Chapter 2 Units & Measurement

8/10/2019 Chapter 2 Units & Measurement

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Units And Measurements (Physics)

Question 2.1:

Fill in the blanks

The volume of a cube of side 1 cm is equal to.....m3

The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...

(mm)2

A vehicle moving with a speed of 18 km h1

covers....m in 1 s

The relative density of lead is 11.3. Its density is ....g cm3

or . ...kg m3

.

Answer

1 cm =

Volume of the cube = 1 cm3

But, 1 cm3 = 1 cm 1 cm 1 cm =

1 cm3

= 106

m3

Hence, the volume of a cube of side 1 cm is equal to 106

m3.

The total surface area of a cylinder of radius rand height h is

S= 2r(r+ h).

Given that,

r= 2 cm = 2 1 cm = 2 10 mm = 20 mm

h = 10 cm = 10 10 mm = 100 mm

= 15072 = 1.5 104

mm2


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Using the conversion,

1 km/h =

Therefore, distance can be o

Distance = Speed Time = 5

Hence, the vehicle covers 5

Relative density of a substan

Relative density =

Density of water = 1 g/cm3

Again, 1g =

1 cm3

= 106

m3

1 g/cm3

=

11.3 g/cm3

= 11.3 103

kg

Question 2.2:

Fill in the blanks by suitable

1 kg m2s2

= ....g cm2

s2

1 m =..... ly

tained using the relation:

1 = 5 m

in 1 s.

ce is given by the relation,

/m3

conversion of units:


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3.0 m s

=.... km h

G= 6.67 1011

N m2

(kg)2

=.... (cm)3s2

g1

.

Answer

1 kg = 103

g

1 m2

= 104

cm2

1 kg m2

s2

= 1 kg 1 m2

1 s2

=103

g 104

cm2

1 s2

= 107

g cm2

s2

Light year is the total distance travelled by light in one year.

1 ly = Speed of light One year

= (3 108

m/s) (365 24 60 60 s)

= 9.46 1015

m

1 m = 103

km

Again, 1 s =

1 s1

= 3600 h1

1 s2

= (3600)2

h2

3 m s2

= (3 103

km) ((3600)2

h2

) = 3.88 104

km h2

1 N = 1 kg m s

2

1 kg = 103

g1

1 m3

= 106

cm3

6.67 1011

N m2

kg2

= 6.67 1011

(1 kg m s2

) (1 m2) (1 s

2)


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= 6.67 10

(1 kg 1 m

= 6.67 1011

(103

g1

)

= 6.67 108

cm3

s2

g1

Question 2.3:

A calorie is a unit of heat or

Suppose we employ a syste

length equals m, the unit o

2

in terms of the new units.

Answer

Given that,

1 calorie = 4.2 (1 kg) (1 m2)

New unit of mass = kg

Hence, in terms of the new u

In terms of the new unit of le

And, in terms of the new uni

1 calorie = 4.2 (1 1

) (1

1 s )

(106

cm3) (1 s

2)

nergy and it equals about 4.2 J where 1J = 1 kg

of units in which the unit of mass equals kg,

time is s. Show that a calorie has a magnitud

1 s2

)

it, 1 kg =

ngth,

of time,

) (1 2) = 4.2

1

2

2

m2s2.

the unit of

4.2 1

2


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Question 2.4:

Explain this statement clearl

To call a dimensional quant

standard for comparison. Innecessary:

atoms are very small objects

a jet plane moves with great

the mass of Jupiter is very la

the air inside this room conta

a proton is much more massi

the speed of sound is much s

Answer

The given statement is true b

comparision to some standar

dimensionless. The coefficiefriction, but less than static f

An atom is a very small obje

A jet plane moves with a spe

Mass of Jupiter is very large

The air inside this room cont

in a geometry box.

A proton is more massive tha

Speed of sound is less than t

Question 2.5:

:

ity large or small is meaningless without spe

view of this, reframe the following statements

speed

ge

ins a large number of molecules

e than an electron

aller than the speed of light.

ecause a dimensionless quantity may be large o

reference. For example, the coefficient of frict

t of sliding friction is greater than the coefficieiction.

ct in comparison to a soccer ball.

ed greater than that of a bicycle.

as compared to the mass of a cricket ball.

ains a large number of molecules as compared t

n an electron.

e speed of light.

cifying a

herever

small in

ion is

nt of rolling

o that present


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A new unit of length is chos

distance between the Sun an

20 s to cover this distance?

Answer

Distance between the Sun an

= Speed of light Time take

Given that in the new unit, s

Time taken, t= 8 min 20 s =

Distance between the Sun a

Question 2.6:

Which of the following is th

a vernier callipers with 20 di

a screw gauge of pitch 1 mm

an optical instrument that ca

Answer

A device with minimum cou

Least count of vernier callip

= 1 standard division (SD)

n such that the speed of light in vacuum is unit

the Earth in terms of the new unit if light takes

d the Earth:

by light to cover the distance

eed of light = 1 unit

500 s

nd the Earth = 1 500 = 500 units

most precise device for measuring length:

isions on the sliding scale

and 100 divisions on the circular scale

measure length to within a wavelength of light

t is the most suitable to measure length.

rs

1 vernier division (VD)

. What is the

8 min and

?


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Least count of screw gauge

Least count of an optical dev

= 0.00001 cm

Hence, it can be inferred that

length.

Question 2.7:

A student measures the thick

of magnification 100. He ma

hair in the field of view of th

thickness of hair?

Answer

Magnification of the microsc

Average width of the hair in

Actual thickness of the hair

Question 2.8:

Answer the following:

You are given a thread and a

thread?

A screw gauge has a pitch of

think it is possible to increas

ice = Wavelength of light 105

cm

an optical instrument is the most suitable devic

ness of a human hair by looking at it through a

es 20 observations and finds that the average

e microscope is 3.5 mm. What is the estimate o

ope = 100

he field of view of the microscope = 3.5 mm

is = 0.035 mm.

metre scale. How will you estimate the diamete

1.0 mm and 200 divisions on the circular scale.

the accuracy of the screw gauge arbitrarily by

e to measure

icroscope

idth of the

the

r of the

Do you

increasing


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the number of divisions on t

The mean diameter of a thin

of 100 measurements of the

of 5 measurements only?

Answer

Wrap the thread on a unifor

very close to each other. Mediameter of the thread is give

It is not possible to increase

divisions of the circular scal

increase its accuracy to a cer

A set of 100 measurements i

errors involved in the former

Question 2.9:

The photograph of a house o

projected on to a screen, and

linear magnification of the p

Answer

Area of the house on the slid

Area of the image of the hou

= 1.55 104

cm2

e circular scale?

rass rod is to be measured by vernier callipers.

iameter expected to yield a more reliable estim

smooth rod in such a way that the coils thus fo

sure the length of the thread using a metre scal

n by the relation,

he accuracy of a screw gauge by increasing the

. Increasing the number divisions of the circula

ain extent only.

more reliable than a set of 5 measurements be

are very less as compared to the latter.

cupies an area of 1.75 cm2on a 35 mm slide. T

the area of the house on the screen is 1.55 m2.

ojector-screen arrangement?

e = 1.75 cm2

se formed on the screen = 1.55 m2

Why is a set

ate than a set

rmed are

. The

number of

scale will

ause random

e slide is

hat is the


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Arial magnification, ma =

Linear magnifications, ml=

Question 2.10:

State the number of significa

0.007 m

2

2.64 1024

kg

0.2370 g cm3

6.320 J

6.032 N m2

0.0006032 m2

Answer

Answer: 1

The given quantity is 0.007

If the number is less than on

the first non-zero) are insigni

not significant. Hence, only

Answer: 3


Here, the power of 10 is irrel

nt figures in the following:

2.

, then all zeros on the right of the decimal poin

ficant. This means that here, two zeros after the

is a significant figure in this quantity.

1024

kg.

evant for the determination of significant figure

(but left to

decimal are

s. Hence, all


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digits i.e., 2, 6 and 4 are sign

Answer: 4


For a number with decimals,and 7, 0 that appears after th

Answer: 4

The given quantity is 6.320 J

For a number with decimals,

appearing in the given quanti

Answer: 4


All zeroes between two non-

Answer: 4


If the number is less than on

the first non-zero) are insigni

significant figures. All zeros

the remaining four digits are

Question 2.11:

The length, breadth and thic

and 2.01 cm respectively. Gi

figures.

Answer

Length of sheet, l = 4.234 m

ificant figures.

g cm3

.

the trailing zeroes are significant. Hence, besiddecimal point is also a significant figure.

.

the trailing zeroes are significant. Hence, all fo

ty are significant figures.

m2.

ero digits are always significant.

32 m2.

, then the zeroes on the right of the decimal poi

ficant. Hence, all three zeroes appearing before

between two non-zero digits are always signific

significant figures.

ness of a rectangular sheet of metal are 4.234

e the area and volume of the sheet to correct si

s digits 2, 3

r digits

nt (but left to

6 are not

ant. Hence,

, 1.005 m,

gnificant


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Breadth of sheet, b = 1.005

Thickness of sheet, h = 2.01

The given table lists the resp

Quantity Number

l 4.234

b 1.005

h 2.01

Hence, area and volume bot

Surface area of the sheet = 2

= 2(4.234 1.005 + 1.005

= 2 (4.25517 + 0.02620 + 0.

= 2 4.360

= 8.72 m2

Volume of the sheet = l b

= 4.234 1.005 0.0201

= 0.0855 m3

This number has only 3 signi

Question 2.12:

The mass of a box measured

masses 20.15 g and 20.17 g athe difference in the masses

Answer

cm = 0.0201 m

ctive significant figures:

Significant Figure

4

4

3

must have least significant figures i.e., 3.

(l b + b h + h l)

0.0201 + 0.0201 4.234)

8510)

h

ficant figures i.e., 8, 5, and 5.

by a grocers balance is 2.300 kg. Two gold pie

re added to the box. What is (a) the total massf the pieces to correct significant figures?

ces of

f the box, (b)


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Mass of grocers box = 2.30

Mass of gold piece I = 20.15

Mass of gold piece II = 20.1

Total mass of the box = 2.3

In addition, the final result s

number with the least decim

Difference in masses = 20.17

In subtraction, the final resul

number with the least decim

Question 2.13:

A physical quantityP is relat

The percentage errors of mearespectively. What is the per

using the above relation turn

result?

Answer

kg

g = 0.02015 kg

g = 0.02017 kg

0.02015 + 0.02017 = 2.34032 kg

ould retain as many decimal places as there are

l places. Hence, the total mass of the box is 2.3

20.15 = 0.02 g

should retain as many decimal places as there

l places.

ed to four observables a, b, c and d as follows:

surement in a, b, c and dare 1%, 3%, 4% and 2entage error in the quantityP? If the value ofP

out to be 3.763, to what value should you roun

in the

kg.

re in the

%,calculated

d off the


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Percentage error inP= 13 %

Value ofP is given as 3.763.

By rounding off the given va

Question 2.14:

A book with many printing eof a particle undergoing a ce

y = a sin vt

(a = maximum displacement

motion). Rule out the wrong

Answer

lue to the first decimal place, we getP= 3.8.

rrors contains four different formulas for the distain periodic motion:

of the particle, v = speed of the particle. T = ti

formulas on dimensional grounds.

placementy

e-period of


14/31

Answer: Correct

Dimension ofy = M0

L1

T0

Dimension of a = M0

L1

T0

Dimension of = M0

L0

T0

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

Hence, the given formula is dimensionally correct.

Answer: Incorrect

y = a sin vt

Dimension ofy = M0

L1

T0

Dimension of a = M0

L1

T0

Dimension of vt= M0

L1

T1

M0

L0

T1

= M0

L1

T0

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.

Answer: Incorrect

Dimension ofy = M0L

1T

0

Dimension of = M0L

1T

1

Dimension of = M0

L1

T1

But the argument of the trigonometric function must be dimensionless, which is not so in

the given case. Hence, the formula is dimensionally incorrect.

Answer: Correct


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Dimension ofy = M0

L1

T0

Dimension of a = M

0

L

1

T

0

Dimension of = M0

L0

T0

Since the argument of the tri

the given case), the dimensio

dimensionally correct.

Question 2.15:

A famous relation in physics

terms of its speed v and the s

special relativity due to Albe

forgets where to put the cons

Answer

Given the relation,

Dimension of m = M1

L0

T0

Dimension of = M1

L0

T0

Dimension of v = M0

L1

T1

onometric function must be dimensionless (wh

ns ofy and a are the same. Hence, the given for

relates moving mass m to the rest mass m0 o

eed of light, c. (This relation first arose as a co

t Einstein). A boy recalls the relation almost co

tant c. He writes:

ich is true in

mula is

f a particle in

nsequence of

rrectly but


16/31

Dimension of v = M L T

Dimension of c = M0

L1

T1

The given formula will be di

same as that of R.H.S. This i

i.e., (1 v2) is dimensionless.

correct relation is

.

Question 2.16:

The unit of length convenien

by . The si

volume in m3

of a mole of h

Answer

Radius of hydrogen atom, r

Volume of hydrogen atom =

1 mole of hydrogen contains

Volume of 1 mole of hydr

= 3.16 107

m3

ensionally correct only when the dimension o

only possible when the factor, is dim

. This is only possible if v2

is divided by c2. Hen

on the atomic scale is known as an angstrom a

e of a hydrogen atom is about what is the

drogen atoms?

0.5 = 0.5 1010

m

6.023 1023 hydrogen atoms.

gen atoms = 6.023 1023

0.524 1030

L.H.S is the

ensionless

ce, the

d is denoted

total atomic


17/31

Question 2.17:

One mole of an ideal gas at svolume). What is the ratio of

(Take the size of hydrogen

Answer

Radius of hydrogen atom, r

Volume of hydrogen atom =

Now, 1 mole of hydrogen co

Volume of 1 mole of hydr

= 3.16 107 m3

Molar volume of 1 mole of h

Vm = 22.4 L = 22.4 103

m

Hence, the molar volume is

reason, the inter-atomic sepa

hydrogen atom.

Question 2.18:

Explain this common observ

tandard temperature and pressure occupies 22.4molar volume to the atomic volume of a mole

olecule to be about 1 ). Why is this ratio so l

0.5 = 0.5 1010

m

tains 6.023 1023

hydrogen atoms.

gen atoms, Va = 6.023 1023

0.524 1030

ydrogen atoms at STP,

.08 104

times higher than the atomic volume.

ation in hydrogen gas is much larger than the s

tion clearly : If you look out of the window of

L (molarf hydrogen?

rge?

For this

ze of a

fast moving


18/31

train, the nearby trees, house

trains motion, but the distan

stationary. (In fact, since you

move with you).

Answer

Line of sight is defined as an

When we observe nearby sta

moving train, they appear to

sight changes very rapidly.

On the other hand, distant ob

large distance. As a result, th

Question 2.19:

The principle of parallax in

very distant stars. The baseli

apart in its orbit around the S

orbit 3 1011

m. However,

baseline, they show parallax

convenient unit of length onshow a parallax of 1 (secon

distance from the Earth to th

Answer

Diameter of Earths orbit = 3

Radius of Earths orbit, r= 1

Let the distance parallax ang

Let the distance of the star b

Parsec is defined as the dista

s etc. seem to move rapidly in a direction oppos

objects (hill tops, the Moon, the stars etc.) see

are aware that you are moving, these distant ob

imaginary line joining an object and an observ

ionary objects such as trees, houses, etc. while

move rapidly in the opposite direction because

ects such as trees, stars, etc. appear stationary

e line of sight does not change its direction rapi

section 2.3.1 is used in the determination of dis

eAB is the line joining the Earths two locatio

un. That is, the baseline is about the diameter o

even the nearest stars are so distant that with su

only of the order of 1 (second) of arc or so. A

he astronomical scale. It is the distance of an o) of arc from opposite ends of a baseline equal

Sun. How much is a parsec in terms of meters

1011

m

.5 1011 m

le be = 4.847 106

rad.

D.

ce at which the average radius of the Earths o

ite to the

to be

jects seem to

rs eye.

sitting in a

he line of

ecause of the

ly.

tances of

s six months

the Earths

ch a long

arsec is a

ject that willto the

bit subtends


19/31

an angle of .

We have

Hence, 1 parsec 3.09 101

Question 2.20:

The nearest star to our solarterms of parsecs? How much

viewed from two locations o

Answer

Distance of the star from the

1 light year is the distance tr

1 light year = Speed of light

= 3 108

365 24 60

4.29 ly = 405868.32 1011

1 parsec = 3.08 1016

m

4.29 ly =

Using the relation,

6m.

ystem is 4.29 light years away. How much is thparallax would this star (named lpha Centaur

the Earth six months apart in its orbit around t

solar system = 4.29 ly

velled by light in one year.

1 year

0 = 94608 1011

m

m

= 1.32 parsec

is distance ini) show when

e Sun?


20/31

But, 1 sec = 4.85 106

rad

Question 2.21:

Precise measurements of phascertain the speed of an airc

at closely separated instants

of radar in World War II. Th

measurements of length, tim

quantitative idea of the preci

Answer

It is indeed very true that pre

development of science. For

used to measure time interva

X-ray spectroscopy is used t

spacing.

The development of mass spprecisely.

Question 2.22:

Just as precise measurements

make rough estimates of qua

sical quantities are a need of science. For examraft, one must have an accurate method to find i

f time. This was the actual motivation behind t

nk of different examples in modern science wh

, mass etc. are needed. Also, wherever you can,

ion needed.

cise measurements of physical quantities are es

example, ultra-shot laser pulses (time interval

ls in several physical and chemical processes.

determine the inter-atomic separation or inter-

ctrometer makes it possible to measure the ma

are necessary in science, it is equally importan

tities using rudimentary ideas and common ob

le, tots positions

e discovery

re precise

give a

ential for the

1015

s) are

laner

s of atoms

to be able to

ervations.


21/31

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):

the total mass of rain-bearing clouds over India during the Monsoon

the mass of an elephant

the wind speed during a storm

the number of strands of hair on your head

the number of air molecules in your classroom.

Answer

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 = 3.3 1012

m2

Hence, volume of rain water, V =A h = 7.09 1012

m3

Density of water, = 1 103

kg m3

Hence, mass of rain water = V= 7.09 1015

kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 1015

kg.

Consider a ship of known base area floating in the sea. Measure its depth in sea (say d1).

Volume of water displaced by the ship, Vb =A d1

Now, move an elephant on the ship and measure the depth of the ship (d2) in this case.

Volume of water displaced by the ship with the elephant on board, Vbe=Ad2

Volume of water displaced by the elephant =Ad2Ad1

Density of water =D

Mass of elephant =AD (d2d1)

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.


22/31


23/31

Density of the Sun =

The density of the Sun is in t

attributed to the intense grav

Sun.

Question 2.24:

When the planet Jupiter is at

angular diameter is measure

Answer

Distance of Jupiter from the

Angular diameter =

Diameter of Jupiter = d

Using the relation,

Question 2.25:

A man walking briskly in rai

angle with the vertical. A s

e density range of solids and liquids. This high

tational attraction of the inner layers on the out

a distance of 824.7 million kilometers from the

to be of arc. Calculate the diameter of

arth,D = 824.7 106 km = 824.7 109 m

with speed v must slant his umbrella forward

udent derives the following relation between

density is

r layer of the

Earth, its

upiter.

aking an

and v: tan =


24/31

v and checks that the relation

assuming there is no strong

Do you think this relation ca

Answer

Answer: Incorrect; on dime

The relation is .

Dimension of R.H.S = M0

L1

Dimension of L.H.S = M0

L0

( The trigonometric functi

Dimension of R.H.S is not e

not correct dimensionally.

To make the given relation c

achieve this is by dividing th

Therefore, the relation reduc

. This relation is d

Question 2.26:

It is claimed that two cesium

disturbance, may differ by o

standard cesium clock in me

Answer

Difference in time of caesiu

has a correct limit: as v 0, 0, as expecte

ind and that the rain falls vertically for a statio

be correct? If not, guess the correct relation.

sional ground

T1

T0

on is considered to be a dimensionless quantity)

ual to the dimension of L.H.S. Hence, the give

rrect, the R.H.S should also be dimensionless.

e R.H.S by the speed of rainfall .

s to

mensionally correct.

clocks, if allowed to run for 100 years, free fro

ly about 0.02 s. What does this imply for the ac

suring a time-interval of 1 s?

clocks = 0.02 s

. (We are

ary man).

relation is

One way to

any

curacy of the


25/31

Time required for this differ

= 100 365 24 60 60

In 3.15 109

s, the caesium

In 1s, the clock will show a t

Hence, the accuracy of a sta

Question 2.27:

Estimate the average mass d

(Use the known values of Av

it with the density of sodium

the same order of magnitude

Answer

Diameter of sodium atom =

Radius of sodium atom, r=

= 1.25 1010

m

Volume of sodium atom, V

According to the Avogadro

and has a mass of 23 g or 23

nce = 100 years

3.15 109

s

lock shows a time difference of 0.02 s.

ime difference of .

dard caesium clock in measuring a time interva

.

nsity of a sodium atom assuming its size to be

ogadros number and the atomic mass of sodiu

in its crystalline phase: 970 kg m3

. Are the tw

If so, why?

ize of sodium atom = 2.5

ypothesis, one mole of sodium contains 6.023

103

kg.

l of 1 s is

bout 2.5 .

). Compare

densities of

1023

atoms


26/31

Mass of one atom =

Density of sodium atom, =

It is given that the density of

Hence, the density of sodiu

not in the same order. This is

inter-atomic separation is ve

Question 2.28:

The unit of length convenien

obey roughly the following e

where r is the radius of the n

1.2 f. Show that the rule imp

nuclei. Estimate the mass de

density of a sodium atom obt

Answer

Radius of nucleus ris given

(i)

= 1.2 f = 1.2 1015 m

Volume of nucleus, V=

sodium in crystalline phase is 970 kg m3

.

atom and the density of sodium in its crystallin

because in solid phase, atoms are closely pack

y small in the crystalline phase.

on the nuclear scale is a fermi : 1 f = 10 15

m.

mpirical relation :

cleus,A its mass number, and r0 is a constant e

ies that nuclear mass density is nearly constant

sity of sodium nucleus. Compare it with the av

ained in Exercise. 2.27.

y the relation,

e phase are

d. Thus, the

uclear sizes

qual to about,

for different

rage mass


27/31

Now, the mass of a nucleiM

M=A amu =A 1.66 102

Density of nucleus,

=

This relation shows that nuclmass densities of all nuclei a

Density of sodium nucleus is

Question 2.29:

A LASER is a source of ver

These properties of a laser li

of the Moon from the Earth

source of light. A laser light

the Moons surface. How m

Answer

Time taken by the laser bea

is equal to its mass number i.e.,

7 kg

ear mass depends only on constant . Hence, te nearly the same.

given by,

intense, monochromatic, and unidirectional be

ht can be exploited to measure long distances.

as been already determined very precisely usin

eamed at the Moon takes 2.56 s to return after

ch is the radius of the lunar orbit around the Ea

to return to Earth after reflection from the Mo

e nuclear

m of light.

he distance

a laser as a

eflection at

th?

n = 2.56 s


28/31

Speed of light = 3 10 m/s

Time taken by the laser bea

Radius of the lunar orbit = D3.84 108

m = 3.84 105

k

Question 2.30:

A SONAR (sound navigatioobjects under water. In a sub

generation of a probe wave a

submarine is found to be 77.

sound in water = 1450 m s1)

Answer

Let the distance between the

Speed of sound in water = 1

Time lag between transmissi

In this time lag, sound waves

ship and the submarine (2S).

Time taken for the sound to

Distance between the ship

Question 2.31:

The farthest objects in our U

light emitted by them takes b

quasars) have many puzzling

What is the distance in km o

to reach Moon =

istance between the Earth and the Moon = 1.28

and ranging) uses ultrasonic waves to detect aarine equipped with a SONAR the time delay

nd the reception of its echo after reflection fro

s. What is the distance of the enemy submarin

.

ship and the enemy submarine be S.

50 m/s

n and reception of Sonar waves = 77 s

travel a distance which is twice the distance be

each the submarine

and the submarine (S) = 1450 38.5 = 55825

iverse discovered by modern astronomers are s

illions of years to reach the Earth. These object

features, which have not yet been satisfactorily

a quasar from which light takes 3.0 billion yea

3 10

8

=

d locatebetween

an enemy

e? (Speed of

tween the

= 55.8 km

o distant that

(known as

explained.

s to reach


29/31

us?

Answer

Time taken by quasar light t

= 3 109 years

= 3 109

365 24 60

Speed of light = 3 108

m/s

Distance between the Earth a

= (3 108) (3 109 365

= 283824 1020 m

= 2.8 1022

km

Question 2.32:

It is a well known fact that d

completely covers the disk o

gather from examples 2.3 an

Answer

The position of the Sun, Mo

figure.

reach Earth = 3 billion years

0 s

nd quasar

24 60 60)

ring a total solar eclipse the disk of the moon a

the Sun. From this fact and from the informati

2.4, determine the approximate diameter of th

n, and Earth during a lunar eclipse is shown in

lmost

n you can

moon.

he given


30/31

Distance of the Moon from t

Distance of the Sun from the

Diameter of the Sun = 1.39

It can be observed that TR

Hence, the diameter of the

Question 2.33:

A great physicist of this cent

Fundamental constants of na

that from the basic constants

and the gravitational constan

Further, it was a very large n

the age of the universe (~15book, try to see if you too ca

can think of). If its coinciden

this imply for the constancy

Answer

One relation consists of som

Where,

e Earth = 3.84 10 m

Earth = 1.496 1011

m

109

m

and TPQ are similar. Hence, it can be writte

oon is 3.57 106

m.

ry (P.A.M. Dirac) loved playing with numerica

ure. This led him to an interesting observation.

of atomic physics (c, e, mass of electron, mass

G, he could arrive at a number with the dimen

mber, its magnitude being close to the present

illion years). From the table of fundamental coconstruct this number (or any other interesting

ce with the age of the universe were significant,

f fundamental constants?

fundamental constants that give the age of the

as:

l values of

Dirac found

f proton)

ion of time.

estimate on

stants in thisnumber you

what would

Universe by:


31/31

t= Age of Universe

e = Charge of electrons = 1.6

= Absolute permittivity

= Mass of protons = 1.67

= Mass of electrons = 9.1

c = Speed of light = 3 108

G = Universal gravitational c

Also, Nm

2

/C

2

Substituting these values in t

1019

C

1027

kg

1031

kg

/s

onstant = 6.67 1011

Nm2

kg2

e equation, we get

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Chapter 2 Units & Measurement