# CBSE Class 11 Physics Chapter 2 Units and Measurements Study Materials

### Class 11 Physics Chapter 2 Units and Measurements

Topics and Subtopics in Class 11 Physics Chapter 2 Units and Measurements:

 Section Name Topic Name 2 Units and Measurements 2.1 Introduction 2.2 The international system of units 2.3 Measurement of length 2.4 Measurement of mass 2.5 Measurement of time 2.6 Accuracy, precision of instruments and errors in measurement 2.7 Significant figures 2.8 Dimensions of physical quantities 2.9 Dimensional formulae and dimensional equations 2.10 Dimensional analysis and its applications

### Units and Measurement Class 11 Notes Physics Chapter 2

1. Measurement
The process of measurement is basically a comparison process. To measure a physical quantity, we have to find out how many times a standard amount of that physical quantity is present in the quantity being measured. The number thus obtained is known as the magnitude and the standard chosen is called the unit of the physical quantity.
2.  Unit
The unit of a physical quantity is an arbitrarily chosen standard which is widely accepted by the society and in terms of which other quantities of similar nature may be measured.
3. Standard
The actual physical embodiment of the unit of a physical quantity is known as a standard of that physical quantity.
• To express any measurement made we need the numerical value (n) and the unit (μ). Measurement of physical quantity = Numerical value x Unit
For example: Length of a rod = 8 m
where 8 is numerical value and m (metre) is unit of length.
4. Fundamental Physical Quantity/Units
It is an elementary physical quantity, which does not require any other physical quantity to express it. It means it cannot be resolved further in terms of any other physical quantity. It is also known as basic physical quantity.
The units of fundamental physical quantities are called fundamental units.
For example, in M. K. S. system, Mass, Length and Time expressed in kilogram, metre and second respectively are fundamental units.
5. Derived Physical Quantity/Units
All those physical quantities, which can be derived from the combination of two or more fundamental quantities or can be expressed in terms of basic physical quantities, are called derived physical quantities.
The units of all other physical quantities, which car. be obtained from fundamental units, are called derived units. For example, units of velocity, density and force are m/s, kg/m3, kg m/s2 respectively and they are examples of derived units.
6. Systems of Units
Earlier three different units systems were used in different countries. These were CGS, FPS and MKS systems. Now-a-days internationally SI system of units is followed. In SI unit system, seven quantities are taken as the base quantities.
(i) CGS System. Centimetre, Gram and Second are used to express length, mass and time respectively.
(ii) FPS System. Foot, pound and second are used to express length, mass and time respectively.
(iii) MKS System. Length is expressed in metre, mass is expressed in kilogram and time is expressed in second. Metre, kilogram and second are used to express length, mass and time respectively.
(iv) SI Units. Length, mass, time, electric current, thermodynamic temperature, Amount of substance and luminous intensity are expressed in metre, kilogram, second, ampere, kelvin, mole and candela respectively.
7. Definitions of Fundamental Units
8. Supplementary Units
Besides the above mentioned seven units,there are two supplementary base units. these are (i) radian (rad) for angle, and (ii) steradian (sr) for solid angle.
9. Advantages of SI Unit System
SI Unit System has following advantages over the other Besides the above mentioned seven units, there are two supplementary base units. These are systems of units:

(i) It is internationally accepted,
(ii) It is a rational unit system,
(iii) It is a coherent unit system,
(iv) It is a metric system,
(v) It is closely related to CGS and MKS systems of units,
(vi) Uses decimal system, hence is more user friendly.
10. Other Important Units of Length
For measuring large distances e.g., distances of planets and stars etc., some bigger units of length such as ‘astronomical unit’, ‘light year’, parsec’ etc. are used.
• The average separation between the Earth and the sun is called one astronomical unit.
1 AU = 1.496 x 1011 m.
• The distance travelled by light in vacuum in one year is called light year.
1 light year = 9.46 x 1015 m.
• The distance at which an arc of length of one astronomical unit subtends an angle of one second at a point is called parsec.
1 parsec = 3.08 x 1016 m
• Size of a tiny nucleus = 1 fermi = If = 10-15 m
• Size of a tiny atom = 1 angstrom = 1A = 10-10 m
11. Parallax Method
This method is used to measure the distance of planets and stars from earth.
Parallax. Hold a pen in front of your eyes and look at the pen by closing the right eye and ‘ then the left eye. What do you observe? The position of the pen changes with respect to the background. This relative shift in the position of the pen (object) w.r.t. background is called parallax.
If a distant object e.g., a planet or a star subtends parallax angle 0 on an arc of radius b (known as basis) on Earth, then distance of that distant object from the basis is given by

• To estimate size of atoms we can use electron microscope and tunneling microscopy technique. Rutherford’s a-particle scattering experiment enables us to estimate size of nuclei of different elements.
• Pendulum clocks, mechanical watches (in which vibrations of a balance wheel are used) and quartz watches are commonly used to measure time. Cesium atomic clocks can be used to measure time with an accuracy of 1 part in 1013 (or to a maximum discrepancy of 3 ps in a year).
• The SI unit of mass is kilogram. While dealing with atoms/ molecules and subatomic particles we define a unit known as “unified atomic mass unit” (1 u), where 1 u = 1.66 x 10-27 kg.
12. Estimation of Molecular Size of Oleic Acid
For this 1 cm3 of oleic acid is dissolved in alcohol to make a solution of 20 cm3. Then 1 cm3 of this solution is taken and diluted to 20 cm3, using alcohol. So, the concentration of the solution is as follows:

After that some lycopodium powder is lightly sprinkled on the surface of water in a large trough and one drop of this solution is put in water. The oleic acid drop spreads into a thin, large and roughly circular film of molecular thickness on water surface. Then, the diameter of the thin film is quickly measured to get its area A. Suppose n drops were put in the water. Initially, the approximate volume of each drop is determined (V cm3).
Volume of n drops of solution = nV cm3
Amount of oleic acid in this solution

The 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 thickness of the film

If we assume that the film has mono-molecular thickness, 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.
13. Dimensions
The dimensions of a physical quantity are the powers to which the fundamental units of mass, length and time must be raised to represent the given physical quantity.
14. Dimensional Formula
The dimensional formula of a physical quantity is an expression telling us how and which of the fundamental quantities enter into the unit of that quantity.
It is customary to express the fundamental quantities by a capital letter, e.g., length (L), mass (AT), time (T), electric current (I), temperature (K) and luminous intensity (C). We write appropriate powers of these capital letters within square brackets to get the dimensional formula of any given physical quantity.
15. Applications of Dimensions
The concept of dimensions and dimensional formulae are put to the following uses:
(i) Checking the results obtained
(ii) Conversion from one system of units to another
(iii) Deriving relationships between physical quantities
(iv) Scaling and studying of models.
The underlying principle for these uses is the principle of homogeneity of dimensions. According to this principle, the ‘net’ dimensions of the various physical quantities on both sides of a permissible physical relation must be the same; also only dimensionally similar quantities can be added to or subtracted from each other.
16. Limitations of Dimensional Analysis
The method of dimensions has the following limitations:
(i) by this method the value of dimensionless constant cannot be calculated.
(ii) by this method the equation containing trigonometric, exponential and logarithmic terms cannot be analyzed.
(iii) if a physical quantity in mechanics depends on more than three factors, then relation among them cannot be established because we can have only three equations by equalizing the powers of M, L and T.
(iv) it doesn’t tell whether the quantity is vector or scalar.
17. Significant Figures
The significant figures are a measure of accuracy of a particular measurement of a physical quantity.
Significant figures in a measurement are those digits in a physical quantity that are known reliably plus the first digit which is uncertain.
18. The Rules for Determining the Number of Significant Figures
(i) All non-zero digits are significant.
(ii) All zeroes between non-zero digits are significant.
(iii) All zeroes to the right of the last non-zero digit are not significant in numbers without decimal point.
(iv) All zeroes to the right of a decimal point and to the left of a non-zero digit are not significant.
(v) All zeroes to the right of a decimal point and to the right of a non-zero digit are significant.
(vi) In addition and subtraction, we should retain the least decimal place among the values operated, in the result.
(vii) In multiplication and division, we should express the result with the least number of significant figures as associated with the least precise number in operation.
(viii) If scientific notation is not used:
(a) For a number greater than 1, without any decimal, the trailing zeroes are not significant.
(b) For a number with a decimal, the trailing zeros are significant.
19.  Error
The measured value of the physical quantity is usually different from its true value. The result of every measurement by any measuring instrument is an approximate number, which contains some uncertainty. This uncertainty is called error. Every calculated quantity, which is based on measured values, also has an error.
20. Causes of Errors in Measurement
Following are the causes of errors in measurement:

Least Count Error. The least count error is the error associated with the resolution of the instrument. Least count may not be sufficiently small. The maximum possible error is equal to the least count.
Instrumental Error. This is due to faulty calibration or change in conditions (e.g., thermal expansion of a measuring scale). An instrument may also have a zero error. A correction has to be applied.
Random Error. This is also called chance error. It makes to give different results for same measurements taken repeatedly. These errors are assumed to follow the Gaussian law of normal distribution.
Accidental Error. This error gives too high or too low results. Measurements involving this error are not included in calculations.
Systematic Error. The systematic errors are those errors that tend to be in one direction, either positive or negative. Errors due to air buoyancy in weighing and radiation loss in calorimetry are systematic errors. They can be eliminated by manipulation. Some of the sources of systematic errors are:
(i) intrumental error
(ii) imperfection in experimental technique or procedure
(iii) personal errors
21. Absolute Error, Relative Error and Percentage Error

1. Combination of Errors
2. IMPORTANT TABLES

### CBSE Class 11 Physics Chapter-2 Important Questions

1 Marks Questions

1.What is the difference between Ao and A.U.?

Ans:Aand A.U. both are the units of distances but 1Ao = 10-10m and 1A.U. = 1.4961011m.

2.Define S.I. unit of solid angle?

Ans:One steradian is defined as the angle made by a spherical plane of area 1 square meter at the centre of a sphere of radius 1m.

3.Name physical quantities whose units are electron volt and pascal?

Ans:Energy and pressure.

4.Fill ups.

(i) 3.0m/s2 = ——————— km/hr2

(ii) 6.67 10-11Nm2/kg2 = ————————- g-1cm3s-2

Ans: (i) 3.0m/s2 =  km/hr2

= 3.9×104 km/hr2

(ii) 6.67 10-11Nm2/kg2 = g-1cm3s-2

5.Write S.I unit of luminous intensity and temperature?

Ans:S.I unit of luminous intensity is candela (cd) and of temperature is Kelvin (k).

6.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. Distance between the Sun and the Earth:

= Speed of light  Time taken by light to cover the distance

Given that in the new unit, speed of light = 1 unit

Time taken, t = 8 min 20 s = 500 s

Distance between the Sun and the Earth = 1  500 = 500 units

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.  Magnification of the microscope = 100

Average width of the hair in the field of view of the microscope = 3.5 mm

∴Actual thickness of the hair is = 0.035 mm.

8.The photograph of a house occupies an area of 1.75on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55. What is the linear magnification of the projector-screen arrangement?

Ans. Linear magnifications, ml =

9. 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.  Time taken by quasar light to reach Earth = 3 billion years

years

Speed of light =  m/s

Distance between the Earth and quasar

m

km

2 Marks Questions

1.When a planet X is at a distance of 824.7 million kilometers from earth its angular diameter is measured to be 35.7211 of arc. Calculate the diameter of ‘X’.

Ans:

Diameter

2.A radar signal is beamed towards a planet from the earth and its echo is received seven minutes later. Calculate the velocity of the signal, if the distance between the planet and the earth is 6.3×1010m?

Ans:

3.Find the dimensions of latent heat and specific heat?

Ans:(1) Latent Heat

Latent Heat =

(2) Specific heat =

4.in Vander Waal’s equation

Ans:

5.E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively. Determine the dimensions of

Ans:

Thus, it is dimension less.

6.Calculate the time taken by the light to pass through a nucleus of diameter 1.56 10-16 m. (speed of light is 3108 m/s)

Ans:

7.Two resistances R1 = 100 and R2 = 200 are connected in series. Then what is the equivalent resistance?

Ans:

8.If velocity, time and force were chosen the basic quantities, find the dimensions of mass?

Ans:

9.A calorie is a unit of heat or energy and it equals about 4.2 J where . Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals  m, the unit of time is y s. Show that a calorie has a magnitude in terms of the new units.

Ans. Given that,

1 calorie = 4.2 (1 kg)

New unit of mass = α kg

Hence, in terms of the new unit, 1 kg =

In terms of the new unit of length,

And, in terms of the new unit of time,

10.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. The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

(a) An atom is a very small object in comparison to a soccer ball.

(b) A jet plane moves with a speed greater than that of a bicycle.

(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.

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

(e) A proton is more massive than an electron.

(f) Speed of sound is less than the speed of light.

11.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 device with minimum count is the most suitable to measure length.

(a) Least count of vernier callipers

= 1 standard division (SD)  1 vernier division (VD)

(b) Least count of screw gauge =

(c) Least count of an optical device = Wavelength of light  cm

= 0.00001 cm

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

12. 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 on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation,

Diameter=

(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

(c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

13. 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?

Ans. Mass of grocer’s box = 2.300 kg

Mass of gold piece I = 20.15g = 0.02015 kg

Mass of gold piece II = 20.17 g = 0.02017 kg

(a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.

(b) Difference in masses = 20.17 – 20.15 = 0.02 g

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

14. A physical quantity P is related to four observables a, b, c and d as follows:

The percentage errors of measurement in abc 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?

Ans.

Percentage error in P = 13 %

Value of P is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

15. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by. The size of a hydrogen atom is aboutwhat is the total atomic volume in of a mole of hydrogen atoms?

Ans.  Radius of hydrogen atom, r = 0.5

Volume of hydrogen atom =

1 mole of hydrogen contains hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms =

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. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Ans. Line of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

17. 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 indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval  s) are used to measure time intervals in several physical and chemical processes.

X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.

The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

18. When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be of arc. Calculate the diameter of Jupiter.

Ans. Distance of Jupiter from the Earth, D =
Angular diameter =
Diameter of Jupiter = d
Using the relation,

19. 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.
Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56 s
Speed of light =  m/s
Time taken by the laser beam to reach Moon =
Radius of the lunar orbit = Distance between the Earth and the Moon =  =
m = 3.84  km

20. 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).
Ans.
Let the distance between the ship and the enemy submarine be.
Speed of sound in water = 1450 m/s
Time lag between transmission and reception of Sonar waves = 77 s
In this time lag, sound waves travel a distance which is twice the distance between the ship
and the submarine (2S).
Time taken for the sound to reach the submarine
∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

3 Marks Questions

1.  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 monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m

Area of country, A =

Hence, volume of rain water, V = A × h =

Density of water, p =

Hence, mass of rain water = p× V =  kg

Hence, the total mass of rain-bearing clouds over India is approximately  kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say).

Volume of water displaced by the ship,

Now, move an elephant on the ship and measure the depth of the ship () in this case.

Volume of water displaced by the ship with the elephant on board,

Volume of water displaced by the elephant =

Density of water = D

Mass of elephant = AD

(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

(d) Area of the head surface carrying hair = A

With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.

∴Area of one hair =

Number of strands of hair

(e) Let the volume of the room be V.

One mole of air at NTP occupies 22.4 l i.e., volume.

Number of molecules in one mole =

∴Number of molecules in room of volume V

=

2.  The unit of length convenient on the nuclear scale is a fermi: 1 f = m. Nuclear sizes obey roughly the following empirical relation:

where r is the radius of the nucleus, A its mass number, and 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 nucleus r is given by the relation,

… (i)

= 1.2 f =  m

Volume of nucleus, V=

Now, the mass of a nuclei M is equal to its mass number i.e.,

M = A amu = kg

Density of nucleus,

p =

This relation shows that nuclear mass depends only on constant. Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus is given by,

3.  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 (ce, 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. One relation consists of some fundamental constants that give the age of the Universe by:

Where,

t = Age of Universe

e = Charge of electrons =

= Absolute permittivity

= Mass of protons = kg

= Mass of electrons = kg

c = Speed of light =  m/s

G = Universal gravitational constant =

Also,

Substituting these values in the equation, we get

4 Marks Questions

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

(a)

(b) y = a sin vt

(c)

(d)

(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. (a) Ans.: Correct

Dimension of y =

Dimension of a =

Dimension of

Dimension of L.H.S = Dimension of R.H.S

Hence, the given formula is dimensionally correct.

(b) Ans.: Incorrect

y = a sin vt

Dimension of y =

Dimension of a =

Dimension of vt =

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

(c) Ans.: Incorrect

Dimension of y =

Dimension of =

Dimension of

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

(d) Ans.: Correct

Dimension of y =

Dimension of a =

Dimension of

Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of y and a are the same. Hence, the given formula is dimensionally correct.

2.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). Why is this ratio so large?

Ans. Radius of hydrogen atom, r = 0.5 = m

Volume of hydrogen atom =

Now, 1 mole of hydrogen contains  hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms,

Molar volume of 1 mole of hydrogen atoms at STP,

Hence, the molar volume is  times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

3.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?

Ans. Distance of the star from the solar system = 4.29 ly

1 light year is the distance travelled by light in one year.

1 light year = Speed of light × 1 year

m

∴4.29 ly =  m

1 parsec =  m

∴4.29 ly = = 1.32 parsec

Using the relation,

Where,

Diameter of Earth’s orbit, d=

Distance of the star from the Earth,D=

But, 1 sec = rad

4.Estimate the average mass density of a sodium atom assuming its size to be about 2.5. (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. Are the two densities of the same order of magnitude? If so, why?

Ans.  Diameter of sodium atom = Size of sodium atom = 2.5

Radius of sodium atom, r =

m

Volume of sodium atom, V =

According to the Avogadro hypothesis, one mole of sodium contains  atoms and has a mass of 23 g or  kg.

∴ Mass of one atom =

Density of sodium atom, p =

It is given that the density of sodium in crystalline phase is 970 kg.

Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.

5. 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.

The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.

Distance of the Moon from the Earth =  m

Distance of the Sun from the Earth = m

Diameter of the Sun =  m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:

Hence, the diameter of the Moon is  m.

5 Marks Questions

1. 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 monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m

Area of country, A =

Hence, volume of rain water, V = A × h =

Density of water, p =

Hence, mass of rain water = p× V =  kg

Hence, the total mass of rain-bearing clouds over India is approximately  kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say).

Volume of water displaced by the ship,

Now, move an elephant on the ship and measure the depth of the ship () in this case.

Volume of water displaced by the ship with the elephant on board,

Volume of water displaced by the elephant =

Density of water = D

Mass of elephant = AD

(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

(d) Area of the head surface carrying hair = A

With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.

∴Area of one hair =

Number of strands of hair

(e) Let the volume of the room be V.

One mole of air at NTP occupies 22.4 l i.e., volume.

Number of molecules in one mole =

∴Number of molecules in room of volume V

=

2.  The unit of length convenient on the nuclear scale is a fermi: 1 f = m. Nuclear sizes obey roughly the following empirical relation:

where r is the radius of the nucleus, A its mass number, and 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 nucleus r is given by the relation,

… (i)

= 1.2 f =  m

Volume of nucleus, V=

Now, the mass of a nuclei M is equal to its mass number i.e.,

M = A amu = kg

Density of nucleus,

p =

This relation shows that nuclear mass depends only on constant. Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus is given by,

3.  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 (ce, 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. One relation consists of some fundamental constants that give the age of the Universe by:

Where,

t = Age of Universe

e = Charge of electrons =

= Absolute permittivity

= Mass of protons = kg

= Mass of electrons = kg

c = Speed of light =  m/s

G = Universal gravitational constant =

Also,

Substituting these values in the equation, we get

### NCERT Solutions for Class 11 Physics Physics Chapter 2 Units and Measurements

NCERT Solutions for Class 11 Physics Physics Chapter 2 Units and Measurements are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

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.
(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
(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

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.
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?
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?
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

Question 8.
(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?
(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.
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
(a) 0 .007 m2 has one significant figures.
(b) 64 x 1024 kg has three significant figures.
(c) 2370 g cm-3 has four significant figures.
(d) 320 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.
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?
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?

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.”
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.
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?

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?

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).
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?

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?
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.
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.
(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.
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.

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.
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 $\frac { v }{ u }$, where u = speed of rainfall. u
Hence, the correct relation becomes: v
tanθ = $\frac { v }{ u }$ 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?
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?
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.
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?
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).
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?
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.