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
The given quantity is 2.64
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
The given quantity is 0.2370
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
The given quantity is 6.032
All zeroes between two non-
Answer: 4
The given quantity is 0.0006
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
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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
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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
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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
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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
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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?
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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.
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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.
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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 =
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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
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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
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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
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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
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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
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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
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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:
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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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