NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 2
Chapter Name : Units And Measurement

Q2.1 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 10.0 cm is equal to _____ (mm)²
(c) A vehicle moving with a speed of 18 km h⁻ⁱ covers _______m in 1 s
(d) The relative density of ead is 11 B. Its density is ______g a-n 30r ______kg m⁻⁵.

Answer. (a)1cm =
1
m
100

Volume of cube = 1 cm³
1 1 1
But, 1cm3 = 1cm × 1cm × 1cm = m × m × m
100 100 100

1cm3 = 10 − 6m3

Hence, the volume of a cube of side 1cm is equal to 10 − 6m3 .

(b) The total surface area of a cylinder of radius and height is

= 2 ( + )

Given that,

= 2cm = 2 × 1cm = 2 × 10mm = 20mm

= 10cm = 10 × 10mm = 100mm

= 2 × 3.14 × 20 × (20 + 100) = 15072 = 1.5 × 104mm2

(c) Using the conversion,

5
1km/h = m/s
18

5
18kmh = 18 × = 5m/s
18

Therefore, distance can be obtained using the relation:

Distance = Speed × Time = 5 × 1 = 5m

Hence, the vehicle covers 5m in 1s .

(d) Relative density of a substance is given by the relation,
Density of substance
Relative density = Density of water

Density of water = 1 g/cm³
Density of lead = Relative density of lead × Density of water
3
= 11.3 × 1 = 11.3g/cm
1
1g = kg
1000

1 cm³ = 10⁻⁶ m³
−3
3 10 3 3 3
1g/cm = −6
kg/m = 10 kg/m
10

11.3 g/cm³ = 11.3 x 10. kg/m³

Page : 35 , Block Name : Exercise

Q2.2 Fill in the blanks by suitable conversion of units:
(a) 1 kg m2² s⁻² = _____g cm2 s⁼²
(b) 3.0 m s⁻²⁰= ______ km h⁻²

(c) G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = ______(cm)³ s⁻² g⁻¹ .

2
2 2
1kgm 1 × 1000 × 10
2 −2 2
( )1kgm s = = gcm
2 2

7 2 −2
= 10 gcm s
1 1 −16
1m = 5
ly ≈ 16
ly = 10 ly
9.46×10 10
−3
3 × 10 km −2
−2 −3
( )3ms = = 3 × 3600 × 3600 × 10 kmh
2
1 2
h
3600

4 −2
= 3.888 × 10 kmh
kgm
−11 2 −2 −11 2 −2
( ) = 6.67 × 10 Nm kg = 6.67 × 10 m kg
2
s
−11 −1 3 −2
= 6.67 × 10 kg m s

3
−11 2
3 6.67 × 10 × 10
m
−11
= 6.67 × 10 =
2 2
kgs 3
10

−8 −3 −2 −1
= 6.67 × 10 cm s g

Page : 35 , Block Name : Exercise

Q2.3 A calorie is a unit of heat or energy 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 a kg, the unit of length equals j8 m, the. unit of time is ys.
Show that a calorie has a magnitude 4.2 −1 −2y2
in terms of the new units.

2 −2
1cal = 4.2kgm s

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

a
=
−1

In terms of the new unit length,
1 −1 2 −2
Im = = or Im =

And, in terms of the new unit of time,
1 −1
1s = =

2 −2
1s =

−2 2
1s =

Calorie = 4.2(1a − 1)(1 − 2)(1y2) = 4.2a − 1 − 2y
2

Page : 35 , Block Name : Exercise

Q2.4 Explain this statement clearly:
"To call a dimensional quantity 'large' or 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. Physical quantities are called large or small depending on the unit (standard) of measurement For
example, the distance between two cities on earth is measured in kilometres but the distance between stars
or intergalactic distances are measured in parsec The later standard parsec is equal to
3.08 × 10
16
m or 3.08 × 10
12
km is certainly larger than metre or kilometre Therefore, the inter-stellar or
intergalactic distances are certainly larger than the distances between two cities on earth.
(a) The size of an atom is much smaller than even the sharp tip of a pin.
(b) A Jet plane moves with a speed greater than that of a super fast train.
(c) The mass of Jupiter is very large compared to that of the earth.
(d) The air inside this room contains more number of molecules than in one mole of air.
(e) This is a correct statement.
(f) This is a correct statement.

Page : 35 , Block Name : Exercise

Q2.5 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?

Distance between Sun and Earth

8
= Speed of light in vacuum time taken by light to travel from Sim to Earth = 3 × 10 m/s × 8

8 8
min 20s = 3 × 10 m/s × 500s = 500 × 3 × 10 m .

8
In the new system, the speed of light in vacuum is unity. So, the new unit of length is 3 × 10 m .

distance between Sun and Earth = 500 new units.

Page : 35 , Block Name : Exercise

Q2.6 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. (a) Least count of vernier callipers = 1/20 = 0.05mm = 5 × 10 −5
m

(b) Least count of screw gauge = Pitch/No. of divisions on circular scale 1 × 10 −3 −5
/100 = 1 × 10

(c) Least count of optical instrument = 6000 A (average wavelength of visible light as 6000A) =

7
6 × 10 m As the least count of optical instrument is least, it is the most precise device out of

three instruments given to us.

Page : 35 , Block Name : Exercise

Q2.7 A student measures the thickness of a human hair by looking at it through a microscope of
magni cation 100. He makes 20 observations and nds that the average width of the hair in the eld of view
of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

As magnification, m = thickness of image of hair/ real thickness of hair = 100

and average width of the image of hair as seen by microscope = 3.5mm

Thickness of hair = 3.5mm/100 = 0.035mm

Page : 35 , Block Name : Exercise

Q2.8 Answer the following:
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to
increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular
scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100
measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements
only?

Answer. (a) Wrap the thread a number of times on a round pencil so as to form a coil having its turns
touching each other closely. Measure the length of this coil, mode by the thread, with a metre scale. If n be
the number of turns of the coil and I be the length of the coil, then the length occupied by each single turn
i.e., the thickness of the thread = l/n This is equal to the diameter of the thread.
(b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on
circular scale is increased, least count is decreased Hence the accuracy is increased. However, this is only a
theoretical idea_Practically speaking, increasing the number of turns would create many dif culties. As an
example, the low resolution of the human eye would make observations dif cult. The nearest divisions
would not clearly be distinguished as separate. Moreover, it would be
technically dif cult to maintain uniformity of the pitch of the screw throughout its length.
(c) Due to random errors, a large number of observation will give a more reliable result than smaller number
of observations. This is due to the fact that the probability (chance) of making a positive random error of a
given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large
number of observations, positive and negative errors are likely to cancel each other. Hence more reliable
result can be obtained

Page : 35 , Block Name : Exercise

Q2.9 The photograph of a house occupies an area of 1.75 cm² on a 35mm 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 magni cation of the
projector-screen arrangement.

2 −4 2
Here area of the house on slide = 1.75cm = 1.75 × 10 m and area of the house of

2
projector-screen = 1.55m

2 −4 −4 2 3
Areal magnification = Area on screen/Area on slide = 1.55m /1.75 × 10 m m = 8.857 × 10

Linear magnification

= Areal magnification

3
= (8.857) × 10

= 94.1

Page : 35 , Block Name : Exercise

Q2.10

State the number of significant figures in the following:

2 4
(a) 0.007m (b)2.64 × 10 kg

3
(c) 0.2370gcm (d)6.320J

−2 2
(e) 6.032Nm (f )0.0006032m

Answer. (a) Answer: 1
The given quantity is 0 007 m². If the number is less than one, then all zeros on the right of the decimal
point (but eft to the rst non-zero) are insigni cant. This means that here, two zeros after the decimal are
not signi cant. Hence, only 7 is a signi cant gure in this quantity.

(b) Answer: 3
The given quantity is 2.64 x 1024 kg. Here, the power of 10 is irrelevant for the determination of signi cant
gures. Hence, all digits i.e., 2, 6 and 4 are signi cant gures.

(c) Answer: 4
The given quantity is 0 2370 g cm⁻³. For a number with decimals, the trailing zeros are signi cant. Hence,
besides digits 2, 3 and 7, 0 that appears after the decimal point is also a signi cant gure.

(d) Answer: 4
The given quantity is 5.320 J. For a number decimals, the trailing zeros are signi cant. Hence, all four digits
appearing in the given quantity are signi cant gures.

(e) Answer: 4
The given quantity is 6 032 Nm⁻³.

(f) Answer: 4
The given quantity is 0 0006032 m². If the number is less than one, then the zeroes on the right of the
decimal point (but left to the rst non-zero) are insigni cant. Hence, all three zeros appearing before 6 are
not signi cant gures. All zeros between two non-zero digits are always signi cant. Hence, the remaining
four digits are signi cant gures.

Page : 35 , Block Name : Exercise

Q2.11 '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 signi cant gures.

Length of sheet, 1 = 4.234m

Breadth of sheet, b = 1.005m

Thickness of sheet, h = 2.01cm = 0.0201m

The given table lists the respective significant figures:

Hence, area and volume both must have least signi cant gures i.e., 3.

Surface area of the sheet = 2 (l x b + b x h + h x l)
= 2(4.234 x 1.005 + 1.005 x 0.0201 + 0.0201 x 4.234)
= 2 x 4.360
0.0855 m³
This number has only 3 signi cant gures i.e., 8, 5, and 5.

Page : 36 , Block Name : Exercise

Q2.12 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 signi cant gures?

(a) Total mass of the box = (2.3 + 0.0217 + 0.0215)kg = 2.3442kg

since the least number of decimal places is 1, therefore, the total mass of the box = 2.3kg .

(b) Difference of mass = 2.17 − 2.15 = 0.02g

since the least number of decimal places is 2 so the difference in masses to the correct

significant figures is 0.02g.

Page : 36 , Block Name : Exercise

Q2.13 A physical quantity P is related to four observables a, b, c and d as follows:
3 2

=
(√ )

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the
percentage error in the quantity P? If the value Of calculated
using the above relation turns out to be 3.763, to what value *Jou d you round off the result?

3 2

=
(√ )

Δ 3Δ 2Δ 1 Δ Δ
= + + +
2

Δ Δ Δ 1 Δ Δ
× 100 % = 3 × × 100 + 2 × × 100 + × × 100 + × 100 %
2

1
= 3 × 1 + 2 × 3 + × 4 + 2
2

= 3 + 6 + 2 + 2 = 13%

Percentage error in p = 13 %
Value of P is given as 3.763.
By rounding off the given value to the rst decimal place, we get P = 3.8.

Page : 36 , Block Name : Exercise

Q2.14 A book with many printing errors contains four different formulas for the displacement y of a particle
undergoing a certain periodic motion:
(a)y = a sin 2

T
t

(b) = sin

(c) y = T
a
sin
t

a

(d) y = a
sin
2

T
t
+ cos
2

T
t

√2

2
(a) = sin

Here, [ L.H.S. ] = [ ] = [ ] = [ ]

2
and [ R.H.S. ] = sin = sin = [ ]

So, the given equation is correct.

(b) = sin

−1
Here, [ ] = [ ] and [ sin ] = sin . = [ sin ]

So, the equation is wrong.

(c) = sin

−1 −1
Here, [ ] = [ ] and sin = sin = sin

So, the equation is wrong.

2 2
(d) = ( √2) sin + cos

Here, [ ] = [ ][ √2] = [ ]

2 2
and sin + cos = sin + cos = dimensionless

So, the given equation is correct.

Page : 36 , Block Name : Exercise

Q2.15 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 rst 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:
0
=
1
2
(1− )2

Answer. Given the relation,
0
=
1
2
(1− )2

0 2
From the given equation, = √1 −

Left hand side is dimensionless.

Therefore, right hand side should also be dimensionless.

2
2
It is possible only when √1 − should be 1 −
2

2
−1/2

Thus, the correct formula is = 0 1 −
2

Page : 36 , Block Name : Exercise

Q2.16 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A: 1 A =
m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in of a mole of
−10 3
10

hydrogen atoms?

Radius of hydrogen atom, = 0.5 = 0.5 × 10 − 10m

4 3
Volume of hydrogen atom =
3

3
4 22 −10
= × × 0.5 × 10
3 7

−30 3
= 0.524 × 10 m

Atomic volume of 1 mole of hydrogen atoms

−31 −7 3
= 6.023 × 1023 × 5.23 × 10 = 3.15 × 10 m

Page : 36 , Block Name : Exercise

Q2.17 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 A.) Why is this ratio so large?

Volume of one mole of ideal gas, Vg

−3 3
= 22.4 litre = 22.4 × 10 m

Radius of hydrogen molecule = 1A/2

−10
= 0.5A = 0.5 × 10 m

3
Volume of hydrogen molecule = 4/3 r

3
−10 3
= 4/3 × 22/7 0.5 × 10 m

−30 3
= 0.5238 × 10 m
23
One mole contains 6.023 × 10 molecules.

Volume of one mole of hydrogen, VH

−30 23 3 −7 3
= 0.5238 × 10 × 6.023 × 10 m = 3.1548 × 10 m

−3 −7 4
Now Vg /NH = 22.4 × 10 /3.1548 × 10 = 7.1 × 10

Hence, the molar volume is 7.08 x 10⁴ times higher than the atomic volume. For this reason, the interatomic
separation in hydrogen gas is much larger than the size of a hydrogen atom.

Page : 36 , Block Name : Exercise

Q2.18 Explain this common observation clearly. If you look out of the window of a fast moving train, the
nearby trees, houses etc., seem to move rapidly in a direction opposite to the triads 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. Line of sight is de ned 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.

Page : 36 , Block Name : Exercise

Q2.19 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 x 10 n m. However, even the nearest stars are so distant
that with such a long baseline, they show parallel 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?

From parallax method we can say

= b/D, where b = baseline, D = distance of distant object or star

′′ 11
since, = 1 (s) and b = 3 × 10 m

11 −6
D = b/20 = 3 × 10 /2 × 4.85 × 10 m

11 −6 16
or D = 3 × 10 /9.7 × 10 m = 30 × 10 /9.7m

16 16
= 3.09 × 10 m = 3 × 10 m

Page : 36 , Block Name : Exercise

Q2.20 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 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

= 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011m

04.29Iy = 405868.32 × 1011m

1 parsec = 3.08 × 1016m
11
405868.32×10
16
= 1.32
3,08×10

=

=

where,

11
Diameter of Earth's orbit, = 3 × 10 m

11
Distance of the star from the Earth, = 405868.32 × 10 m
11
3×10 −6
= = 7.39 × 10 rad
11
405868.32×10

t, 1sec = 4.85 × 10 − 6rad
−6
−6 7.39×10 ′′
7.39 × 10 rad = = 1.52
−6
4.85×10

Page : 37 , Block Name : Exercise

Q2.21 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 nd 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 modem science where precise measurements of length, time, mass etc., are needed. Also,
wherever you can, give a quantitative idea of the precision needed.

Answer. It is indeed very true that precise measurements of physical quantities are essential for the
development of science. For example, ultrashort laser pulses (time interval 0 10—15 s) are used to measure
time intervals in several physical and chemical processes. X-ray spectroscopy is used to determine the

interatomic separation or inter-planar spacing. The development of mass spectrometer makes it possible to
measure the mass of atoms

Page : 37 , Block Name : Exercise

Q2.22 Just as precise measurements are necessary in science, it is equally important to be able to make
rough estimates of quantities using rudimentary ideas and common
observations. Think of ways by which you can estimate the following (where an estimate is dif cult 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) The average rainfall of nearly 100cm or 1m is recorded by meteorologists, during

6
Monsoon, in India. If A is the area of the country, then A = 3.3 million sq. km = 3.3 × 10 (km)2 =

6 6 2 12 2
3.3 × 10 × 10 m = 3.3 × 10 m

Mass of rain-bearing clouds

12 15
= area height density = 3.3 × 10 × 1 × 1000kg = 3.3 × 10 kg

(b) Measure the depth of an empty boat in water. Let it be dl. If A be the base area of the boat, then volume
of water displaced by boat, VI = Ad2
Let d2 be the depth of boat in water when the elephant is moved into the boat. Volume of water displaced by
(boat + elephant), V2 Ad2 Volume of water displaced by elephant,
V = V2-V1 = A(d2 -dl)
If p be the density of water, then mass of elephant = mass of water displaced by it = A(d2 - dl) p.
(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) Let us assume that the man is not partially bald Let us further assume that the hair on the head are
uniformly distributed We can estimate the area of the head. The thickness of a hair can be measured by
using a screw gauge. The number of hair on the head is clearly the ratio of the area of head to the cross-
sectional area of a hair.
Assume that the human head is a circle of radius 0.08 m i.e., 8 cm. Let us further assume that
= Area of the head/Area of cross - section of a hair

−5 −10 6
= (0.08)2/ 5 × 10 = 64 × 10 − 4/25 × 10 = 2.56 × 10

The number of hair on the human head is of the order of one million.

(e) We can determine the volume of the classroom by measuring its length, breadth and height
3
Consider a class room of size 10m × 8m × 4m . Volume of this room is 320m . We know that

−3 3 23
22.41 or 22.4 × 10 m of air has 6.02 × 10 molecules (equal to Avogadro's number).

Number of molecules of air in the classroom

23 −3 27
= 6.02 × 10 /22.4 × 10 × 320 = 8.6 × 10

Page : 37 , Block Name : Exercise

Q2.23 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its
outer surface at a temperature of about 6000 K. At these high
temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density

of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from
the following data: mass of the Sun = 2.0 x 10³⁰ kg, radius of the sun = 7.0 x 10⁸ m.

30 8
Given = 2 × 10 kg, r = 7 × 10 m

3
3 8 27 3
Volume of Sun = 4/3 r × 3.14 × 7 × 10 = 1.437 × 10 m

30 27 −3 3 −3
As p = M/V, : p = 2 × 10 /1.437 × 10 = 1391.8kgm = 1.4 × 10 kgm

Mass density of Sun is in the range of mass densities of solids/liquids and not gases.

Page : 37 , Block Name : Exercise

Q2.24 When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular
diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.

−6
Given angular diameter = 35.72 = 35.72 × 4.85 × 10 rad

−6 −4
= 173.242 × 10 = 1.73 × 10 rad

−4 9
Diameter of Jupiter D = × d = 1.73 × 10 × 824.7 × 10 m

3 8
= 1426.731 × 10 = 1.43 × 10 m

Page : 37 , Block Name : Exercise

Q2.25 A man walking briskly in rain with speed v must slant his umbrella forward making an angle e with
the vertical. A student derives the following relation between e and v: tane = v and checks that the relation
has a correct limit: as v → , → 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.

According to principle of homogenity of dimensional equations,

Dimensions of L.H.S. = Dimensions of R.H.S.

Here, = tan

1 −1
i.e., L T = dimensionless, which is incorrect.

Correcting the L.H.S. we. get

v/u = tan , where u is velocity of rain.

Page : 37 , Block Name : Additional Exercise

Q2.26 It is claimed that two cesium clocks, if allowed to run for 100 veus, free from' any disturbance, may
differ by only out 0.02 s. What does this imply for the accuracy of the stand&d cesium clock in measuring a
time-interval of 1 s ?

Answer. Difference in time of caesium clocks — 0.02 s
Time required for this difference • t 00 years
= 100 x 365 x 24 x 60 x 60 = 3.15 x 109 s
In 3.15 x 109 s, the caesium clock shows e time difference of 0.02 s.
In Is, the clock Rill show a time difference of s
0.02
9
3.15×10

Hence, the accuracy of a standard caesium clock in measuring a time interval of 1 s is

9
3.15×10 9 11
= 157.5 × 10 s ≈ 1.5 × 10 s
0.02

Page : 37 , Block Name : Additional Exercise

Q2.27 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 . Are the two densities of the same order of magnitude? If so, Why ?
3−

−10
Answer: It is given that radius of sodium atom, = 2.5A = 2.5 × 10 m

3
Volume of one mole atom of sodium, V = NA.4/3 R
3
23 −10 3
= 6.023 × 10 × −4/3 × 3.14 × 2.5 × 10 m and mass of one mole atom of

−3
sodium, M = 23g = 23 × 10 kg

Average mass density of sodium atom, p = M/V

−3 23 −10
= 23 × 10 /6.023 × 10 × 4/3 × 3.14 × 2.5 × 10

2 −3 −3 −3
= 6.96 × 10 kgm = 0.7 × 10 kgm
−3
The density of sodium in its crystalline phase = 970kgm

3 −3
= 0.97 × 10 kgm
3 −3
Obviously the two densities are of the same order of magnitude = 10 kgm . It

is on account of the fact that in solid phase atoms are tightly packed and so the

atomic mass density is close to the mass density of solid.

Page : 37 , Block Name : Additional Exercise

Q2.28
The unit of length convenient on the nuclear scale is a fermi:

where r is the radius of the
−15
f = 10 m . Nuclear sizes obey roughly the following empirical relation:

1/3
r = r0A

nucleus, A its mass number, and ro is a constant equal to about,l .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 Q2.27.

Answer: Assume that the nucleus is spherical. Volume of nucleus

3
3 1/3 3
= 4/3 = 4/3 0 = 4/3
0

Mass of nucleus =

Nuclear mass density = Mass of nucleus/Volume of nucleus

3 3
= A/ 4/3 r A = 3/4 r
0 0

Since ro is a constant therefore the right hand side is a constant So, the nuclear mass density is independent
of mass number Thus, nuclear mass density is constant for different nuclei.
For sodium, A = 23
radius of sodium nucleus,
−15 1/3 −15 −15
= 1.2 × 10 (23) m = 1.2 × 2.844 × 10 m = 3.4128 × 10

4 3
Volume of nucleus =
3

4 22 3
−15 3 −43 3
= × 3.4128 × 10 m = 1.66 × 10 m
3 7

If we neglect the mass Of electrons Of a sodium atom, then the mass Of its nucleus can be taken to the mass
of its atom.
−26
Mass of sodium nucleus = 3.82 × 10 kg

( Refer to Q.2.27)

Mass density of sodium nucleus

Mass of nucleus
=
Volume of nucleus
−26
3.82 × 10 −3 17 −3
= kgm = 2.3 × 10 kgm
−43
1.66 × 10
3 −3
Mass density of sodium atom = 4.67 × 10 kgm

(Refer to Q. 2.27

The ratio of the mass density of sodium nucleus to the average mass density of a sodium

atom is
17
2.3×10 13
3
i.e., 4.92 × 10 .
4.67×10

Page : 37 , Block Name : Additional Exercise

Q2.29 A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These
properties of a light can be exploited to measure long distances The distance Of the Moon from the Earth
has been already determined very precisely a laser as a source of light. A laser light burned at the Moon
takes 2.56 s to return after re ection at the Moon's surface. How much is the radius of the lunar orbit around
the Earth?

Answer. We known that speed of laser light = C = 3 x 10 m/s. If d be the distance of Moon from the earth,
8

the time taken by laser signal to return after re ection at the Moon's surface
2 2
= 2.56s = =
8 −1
3 × 10 ms

1
8 8
= × 2.56 × 3 × 10 m = 3.84 × 10 m
2

Page : 37 , Block Name : Additional Exercise

Q2.30
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In
a submarine equipped with a
SONAR the time delay between generation of a probe wave and the reception of its echo after re ection
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 ). −1

−1
Here speed of sound in water = 1450ms and time of echot = 77.0

s.

If distance of enemy submarine be d, then t = 2d/v

3
d = vt/2 = 1450 × 77.0/2 = 55825m = 55.8 × 10 m or 55.8km.

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Q2.31 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: The time taken by light from the quasar to the observer

9 15
= 3.0 billion years = 3.0 × 10 years As 1 ∣
∣ = 9.46 × 10 m

9 15
Distance of quasar from the observer = 3.0 × 10 × 9.46 × 10 m
24 25 22
= 28.38 × 10 m = 2.8 × 10 m or 2.8 × 10 km

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Q2.32 It is a well known fact that during a total solar eclipse the disk of the Moon almost completely covers
the disk of the Sun. From this fact and
from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the
Moon.

′′ 8
From examples 2.3 and 2.4, we get = 1920 and = 3.8452 × 10 m .

During the total solipse, the disc of the moon completely covers the disc of

the sun, so the angular diameter of both the sun and the moon must be equal.

Angular diameter of the moon, = Angular diameter of the sun
8
The earth-moon distance, = 3.8452 × 10 m : The diameter of the moon, D =

×S

6 8 2 3
= 1920 × 4.85 × 10 × 3.8452 × 10 m = 3506.5024 × 10 m = 3581 × 10 m

3581km.

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Q2.33 A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of fundamental
constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of
atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a
number with the dimension of time. Further, it was a very large number, its magnitude being close to the
present estimate on the age of the universe (-15 billion years). From the table of fundamental constants in
this book, try to see if you too can construct this number (or any other interesting number you can think of).
If its coincidence with the age of the universe were signi cant, what would this imply for the constancy of
fundamental constants?

−19
Charge on an electron, = 1.6 × 10 C
−31
Mass of an electron, = 9.1 × 10 kg

−27
Mass of a proton, = 1.67 × 10 kg

8
Speed of light, = 3 × 10 m/s

−11 2 −2
Gravitational constant, = 6.67 × 10 Nm kg

1 2 −2
9
= 9 × 10 Nm C
4 0

We have to try to make permutations and combinations of the universal constants and see if there can be
any such combination whose dimensions
come out to be the dimensions of time. One such combination is:

2
2
1
⋅
2
4 0
3

of electrostatics,

1 ( )( )
=
4 0
2

2
2 2 4
1 1
= or =
4 2 4 4
0 0

According to Newton's law of gravitation,
2
1 2
= or =
2
1 2

4 2 4
1
4 1 2
Now, =
2 2 4 2 3 2
3
(4 0)

2 −2 2

= = = [ ]
3 3 −3

Clearly, the quantity under discussion has the dimensions of time. Substituting values in the quantity under
discussion, we get
−19 4 9 2
1.6×10 9×10

−27 −31 2 8 3 −11
1.69×10 9.1×10 3×10 6.67×10

16
= 2.1 × 10 second
16
2.1 × 10
= years
60 × 60 × 24 × 365.25
8
= 6.65 × 10 years

9
= 10 years

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