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Units And Measurements (Physics)

Question 2
...
m3
The surface area of a solid cylinder of radius 2
...
0 cm is equal to
...
m in 1 s
The relative density of lead is 11
...
Its density is
...
kg m–3
...

The total surface area of a cylinder of radius r and height h is
S = 2πr (r + h)
...
5 × 104 mm2

Using the conversion,

1 km/h =

Therefore, distance can be obtained using the relation:
Distance = Speed × Time = 5 × 1 = 5 m
Hence, the vehicle covers 5 m in 1 s
...
3 g/cm3 = 11
...
2:
Fill in the blanks by suitable conversion of units:
1 kg m2s–2=
...
ly

3
...
km h–2
G= 6
...
(cm)3s–2 g–1
...

1 ly = Speed of light × One year
= (3 × 108 m/s) × (365 × 24 × 60 × 60 s)
= 9
...
88 × 10–4 km h–2
1 N = 1 kg m s–2
1 kg = 10–3 g–1
1 m3 = 106 cm3

∴ 6
...
67 × 10–11 × (1 kg m s–2) (1 m2) (1 s–2)

= 6
...
67 × 10–11 × (10–3 g–1) × (106 cm3) × (1 s–2)
= 6
...
3:
A calorie is a unit of heat or energy and it equals about 4
...

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 is γ s
...
2 α–1 β–2
γ2 in terms of the new units
...
2 (1 kg) (1 m2) (1 s–2)
New unit of mass = α kg

Hence, in terms of the new unit, 1 kg =
ew
In terms of the new unit of length,

And, in terms of the new unit of time,

2
∴1 calorie = 4
...
2 α–1 β–2 γ2

Question 2
...
In view of this, reframe the following statements wherever
necessary:
atoms are very small objects
a jet plane moves with great speed
the mass of Jupiter is very large
the air inside this room contains a large number of molecules
s
a proton is much more massive than an electron
the speed of sound is much smaller than the speed of light
...
For example, the coefficient of friction is
dimensionless
...

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

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

A proton is more massive than an electron
...


Question 2
...
What is the
distance between the Sun and the Earth in terms of the new unit if light takes 8 min and
Earth
20 s to cover this distance?
Answer

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
Distance

Question 2
...

Least count of vernier callipers
= 1 standard division (SD) – 1 vernier division (VD)

Least count of screw gauge =

Least count of an optical device = Wavelength of light ∼ 10–5 cm
= 0
...


Question 2
...
He makes 20 observations and finds that the average width of the
hair in the field of view of the microscope is 3
...
What is the estimate on the
thickness of hair?
Answer

Magnification of the microscope = 100
Average width of the hair in the field of view of the microscope = 3
...
035 mm
...
8:
Answer the following:
You are given a thread and a metre scale
...
0 mm and 200 divisions on the circular scale
...
Why is a set
of 100 measurements of the diameter expected to yield a more reliable estimate t
than a set
of 5 measurements only?
Answer

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

It is not possible to increase the accuracy of a screw gauge by increasing the number of
divisions of the circular scale
...

A set of 100 measurements is more reliable than a set of 5 measurements because random
asurements
errors involved in the former are very less as compared to the latter
...
9:
The photograph of a house occupies an area of 1
...
The slide is
projected on to a screen, and the area of the house on the screen is 1
...
What is the
linear magnification of the projector-screen arrangement?
projector
Answer

Area of the house on the slide = 1
...
55 m2
= 1
...
10:
State the number of significant figures in the following:
0
...
64 × 1024 kg
0
...
320 J
6
...
0006032 m2
Answer

Answer: 1
The given quantity is 0
...

If the number is less than one, then all zeros on the right of the decimal point (but left to
the first non-zero) are insignificant
...
Hence, only 7 is a significant figure in this quantity
...
64 × 1024 kg
...
Hence, all

digits i
...
, 2, 6 and 4 are significant figures
...
2370 g cm–3
...
Hence, besides digits 2, 3
and 7, 0 that appears after the decimal point is also a significant figure
...
320 J
...
Hence, all four digits
cimals,
appearing in the given quantity are significant figures
...
032 Nm–2
...

zero
Answer: 4
The given quantity is 0
...

If the number is less than one, then the zeroes on the right of the decimal point (but left to
the first non-zero) are insignificant
...
All zeros between two non-zero digits are always significant
...


Question 2
...
234 m, 1
...
01 cm respectively
...

Answer

Length of sheet, l = 4
...
005 m
Thickness of sheet, h = 2
...
0201 m
The given table lists the respective significant figures:
Quantity

Number

Significant Figure

l

4
...
005

4

h

2
...
e
...

Surface area of the sheet = 2 (l × b + b × h + h × l)
(
= 2(4
...
005 + 1
...
0201 + 0
...
234)
= 2 (4
...
02620 + 0
...
360
= 8
...
234 × 1
...
0201
= 0
...
e
...


Question 2
...
300 kg
...
15 g and 20
...
What is (a) the total mass of the box, (b)
the difference in the masses of the pieces to correct significant figures?
Answer

Mass of grocer’s box = 2
...
15g = 0
...
17 g = 0
...
3 + 0
...
02017 = 2
...
Hence, the total mass of the box is 2
...

Difference in masses = 20
...
15 = 0
...


Question 2
...
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
...
8
...
14:
A book with many printing errors contains four different formulas for the displacement y
of a particle undergoing a certain periodic motion:

y = a sin vt

(a = maximum displacement of the particle, v = speed of the particle
...
Rule out the wrong formulas on dimensional grounds
...
H
...
H
...

Answer: Incorrect
y = a sin vt
Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0
Dimension of vt = M0 L1 T–1 × M0 L0 T1 = M0 L1 T0
But the argument of the trigonometric function must be dimensionless, which is not so in
the given case
...

Answer: Incorrect

Dimension of y = M0L1T0

Dimension of

= M0L1T–1

Dimension of

= M0 L–1 T1

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

Answer: Correct

Dimension of y = M0 L1 T0
Dimension of a = M0 L1 T0

Dimension of

= M0 L0 T0

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
...


Question 2
...
(This relation first arose as a consequence of
special relativity due to Albert Einstein)
...
He writes:

Answer

Given the relation,

Dimension of m = M1 L0 T0
Dimension of

= M1 L0 T0

Dimension of v = M0 L1 T–1

Dimension of v2 = M0 L2 T–2
Dimension of c = M0 L1 T–1
The given formula will be dimensionally correct only when the dimension of L
...
S is the
same as that of R
...
S
...
e
...
This is only possible if v2 is divided by c2
...


Question 2
...
The size of a hydrogen atom is about
volume in m3 of a mole of hydrogen atoms?

what is the total atomic

Answer

Radius of hydrogen atom, r = 0
...
5 × 10–10 m

Volume of hydrogen atom =

1 mole of hydrogen contains 6
...


∴ Volume of 1 mole of hydrogen atoms = 6
...
524 × 10–30
= 3
...
17:
One mole of an ideal gas at standard temperature and pressure occupies 22
...
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

)
...
5

= 0
...
023 × 1023 hydrogen atoms
...
023 × 1023 × 0
...
16 × 10–7 m3

Molar volume of 1 mole of hydrogen atoms at STP,
Vm = 22
...
4 × 10–3 m3

Hence, the molar volume is 7
...
For this
reason, the inter-atomic separation in hydrogen gas is much larger than the size of a
atomic
hydrogen atom
...
18:
Explain this common observation clearly : If you look out of the window of a fast moving
mo

train, the nearby trees, houses etc
...
) seem to be
stationary
...

Answer

Line of sight is defined as an imaginary line joining an object and an observer’s eye
...
while sitting in a
moving train, they appear to move rapidly in the opposite direction because the line of
sight changes very rapidly
...
appear stationary because of the
large distance
...

rapidly
...
19:
The principle of ‘parallax’ in section 2
...
1 is used in the determination of distances of
very distant stars
...
That is, the baseline is about the diameter of the Earth’s
orbit ≈ 3 × 1011m
...
A parsec is a
convenient unit of length on the astronomical scale
...
How much is a parsec in terms of meters?
Answer

Diameter of Earth’s orbit = 3 × 1011 m
arth’s
Radius of Earth’s orbit, r = 1
...
847 × 10–6 rad
...

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends

an angle of


...
09 × 1016 m
...
20:
The nearest star to our solar system is 4
...
How much is this distance in
terms of parsecs? How much parallax would this star (named Alpha Centauri) show when
Centauri
viewed from two locations of the Earth six months apart in its orbit around the Sun?
Answer

Distance of the star from the solar system = 4
...

1 light year = Speed of light × 1 year
= 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011 m
∴4
...
32 × 1011 m
1 parsec = 3
...
29 ly =

Using the relation,

= 1
...
85 × 10–6 rad
∴

Question 2
...
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
...
Think of different examples in modern science where precise
measurements of length, time, mass etc
...
Also, wherever you can, give a
quantitative idea of the precision needed
...
For example, ultra-shot laser pulses (time interval ∼ 10–15 s) are
ultra
used to measure time intervals in several physical and chemical processes
...

The development of mass spectrometer makes it possible to measure the mass of atoms
precisely
...
22:
Just as precise measurements are necessary in science, it is equally important to be able to
equally
make rough estimates of quantities using rudimentary ideas and common observations
...

Answer

During monsoons, a metrologist records about 215 cm of rainfall in India i
...
, the height
of water column, h = 215 cm = 2
...
3 × 1012 m2
Hence, volume of rain water, V = A × h = 7
...
09 × 1015 kg
Hence, the total mass of rain-bearing clouds over India is approximately 7
...

Consider a ship of known base area floating in the sea
...

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
...
As wind blows, it rotates
...


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
...

∴Area of one hair = πr2

Number of strands of hair
Let the volume of the room be V
...
4 l i
...
, 22
...

Number of molecules in one mole = 6
...
915 × 1026 V

= 1
...
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
...
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
gases?
your guess is correct from the following data: mass of the Sun = 2
...
0 × 108 m
...
0 × 1030 kg
Radius of the Sun, R = 7
...
This high density is
attributed to the intense gravitational attraction of the inner layers on the outer layer of the
Sun
...
24:
When the planet Jupiter is at a distance of 824
...
Calculate the diameter of Jupiter
...
7 × 106 km = 824
...
25:
A man walking briskly in rain with speed v must slant his umbrella forward making an
angle θ with the vertical
...
(We are
assuming there is no strong wind and that the rain falls vertically for a stationary man)
...

Answer

Answer: Incorrect; on dimensional ground
The relation is


...
H
...
H
...
H
...
H
...
Hence, the given relation is
not correct dimensionally
...
H
...
One way to
should
achieve this is by dividing the R
...
S by the speed of rainfall
...
This relation is dimensionally correct
...
26:
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any
disturbance, may differ by only about 0
...
What does this imply for the accuracy of the
standard cesium clock in measuring a time-interval of 1 s?
time
Answer

Difference in time of caesium clocks = 0
...
15 × 109 s
In 3
...
02 s
...


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

...
27:
Estimate the average mass density of a sodium atom assuming its size to be about 2
...

(Use the known values of Avogadro’s number and the atomic mass of sodium)
...
Are the two densities of
the same order of magnitude? If so, why?
Answer

Diameter of sodium atom = Size of sodium atom = 2
...
25 × 10–10 m

Volume of sodium atom, V =

According to the Avogadro hypothesis, one mole of sodium contains 6
...


∴ Mass of one atom =

Density of sodium atom, ρ =
It is given that the density of sodium in crystalline phase is 970 kg m–3
...
This is because in solid phase, atoms are closely packed
...

atomic

Question 2
...
Nuclear sizes
obey roughly the following empirical relation :
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to about,
1
...
Show that the rule implies that nuclear mass density is nearly constant for different
nuclei
...
Compare it with the average mass
density of a sodium atom obtained in Exercise
...
27
...
2 f = 1
...
e
...
66 × 10–27 kg

Density of nucleus,

ρ=

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

Density of sodium nucleus is given by,

Question 2
...

These properties of a laser light can be exploited to measure long distances
...
A laser light beamed at the Moon takes 2
...
How much is the radius of the lunar orbit around the Earth?
Answer

Time taken by the laser beam to return to Earth after reflection from the Moon = 2
...
28 × 3 × 108 =
3
...
84 × 105 km

Question 2
...
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
...
What is the distance of the enemy submarine? (Speed of
sound in water = 1450 m s–1)
...

‘
Speed of sound in water = 1450 m/s
Time lag between transmission and rece
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)
...
5 = 55825 m = 55
...
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
...

features,
What is the distance in km of a quasar from which light takes 3
...
8 × 1022 km

Question 2
...
From this fact and from the information you can
gather from examples 2
...
4, determine the approximate diameter of the moon
...


Distance of the Moon from the Earth = 3
...
496 × 1011 m
Diameter of the Sun = 1
...
Hence, it can be written as:
written

Hence, the diameter of the Moon is 3
...


Question 2
...
A
...
Dirac) loved playing with numerical values of
Fundamental constants of nature
...
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
...
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)
...
6 ×10–19 C
×
= Absolute permittivity
= Mass of protons = 1
...
1 × 10–31 kg
c = Speed of light = 3 × 108 m/s
G = Universal gravitational constant = 6

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