RBSE Solutions for Class 11 Physics Chapter 2 Units and Measurements

Rajasthan Board RBSE Solutions for Class 11 Physics Chapter 2 Units and Measurements Textbook Exercise Questions and Answers.

RBSE Class 11 Physics Solutions Chapter 2 Units and Measurements

RBSE Class 11 Physics Units and Measurements Textbook Questions and Answers

Question 2.1.
Fill in the blanks:
(a) The volume of the cube of side 1 cm is equal to ................................. m³.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ................................. (mm)².
(c) A vehicle moving with speed of 18 km h^-1 covers ................................. m in 1 s.
(d) The relative density of lead is 11.3. Its density is ................................. gcm^-3 or ................................. kg.m^-3.
Answer:
(a) Volume V = l³ = (1 cm)³ = 1 cm³
= 1 x 10^-6 m³
∴ Answer is 10^-6 m^-3

(b) Total surface area of a cylinder,
A = 2πrh + 2 x πr²
= 2πr (h + r)
= 2 x \(\frac{22}{7}\)  x 2(10 + 2) cm²
= \(\frac{44 \times 2 \times 12}{7}\) cm²
= 150.85 cm² = 1.5 x 10² cm²
= 1.5 x 10² cm² = 1.5 x 10² x 10² mm²
= 1.5 x 10⁴ mm²
∴ Answer is 1.5 x 10⁴

(c) Speed v = 18 kmh^-1 = 18 x \(\frac{5}{18}\) ms^-1 = 5 ms^-1
∴ Distance covered d = vt = 5 x 1 = 5 m,
∴ Answer = 5m.

(d) ∵ Relative density of a substance = \(\frac{\text { Density of substance }}{\text { Density of water }}\)
and density of water is 1 gcm^-3.
= 11.3 x \(\frac{10^{-3} \mathrm{~kg}}{10^{-6} \mathrm{~m}^3}\)
= 11.3 x 10³ kgm^-3
∴ Answer is 11.3 g cm^-3 and 11.3 x 10³ or 1.13 x 10⁴.

Question 2.2.
Fill in the blanks by suitable conversion of units:
(a) 1 kgm²s^-2 = ......................... gcm²s^-2
(b) 1 m = ......................... ly
(c) 3.0 ms^-2 = ......................... kmh^-2
(d) G = 6.67 x 10^-11 Nm² kg^-2 = ......................... (cm)³ s^-2 g^-1
Answer:
(a) 1 kg. m².s^-2 = 1 x 10³ g x 10⁴ cm² s^-2 = 1 x 10⁷g cm². s^-2
∴ Answer is 10⁷ g cm²s^-2.

(b) 1 ly = 9.46 x 10¹⁵ m
∴ 1 m = \(\frac{1}{9.46 \times 10^1}\)ly = 0.1057 x 10^-15 ly
= 1.05 x 10^-16 = 10^-16 ly
∴ Answer is 10^-16 ly.

(c) 3.0 ms^-2 = 3.0\(\frac{\mathrm{m}}{\mathrm{s}^2}=\frac{3.0 \times 10^{-3} \mathrm{~km}}{\frac{1}{3600} \times \frac{1}{3600} \mathrm{~h}^2}\)
= 3.0 x 10^-3 km x 3600 x 3600 h^-2
= 3.0 x 36 x 360 km.h^-2
= 38880 kmh^-2
= 3.888 x 10⁴ kmh^-2
= 3.9 x 10⁴ km-h^-2
∴ Answer is 3.9 x 10⁴ km-h^-2

(d) G = 6.67 x 10^-11 Nm² kg^-2
= 6.67 x 10^-11 kg ms^-2. m² kg^-2
= 6.67 x 10^-11 s^-2 m³ kg^-1
= 6.67 x 10^-11 s^-2 x \(\frac{10^6 \mathrm{~cm}^3}{1000 \mathrm{~g}}\)
= 6.67 x 10^-8 cm³. s^-2 g^-1
∴ Answer is 6.67 x 10^-8

Question 2.3.
A calorie is a unit of heat (energy in transit) a.nd it equals about 4.2 J, where 1J = 1kg m²s^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals ß m and unit of time is γ s. Show that a calorie has a magnitude of 4.2 α^-1ß^-2γ² in terms of the new units.
Answer:
1 cal = 4.2 kg. m² s^-2
Formula used, n₂ = n₁\(\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^a\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^b\left[\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right]^c\)
Dimensional formula of energy = [M¹L²T^-2]
∴ a = 1, b = 2, c = -2

SI System	New System
n1 = 4.2	n2 = ?
M1 = 1 kg	M2 = α kg
L1 = 1 m	L2 = ß
T1 = 1 s	T2 = γ

∴ n₂ = 4.2\(\left[\frac{\mathrm{kg}}{\alpha \mathrm{kg}}\right]^1\left[\frac{\mathrm{m}}{\beta \mathrm{m}}\right]^2\left[\frac{\mathrm{s}}{\gamma \cdot \mathrm{s}}\right]^{-2}\)
= 4.2 x \(\frac{1}{\alpha} \times \frac{1}{\beta^2} \times \frac{1}{\gamma^{-2}}\)
or n₂ = 4.2 α^-1ß^-2γ²

Question 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 re-frame the following statements whenever 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.
Answer:
Without specifying an appropriate standard for comparison, we can not call a dimensional quantity large or small. As an illustration, the height of a man is very small as compared to the height of a mountain. However, height of a man is very large as compared to the diameter of atom. Morevoer, we must choose appropriate units for measurement of quantities. As an illustration, it would not be appropriate to measure height of a man in light year (ly) or angstroms.
(a) Atoms are very small objects as compared to objects of daily life.
(b) A jet plane moves with high speed compared to the speed of a pedestrian.
(c) The mass of Jupiter is very large compared to the mass of Moon.
(d) The air inside this room contains a large number of molecules compared to the number of air molecules in a small empty bottle.
(e) The statement is already correct.
(f) The statement is already correct.

Question 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?
Answer:
Given: New unit of length = 3 x 10⁸ m
Time taken by light from sun to Earth,
t = 8 min + 20 s
= 8 x 60s + 20s = (480 + 20)s
= 500 s
∴ Distance between Sun and Earth
d = v x t = c x t
= 3 x 10⁸ x 500 m
= 500 x \(\frac{3 \times 10^8}{3 \times 10^8}\) new unit
= 500 new unit

Question 2.6.
Which of the following is 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 circular scale.
(c) An optical instrument that can measure length to with in a wavelength of light.
Answer:
The device having least value of least count is the most precise device.
(a) Least count of vernier calliperes
= 1 MSD - 1 VSD
= 1 MSD - \(\frac{19}{20}\) MSD
= \(\frac{1}{20}\) MSD = 0.05 mm
= 0.005 cm

(b) Least count of screw gauge = \(\frac{\text { Pitch }}{\text { Number of division on circular scale }}\)
= \(\frac{1 \mathrm{~mm}}{100}\) = 0.01 mm
= 0.001 cm

(c) Least count of optical instrument = wavelength of light
= 10^-5 cm
= 0.00001 cm
Since the least count of optical instrument is the lest, therefore it is the most precise apparatus.

Question 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 of the thickness of hair?
Answer:
Magnification = \(\frac{\text { Observed width }}{\text { Real width }}\)
∴ Real width = \(\frac{\text { Observed width }}{\text { Magnification }}=\frac{3.5 \mathrm{~mm}}{100}\)
= 0.035 mm

Question 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 calliper. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than 5 measurements only?
Answer:
(a) Wrap the thread a number of times on a round pencil so as to form a coil having its turns touching each other closely. Measure the length of this coil, made by the thread, with a meter scale. If n be the number of turns of the coil and l be the length of the coil, then the length occupied by each single turn i. e., the thickness of the thread = \(\frac{l}{n}\). This is equal to the diameter of the thread.

(b) Least count of screw gauge is given by following relation
Least count = \(\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}\)
It is clear that on increasing the number of divisions on circular scale, the least count will decrease. Hence the accuracy is increased. However, this is only theoretical idea. Practically, on increasing the number of division circular scale, many difficulties may create. For example the low rsolution of the human eye would make observation difficult. The nearest divisions would not be clearly distinguidshed as separate. Moreover, it could be technically difficult to maintain uniformity of the pitch of the screw throughout its length.

(c) Due to random errors, a large number of observations will give a more reliable result than smaller nilmber of observations. This is due to the fact that the probability (chance) of making a positive random error of a given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of observations, positive and negative errors are likely to cancel each other. Hence more reliable result can be obtained.

Question 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?
Answer:
Areal magnification = \(\frac{\text { Area of image }}{\text { Area of object }} \)
= \(\frac{1.55 \mathrm{~m}^2}{1.75 \mathrm{~cm}^2}=\frac{1.55 \times 10^4 \mathrm{~cm}^2}{1.75 \mathrm{~cm}^2}\)
= 8857.1
∴ Linear magnification = \(\sqrt{8857.1}\) = 94%

Question 2.10.
State the number of significant figures in the following:
(a) 0.007 m²
(b) 2.64 x 10²⁴ kg
(c) 0.2370 gcm^-3
(d) 6.320 J
(f) 0.0006032 m²
Answer:
(a) 1
(b) 3
(c) 4
(d) 4
(e) 4
(f) 4

Question 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.
Answer:
Given: Length of sheet l = 4.234 m
Breadth of sheet b = 1.005 m
and thickness of sheet t = 2.01 cm
= 2.01 x 10^-2 m.
∴ Volume of the sheet
V = l x b x t = 4.234 x 1.005 x 2.01 x 10^-2 m³
= 8.55289 x 10^-2 m³
= 0.0855 m³
Surface area of the sheet
A = 2[l x b + b x t + l x t]
= 2 [4.234 x 1.005 + 1.005 x 2.01 x 10^-2 +4.234 x 2.01 x 10^-2]m²
= 2[4.25517 + 0.0202005 + 0.0851034]m²
= 2 x [4.3604739] = 8.7209478 m²
= 8.72 m²

Question 2.12
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) total mass of the box; (b) the difference in the masses of the pieces to correct significant figures?
Answer:
(a) Total mass of the box
= [2.3 + 0.02017 + 0.02015] kg
= 2.3442 kg
= 2.3 kg (because minimum number of significant figures is 2)

(b) Difference between masses of the pieces = 20.17 - 20.15
= 0.02 gram

Question 2.13.
A physical quantity P is related to four observables a, b, c and d as follows:
p = \(\frac{a^3 b^2}{\sqrt{c} d}\)
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 is calculated using the above relation turns out to be 3.763, to what value should you round off the result?
Answer:
The given formula
P = \(\frac{a^3 b^2}{\sqrt{c} d}\)
∴ Maximum relative error in value of P,

There are only two significant figures in percentage error of P. Therefore, there must be only two significant figures in the value of P. 
∵ The calculated value of P = 3.763
∴ The value of P = 3.8

Question 2.14.
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing certain periodic motion:
(a) y = a sin\(\frac{2 \pi t}{T}\)
(b) y = a sin vt
(c) y = \(\left(\frac{a}{T}\right) \sin \frac{t}{a}\)
(d)  y = (a\(\sqrt{2}\)) (sin \(\frac{2 \pi t}{T}\) + cos \(\frac{2 \pi t}{T}\))
Where (a = maximum displacement of the particle; v = speed of the particle; T = time-period of motion).
Rule out the wrong formula on dimensional grounds.
Answer:
Note: The argument of a trigonometric function i.e, angle is always dimensionless.
(a) Angle [\(\frac{2 \pi t}{T}\)] = [M⁰L⁰T⁰]
∴ |y| = |a|
Hence this equation is correct

(b) ∵ vt is not dimensionaless
Therefore this relation is not correct

(c) \(\frac{t}{a}\)  is not dimensionaless.
∴ This equation is also not correct.

(d) ∵|\(\frac{2 \pi t}{T}\)| = [M⁰L⁰T⁰] ∴ y = a\(\sqrt{2}\)
Hence this equation is correct.

Question 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 = \(\frac{m_0}{\left(1-v^2\right)^{1 / 2}}\)
Guess where to put the missing c.
Answer:
The given relation can be written as:
\(\frac{m_0}{m}=\sqrt{1-v^2}\)
L.H.S. of this relation \(\frac{m_0}{m}\) is dimensionless, therefore R.H.S. of the relation (i.e, \(\sqrt{1-v^2}\)) should also be dimensionless for validity of the relation and this is possible only when we write R.H.S. as \(\sqrt{1-\frac{v^2}{c^2}}\).
Thus the correct formula will be as given below:

Question 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 hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Answer:
Given: Diameter of hydrogen atom.
D = 1 x 10^-10 m
∴ Radius of hydrogen atom r = \(\frac{D}{2}\) = 0.5 x 10^-10 m
Therefore volume of one hydrogen atom
V₁ = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) x 3.14 x (0.5 x 10^-10)³ m³
= 5.236 x 10^-31 m³
Number of hydrogen atoms in 1 mole of hydrogen = N (Avogadro's number)
= 6.023 x 10²³
∴ Volume of 1 mole of hydrogen atoms.
V = NV₁ = 6.023 x 10²³ x 5.236 x 10^-31
or V = 3.15 x 10^-7 m³

Question 2.17.
One mole of an ideal gas at standard temprature 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 lA). Why is this ratio so large?
Answer:
Vmole = 22.4L = 22.4 x 10^-3 m³
Radius of hydrogen molecule
r = \(\frac{1}{2}\) Å = 0.5 x 10^-10 m
∴ Volume of one molecule of hydrogen
V₁ = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) x 3.14 x (0.5 x 10^-10)³
= 0.5236 x 10^-30 m³
∴ Volume of 1 mole molecule of hydrogen
Vmolecule = N.V₁ = 6.023 x 10²³ x 0.5236 x 10^-30 m³
= 3.153 x 10^-7 m³
∴ \(\frac{V_{\text {mole }}}{V_{\text {molecule }}}=\frac{22.4 \times 10^{-3}}{3.153 \times 10^{-7}}\) = 7.1 x 10³ ≈ 10⁴
The reason of this ratio to be so high is that the intermolecular space is very much large in comparison to size of molecule.

Question 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 direction opposite to the trains motion, but the distant objects (hill tops, the moon, the stars etc) seem to be stationary. (In fact since you aware that you are moving, these distant objects seem to move with you).
Answer:
The visual angle subtended by near objects at the eyes of observer is more than that by distant objects. When observer will move, the change in visual angle by near objects will be more than that by distant objects. This is why the near objects seen to move in opposite direction to that of observer’s motion while for or distant objects seen to move with observer.

Question 2.19.
The principle of ‘parallax’ 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 x 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 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 base line equal to the distance from the Earth to the sun. How much is a parsec in terms of meters?
Answer:
Given: θ = 1" = \(\left(\frac{1}{60}\right)^1\)
= \(\left(\frac{1}{60 \times 60}\right)^{\circ}=\left(\frac{1}{3600}\right)^{\circ}\)

= \(\frac{\pi}{3600 \times 180}\) rad
Diameter AB of orbit = 3 x 10¹¹ m.
∴ Radius l of the orbit = \(\frac{3}{2}\) x 10¹¹ m
Now parallax angle
θ = \(\frac{l}{r}\) or r = \(\frac{l}{\theta}\)
∴ r = \(\frac{3 / 2 \times 10^{11} \times 3600 \times 180}{\pi}\) m
= \(\frac{3 \times 10^{11} \times 3600 \times 180}{2 \times 3.14}\)
= 3.09 x 10¹⁶ m
⇒ 1 parsec = 3.09 x 10¹⁶ m

Question 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?
Answer:
(a) We know that : 1 ly = 9.46 x 10¹⁵ m
∴ 4.29 ly = 4.29 x 9.46 x 10¹⁵ m= 4.058 x 10¹⁶ m
∵ 1 Parsec = 3.08 x 10¹⁶ m
∴ 4.29 ly = \(\frac{4.058 \times 10^{16}}{3.08 \times 10^{16}}\) = 1.318 Parsec
= 1.32 Parsec

(b) The distance between two observations after time interval of 6 months, is equal to diameter of the orbit of Earth around the Sun.
∴ Diameter of the orbit b = 2 x radius of orbit
or b = 2 x 1AU = 2AU
= 2 x 1.496 x 10¹¹ m
Distance of star from Earth
r = 4.058 x 10¹⁶ m
∴ Parallax angle, θ = \(\frac{b}{r}\)
or θ = \(\frac{2 \times 1.496 \times 10^{11}}{4.058 \times 10^{16}}\) = 7.37 x 10^-6 rad
= \(\frac{7.37 \times 10^{-6} \times 180 \times 3600}{3.14}\) = 1.5" (second)
or Q = 1.5" (second)

Question 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.
Answer:
Precise measurement of physical quantities like length, mass and time are basic requirements for development of quantitative laws of Physics or any other science. The following examples illustrate this point.

1. In measurement of distance of Moon from the Earth by laser beam, a very accurate measurement of time is required. This accuracy should be of the order of 10^-9 s.
2. For radio and laser therapy, the precise location,size and mass of a tumor must be known. The precision should be the order of 10^-9 m.
3. the precise measurement of the mass of an atom is made by mass spectrometer.
4. This has made it possible to verify one prediction of the theory of relativity i. e. the variation of mass of a particle with velocity.
5. The thickness of the various types of materials used in semiconductor technology should be known very precisely.

Question 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.
Answer:
(a) First of all we will count the total rain in India, after knowing this we will able to know the mass of the water. Altimately we can guess the mass of clouds.

(b) To guess the mass of elephant, we take a boat of known base-area A Measure the portion of boat' in-side the water. Suppose this distance is The volume of displaced water V₁ = Ax₁. Now the elephant is loaded on the boat. As a result the boat will immerse in water more: Now again measure the part of boat in water, let it be x₂.
∴ Volume of water displaced by boat with elephant
V₂ = Ax₂
∴ The volume of water displaced by elephant
V = V₂ - V₁ = Ax₂ - Ax₁ = A(x₂ - x₁)
∴ Mass of elephant = V.ρ = A (x₂ - x₁ ).ρ
where ρ is density of water.

(c) Speed of air can be guessed by pressure exertedby air in storm.

(d) To know the number of hairs on our head, we gues the number of hairs per unit area of our head. By guessing the total area of our head, we can estimate the total numebr of hairs on our head.

(e) We can know the density of air, therefore we can judge the number of air molecules per cm3. In this way we can guess the number of air molecules in our class-room by measuring the volume of class-room.

Question 2.23.
The Sun is a hot plasma (ionized matter) with its inner core at 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 liquid's or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0 x 10³⁰ kg, radius of the Sun = 7.0x10 s m.
Answer:
Given: M = 2 x 10³⁰ kg, r = 7x 10⁸ m,
Volume of the Sun,
V = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) x 3.14 x (7 x 10⁸)³ m³
= 1.437 x 10²⁷ m³
∴ Density of Sun, ρ = \(\frac{M}{V}=\frac{2 \times 10^{30}}{1.437 \times 10^{27}}\)
or ρ = 1391.8 kgm^-3 = 1.4 x 10³ kg-m^-3
Therefore the density of Sun is in the density range of solids and liquids. It does not come in the density range of gases because due to inner layer of Sun gravitational attraction is effective on outer layers, therefore inner density becomes greater than outer density, i.e., the density of plasma increases.

Question 2.24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angualr diameter is measured to be 35.72"of arc. Calculate the diameter of Jupiter.
Answer:
Given: Angular measurement θ = 35.72"
= 35.72 x 4.86 x 10^-6 rad.
= 1.73 x 10^-4 rad
Distance of Jupiter from Earth,
d = 824.7 x 10⁶ km = 824.7 x 10⁹ m
∴ Diameter of Jupiter,
D = θ x d = 1.73 x 10^-4 x 824.7 x 10⁹ m
= 1426.731 x 10⁵ m
= 1.43 x 10⁸ m

Additional Exercises

Question 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: tanθ = v and checks that the relation has a correct limit: 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.
Answer:
The given relation tanθ = v is wrong dimensionally because the L.H.S. is dimensionless while the dimensions of R.H.S. are [M⁰L¹T^-1] i.e. it does not obey the principle of homogeneity of dimensions.
If the R.H.S. of equation tanθ = v is corrected as tanθ = \(\frac{v}{u}\), where u is the speed of rain fall, then it will become dimensionally corrected.

Question 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?
Answer:
Total time = 100 years = 100 x 24 x 3600 s
∴ Error in 1 s = \(\frac{0.02}{100 \times 24 \times 3600}\) = 6.36 x 10^-12 s
Therefore accuracy in one time interval is from 10¹¹ to 10¹².

Question 2.27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (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 kgm-3. Are the two densities of the same order of magnitude? If so, why?
Answer:
Volume of sodium atom,
V = \(\frac{4}{3}\)πr³ = \(\frac{4}{3}\) x 3.14 x (2.5 x 10^-10)³
= 65.5 x 10^-30 m³
Number of sodium atoms in one mole (i.e. 23 x 10^-3 kg)
= N = 6.023 x 10²³
∴ Mass of one sodium atom
= M = \(\frac{23 \times 10^{-3}}{6.023 \times 10^{23}}\) = 3.82 x 10^-26 kg
∴ Density of sodium atom,
ρ = \(\frac{M}{V}=\frac{3.62 \times 10^{-26}}{65.5 \times 10^{-30}}\) = 0.6 x 10³ kg.m^-3
= 600 kg.m^-3
Density of crystalline sodium = 970 kgm^-3
= 0.970 x 10³ kgm^-3
Thus the order of magnitude of both densities is same equal to 3.
In the solid phase, the atoms are tightly packed. So, the atomic mass density is close to the mass density of the solid.

Question 2.28.
The unit of length convenient on the nuclear sclae is a fermi: 1 f = 10^-15 m. Nuclear size obey roughly the following empirical relation: r = r₀A^1/3
where r is the radius of the nucleus, A its mass number and r₀ is a constant equal to about 1.2 fermi. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of a sodium atom obtained in exercise 2.27.
Answer:
Mass of nucleus in terms of mass number A = (1.66 x 10^-27)Akg
Radius r = r₀A^1/3 = (1.2 x 10^-15)A^1/3
∴ Volume of the nucleus, V = \(\frac{4}{3}\)πr³
or V = \(\frac{4}{3}\) x 3.14 x [(1.2 x 10^-15) A^1/3]³ m³
= 7.24 x 10^-45 A.m³
∴ Density of the nucleus;
ρ = \(\frac{M}{V}=\frac{1.66 \times 10^{-27} \mathrm{~A}}{7.24 \times 10^{-45} \mathrm{~A}}\)
= 2.29 x 10¹⁷ kgm^-3 = 0.3 x 10¹⁸ kg.m^-3
The ratio of the mass density of sodium nucleus to the average mass density of sodium atom.
= \(\frac{2.3 \times 10^{17}}{4.67 \times 10^3}\) = 4.92 x 10¹³
So, the nuclear mass density is nearly 50 million times more than the atomic mass density for a sodium atom.

Question 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?
Answer:
Radius of lunar orbit
d = \(\frac{1}{2}\) v x t = \(\frac{3 \times 10^8 \times 2.56}{2}\) = 3.84 x 10⁸ m

Question 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 ms^-1).
Answer:
The distance of energy submarine,
d = \(\frac{1}{2}\) v x t
Given: v = 1450 ms^-1 and t = 77.0 s
∴ d = \(\frac{1450 \times 77.0}{2}\) m
or d = 55825 m = 55.825 km
or d = 55.8 km

Question 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? 
Answer:
Given: t = 3 billion years
= 3 x 10⁹ x 365.25 x 24 x 3600 s
Velocity of hight, c = 3 x 10⁸ ms^-1
∴ Distance of quasar,
d = c x t = 3 x 10⁸ x 3 x 10⁹ x 365.25 x 24 x 3600
= 2840184 x 10¹⁹ m = 2840184 x 10¹⁶ km
= 2.84 x 10²² km

Question 2.32.
It is a well known fact that during 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.
Answer:

According to question, AB is diameter of Moon and CD is the diameter of Sun, E is the situation of Earth.
∴ \(\frac{A B}{C D}=\frac{B E}{D E} \Rightarrow A B=\frac{B E}{D E} \times C D\)
But BE = 3.8452 x 10⁸m; DE = 1.496 x 10¹¹ m and CD = 13.92 x 10⁸ m
∴ AB = \(\frac{3.8452 \times 10^8 \times 13.92 \times 10^8}{1.496 \times 10^{11}}\)
= 3.58x 10³ km

Question 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 (i.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?
Answer:
In Atomic Physics a quantity is obtained from fundamental constants (c = speed of light, e = electronic charge, mass of electron me, mass of proton m_p) and gravitational constnat G. This quantity has the dimensions of time, this quantity is:
t = \(\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^2 \times \frac{1}{m_p \times m_e^2 c^3 G}\)
= \(\frac{e^4}{\left(4 \pi \varepsilon_0\right)^2} \times \frac{1}{m_p m_e^2 c^3 G}\)
∵ e = 1.6 x 10^-19 C; \(\frac{1}{4 \pi \varepsilon_0}\) = 9 x 10⁹
c = 3 x 10⁸ ms^-1; G = 6.67 x 10^-11 Nm² kg^-2
∴ t = \(\begin{gathered} \frac{\left(1.6 \times 10^{-19}\right)^4 \times\left(9 \times 10^9\right)^2}{\left(1.67 \times 10^{-27}\right) \times\left(9 \times 10^{-31}\right)^2 \times\left(3 \times 10^8\right)^3} \\ \times\left(6.67 \times 10^{-11}\right) \end{gathered}\)
= 2.18 x 10¹⁶ second
This value of t is of order of age of Earth.