} $$
or n2 = 4.2 α–1 β–2 γ2.
2.4. Explain this statement clearly :
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.
Ans. Depending on the unit (standard) of measurement, physical amounts are referred to as large or little. The distance between the two towns on Earth, for example, is measured in kilometers, whereas the distance between stars, or inter-galactic distances, is measured in parsec. The subsequent standard parsec, which is 3.08 × 1016 m or 3.08 × 1013 km, is unquestionably greater than a meter or a kilometer. As a result, interstellar or intergalactic distances are unquestionably greater than those between two cities on Earth.
(a) The size of an atom is much smaller than even the sharp tip of a pin.
(b) A Jetplane moves with a speed greater than that of a super fast train.
(c) The mass of Jupiter is very large compared to that of the earth.
(d) The air inside this room contains more molecules than in one mole of air.
(e) This is a correct statement.
(f) This is a correct statement.
2.5. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?
Ans. Dstance between the Sun and the Earth
= Speed of light in vacuum × time it takes for light to travel from Sun to Earth
= 3 × 108 m/s × 8 min 20 s
= 3 × 108 m/s × 500 s
= 500 × 3 × 108 m
The speed of light in vacuum equals unity in the new system. As a result, the new length unit is
3 × 108 m.
500 new units for the distance between the Sun and the Earth.
2.6. Which of the following is the most precise device for measuring length :
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light ?
Ans. (a) Least count of vernier callipers
$$ = \frac{1}{20} = 0.05\space \text{mm = 5 ×} 10^{-5} m \\ \text{(b) Least count of screw gauge = }\frac{\text{Pitch}}{\text{No. of divisions on circular scale}}\\ = \frac{1×10^{-3}}{100} = 1 × 10^{–5} m $$
(c) Least count of optical instruments = 6000Å (average wavelength of visible light as 6000Å) = 6 × 10-7 m. As the least count of optical instrument is least, it is the most precise device out of three instruments given to us.
Those instruments which have the minimum least count are the most precise one.
2.7. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?
Ans. As magnification,
$$ \text{m =} \frac{\text{thickness of image of hair}}{\text{real thickness of hair}} = 100 $$
and average width of the image of hair as seen by microscope = 3.5 mm
Thickness of hair = 3.5 mm/100 = 0.035 mm
2.8. Answer the following :
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread ?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?
Ans. (a) Wrap the thread around a circular pencil several times to make a coil with turns that are close together. With a metre scale, measure the length of this coil, made by thread. If n is the number of coil turns and l is the coil length, then the length occupied by each single turn, i.e. the thread thickness, is 1/n.
This is the same as the thread's diameter.
(b) On a circular scale, the least count is the pitch divided by the number of divisions. As the number of divisions on the circular scale grows, the least count decreases. As a result, precision increases. This, however, is only a hypothesis. Increasing the amount of 'turns would result in a host of issues in practice.
The poor resolution of the human eye, for example, would make observations problematic. It would be difficult to tell the closest divisions apart. Furthermore, maintaining pitch uniformity along the length of the screw would be technically challenging.
(c) A large number of observations will offer a more dependable result than a smaller number of observations due to random mistakes. This is because making a positive random error of a given size has the same probability (chance) as making a negative random error of the same magnitude. Positive and negative mistakes are likely to cancel each other out over a large number of observations. As a result, more trustworthy results can be achieved.
2.9. The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.
Ans. Here area of the house on slide = 1.75 cm2 = 1.75 × 10-4 m2 and area of the house of projector-screen = 1.55 m2
$$ \text{Area magnification =} \frac{\text{Area of screen}}{\text{Area of slide}} \\ = \frac{1.55\space m^2}{1.75×10^{-4} m^2} = 8.857× 10^3 \\ \text{Linear magnification = }\sqrt{\text{Area magnification}} \\ = \sqrt{(8.857)×10^3} \\ =94.1 $$
2.10. State the number of significant figures in the following :
(a) 0.007 m2
(b) 2.64 × 1024 kg
(c) 0.2370 g cm–3
(d) 6.320 J
(e) 6.032 N m–2
(f) 0.0006032 m2
Ans. (a) 1
(b) 3
(c) 4
(d) 4
(e) 4
(f) 4.
2.11. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Ans. As we know 4.234 × 1.005 = 4.255 m2 = 4.3 m2
Volume = (4.234 × 1.005) × (2.01 × 10-2)
= 8.55289 × 10-2 = 0.0855 m3.
2.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box.
What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?
Ans. (a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg
Since the least number of decimal places is 1, therefore, the total mass of the box = 2.3 kg.
(b) Difference of mass = 2.17 – 2.15 = 0.02 g
Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is 0.02 g.
2.13. A physical quantity P is related to four observables a, b, c and d as follows :
Ans.
$$P = \frac{a^3b^2}{\sqrt{c}d}$$
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity
P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
$$ \textbf{Ans. } As \space P = \frac{a^3b^2}{\sqrt{c}d}= a^3b^2c^{-1/2}d^{-1} $$
∴ Maximum fractional error in the measurement
$$ \frac{\Delta P}{P} = 3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+ \frac{\Delta d}{d}\\ \text{As} \frac{\Delta a}{a}= 1\%, \frac{\Delta b}{b} = 3\%, \frac{\Delta c}{c} = 4\% \space and \space \frac{\Delta d}{d} = 2\% $$
∴ Maximum fractional error in the measurement
$$\frac{\Delta P}{P} = 3 × 1\% + 2 × 3\% - \frac{1}{2} × 4\% + 2\%$$
= 3% + 6% – 2% + 2% = 11%
If P = 3.763, then DP = 13% of P
$$\frac{11P}{100} = \frac{11 × 3.763}{100} = 0.376 $$
As the error lies in first decimal place, the answer should be rounded off to first decimal place. Hence, we shall express the value of P after rounding it off as P = 3.8.
2.14. A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
(a) y = a sin 2π t/T
(b) y = a sin vt
(c) y = (a/T) sin t/a
(d) y = a (sin 2πt / T + cos 2πt / T)
(sin 2πt / T + cos 2πt / T)(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.
Ans. According to dimensional analysis an equation must be dimensionally homogeneous.
$$(a)\space y = a\space sin \frac{2\pi t}{T}\\ \text{Here, [L.H.S.] = [y] = [L]} \\ \text{and [R.H.S.]} = \begin{bmatrix}\text{a sin} \frac{2\pi t}{T}\end{bmatrix}\\ =\begin{bmatrix}\text{L sin}\frac{T}{T} \end{bmatrix}= [L] \\ \text{So, it is correct.}$$
(b) y = a sin vt
Here, [y] = [L] and [a sin vt] = [L sin (LT–1. T)]
= [L sin L]
So, the equation is wrong
$$ (c)\space y = \left(\frac{a}{T}\right)\space sin \frac{t}{a} \\ \text{Here},\space \space [y] = [L] \\ and\space \space \begin{bmatrix} \left(\frac{a}{T}\right)\space sin\frac{t}{a}\end{bmatrix} = \begin{bmatrix}\frac{L}{T} sin\frac{T}{L}\end{bmatrix} \\ = [LT^{–1} \text{sin}\space TL^{–1}]$$
So, the equation is wrong.
$$ (d)\space\space y = (a\sqrt{2}) \left(sin\frac{2\pi T}{T}+cos\frac{2\pi t}{T}\right) \\ Here, [y] = [L],\space [a\sqrt{2}] = [L] \\ and \space \begin{bmatrix}sin\frac{2\pi t}{T} + cos\frac{2\pi t}{T}\end{bmatrix} = \begin{bmatrix}sin \frac{T}{T} + cos \frac{T}{T} \end{bmatrix}$$
= dimensionless
So, the equation is correct.
2.15. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m0 of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :
$$m = \frac{m_0}{(1-v^2)^{1/2}}$$
Guess where to put the missing c.
Ans. From the given equation,
$$\frac{m_0}{m} = \sqrt{1-v^2}$$
Left hand side is dimensionless
Therefore, right hand side should also be dimensionless.
$$\text{It is possible only when} \space \sqrt{1-v^2} \text{ should be } \sqrt{1-\frac{v^2}{c^2}}. \\ \text{Thus, the correct formula is m = } m_0\left(1-\frac{v^2}{c^2}\right)^{-1/2}$$
2.16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å : 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ?
Ans. Volume of one hydrogen atom
= 4/3 πr3 (volume of sphere)
= 4/3 × 3.14 × (0.5 × 10-10)–3 m3
= 5.23 × 10-31 m3
According to Avagadro’s hypothesis, one mole of hydrogen contains 6.023 × 1023 atoms.
Atomic volume of 1 mole of hydrogen atoms
= 6.023 × 1023 × 5.23 × 10-31
= 3.15 × 10-7m3.
2.17. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large ?
Ans. Volume of one mole of ideal gas, Vg
= 22.4 litre = 22.4 × 10-3 m3
Radius of hydrogen molecule
$$= \frac{1}{2}A$$
= 0.5 A = 0.5 × 10-10 m
Volume of hydrogen molecule
= 4/3 πr3
= 4/3 × 22/7 (0.5 × 10-10)3 m3
= 0.5238 × 10-30 m3
One mole contains 6.023 × 1023 molecules.
Volume of one mole of hydrogen,
VH = 0.5238 × 10-30 × 6.023 × 1023 m3
= 3.1548 × 10-7 m3
$$\text{Now, } \space \space \frac{V_g}{V_H} = \frac{22.4 × 10^{-3}}{3.1548 × 10^{-7}} = 7.1 × 10^4 $$
The ratio is very large. This is because the interatomic separation in the gas is very large compared to the size of a hydrogen molecule.
2.18. Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the motion of the train, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
Ans. The line of sight is the path that connects a specific item to our eye. When a train moves quickly, a passenger's line of sight for neighbouring trees changes direction quickly. As a result, adjacent trees and other things appear to be moving in the opposite direction of the train. The line of sight of distant and large-scale objects, such as hilltops, the Moon, and stars, however, nearly remains unaltered (or changes by an extremely small angle). As a result, the faraway object appears to be still.
2.19. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit
≈ 3 × 1011 m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?
Ans. From parallax method we can say θ = b/D, where b = baseline , D = distance of distant object or star
Since, θ = 1″ (s) and b = 3 × 1011 m
D = b/20 = 3 × 1011/2 × 4.85 × 10-6 m
or
D = 3 × 1011/9.7 × 10-6 m
= 30 × 1016/9.7 m
= 3.09 × 1016 m = 3 × 1016 m.
2.20. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsec? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
Ans. As we know,
1 light year = 9.46 × 1015 m
4.29 light years = 4.29 × 9.46 × 1015
= 4.058 × 1016 m
Also, 1 parsec = 3.08 × 1016 m
4.29 light years = 4.508 × 1016/3.80 × 1016
= 1.318 parsec = 1.32 parsec.
As a parsec distance subtends a parallax angle of 1″ for a basis of radius of Earth’s orbit around the Sun (r). In present problem base is the distance between two locations of the Earth six months apart in its orbit around the Sun = diameter of Earth’s orbit (b = 2r).
∴ Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec).
∴ Parallax angle subtended by the star Alpha Centauri at the given basis θ = 1.32 × 2 = 2.64″.
2.21. Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Ans. Modern science necessitates extremely accurate measurements. For example, we must monitor time to a precision of 1 microsecond when launching a satellite utilizing a space launch rocket system. Again working with lasers we require length measurements to an angstrom unit (1Å = 10–10 m) or even a fraction of it. For estimating nuclear sizes we require a precision of 10-15 m. To measure atomic masses using mass spectrograph we require a precision of 10-30kg and so on.
2.22. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.
Ans. (a) The average rainfall of nearly 100 cm or 1 m is recorded by meteorologists, during Monsoon, in India. If A is the area of the country, then A = 3.3 million sq. km = 3.3 × 106 (km)2 = 3.3 × 106 × 106 m2 = 3.3 × 1012 m2
Mass of rain-bearing clouds
= area × height × density
= 3.3 × 1012 × 1 × 1000 kg = 3.3 × 1015 kg.
(b) Measure the depth of an empty boat in water. Let it be d1. If A be the base area of the boat, then volume of water displaced by boat V1 = Ad2
Let d2 be the depth of the boat in water when the elephant is moved into the boat. Volume of water displaced by (boat + elephant), V2 = Ad2 Volume of water displaced by elephant,
V = V2 – V1 = A(d2 – d1)
If ρ be the density of water, then mass of elephant = mass of water displaced by it = A(d2 – d1)ρ.
(c) Wind speed can be estimated by floating a gas-filled balloon in air at a known height h. When there is no wind, the balloon is at A. Suppose the wind starts blowing to the right such that the balloon drifts to position B in 1 second.
Now, AB = d = hθ.
The value of d directly gives the wind speed.
(d) Let us assume that the man is not partially bald. Let us further assume that the hair on the head is uniformly distributed. We can estimate the area of the head. The thickness of a hair can be measured by using a screw gauge. The number of hair on the head is clearly the ratio of the area of head to the cross-sectional area of a hair.
Assume that the human head is a circle of radius 0.08 m i.e., 8 cm. Let us further assume that the thickness of a human air is 5 × 10-5 m.
Number of hair on the head
$$=\frac{\text{Area of the head}}{\text{Area of cross-section of hair}}$$
= π(0.08)2/π(5 × 10–5)2
= 64 × 10-4/25 × 10–10
= 2.56 × 106
The number of hair on the human head is of the order of one million.
(e) We can determine the volume of the class-room by measuring its length, breadth and height. Consider a class room of size 10 m × 8 m × 4 m. Volume of this room is 320 m3. We know that 22.4l or 22.4 × 10–3 m3 of air has 6.02 × 1023 molecules (equal to Avogadro’s number).
Number of molecules of air in the classroom
= (6.02 × 1023 /22.4 × 10–3 ) × 320
= 8.6 × 1027
2.23. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m.
Ans. Given M = 2 × 1030 kg, r = 7 × 108 m
$$\therefore \text{ Volume of Sun =} 4/3πr^3 = \frac{4}{3} × 3.14 × (7 × 10^8)^3 \\= 1.437 × 10^{27} m^3 \\ \text{As } \rho = M/V, \\ \therefore \rho = \frac{2× 10^{30}}{1.437 × 10^{27}}$$
= 1391.8 kg m-3
= 1.4 × 103 kg m-3
Mass density of the Sun is in the range of mass densities of solids/liquids and not gases.
2.24. When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.
Ans. Given angular diameter
θ = 35.72 = 35.72 × 4.85 × 10–6 rad
= 173.242 × 10–6 = 1.73 × 10–4 rad
Diameter of Jupiter
D = θ × d
= 1.73 × 10-4 × 824.7 × 109 m
= 1426.731 × 103
= 1.43 × 105 km
2.25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v : tan θ = v and checks that the relation has a correct limit: as v → 0, θ→ 0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.
Ans. According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S = Dimensions of R.H.S
In relation v = tan θ, tan θ is a trigonometric function and it is dimensionless. The dimension of v is [LT-1]. Therefore, this relation is incorrect.
To make the relation correct, the L.H.S must be divided by the velocity of rain, u.
Therefore, the relation becomes
$$\frac{v}{u} = \text{tan θ}$$
This relation is correct dimensionally.
2.26. It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
Ans. Total time = 100 years
= 100 × 365 × 24 × 60 × 60 s
Error in 1 second = 0.02/100 × 365 × 24 × 60 × 60
= 6.34 × 10–12 s
Accuracy of the standard cesium clock in measuring a time-interval of 1 s is 10-12 s.
2.27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : 970 kg m–3. Are the two densities of the same order of magnitude ? If so, why ?
Ans. The diameter of sodium = 2.5 A = 2.5 × 10–10 m
Therefore, the radius is 1.25 x 10-10 m
Volume of sodium atom,
V = (4/3)πr3
= (4/3) × (22/7) × (1.25 × 10–10)3
= 8.177 × 10–30 m3
Mass of one mole atom of sodium
= 23 g = 23 × 10–3 kg
1 mole of sodium contains 6.023 × 1023 atoms
Therefore, the mass of one sodium atom,
M = 23 × 10–3/6.023 × 1023
= 3.818 × 10-26 kg
Atomic mass density of sodium,
r = M/V
= 3.818 × 10-26/8.177 × 10-30
= 0.46692 × 104
= 4669.2 kg m-3
The density of sodium in the crystalline phase is stated as 970 kg m–3.
As a result, the sodium densities in two phases are not of the same order. Atoms are closely packed in the solid state. As a result, in the crystalline phase, the interatomic separation is exceedingly small.
2.28. The unit of length convenient on the nuclear scale is a fermi : 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
Ans. Radius of the nucleus
r = r0 A1/3
r0 = 1.2 f = 1.2 × 10–15 m
Considering the nucleus is spherical. Volume of nucleus
= 4/3 πr3 = 4/3 π [r0 A1/3]3
= 4/3 πr03A
Mass of nucleus = mA
m is the average mass of the nucleon
A is the number of nucleons
Nuclear mass density
$$= \frac{\text{ Mass of nucleus}}{\text{Volume of nucleus}}$$
= mA/(4/3πr3) = 3 mA/4πr3 = 3 mA/4πr03A
= 3 m/4πr03
Using m = 1.66 × 10–27 kg and r0 = 1.2 f = 1.2 × 10–15 m in the above equation
= 3 × 1.66 × 10–27 /4 × 3.14 × (1.2 × 10–15)3
= 4.98 × 10–27/21. 703 x 10–45
= 2.29 × 1017 kg/m3
So, the nuclear mass density is much larger than atomic mass density for a sodium atom we got in 2.27.
2.29. A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth ?
Ans. We know that the speed of laser light = c = 3 × 108 m/s. If d be the distance of Moon from the earth, the time taken by laser signal to return after reflection at the Moon’s surface
$$ \text{t = 2.56 s = }\frac{2d}{c} = \frac{2d}{3× 10^8 ms^{\normalsize-1}} \\ \Rightarrow \space \space d = \frac{1}{2}×2.56×3×10^8 m$$
= 3.84 × 108 m.
2.30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s–1).
Ans. Here speed of sound in water v = 1450 m s–1 and time of echo t = 77.0 s.
If distance of enemy submarine be ‘d’ ,
then t = 2d/v
$$\therefore d = \frac{vt}{2} = \frac{1450 × 77}{2} \text{= 55825 m}$$
= 55.8 × 103 m or 55.8 km.
2.31. The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?
Ans. The time taken by light from the quasar to the observer t = 3.0 billion years = 3.0 × 109 years As 1 ly = 9.46 × 1015 m
∴ Distance of quasar from the observer
d = 3.0 × 109 × 9.46 × 1015 m
= 28.38 × 1024 m
= 2.8 × 1025 m or 2.8 × 1022 km.
2.32. It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.
Ans. From examples 2.3 and 2.4, we get θ = 1920″ and S = 3.8452 × 108 m. During the total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal. Angular
diameter of the moon,
θ = Angular diameter of the sun
= 1920″ = 1920 × 4.85 x 10–6 rad
[1″ = 4.85 × 10-6 rad]
The earth-moon distance,
S = 3.8452 × 108 m
∴ The diameter of the moon,
D = θ × S
= 1920 x 4.85 x 10-6 x 3.8452 x 108 m
= 35806.5024 x 102 m
= 3581 x 103 m 3581 km.
2.33. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Ans. The values of different fundamental constants are given below :
Charge on an electron,
e = 1.6 × 10–19 C
Mass of an electron,
me = 9.1 × 10–31 kg
Mass of a proton,
mp = 1.67 × 10–27 kg
Speed of light,
c = 3 × 108 m/s
Gravitational constant,
G = 6.67 × 10–11 Nm2 kg–2
$$\frac{1}{4\pi \varepsilon_0} = 9 × 10^9 Nm^2 C^{-2}$$
We have to try to make permutations and combinations of the universal constants and see if there can be any such combination whose dimensions come out to be the dimensions of time. One such combination is:
$$\left(\frac{e^2}{4\pi \varepsilon_0} \right)^2 \cdot \frac{1}{m_pm^2_ec^3G}$$
According to Coulomb's law of electrostatics,
$$F = \frac{1}{4\pi \varepsilon_0}\frac{(e)(e)}{r^2} \\ or, \space \space \frac{1}{4\pi \varepsilon_0} = \frac{Fr^2}{e^2}\\ or \space \space \left(\frac{1}{4\pi \varepsilon_0} \right)^2 = \frac{F^2r^4}{e^4}$$
According to Newton's law of gravitation,
$$ F = G\frac{m_1 m_2}{r^1}\\ or \space \space G = \frac{Fr^2}{m_1m_2} \\ \text{Now,} \space \begin{bmatrix}\frac{e^4}{(4\pi \varepsilon_0)^2 m_p m^2_e C^3G} \end{bmatrix} \\ \text{Now,} \space \begin{bmatrix}\frac{e^4}{(4\pi \varepsilon_0)^2 m_p m^2_e C^3G} \end{bmatrix} \\ = \begin{bmatrix} e^4\left(\frac{F^2r^4}{e^4}\right) \frac{1}{m_pm^2_ec^3} \frac{m_1m_2}{Fr^2} \end{bmatrix} \\ =\begin{bmatrix} \frac{Fr^2}{mc^3}\end{bmatrix} = \begin{bmatrix} \frac{MLT^{-2}L^2}{ML^3T^{-3}}\end{bmatrix} = [T]$$
Clearly, the quantity under discussion has the dimensions of time.
Substituting values in the quantity under discussion, we get
$$\frac{(1.6 × 10^{\normalsize-19})^4(9×10^9)^2}{(1.69×10^{\normalsize-27})(9.1 ×10^{\normalsize-31})^2(3×10^8)^3(6.67 ×10^{\normalsize-11})}\\ = 2.1 ×10^{16} \text{second} \\ = \frac{2.1 × 10^{16}}{60×60×24×365.25}\space\text{year} \\ $$
= 6.65 × 108 years
= 109 years
The estimated time is nearly one billion years.
So, the calculated age of the universe is 6 billion years. It is observed that the calculated age of the universe does not exactly coincide with the actual age of the universe, although it is very close.
NCERT Solutions for Class 11 Physics Chapter 2 Free PDF Download
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