Class 11 Physics Chapter 2 Units and Measurements

NOTES ON Units and Measurements

Physical quantities:

Quantities that can be measured also, in term of which laws of Physics can be described are called physical quantities. Example: mass, length etc.

Unit:

Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard known as unit.

Fundamental (or Base) units & Derived units:

The units for the base or fundamental quantities are called base or fundamental units.

There are 7 fundamental quantities and they are:

• Length

• Mass

• Time

• Electric Current

• Thermodynamic Temperature

• Amount of Substance

• Luminous Intensity

The units of all other physical quantities can be expressed as combinations of the base units. Such units obtained for the derived quantities are called derived units.

System of Units:

A complete set of these units, both the base units and derived units, is known as the system of units.

The international system of units:

Three common systems of units frequently used in Physic are:

• CGS system (In CGS system they were centimetre, gram and second respectively)

• FPS (or British) system (In FPS system they were foot, pound and second respectively)

• MKS system (In MKS system they were metre, kilogram and second respectively)

• SI System of Units

SI System of Units:

The system of units which is at present internationally accepted for measurement is the Système Internationale d’ Unites (French for International System of Units), abbreviated as SI.

The SI, with standard scheme of symbols, units and abbreviations, was developed and recommended by General Conference on Weights and Measures in 1971 for international usage in scientific, technical, industrial and commercial work.

SI units used decimal system, conversions within the system are quite simple and convenient.

SI units of 7 Fundamental & 2 Supplementary Physical Quantities (and their definitions):

SI units of 7 Fundamental Physical Quantities
Base quantity	Name (& Symbol) of SI Units	Definition
Length	metre (m)	The metre is the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second. (1983).
Mass	kilogram (kg)	The kilogram is equal to the mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at international Bureau of Weights and Measures, at Sevres, near Paris, France. (1889)
Time	second (s)	The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom. (1967)
Electric current	ampere (a)	The ampere is that constant current which, if maintained in current two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2×10–7 newton per metre of length. (1948)
Thermodynamic Temperature	kelvin (k)	The kelvin, is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. (1967)
Amount of substance	mole (mol)	The mole is the amount of substance of a system, which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon - 12. (1971)
Luminous intensity	candela (cd)	The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian. (1979)
SI units of 2 Supplementary Physical Quantities
Plane angle	Radian (rad)
Solid angle	Sterdian (sr)

Plane angle (dθ): as the ratio of length of arc ds to the radius r.

Solid angle (dΩ): as the ratio of the intercepted area dA of the spherical surface, described about the apex O as the centre, to the square of its radius r.

Order of Magnitude:

Order of magnitude gives us the value nearest to the actual value, in terms of suitable powers of 10. (It does not tell us the absolute value of the quantity).

 To express a number in nearest power of 10, a number less than 5 is treated as 1 and a number between 5 and 10 is treated as 10

Example: Radius of earth is 6.4 ×10⁶ m. Taking 6.4 as 10, size of the earth is of the order of 10⁷m.

Units and Measurement

Measurement of length:

There are two ways to measure length or distances

(i) Direct method (ii) Indirect method

Direct Method:

• A metre scale for distances from from 10^‒3 m to 10² m

• A verner calipers for distance upto 10^‒4 m

• A screw gauge and a spherometer for distances upto 10^‒5 m

Indirect method for measuring large distances:

Some indirect methods to measure large distances are:

Parallax method

This method is used to measure distances of planets which are very far away from earth.

When a man holds a pen in front of him against some specific point on the background (a wall) and look at the pen first through his left eye A (closing the right eye) and then look at the pen through his right eye B (closing the left eye), he would observe that the position of the pen seems to change with respect to the point on the wall. This is called parallax.

The distance between the two points of observation is called the basis. In this example, the basis is the distance between the eyes.

The situation mentioned above is shown in the figure given below

As the planet is very far away, (b/D) <<  1, and therefore, θ is very small.

Then we approximately take AB as an arc of length b of a circle with centre at S and the distance D as

The radius AS = BS so that AB = b = D θ where θ is in radians or we can write D = b/ θ

Some other indirect methods used to measure long distances are:

Echo method:

An echo is the phenomenon of repetition of sound on reflection from an obstacle.

Here, a sound wave is sent to the obstacle and we observe the time it takes to return.

Suppose, distance between source producing the sound and the object is x, then, the total distance sound wave covered is going and coming back will be 2 x.

As we know, speed = (distance covered) / (Time taken) or speed of sound = (2x) / Time taken.

Other methods for measurement of distances, based on the same principle are: Laser method, RADAR method, SONAR method.

Indirect method for measuring very small distances or length:

A simple method for estimating the molecular size of oleic acid is given below. Oleic acid is a soapy liquid with large molecular size of the order of 10^–9 m.

The idea is to first form mono-molecular layer of oleic acid on water surface.

• We dissolve 1 cm³ of oleic acid in alcohol to make a solution of 20 cm³.

• Then, we take 1 cm³ of this solution and dilute it to 20 cm³, using alcohol.

• So, the concentration of the solution is equal to [1/(20×20)] cm³ of oleic acid/cm³ of solution.

• Next we lightly sprinkle some lycopodium powder on the surface of water in a large trough and we put one drop of this solution in the water.

• The oleic acid drop spreads into a thin, large and roughly circular film of molecular thickness on water surface.

• Then, we quickly measure the diameter of the thin film to get its area A.

• Suppose we have dropped n drops in the water.

• Initially, we determine the approximate volume of each drop (V cm³).

• Volume of n drops of solution = nV cm³

• Amount of oleic acid in this solution = [nV] [1/(20×20)] cm³

• This solution of oleic acid spreads very fast on the surface of water and forms a very thin layer of thickness t. If this spreads to form a film of area A cm², then the thickness of the film

t = (volume of the film) / (area of the film) or t = [nV/(20×20A)] cm

If we assume that the film has mono-molecular thickness, then this becomes the size or diameter of a molecule of oleic acid. The value of this thickness comes out to be of the order of 10^–9 m.

Range of variations of length:

The sizes of the objects in the universe vary over a very wide range. Some examples are given in table given below:

Measurement of Mass

Mass is a basic property of matter. It does not depend on the temperature, pressure or location of the object in space. The SI unit of mass is kilogram (kg).

While dealing with atoms and molecules, we use another important standard unit of mass, called the unified atomic mass unit (u), which has been established for expressing the mass of atoms as

1 unified atomic mass unit = 1u = (1/12) of the mass of an atom of carbon-12 isotope (¹²C₆) including the mass of electrons = 1.66 × 10^–27 kg

Small masses of commonly available objects can be determined by a common balance (the one used in shops).

Large masses in the universe like planets, stars, etc., based on Newton’s law of gravitation can be measured by using gravitational method (will be discussed in further chapters).

Measurement of small masses of atomic/subatomic particles etc., is done by mass spectrograph technique in which radius of the trajectory is proportional to the mass of a charged particle moving in uniform electric and magnetic field.

Range of Masses

The masses of the objects, we come across in the universe, vary over a very wide range. Some examples are given below in the table

Measurement of Time

For measurement of any time interval, we need a clock. Any phenomenon which repeats itself after a fixed interval can serve the purpose of a clock. 

Atomic clock: We now use an atomic standard of time, which is based on the periodic vibrations produced in a cesium atom. This is the basis of the cesium clock, sometimes called atomic clock, used in the national standards. It is highly accurate.

Other clocks, for example, wrist watch, pendulum clock etc., loose and gain time due to the effect of temperature, pressure and variation of acceleration due to gravity, so they are not considered as accurate.

In view of the tremendous accuracy in time measurement, the SI unit of length has been expressed in terms the path length light travels in certain interval of time (1/299, 792, 458 of a second).

NCERT Exercises

Question 1.
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 cm is equal to ….(mm)2.
(c) A vehicle moving with a speed of 18 km h-1 … m in 1 s.
(d) The relative density of lead is 11.3. Its density is ….g cm-3 or …kg m-3.
Answer:
(a) The volume of a cube of side 1 cm is
given by, V = (1 cm)3
or V= (10-2m)3 = Kb6 m3.
(b) The surface area of a solid cylinder of radius r and height h is given by :
A = Area of two caps + curved surface area
= 2πr2 + 2πrh = 2πr(r + h)
here r = 2 cm = 20 mm, h = 10 cm = 100 mm

∴ density of lead = relative density of lead x density of water
= 11.3 x 1 g cm-3 = 11.3 g cm-3
Also in S.I. system density of water = 103 kg m-3
density of lead = 11.3 x 103 kg m-3
= 1.13 x 104 kg m-3

Question 2.
Fill in the blanks by suitable conversion of units
(a) 1 kg m2 s-2 = ….g cm2 s-2
(b) 1 m =… ly
(c) 0 m s-2 = …km h-2
(d) G = 6.67 x 10-11 N m2 (kg)-2 =…. (cm)3 s-2
Answer:

(a) 1 kg m2s-2 = 1 x 103 g (102 cm)2 s-2 = 107 g cm2 s-2

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 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 isϒ s. Show that a calorie has a magnitude 4.2 α-1 β-2 ϒ-2 in terms of the new units
Answer:


Question 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.
Answer:
The given statement is correct. Measurement is basically a comparison process. Without specifying a standard of comparison, it is not possible to get an exact idea about the magnitude of a dimensional quantity. For example, the statement that the mass of the earth is very large, is meaningless. To correct it, we can say that the mass of the earth is large in comparison to any object lying on its surface.
(a) The size of an atom is much smaller than the sharp tip of a pin.
(b) A jet plane moves with a much larger speed than a superfast train.
(c) The mass of Jupiter is very large as compared to that earth.
(d) The air inside this room contains a very large number molecules as compared to that in a balloon.
(e) The given statement is correct.
(f) The given statement is correct.

Question 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:
According to problem, speed of light in vacuum, c = 1 new unit of length s-1.
Time taken by light to cover distance between sun and the earth.
t = 8 min 20 s = 500 s.
∴ Distance between sun and earth
= c x t = 1 new unit of length x 500 s
= 500 new units of length

Question 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?
Answer:
The most precise device is that whose least count is minimum.
Now:

Question 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
Answer:

Question 8.
Answer the following:
(a) You are given a thread and a meter 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?
Answer:
(a) The diameter of a thread is so small that it cannot be measured using a meter scale. We wind a number of turns of the thread on the meter scale so that the turns are closely touching one another. Measure the length (Z) of the windings on the scale which contains n number of turns
Diameter of thread =1/n
(b)

∴ theoretically speaking, least count decreases on increasing the number of divisions on the circular scale. Hence, accuracy would increase. Practically, it may not be possible to take the reading precisely due to low resolution of human eye.
(c) A large number of observations (say, 100) will give more reliable result than smaller number of observations (say, 5). This is because larger the number of readings, closer is the arithmetic mean to the true value and hence smaller the random error.

Question 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:
Here, size of an object = area of object
= 1.75 cm2 = 1.75 x 10-4 m2
Size of the image = area of the image = 1.55 m2

Question 10.
State the number of significant figures in the following:
(a) 007 m2
(b) 2.64x 1024kg
(c) 0.2370gem3
(d) 6.320J
(e) 032 Nm2
(f) 0.0006032 m2
Answer:
(a) 0 .007 m2 has one significant figures.
(b)2.64 x 1024 kg has three significant figures.
(c) 2370 g cm-3 has four significant figures.
(d) 6320 J has four significant figures.
(e) 032 N nr2 has four significant figures.
(f) 0006032 m2 has four significant figures.
(g) The length, breadth and thickness of a

Question 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, (Z) = 4.234 m,
breadth (b) = 1.005 m
thickness, d = 2.01 cm = 2.01 x 10-2 m
Area of sheet = 2 (lb + bd + dl)
= 2(4.234 x 1.005 +1.005 x 0.0201 + 0.0201 x 4.234)
= 2(4.3604739) = 8.7209478 m2
As the least number of significant figure in thickness is 3. Therefore, area has 3 significant
figure, Area = 8.72 m2
volume of metal sheet = Z x b x d
= 4.234 x 1.005 x 0.0201 m3 = 0.085528917 m3
After rounding off = 0.0855 m3

Question 12.
The mass of a box measured by a grocer’s balance is 2.300 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?
Answer:
Here, mass of the box, m = 2.3 kg
Mass of one gold piece,m1= 20.15 g = 0.02015 kg
Mass of other gold piece, m2 = 20.17 g = 0.02017 kg
(a) Total mass = m + m1 + m2
= 2.3 + 0.02015 + 0.02017 = 2.34032 kg As the result is correct only upto one place of decimal, therefore, on rounding off total mass = 2.3 kg
(b) Difference in masses = m2– m1
= 20.17-20.15 = 0.02 g
(correct upto two places of decimal).

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows:
P = a3b2l (√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 PI 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?
Answer:

Question 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 = asin2πt/T
(b) y = asinvt
(c) y = (a/T) sin t/a
(d) y = (a√2) (sin2πt/T+ cos2πt/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.”
Answer:
The argument of a trigonometrical function, i.e. angle is dimensionless. Now using the principle of homogeneity of dimensions.

Question 15.
Famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ ma of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special theory of relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

Guess where to put the missing c.
Answer:
From principle of homogenetity of dimensions both sides of above formula must be same dimensions. For this, (1 – υ2)1/2 must be dimensionless.
Therefore, instead of (1 – υ2)112, it will be (1 – υ2/c2)112.
Hence relation should be

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by
A: 1 A = 10-10 The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m3 of a mole of hydrogen atoms?
Answer:

According to Avogadro’s hypothesis, one mole of hydrogen contains :
N = 6.023 x 1023 atoms
∴ Atomic volume of 1 mole of hydrogen atoms,
V=NV1,
or V= 6.023 x 1023 x 5.233 x 10-3
= 3.152 x 10-7m3 ≅ 3 x 10-7 m3

Question 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). What is this ratio so large?
Answer:

The large value of ratio shows that the inter molecular separation in a gas is much larger than the size of a molecule

Question 18.
Explain this common observation clearly: If you look out of the window of a fast moving train, the near by 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. (In fact, since you are aware that you are moving,, these distant objects seem to move with you).
Answer:
The line joining the object to the eye is called the line of sight. When a train moves rapidly, the line of sight of a nearby tree changes its direction of motion rapidly i.e. near objects make greater angle than distant objects. Therefore the trees appear to run in opposite direction.

On the other hand, the angular change i.e. the line of sight of far off objects (hill tops, the moon, the stars etc.) changes its direction extremely slowly and hence the relative shift in their position is negligible. Hence they appear to be stationary i.e. move in the direction of the train i.e. appear to move with the observer in the train.

Question 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 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 meters?
Answer:


Question 20.
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? 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:
Distance = 4.29 light year
= 4.29 x 9.46 x 1015m
( 1 light year = 9.46 x 10-5m)

Question 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:
Some of the examples of modem science, where precise measurements play an important role, are as follows :

1. Electron microscope uses an electron beam of wavelength 0.2 A to study very minute objects like viruses, microbes and the crystal structure of solids.
2. The successful launching of artificial satellites has been made possible only due to the precise technique available for accurate measurement of time-intervals.
3. The precision with which the distances are measured in Michelson-Morley Interferometer helped in discarding the idea of hypothetical medium ether and in developing the Theory of Relativity by Einstein.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able I       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) During Monsoon in India, the average rain fall is about 100 cm i.e. 1 m over , the area of the country, which is about
A = 3.3 x 106 km2 = 3.3 x 106 x 106 = 3.3 x 1012 m2
Therefore, volume of the rain water,
V = A h = 3.3 x 1012 x 1 = 3.3 x 1012 m3
Now, density of water, p = 103 kg m-3
Hence, the total mass of rain-bearing clouds over India,
m = V ρ = 3.3 x 1012 x 103 = 3.3 x 1015 kg
(b) To estimate the mass of an elephant, consider a boat having base area A in a river. Mark a point on the boat upto which it is inside the water.
Now, move the elephant into the boat and again mark a point on the boat upto which it is inside the water. If h is the distance between the two marks, then
Volume of the water displaced by the elephant, V = Ah
According to Archimedes’ principle, mass of the elephant,
M = mass of the water displaced by the elephant
If ρ (= 103 kg m-3) density of the water, then M = Vρ= Ah ρ
(c) The wind speed during a storm can be found by measuring the angle of drift of an air balloon in a known time. Consider that an air balloon is at the point A at a vertical height h above the observation point O on the ground, when there is no wind Storm.
During the storm, suppose that the balloon moves to the point B in an extremely small time t as shown in figure.

If θ is the angle of drift of the balloon, then from the right angled ΔOAB, we have

(d) Let the area of the hair-bearing head be equal to A. With a fine micrometer, measure the thickness d (diameter) of the hair. Then, area of cross-section of the hair, a = π d2/4
If we ignore the interspacing between the hair, then the number of strands of hair on the head,

(e) At N.T.P., one mole of air occupies a volume of 22.4 litres e. 22.4 x10-3 m3 and contains molecules equal to Avogadro’s number (= 6.023 x1023).
Therefore, number of air molecules per m3,

Suppose that the dimensions of the classroom are 7m x 5m x 4m.
Therefore, volume of the classroom, y=7m x 5m x 4m = 140 m3
Therefore, number of air molecules in the classroom,
N = V.n = 140 x 2.69 x 1025 = 3.77 x 1027

Question 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 600 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 x 1030 kg,
radius of the Sun = 7.0 x 108 m.
Answer:
Here M=2.0 x 1030 kg R=7.0 x 108 m.

This is the order of density of solids and liquids; and not gases. The high density of sun is due to inward gravitational attraction on outer layers, due to the inner layers of the sun.

Question 24.
When the planet Jupiter is at a distance of 7 million kilometers from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.
Answer:

Question 25.
A man walking briskly in rain with speed v must slant his umbrella forward making an angle 0 with the vertical. A student derives the following relation between 0 and v : tanθ = 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.
Answer:
Here, given relation is tanθ = v No, this relation is not correct.
Since the left hand side of this relation is a trigonometrical function which is dimensionless, so R.H.S. must also be dimensionless. So v must
be , where u = speed of rainfall. u
Hence, the correct relation becomes: v
tanθ =  u

Question 26.
lt 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:
Here, the difference shown by two clocks in 100 years = 0.02 s Therefore, the difference, the two clocks will show in 1s


Question 27.
Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (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 kg m3. Are the two densities of the same order of magnitude? If so, why?
Answer:
Here, average radius of sodium atom,


Yes, both densities are of the same order of magnitude, i.e. of the order of 103. This is because in the solid phase atoms are tightly packed.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10-15m. Nuclear sizes obey roughly the following empirical relation:
r = r0A1/3
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 question 27.
Answer:
Let m be the average mass of a nucleon (neutron or proton).
As the nucleus contains A nucleons,
mass of nucleus M = mA
radius of nucleus r = r0 A1/3

As m and r0 are constant, therefore, nuclear density is constant for all nuclei.
Using m = 1.66 x 10-27 kg and


Question 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:
Here, t = 2.56 s
velocity of laser light in vacuum,
c = 3 x 108 m/s
The radius of lunar orbit is the distance of moon from earth. Let it be x

Question 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 s1).
Answer:
Here, υ= 1450 m s_1; t = 77.0 s
The required distance,

Question 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:
Time taken, t = 3 x 109 years
= 3 x 109 x 365 x 24 – 60 x 60 s
Velocity of light, c = 3 x 108 m s_1
Distance of quasar from earth = ct
= 3 x 108 x 3 x 109 x 365 x 24 x 3600 m
= 2.84 x 1025 m = 2.84 x 1022 km.

Question 32.
lt is a well known fact that during a total solar eclipse the disc of the moon almost completely covers the disc of the Sun. From this fact determine the approximate diameter of the moon.

Answer:
Distance of moon from earth,
ME = 3.84 x 108 m
Distance of sun from earth,
SE= 1.496 x 1011m.
Diameter of sun AB = 1.39 x 109 m.
The situation during total solar eclipse is shown in figure

Question 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 the NCERT 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:
The basic constants of atomic physics namely c-speed of light, e-charge on electron, mc-mass of electron and mp-mass of proton; and the gravitational constant G give rise to the quantity.