Download 1. Estimate the average mass density of a sodium atom assuming its

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1. Estimate the average mass density of a sodium atom assuming its size to be about 2.5
Å. (Use the known values of Avagadro number and the atomic weight of sodium). Compare
it with the density of sodium in its crystalline phase: 970 kg m-3. Are the two densities of
the same order of magnitude? If so, why?
Solution:
09
= 1.25 × 10-10 m3
Average radius of the sodium atom, R =
Therefore, volume of one sodium atom, V =
-3
Now, one mole i.e., 23 × 10
× R3
=
π€× (1.25 ( 10-10)3 = 8.18 × 10-30 m3
kg of sodium= 6.0225 (1023 atoms
Therefore, average mass of one sodium atom, m =
= 3.82 × 10-26 kg
Average density of sodium atom,
= 4.67 × 103 kg m-3
-3
Now, density of sodium in its crystalline state, pc = 970 kg m
Since 4.67 x 103 » 103 and 970 » 103, the two densities are of the same order.
2. One mole of an ideal gas at standard temperature and pressure occupies 22.4 litre
(molar volume). What is the ratio of molar volume to the atomic volume of hydrogen? (Take
the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?
Volume of one hydrogen molecule =
πR3
ca
Solution:
Volume of one mole of an ideal gas, Vgas = 22.4 litre
= 22.4 × 10-3 m3
Radius of hydrogen molecule, R = 1Å /2 = 0.5 × 10-10m
ti
= π × 0.5 (10-10)3 = 0.5236 × 10-30 m3
If N is the Avagadro number, then volume of one mole of hydrogen molecules, is given by
VH = 0.5236 × 10-30 × N
= 0.5236 × 10-30 × 6.0225 × 1023
= 3.154 × 10-7 m3
104
.in
on
Therefore
The ratio
is so large due to the fact that the intermolecular separation in a gas is much larger
than the size of the hydrogen molecule.
3. A calorie is a unit of heat energy and it is equal to about 4.2 J, where 1 J = 1 kg m2 s-2.
Suppose we employ a system of units in which the unit of mass equals a kg, the unit of
length equals b m, the unit of time is g s. Show that a calorie has a magnitude 4.2 a-1b2 2
g in terms of the new units.
1
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Solution:
1 calorie = 4.2 J = 4.2 kg m2 s-2
α-1 ( do not use / for division
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As new unit of mass = ( kg, hence in terms of new unit 1 kg = 1/α =
only horizontal proper sign for division is to be used )
Similarly, in terms of the new unit of length
β-1
= β-2
1m=
or
2
1m
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In terms of the new unit of time, 1 s = γ-1
Therefore 1s-2 = (γ-1)-2 = γ2
Hence, 1 calorie = 4.18 kg m2 s-2
= 4.18 (α-1)(β2)(γ2) = 4.18
αβ2γ2 kg m/s.
4. Explain this statement clearly: "To call a dimensional quantity 'large' or 'small' is
meaningless without specifying a standard for comparison". In view of this, reword the
following statements wherever necessary:
a. Atoms are very small objects.
b. A jet plane moves with very 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
Solution:
a. Atoms are very small objects of the order of 10-10 m.
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b. A jet plane moves with very great speed as compared to the speed of a car or any two wheeler.
c. The mass of Jupiter is very large compared to the mass of Earth or moon.
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d. The air inside this room contains a larger number of molecules than that inside a box or a balloon.
f. The speed of sound is much smaller than the speed of light.
.in
e and f are correct statements and need no change.
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e. A proton is much more massive than an electron.
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 the light takes 8
minutes and 20 seconds to cover this distance?
Solution:
The new unit of distance = Speed of light in vacuum = distance traveled by light in 1s.
or New unit of distance = 3 × 108 m
Time taken by light to cover the distance between Sun and Earth = 8 min and 20 = 500 s
2
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The distance between the sun and the earth = 3 × 108 × 500 m.
= 500
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The distance between the sun and the earth in terms of the new units =
new units.
09
6. Which of the following is the most precise device for measuring length:
(i) a vernier calipers with 20 divisions on the sliding scale.
(ii) a screw gauge of pitch 1 mm and 100 divisions on circular scale.
(iii) an optical instrument that can measure length to within a wavelength of light?
Solution:
(iii) optical instrument that can measure length to within a wavelength of light is more accurate
device for measuring length:
7. A student measures the thickness of human hair by looking at it through a microscope
of magnification 100. He makes 20 observations and finds that the average width of the
hair in the field of view of the microscope is 3.5 mm. What is the estimate on thickness of
the hair?
Solution:
Actual width =
8. Answer the following:
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Diameter of the hair 1mm =
= 0.035 mm
Thickness of the hair
= 0.035 mm.
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(a) You are given a thread and a metre scale. How will you estimate the diameter of the
thread?
(b) A screw guage has a pitch of 1.0 mm and 200 divisions on the main scale. Do you think
it is possible to increase the accuracy of the gauge arbitrarily by increasing the number of
divisions on the circular scale?
.in
(c) The mean diameter of a thin brass rod is to be measured by a vernier calipers. Why is a
set of 100 measurements expected to yield a more reliable estimate than a set of 5
measurements only?
Solution:
(a) First of all wrap the thread a number of times on a metre scale very close to each other to form a
coil.
Let the length of this thread coil be 'l' cm.
Let the number of turns in the coil be 'n'
3
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Thickness (diameter) of the thread t= l/n cm,
(b) Yes, the accuracy can be increased to a greater extent by increasing the number of divisions on
the circular scale because the least count of the gauge would become less. Least count = pitch/ no.
of divisions on circular scale
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(c) By taking a large number of measurements of the same quantity, it is quite possible that the
majority of these measurements will have small errors, which might be positive or negative. While
taking an average, these positive and negative errors are likely to cancel each other. Hence, a set of
about 100 measurements of the diameter is expected to produce a more reliable estimate than the
set of 5 measurements only.
9. The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is
projected on to a screen and the area of the house on the screen is 1.55 m2. What is the
linear magnification of the projector-screen arrangement?
Solution:
Linear magnification
=
=
=
=
= 94.11
10. State the number of significant figures in the following:
(i) 0.007 m2
(ii) 2.64 × 1024 kg
(iii) 0.2370 gcm-3
(v) 6.032 Nm-2
(iii) 0.2370 g cm-3 = 2.370 × 10-1 g cm-3
The significant figures are 2, 3, 7 and 0 and their number is 4.
.in
(ii) 2.64 × 1024 kg
The significant figures are 2, 6 and 4 and their number is 3.
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Solution:
(i) 0.007 m2 = 7 × 10-3 m2
the number of significant figure is only one.
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(vi) 0.0006032 m2
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(iv) 6.320 J
(iv) 6.320 J
The significant figures are 6, 3, 2 and 0 and their number is 4.
(v) 6.023 Nm-2
The significant figures are 6, 0, 2 and 3, i.e. their number is 4.
4
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(vi) 0.0006032 = 6.032 × 10-4
The significant figures are 6, 0, 3 and 2 i.e., their number is 4.
Solution:
[The number of significant figures in a calculated figure is the same as that of the number with the
lowest significant digit in the constituting numbers.]
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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 to the correct significant figures?
Area = 2 (lb + bh + hl)
= 2 (4.234 (1.005 + 1.005 0.0201 + 0.0201 × 40234)
= 8.7209468 m2 3 significant figures.
Since the least accuracy in the given data is upto three places of decimal,
Area = 8.720 m2
Volume = length × breadth × thickness
= (4.234 (1.005 (0.0201))) m3
= 0.0855 m2.
12. The mass of a box measured by a grocer’s balance is 2.3 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 masses of the pieces to the correct significant figures?
•
Solution:
a. Total mass of the box = 2.3 + 0.0215 + 0.0217 kg = 2.34032 kg
= 2.3 kg (rounded off to two significant figures as mass of box has two
significant figures only)
ca
b. The difference in masses of the pieces added = 20.17 - 20.15 g = 0.02 g.
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13. A physical quantity P is related to four observables a, b, c and d as follows:
P = a3 b2/√
√(cd). 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 P
calculated using the above relation turns out to be 3.763, to what value should you round
off the result.
.in
Solution:
percentage error in P = ((P/P) (100))
(∆P/P) × 100 = [3 × (Da/a) + 2 × (Db/b) + 1/2 x (Dc/c) + (Dd/d)] × 100
(∆a/a) × 100 = 1%
(∆b/b) × 100 = 3%
(∆c/c) × 100 = 4%
(∆d/d) × 100 = 2%
(∆P/P)€× 100 = [3 × 1% + 2 × 3% + ½ × 4% + 2%]
= 13 %
Observed value P = 3.763 ± 3.763 × 12% = 3.763 ± 0.451.
on
•
Here only the first digit is reliable. We round off the value to one additional digit.
Rounded value P = 3.8. The third digit is 6 which is greater than 5.
5
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14. A book with many printing errors contain four different formulas for displacement of
y for a particle undergoing certain periodic motion:
(i) y = a sin 2pt / T
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(ii) y = a sin vt
(iii) y = (a/T) sin t/a
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(iv) y = (a/√
√2) ( sin 2pt/T + cos 2pt/T)
(a = maximum displacement of the particle, v = speed of the particle, T = time-period of
motion). Rule out the wrong formulae on dimensional grounds.
•
Solution:
RHS and LHS of an equation should have same dimensions, in order to be a correct formula.
(i) [L] = [L] Equation is dimensionally correct.
(ii) The argument of a Trigonometric function must always be dimensionless.
Lsin( LT-1 .T) = L sin (L)
The Equation is dimensionally correct
(iii) [L] = [T]-1 Equation is dimensionally wrong. Also, The in the Trigonometric function the equation
is dimensionally wrong since t/a is not dimensionless, so it is not an angle.
(iv) Equation is dimensionally correct.
•
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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 first 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 m = mo /( 1 - v2)1/2. Guess
where to put the missing c.
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Solution:
Since m and m0 are of same dimension [M], the denominator v2/c2 should be a dimensionless
number.
For this, v2 should be divided by a (velocity factor)2 i.e. c2. The formula is:
m = m0/{1 - (v2 / c2)}1/2
The denominator must be dimensionless , V2/C2 = (LT –1) 2 /(LT-1)2 = L0 T0.
The correct equation is m = mo /( 1 - v2 / c2)1/2.
on
.in
16. A man walking briskly in rain with speed v must slant his umbrella making an
angle θ with the vertical. A student derives the following relation between q and v: tan θ = v
and checks that the relation has a correct limit: as v → 0, ???? check symbol θ → 0, as
expected. (We are assuming that there is no strong wind and that the rain falls vertically
for a stationary man). Do you think that the relation is correct? If not, guess the correct
relation
•
Solution:
6
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Trigonometric function is a dimensionless factor. Hence the formula should be:
tan
θ€= v/ur.
The given relation is not correct because L.H.S has no dimensions. Where u is the speed with which
the rain is falling.
17. It is claimed that two cesium clocks, if allowed to run for 100 years, free from any
disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the
standard cesium clock in measuring a time-interval of 1 s?
•
Solution:
While measuring a time interval of 1 s, it implies an accuracy of
=
= 6.35 × 10-10
=
.
Solution:
Size of the hydrogen atom, i.e. R = 0.5Å = 0.5 × 10-10m
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•
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18. The unit of length convenient on the atomic scale as an angstrom and is denoted by
A: 1Å = 10-10m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume
in m3 of a mole of hydrogen atoms?
Therefore volume of the hydrogen atom can be found using the given formula,
One mole of hydrogen contains atoms equal to Avagadro's number i.e.
.in
× 3.14 × (r)3 =
× 3.14 × 0.5 (10-10)3 change alignment
= 5.236 × 10-31m3
on
1 mole = 6.02 × 1023
1 mole = 6.0225 × 1023 atoms,
Therefore the atomic volume of 1 mole of hydrogen atoms is
= 5.236 ×€10-31 × 6.0225 × 1023 m3
= 3.154 × 10-7 m3.
7
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19. The unit of length convenient on the nuclear scale is a fermi: 1F = 10-15m. Nuclear
sizes obey roughly the following empherical relation: r = r0A1/3; where r is the radius of the
nucleus, A its mass number and r0 is a constant equal to about 1.2 F. Show that the rule
implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass
density of sodium nucleus. Compare it with the average mass density of a sodium atom
obtained in problem 20?
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Solution:
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Mass of the nucleus of mass number A, m =
a.m.u
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=
1/3
1.66 × 10-27 kg
Radius of the nucleus, r = r0 A
= 1.2 × 10-15 A1/3
Therefore, volume of the nucleus, V =
× 3.14r3
=
× 3.142 × (1.2 × 10-15 A1/3)3
= 7.24 × 10-45 A
Therefore, density of the nucleus, p =
=
= 2.29 × 1017 kg m-3
As density of the nucleus is independent of A, it may be concluded that nuclear density is same for
different nuclei. Therefore, density of sodium nucleus pnuc = 2.29 × 1017 kg m-3
Density of sodium atom patom = 4.67 × 103 kg m-3
Therefore
=
= 4.9 × 1013.
•
Solution:
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20. 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 in a direction opposite to the train’s motion, but the distant objects (hilltops, 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)?
.in
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There is a relative motion between the observer in the train and outside the objects, which are
stationary. But in case of objects like nearby tree, houses etc, the angle subtended by them with the
line of vision is large and hence changes rapidly with time. On the other hand, for far-off objects, like
the hill tops or so, the subtended by them is small, the change in angle is also small .hence, they
appear to be stationary.
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21. The principle of ‘parallax’ is used in the determination of distances of very distant
stars. The baseline AB is the line joining the Earth’s two locations six months apart in its
orbit around the Sun. That is, the baseline is about the diameter of the Earth’s
orbit ∼ 3 ×1011m. However, even the nearest stars are so distant that with such a long
baseline they show parallax only of the order of 1" (second) of arc or so. A ‘parsec’ is a
convenient unit of length on the astronomical scale. It is the distance of an object that will
show a parallax of 1" (second) of arc from the opposite ends of a baseline equal to the
distance from the Earth to the Sun. How much is a parsec in terms of metres?
Solution:
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1 s of an arc = 4.9 × 10-6 rad.
Diameter of earth's orbit = 3 × 1011 m
... Radius of the earth's orbit = 1.5 × 1011 m
Distance of the object = 1.5 × 1011 m
4.9 × 10-6
One parsec = distance at which an arc length of the distance between between the earth and Sun
subtends one second angle.
= 1.5 × 1011 m/ (57.3o/3600)
= 3.1 × 1016 m, As a unit of length 1 parsec is defined to be equal to 3.05 × 1016 m.
.in
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22. 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 on Earth six months apart in its orbit around the
Sun?
Solution:
1 light year = 9.46 × 1015 m
1 parsec = 3.08 × 1016 m
4.29 light years S =
= 1.31 parsec
1.In the orbit of the Sun, the distance between the two locations of the earth six months apart will
be equal to diameter of the orbit of earth i.e. l = 2 AU = 2 × 1.496 (1011 m. Therefore, the star will
show parallax.)
9
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Arc subtended = base line in earth's orbit /distance to the star = l/s
=
= 7.34 (1016 radian)
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= 1.54 seconds of an arc.
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23. A LASER is a source of very intense, monochromatic, and unidirectional beam of
light. These properties of a laser light can be exploited to measure long distances. The
distance of the Moon from the Earth has been already determined very precisely using a
laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after
reflection at the Moon’s surface. How much is the radius of the lunar orbit around the
Earth?
Solution:
Velocity of the laser light = c = 3 × 108 m/s
Time taken for laser light to travel from the earth to moon and back =2 t = (2 .56s) / (2) = 1.28 s
... Required distance = Speed of laser light × t = 3 × 108 × ( 2.56) /2
= 3.84 × 108 m.
24. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate
objects under water. In a submarine equipped with a SONAR the time delay between
generation of a probe wave and the reception of its echo after reflection from an enemy
submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of
sound in water = 1450 ms-1)
ca
Solution:
The time taken by ultrasonic (SONAR) to travel to and fro of the submarine 2t = 77.0 s,
t = 77.0 = 38.5 s
2
speed of sound in water = 1450 ms-1
... Required distance = 1450 × 38.5 = 55825 m
= 5.583 × 104 m.
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25. 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?
on
.in
Solution:
The time taken by the light to reach us = t = 3.0 billion years.
One light year = 9.46 × 1015 m = 9.46 × 1012 km
= 3.0 × 109 × 365 × 24 × 60 × 60 s
We know that the velocity of the light c = 3 × 108 m/s
Therefore, distance of the quasar from us, S = 3 × 108 × 3.0 x 1015 × 365 × 24 × 60 × 60
= 2.838 × 1025 m.
26. Precise measurements of physical quantities are a need of modern times. For
example, to ascertain the speed of an enemy fighter plane, one must have an accurate
method to find its positions at closely separated instants of time. Only then we can hope to
shell by an anti-aircraft gun. This was the actual motivation behind the discovery of radar in
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World war II. Think of different examples in modern science where precision
measurements of length, time, mass etc. are needed. Also, wherever you can, give a
quantitative idea of the precision needed.
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Solution:
(i) Measurement of length
Precision measurement of length is needed to determine the interatomic distance in crystals in the
field of crystallography.
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(ii) Measurement of time
In Michelson-Morley experiment, precision measurement of time was needed. The time interval was
measured in terms of shift of the interference pattern of light. This measurement is responsible for
the formulation of the theory of relativity by Einstein.
(iii) Measurement of mass
Precision measurement of mass of an atom is made by a mass spectrometer. This has verified the
relation
m=
where m is the mass of a particle moving with velocity v and m0 is its rest mass.
27. Just as precise measurements are necessary in science, it is equally important to
make rough estimates of quantities using rudimentary ideas and common observations.
Think of the ways by which you can estimate the following: (Where estimate is difficult to
obtain, try to get an upper bound of the quantity)
a. The total quantity of rain-bearing clouds over India during monsoon.
b. The mass of an elephant
d. The number of strands of hair on your head.
ti
e. The number of air molecules in your classroom.
ca
c. The wind-speed during a storm
.in
on
Solution:
a. Rainfall is measured extensively covering the entire nation along with an approximate demarcation
of the areas involved. This will give the data for computing the total volume of rainfall in the country.
The projection of the trend for future can be forecast in general based on the statistical data
collected over years. An average rainfall of about 100 cm during monsoon in India is already
recorded by the meteorologist.
b. The elephant can be made to stand in the middle of a steel bridge and the deflection of the bridge
can be measured by a dial gauge. Afterwards the bridge can be loaded by steel billets to give the
same deflection. The billets can be weighed individually and added to give the total weight of the
elephant. A traditional method is to make the elephant stand on a barge floating in water and then
afterwards add bricks to sink the barge to the same extent. Mass of an elephant can be estimated by
making it stand at the edge of long and strong wooden beam and then applying the principle of
lever.
c. If the wind has been sudden and the direction of movement can be reasonably judged, one can
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deduce an idea of the speed of the wind. Otherwise the extent of bending of steel transmission
towers, electric posts etc can give an idea about the speed in the local area.
Wind speed can be estimated by floating a gas filled balloon in air at a known height h. At the onset
of the strom, the angle of drift of the balloon in one second can be estimated. Knowing the height h
and the angle of drift, the wind speed can be calculated.
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d. The first assumption is that the man is not partially bald. Area of the head is estimated and the
thickness of the hair is measured using a screw gauge. Assuming the hair on the head is distributed
unformly, the number of hair on the head can be estimated as the ratio of area of head to the cross
section area of hair.
28. The Sun is a hot plasmo (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 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 × 1030 kg,
radius of the Sun = 7 .0 × 108 m ?
Solution:
Density = (2 × (1030) /{(4/3) ((π) (73)1024} kg/m3= 1.4 × 103 kg/m3.
The mass density of the Sun is in the range of density of liquids and solutions not gases.
This high density of the hot plasma arises due to inward gravitational attraction on outer layers due
to inner layers of the Sun, and the gases inside the sun are subjected to enormous hydrostatic
pressure of gases themselves.
29. When the planet Jupiter is at a distance of 824.7 million kilometers from the earth,
its angular diameter is measured to be 35.72 second of arc. Calculate the diameter of
Jupiter.
35.72 second of arc
=
(in radians)
d = 35.72 × 4.85 × 10-6 × 824.7 × 10-6 km
= 1.429 × 105 km.
.in
∴ 35.72 × 4.85 × 10-6 =
on
=
ti
Angular diameter q
ca
Solution:
30. Assuming that the orbit of the planet Mercury around the sun to be a circle,
Copernicus determine the orbital radius to be 0.38 AU. From this, determine the angle of
maximum elongation for Mercury and its distance from the earth when the elongation is
maximum.
Solution:
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We know that sin Î =
where rps is the distance of the planet from the sun and res is the distance of the earth from the sun.
rps
= sin Î res
res
= 1 AU
sin Î
= rps (in AU)
sin Î
= 0.38
(... rps = 0.38 AU)
-1
€€∴€
Î
= sin (0.38)
= 22.33°
Now,
cos Î =
rpe = cos Î
res = cos 22.33° AU
= 0.925 AU
= 0.925 × 1.496 × 1011 m
= 1.384 × 1011 m
= 1.384 × 108 km
ca
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31. If, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius
of the orbit is 4.22 x 108 m. Show that the mass of Jupiter is about one thousandth that of
the sun. NOTE to be shifted to chapter gravitation
.in
∴G
where
ms = mass of satellite
mj = mass of Jupiter
r = distance of satellite from Jupiter
u = velocity of satellite
on
Solution:
The gravitational force of attraction between the satellite and the planet provides the necessary
centripetal force.
Now,
u=
where T is the period of revolution of the satellite
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=
or
mj =
w
mj =
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mj =
kg
=
= 1901.15 ×€1024 kg
= 1.9 × 1027 kg
Mass of sun = 2 × 1030 kg
kg
= 9.5 ×€10-4 kg
32. It is a well-known fact that during a total solar eclipse the disc of the moon almost
completely covers the disc of the sun. From this fact determine the approximate diameter of
the moon.
(Given: earth-moon distance = 3.8452 × 108 m, sun-earth distance = 1.496 × 10 11 m and
diameter of the sun = 13.92 × 108 m).
Solution:
ti
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.in
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In the figure, CD represents the diameter of the sun, AB represents the diameter of the moon and E
represents the earth.
As the triangles CDE and ABE are similar
or
We know that
AB =
× CD
BE = 3.8452 × 108 m
DE = 1.496 × 1011 m
CD = 13.92 × 108 m
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∴ AB =
= 3.58 × 103 km.
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33. Suppose the sun shrank from its present size so that its radius is halved. What would
be the change in its gravitational potential energy? (Calculate the actual number in joules)
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Solution:
Initial gravitational potential energy of sun =
Final gravitational potential energy of sun =
(i.e., energy after shrinking)
Change in gravitational potential energy
=
=
Gravitational P.E. =
= 2.3 × 1041 joule
joule
34. Let us assume that our galaxy consists of 2.5 x 1011 stars each of one solar mass.
How long will a star at a distance of 50,000 light year from the galactic centre take to
complete one revolution?
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Solution:
Number of stars in our galaxy = 2.5 × 1011
Mass of each star
= solar mass = 2 × 1030 kg
Distance from galactic centre r = 50,000 light years
= 50,000 × 9.46 × 1015 m
Let us assume that the centripetal force required by the star for its circular motion about galactic
centre is provided by the gravitational force of attraction between star and galaxy.
ti
Then,
Or
T=
T=
.in
Period of revolution of star,
on
Or
where m is the mass of the star, u is the velocity of the star, M is the mass of the galaxy and r is the
distance of the star from the galactic centre
= 2p
Or
T = 2 × 3.14
= 1.12 × 1016 s
= 3.55 × 108 year.
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09
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35. A great physicist of this century (P.A.M. Dirac) loved playing with the numerical
values of Fundamental constants of nature. This led him to an interesting observation. Dirac
found that from 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 dimensions 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 is significant, what would this
imply for the constancy of fundamental constants?
Solution:
The quantity
has the dimension of time.
=
=
The constant called the fine structure constant determines the fine structure of spectral lines of
hydrogen atoms. Since the present number arrived at by Dirac agreed with the age of the universe,
the basic constants like mass of the electron, mass of the proton, and G should change with time as
the present age of the universe will increase with time.
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