NCERT Solutions for Class 11 Physics Chapter 2

Q.1 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 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.

One such quantity consisting of some fundamental constants and having dimensions of time is given as: t = e 2 4πε 0 2 × 1 m p m e 2 c 3 G Here,  ε 0  represents absolute permittivity e representing charge of electrons = 1.6 ×  10 -19  C m e  representing mass of electrons = 9 .1×10 -31  kg m p  representing mass of protons = 1 .67×10 -27  kg 1 4πε 0 = 9×10 9   Nm 2 C -2 c representing speed of light = 3×  10 8   ms -1  and G = 6 .67×10 -11   Nm 2 kg -2 Putting the above values in the given relation, we obtain: t = 1 .6×10 -19 4 × 9×10 9 2 9 .1×10 -31 2 ×1 .67×10 -27 × 3×  10 8 3 ×6 .67×10 -11  = 2 .18×10 16  s

Q.2 It is a well known fact that during a 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.

The given figure shows the positions of Sun, Moon and Earth  at the time of a lunar eclipse. Here, distance between Moon and Earth, UE = 3 .84×10 8  m Distance between Sun and Earth, SE = 1 .496×10 11  m Diameter of Sun, AB = 1 .39×10 9  m ∵ ΔEMN and ΔEAB are similar ∴ AB MN   = SE UE 1 .39×10 9 m MN = 1 .496×10 11 m 3 .84×10 8 m ∴MN = 1.39×3.84 1.496 ×10 6 m = 3 .57×10 6  m ∴Diameter of Moon  =  3 .57×10 6  m

Q.3 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 yet not been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Ans.

Given, time taken by quasar light to reach Earth, t = 3 billion years t = 3×10 9 ×365×24×60×60 s Velocity of light in vacuum,  c = 3×10 8   ms -1 As, distance = velocity×time ∴Distance between Earth and quasar, d = ct d = 3×10 8 ms -1 ×3×10 9 ×365×24×60×60 s = 2 .84×10 22  km

Q.4 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).

Ans.

Let distance between the ship and the enemy submarine = s Speed of sound in water, v = 1450  ms -1 Time delay between transmission and reception of SONAR waves, t = 77 s From the relation: s = v×t 2 ∴Distance between ship and enemy submarines is given as: s = 1450 ms -1 ×77 s 2 = 55825 m

Q.5 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.

Given, time taken by the LASER beam to return earth , t = 2.56 s Velocity of light in vacuum,  c = 3×10 8   ms -1 Radius of lunar orbit = Separation between Earth and Moon Let radius of lunar orbit be x From the relation: x = c×t 2 x = 3×10 8 ms -1 ×2.56 s 2 = 3 .84×10 8  m

Q.6 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 = 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 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.

Given, radius of nucleus,  r = r 0 A 1/3 ∴Volume of nucleus, V = 4 3 π r 3 = 4 3 π r 0 A 1/3 3 = 4 3 π r 0 3 A   ∵ the mass of the nucleus = Mass number ∴ M = A amu = A × 1 .66×10 -27  kg ∴Density of sodium nucleus, ρ = 3×1 .66×10 -27 kg 4×3.14× 1 .2×10 15  m 3 ρ = 4.98 21.71 ×10 18 kgm -3 = 2 .29×10 17   kgm -3 Density of sodium atom, ρ’ = 4 .67×10 3   kgm -3 ∴ ρ ρ’ = 2 .29×10 17 kgm -3 4 .67×10 3   kgm -3 = 4 .9×10 13

Q.7 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 density of sodium in its crystalline phase: 970 kgm^-3. Are the two densities of the same order of magnitude? If so, why?

Ans.

Given, size of sodium atom = Diameter of sodium atom = 2.5 Å ∴Radius of sodium atom, r = 1 2 ×2.5 Å = 1.25 Å = 1.25 ×  10 -10  m Volume of sodium atom, V = 4 3 πr 3  = 4 3 × 22 7 × 1 .25×10 -10 m 3 As per Avogadro hypothesis, Number of atoms in one mole of sodium = 6 .023×10 23 Mass of one mole of sodium = 23  g = 23×10 -3  kg ∴Mass of one atom of sodium = 23×10 -3 6 .023×10 23 kg ∴Density of sodium atom = mass volume Density of sodium atom = 23×10 -3 6 .023×10 23 kg 4 3 × 22 7 × 1 .25×10 -10 m 3  = 4 .67×10 3   kgm -3 Due to the inter-atomic separation in the crystalline phase, the two densities are not of the same order.

Q.8 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.

Given, error in time of cesium clocks = 0.02 s Time taken = 100 years = 100×365 1 4 ×24×60×60 s Error in time of cesium clocks in 100 years = 0.02 s ∴ Error in time of cesium clocks in 1 s, Δt = 1 s× 0.02 s  100×365 1 4 ×24×60×60 s Δt = 6 .337×10 -12  s ≈ 10 -11 s ∴Accuracy of a standard cesium clock for measuring the time interval of 1 s =   10 -11  s

Q.9 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.

The given relation is incorrect dimensionally. One way to make the given relation correct is by dividing the R.H.S of the given relation by the speed of rainfall v’. The given relation becomes: tanθ = v v’ This relation is dimensionally correct.

Q.10 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, distance between Jupiter and Earth, r = 824 .7×10 6  km Angular diameter of Jupiter, θ = 35.72″ θ = 35.72 60×60 × π 180 rad Let diameter of Jupiter be l. From the relation: θ = l r ∴l = θr l = 35.72 60×60 × π 180 rad×824 .7×10 6  km = 1 .429×10 5  km

Q.11 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,Mass of the Sun, M = 2 .0×10 30  kg Radius of the Sun, R = 7 .0×10 8  m ∴ Volume of the Sun, V = 4 3 πR 3 ∴V = 4 3 × 22 7 × 7 .0×10 8 m  = 1 .4377×10 27   m 3 ∴Density of Sun = Mass Volume Density of Sun = 2 .0×10 30 kg 1 .4377×10 27 m 3 = 1 .3911×10 3   kgm -3 The density of sun is of order of the density of solids and liquids. The enormous gravitational attraction of the inner layers on the outerlayer of Sun is responsible for the high density of Sun.

Q.12 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) During the monsoon season, a metrologist records nearly 215 cm of rainfall in India. ∴Height of water column, h = 215 cm = 2.15 m Area of the Country, A = 3.3 ×  10 12   m 2 ∴Volume of rain water, V = A × h = 7.09 ×  10 12   m 3 ∵ density of rain water, ρ = 1 ×  10 3   kgm -3 ∴Mass of rain water = ρV = 7.09 ×  10 15  kg

(b) Let us assume a boat of known base area A. Let depth of boat in sea be  d 1 Volume of water displaced by boat,  V 1 = Ad 1 Now move the elephant on the boat Let depth of boat when elephant is moved on the  boat be d 2 ∴Volume of water displaced by boat and elephant,  V 2 = Ad 2 ∴Volume of water displaced by elephant,  V = V 2 – V 1 = A d 2 – d 1 Let density of water be ρ Mass of elephant = Mass of water displaced by elephant = Vρ = A d 2 -d 1 ρ

(c) With the help of an anemometer we can measure the wind speed. The wind speed is given by the rotation made by the anemometer in one second.

(d) Let area of head carrying hair be A. The radius of a hair can be measured with help of a screw gauge. Let radius of a hair be r ∴Area of one hair = π r 2 Considering that the uniform distribution of hair over the head, Number of strands of hair = Total surface area Area of one hair = A π r 2

(e) Volume occupied by one mole of air at NTP = 22.4 l = 22 .4×10 -3   m 3 Number of molecules in one mole of air = 6 .023×10 23   Let volume of the room be V ∴Number of air molecules in room of volume V = 6 .023×10 23 22 .4×10 -3 m 3 ×V = V×1 .35×10 28   m -3

Q.13 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.

It is true that precise measurements of physical quantities are essential for the development of laws of Physics or any other Science. For example:

(a) In various physical and chemical processes, ultra-shot LASER pulses are used to determine small time intervals.

(b) Mass spectrometer can be used to measure the mass of atoms precisely.

(c) The inter-atomic separation can be determined by using X-ray spectroscopy.

Q.14 The nearest star to our solar system is 4.29 light years away. How much is the 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.

Distance between the star and the solar system=4.29 ly 1 light year = speed of light × 1 year 1 light  year = 3×10 8 ms -1 ×365×24×60×60 s = 94608×10 11  m ∴4.29 ly = 4 .29×94608×10 11  m = 405868 .32×10 11  m ∵1 parsec = 3 .08×10 16  m ∴4.29 ly = 405868 .32×10 11 3 .08×10 16 = 1.32 parsec Here, diameter of Earth’s orbit,  d = 3×10 11  m Distance of Star from the Earth, D = 405868.32 ×  10 11  m From the relation: θ = d D ∴θ = 3×10 11 405868 .32×10 11 = 7 .39×10 -6  rad ∵ 1 sec = 4 .85×10 -6  rad ∴7 .39×10 -6  rad = 7 .39×10 -6 4 .85×10 -6 = 1.52″

Q.15 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.

Given, dameter of Earth’s  orbit = 3×10 11  m ∴Radius of Earth’s orbit,r = 1 .5×10 11  m Given that, parallax angle, θ = 1″ θ = 1′ 60 = 1° 60×60 = π 180 × 1 60×60 rad = 4 .847×10 -6  rad Let the distance of star be D From the relation: θ = r D ∴D = r θ D = 1 .5×10 11 4 .847×10 -6 = 3 .09×10 16  m ∴ 1 parsec≈3 .09×10 16  m

Q.16 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 train’s motion, but the distant objects (hill tops, the moon, the stars etc.) seem to be stationary.

Ans.

The imaginary line joining an object to the eye is called the line of sight. When a train is moving rapidly, the line of sight of a nearby house changes its direction very rapidly. Therefore, the house seems to move rapidly in a direction opposite to the train’s motion.

Since distant objects such as the hill top, the moon, the stars, etc. are extremely large distances apart, therefore, the line of sight does not change its direction rapidly and they appear to be stationary.

Q.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.

Given, size of hydrogen molecule = 1 Å ∴Atomic radius, r = 0 .5×10 -10  m Volume of each H atom, V = 4 3 πR 3 V = 4 3 ×3.14× 0 .5×10 -10  m 3 = 5 .236×10 -31   m 3 Since number of hydrogen atoms in 1 g mole of hydrogen, Avogadro’s Number,  N 0 = 6 .023×10 23 ∴Atomic volume of 1 g mole of H  atom = N 0 ×V = 5 .236×10 -31 m 3 ×6 .023×10 23 = 3 .154×10 -7   m 3 Molar volume = 22.4 L = 22 .4×10 -3   m 3 Molar volume Atomic volume = 22 .4×10 -3   m 3 3 .154×10 -7   m 3 = 7 .1×10 4 This ratio is large because of large intermolecular separations.

Q.18 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo 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 = m 0 1 – v 2 1/2 Guess where to put the missing c.

Ans.

∵ As per the principle of Homogeneity of dimensions, dimensions of M, L, and T on one side of dimensional physical relation should be equal to their respective dimensions on the other side of the relation. ∴On RHS, the denominator  1 – v 2 1/2  must be dimensionless. ∴In place of  1 – v 2 1/2 , there should be  1 – v 2 c 2 1/2     ∴The correct relation is: m = m 0 1 – v 2 c 2 1/2 Sr19 (1702789) Instruction Question

Q.19 A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion:

(i) y = a sin 2πt T

(ii) y = a sin vt 

(iii) y = a T sin t a  

(iv) y = a 2 [ sin  2πt T + cos 2πt T ]

(a = maximum displacement of particle, v = speed of particle, T = time-period of motion). Rule out the wrong formulae on dimensional grounds.

Ans.

Angle is a dimensionless physical quantity. (i)  2πt T = T T = 1 = M 0 L 0 T 0     The above formula is dimensionless. (ii) vt = LT -1 T   = L = M 0 L 1 T 0 The above formula is not dimensionless. (iii)  t a = T L = L -1 T 1 (iv)  2πt T = T T = 1 = M 0 L 0 T 0    The above formula is not dimensionless. ∴Formulae (ii) and (iii) are wrong. Sr20 (1702761) Instruction Question

Q.20 A physical quantity P is related to four observables a, b, c and d as follows:P = a 3 b 2 ( 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 is 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?

Ans.

Given, P = a 3 b 2 c d   ∵ Fractional error in a quantity raised to power n  is n times the fractional error in the individual quantity, ∴Maximum fractional error in P = ΔP P ΔP P = 3 Δa a + 2 Δb b + 1 2 Δc c + Δd d ΔP P = 3 1 100 + 2 3 100 + 1 2 4 100 + 2 100 = 13 100 = 0.13 ∴Percentage error in P = ΔP P ×100 = 0.13×100 = 13% Since the result has two significant figures, ∴ The result, P = 3.763 would be rounded off to 3.8. Sr21 (1702746) Instruction Question

Q.21 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.

Given, mass of grocer’s box, m = 2.3 kg Mass of first gold piece,  m 1 = 20.15 g = 0.02015 kg Mass of the second gold piece,  m 2 = 20.17 g = 0.02017 kg (a) Total mass of box,  m T = m + m 1 + m 2 m T = 2.3 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg Since the result is accurate only up to one place of decimal, ∴On rounding off, we obtain: Total mass of box = 2.3 kg (b) Difference in masses of the gold  pieces = m 2 – m 1 = 20.17 g – 20.15 g = 0.02 g

Q.22 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.

Given, radius of each hydrogen atom, r = 0.5 Å = 0 .5×10 -10  m ∴Volume of each hydrogen atom, V = 4 3 πR 3 V = 4 3 ×3.14 0 .5×10 -10 3 = 5 .236×10 -31   m 3 ∵ Number of H atoms in 1 g mole of hydrogen = 6 .023×10 23 ∴Atomic volume of 1 g mole of H atom = 6 .023×10 23 ×V = 6 .023×10 23 ×5 .236×10 -31   m 3 = 3 .154×10 -7   m 3

Q.23  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.

Given, length of the metal sheet, l = 4.234 m Breadth of the metal sheet, b = 1.005 m Thickness of the metal sheet, t = 2.01 cm = 2 .01×10 -2 m Area of the metal sheet, A = 2 l×b + b×t + t×l A = 2 4.234 m×1.005 m + 1.005 m×2 .01×10 -2 m + 2 .01×10 -2 m×4.234 m  A = 2 4 .3604739 m 2 = 8.7209478  m 2 Since area and volume can contain a maximum of three significant figures,  ∴ On rounding off, we obtain Area = 8.72  m 2 Volume of metal sheet, V = l×b×t V = 4.234 m×1.005 m×0.0201 m = 0.0855289 = 0.0855  m 3 Sr24 (1702663) Instruction Question

Q.24 State the number of significant figures in the following:
(a) 0.007 m²
(b) 2.64 × 10²⁴ kg
(c) 0.2370 gcm^-3
(d) 6.320 J
(e) 6.032 Nm^-2
(f) 0.0006032 m²

Ans.

As per the rules of finding significant figures, the number of significant figures is given below:

(a) one

(b) three

(c) four

(d) four

(e) four

(f) four

Q.25 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.

Given, area of the object = 1.75  cm 2  area of the image = 1.55  m 2 = 1 .55×10 4 cm 2 Areal magnification is given as:  m A = area of image area of object   = 1 .55×10 4 cm 2 1 .75 cm 2   = 8857 Linear magnification of projector screen arrangement = m A = 8857   = 94.1

Q.26 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) Since the diameter of a thread is very small, therefore, it cannot be measured by using a metre scale. Wind a number of turns of the thread on the metre such that the turns are very closely touching each other. Now measure the length (L) of the windings of thread on the scale having n number of turns.

∴  diameter of thread= Length of thread Number of turns = l n

(b) Least count of screw gauge = Pitch of screw gauge Number of divisions on the circular scale

Since on increasing the number of divisions on the circular scale, the least count decreases, therefore, accuracy increases.

However, due to the low resolution of the human eye, it may not be possible to take the reading precisely.

(c) Since the probability of making a positive random error of definite magnitude is equal to that of making a negative random error of the same magnitude, therefore, when the number of observations is large, random errors are probable to cancel and the result may become more accurate. Therefore, a large number of observations (100) will give more accurate results than a smaller number of observations.

Q.27 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.

Here, magnification of microscope = 100

Observed width of hair, y = 3.5 mm Let real width of hair be x Magnification of microscope is given as:

m = observed width real width ∴m = y x ∴Real width, x = y m x = 3.5 mm 100 = 0.035 mm Sr28 (1702657) Instruction Question

Which of the following is the most precise device for measuring length :
(a) a vernier calipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1mm 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 the given vernier callipers    = Value of one main scale division Total number of divisions on the sliding scale = 1 mm 20 = 0.05 mm = 0.005 cm

(b) Least count of screw gauge = Pitch of screw gauge Total no. of divisions on circular scale = 1 100 mm = 0.01 mm = 0.001 cm

(c) Here, wavelength of light, λ≈ 10 -5 cm = 0.00001 cm The most precise device is that whose least count is minimum. ∴An optical instrument is the most precise device.

Q.28 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 of light takes 8 min and 20 s to cover this distance?

Ans.

Given, velocity of light in vacuum, c = 1 new unit of length  s -1 Time taken by sun light to reach the Earth, t = 8 min 20 s t = 8×60 s + 20 s = 500 s ∴ Distance between Sun and Earth is given as: x = c×t = 1 new unit of length  s -1 ×500 s ∴Distance between Sun and Earth = 500 new units of length Sr30 (1702654) Instruction Question

Q.29 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.

(i) Since a dimensionless quantity can be large or small in comparison to some standard reference, therefore, the given statement is true. For example, the angle is a dimensionless quantity.

∠θ = 60° is greater than ∠θ = 30° , but smaller than ∠θ = 90°

(ii) (a) An atom is smaller as compared to the sharp tip of a pin.

(b) A jet plane moves with a speed greater than a superfast train.

(c) The mass of Jupiter is very large as compared to the mass of Earth.

(d) The air inside this room contains a large number of molecules as compared to that present in one mole of air.

(e) The given statement is already correct.

(f) The given statement is already correct.

Q.30 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kgm² 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^–1β^–2γ² in terms of the new units.

Ans.

Given, 1 calorie = 4.2 J = 4.2 kg  m 2   s -2 Since new unit of mass = α kg ∴1 kg = 1 α  new unit of mass =  α -1  new unit of mass In the same way, 1 m =  β -1  new unit of length  1 s = γ -1  new unit of time Substituting these values in (i), we obtain: 1 calorie = 4.2 α -1  new unit of mass β -1  new unit of length 2 γ -1  new unit of time -2 1 calorie = 4.2 α -1 β -2 γ 2  new unit of energy

Q.31 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 × 10^–11 Nm²(kg)^–2 = …. (cm)³ s^–2g^–1

Ans.

(a) 1  kgm 2 s -2 = 1×1000 g 100 cm 2 s -2 = 10 7   gcm 2 s -2

(b) Light year is the distance travelled by light in one year. 1 light year = 9 .46×10 15  m ∴ 1 m = 1 9 .46×10 15  light year = 1 .053×10 -16  light year

(c) ∵1  m = 10 -3  km 1 s = 1 3600  h ∴1  s -1 = 3600  h -1 ∴3  ms -2 = 3×10 -3  km 3600 h -2 3  ms -2 = 3×10 -3 ×3600×3600  kmh -2 = 3 .888×10 4   kmh -2

(d) G = 6 .67×10 -11   Nm 2 kg -2 = 6 .67×10 -11   kgms -2 m 2 kg -2 G = 6 .67×10 -11   m 3 s -2 kg -1 = 6 .67×10 -8   cm 3 s -2 g -1

Q.32 Fill in the blanks
(a) The volume of a 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 a speed of 18 kmh^–1 covers….m in 1 s.
(d) The relative density of lead is 11.3. Its density is ….gcm^–3 or ….kgm^–3.

Ans.

(a) Given, length of side of  cube, L = 1 cm = 10 -2  m   ∴ Volume of cube = L 3 = (10 -2 m) 3 = 10 -6   m 3

(b) Given, radius of cylinder, r = 2.0 cm = 20 mm    Height of cylinder, h = 10.0 cm = 100 mm Surface area of solid cylinder is given as: A = (2πr)h = 2× 22 7 ×20×100  mm 2 Surface area of solid cylinder = 1 .26×10 4   mm 2

(c)   Given, speed of vehicle, v = 18  kmh -1 v = 18×1000 m 60×60 s = 5  ms -1