Q

Q 2.33: A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of ). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

D Devendra Khairwa

$T = \frac{e^4}{16\ \pi^2 \ \epsilon _{0}^{2}\ m_{p} m_{e}^2\ c^3\ G } \ s$

The above equation consisting of basic constants of atomic physics and the gravitational constant G has the dimension of time.

e = charge of Electron = $-1.6\times10^{-19} C$

$\epsilon _{0}$ = absolute permittivity = $8.85\times10^{-12} F/m$

$m_{p}$ = Mass of the Proton = $1.67\times10^{-27} kg$

$m_{e}$ = Mass of the Electron = $9.1\times10^{-31} kg$
c = Speed of light in vacuum = $3\times10^8 m/s$
G = Universal Gravitational constant = $6.67\times10^{11} Nm^2kg^{-2}$

Considering T as the age of the universe and putting the values of the constants, we get:

$\therefore T \approx 6\times10^9\ years$

The age of the universe  ≈ 6 billion years.!

Q 2.32: It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

S safeer

From Examples 2.3 and 2.4 , we have,

Diameter of the Earth = $1.276 \times 10^7 m$
Distance between the Moon and the Earth,$D_{moon}$ = $3.84\times10^8 m$

Distance between the Sun and the Earth, $D_{sun}$ = $1.496\times10^{11} m$

Diameter of the Sun, $d_{sun}$ = $1.39\times10^9 m$

Let, Diameter of the Moon be  $d_{moon}$

Now, During Solar eclipse, angle subtended by Sun's diameter on Earth = angle subtended by moon's diameter

$\frac{1.39\times10^9}{1.496\times10^{11}} = \frac{d_{moon}}{3.83\times10^8}$            $(\because \Theta = d/D)$

$\Rightarrow d_{moon} = \frac{1.39\times10^9}{1.496\times10^{11}}\times 3.83\times10^8 = 3.56\times10^6 m$

Therefore, Diameter of moon = $3.56\times10^3 km$

Q 2.31: The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?

S safeer

Let the distance of a quasar from Earth be D km.

We know , Speed of light = $3\times10^8 ms^{-1}$

And, time taken by light to reach us , t = 3.0 billion years = $3\times 10^{9} years = 3\times 10^{9} \times365\times24\times60\times60 \ s$

$\therefore$ D =Speed of Light x t

= $(3 \times 10^8 ms^{-1}) \times(3\times 10^{9} \times365\times24\times60\times60 \ s)$

= $2.8\times 10^{22}\ m$

Q 2.30: 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 = $1450\ m s^{-1}$).

S safeer

Given,

77.0 s is the total time between generation of a probe wave and the reception of its echo after reflection.

$\therefore$ Time taken by sound to reach the enemy submarine = Half of total time = $\frac{1}{2}\times 77 s = 38.5 s$

Distance of enemy ship = Speed of sound x Time taken to reach the submarine

$1450\ ms^{-1} \times 38.5\ s = 55825\ m = 55.8\ km$

Q 2.29: 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 ?

S safeer

2.56 s is the total time taken by the LASER to reach Moon and again back Earth.

$\therefore$ Time taken by LASER to reach Moon = $\frac{1}{2}\times 2.56 s = 1.28 s$

We know , Speed of light $\approx 3\times10^8 ms^{-1}$

$\therefore$ The radius of the lunar orbit around the Earth = Distance between Earth and Moon =

= Speed of light x Time taken by laser one-way = $3\times10^8\ ms^{-1}\times1.28\ s = 3.84\times 10^8 \ m$

Q 2.28: The unit of length convenient on the nuclear scale is a fermi : $1 f = 10^{-15} m$. Nuclear sizes obey roughly the following empirical relation :$r = r_{o} A^{1/3}$ where r is the radius of the nucleus, A its mass number, and $r_{o}$ 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 Exercise. 2.27.

S safeer

The equation for the radius of the nucleus is given by,

$r = r_{0} A^{1/3}$

Volume of the nucleus using above relation, $\small V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (r_{0}A^{1/3})^3 = \frac{4}{3}\pi Ar_{0}^3$

We know,

Mass = Mass number× Mass of single Nucleus

= $\small A\times1.67\times10^{-27} kg$ (given)

$\small \therefore$ Nuclear mass Density = Mass of nucleus/ Volume of nucleus =

$\small \frac{A\times1.67\times10^{-27} kg}{\frac{4}{3}\pi Ar_{0}^3\ m^3} =\frac{3\times1.67\times10^{-27}}{4\pi r_{0}^3\ }\ kgm^{-3}$

The derived density formula contains only one variable,$\small r_{0}$ and is independent of mass number A. Since $\small r_{0}$ is constant, hence nuclear mass density is nearly constant for different nuclei.

∴ Density of sodium atom nucleus = $\small 2.29\times 10^{17} kgm^{-3} \approx 0.3\times 10^{18} kgm^{-3}$               $\small (Putting\ r_{0} = 1.2 f = 1.2\times 10^{-15} m )$

Comparing it with the average mass density of a sodium atom obtained in Q 2.27. (Density of the order $\small 10^3 kgm^{-3}$)

Nuclear density is typically $\small 10^{15}$ times atomic density of matter!

Q 2.27: Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : $970 \ kg\ m^{-3}$. Are the two densities of the same order of magnitude ? If so, why ?

S safeer

Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : . Are the two densities of the same order of magnitude ? If so, why ?Given, Diameter of sodium = 2.5 $\AA$

Radius, r = $0.5 \times 2.5 � = 1.25 \times 10^{-10}\ m$

Now, Volume occupied by each atom  $V = (4/3)\pi r^3 = (4/3)\times (22/7)\times(1.25\times 10^{-10})^3$

$= 8.18 \times 10^{-30} m^3$

We know, One mole of sodium has $6.023 \times 10^{23}$ atoms and has a mass of  $23 \times 10^{-3} kg$

$\therefore$ Mass of each Sodium atom $= \frac{23\times10^{-3}}{6.023\times10^{23}} kg$

$\therefore$ Density  = Mass/ Volume $= (\frac{23\times10^{-3}}{6.023\times10^{23}}\ kg )/ (8.18 \times 10^{-30} m^3) = 466.8 \times 10^{1}\ kgm^{-3}$

But , the mass density of sodium in its crystalline phase =  $970 \ kg\ m^{-3}$

$\therefore$ The densities are almost of same order. In the solid phase atoms are tightly packed and thus interatomic space is very small.

Q 2.26: It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?

D Devendra Khairwa

In terms of seconds , 100 years = $100 \times 365\times 24\times 60\times60 = 3.154 \times10^{9}\ s$

Given, Difference between the two clocks after 100 years = 0.02 s

$\therefore$ In 1 s,  the time difference  $= \frac{0.02}{3.15\times10^{9}} = 6.35 \times 10^{-12} s$

$\therefore$ Accuracy in measuring a time interval of 1 s =

$\frac{1}{6.35 \times 10^{-12}} = 1.57 \times 10^{11} \approx 10^{11}$

$\therefore$ Accuracy of 1 part in  $10^{11}\ to\ 10^{12}$

Q 2.24: 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" of arc. Calculate the diameter of Jupiter.

S safeer

Given,

Distance of Jupiter, D = $\dpi{100} 824.7 \times10^6\ km$

Angular diameter, $\dpi{100} \Theta = 35.72''$$\dpi{100} = 35.72 \times 4.848 \times 10^{-6} rad$         ($\dpi{100} \because 1'' = 4.848 \times 10^{-6} rad$)

Let diameter of Jupiter = d km

$\dpi{100} \\ \therefore d = \theta \times D = 824.7 \times 10^6 \times 35.72 \times 4.848 \times 10^{-6} \\ = 1.428 \times 10^{5}\ km$    $\dpi{100} (\because \theta = \frac{d}{D})$

Q 2.23: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7 K$, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = $2.0 \times 10^{30} kg$, radius of the Sun = $7.0 \times 10^8 m$

S safeer

Given,

Mass of the Sun, m = $3 \times10^{30}\ kg$

Radius of the Sun, r = $8 \times10^8\ m$

$\therefore$ Volume V = $\frac{4}{3}\pi r^3$

$\frac{4}{3} \times \frac{22}{7} \times (8 \times 10^8)^3 = 2145.52 \times10^{24} m^3$

Now,  Density = Mass/Volume = $\frac{3\times10^{30}}{2145.52\times10^{24}} = 1.39 \times 10^3\ kgm^{-3}$

Therefore, the density of the sun is in the range of solids and liquids and not gases.This high density arises due to inward gravitational attraction on outer layers due to inner layers of the Sun. (Imagine layers and layers of gases stacking up like a pile!)

Q 2.22: Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :

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

(b) the mass of an elephant

(c) the wind speed during a storm

(e) the number of air molecules in your classroom.

D Devendra Khairwa

(a) Height of water column during monsoon is recorded as 215 cm.

H = 215 cm = 2.15 m

Area of the country, $A = 3.3 \times 10^{12} m^2$

Volume of water column, V = AH

V = $3.3 \times 10^{12} m^2 \times 2.15 m = 7.1 \times 10^{12} m^3$

Mass of the rain-bearing clouds over India during the Monsoon, m = Volume x Density

m = $7.1 \times 10^{12} m^3 \times 10^3 kg m^{-3}$ =  $7.1 \times 10^{15} kg$       (Density of water = 103 kg m-3 )

b) Consider a large solid cube of known density having density less than water .

Measure volume of water displaced when it immersed in water = v

Measure the volume again when elephant is kept on the cube = V

Volume of water displaced by elephant, V' = V – v

The mass of this volume of water is equal to the mass of the elephant.

Mass of water displaced by elephant, m = V' x Density of water

This gives approximate mass of the elephant.

(c) A rotating device can be used to determine speed of wind. As the wind blows, the number of rotations per second will give the wind speed.

(d) Let A be the area of the head covered with hair.

If r is the radius of the root of hair, the area of the hair strand, $a = \pi r^2$

So, number of hair , $n = A/a = A/\pi r^2$

(e) Let l, b and h be the length, breadth and height of the classroom, $\therefore$ Volume of the room, v = lbh.

Volume of the air molecule,  $v' = (4/3)\pi r^3$  (r is the radius of an air molecule)

So, number of air molecules in the classroom, $n = v'/v = 4\pi r^3/3lbh$

Q 2.21: Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

S safeer

The statement "Precise measurements of physical quantities are a need of science" is indeed true. In Space explorations, very precise measurement of time in microsecond range is needed. In determining half life of radioactive material, very precise value of mass of nuclear particles are required. Similarly, in Spectroscopy precise value of length in Angstroms is required.

Q 2.20: The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?

S safeer

Given, Distance of the star from the solar system = 4.29 ly (light years)
1 light year is the distance travelled by light in one year.

(Note: Light year is a measurement of distance and not time!)
(a) $1 ly = (3 \times 10^8)ms^{-1} \times (365 \times 24 \times 60 \times 60)s = 94608 \times 10^{11} m$
4.29 ly = $405868.32 \times 10^{11} m$
We know ,1 parsec = $3.08 \times 10^{16} m$

$\therefore$ 4.29 ly = $\frac{405868.32\times10^{11}}{3.08\times10^{16}}$= 1.32 parsec

Now,

(b) $%u03B8 = d/D$$\Theta = d/D$

& d = $3 \times 10^{11} m$; D = $405868.32 \times 10^{11} m$

$\theta = (3\times10^{11})/(405868.32\times10^{11} )= 7.39\times10^{-6} rad$

Also, We know $1\ sec = 4.85 \times10^{-6} rad$

$\therefore 7.39\times10^{-6} rad = \frac{7.39\times10^{-6}}{4.85 \times10^{-6}} = 1.52''$

Q 2.25: A man walking briskly in rain with speed v must slant his umbrella forward making an angle $\Theta$ with the vertical. A student derives the following relation between $\Theta$ and $v: tan\Theta = v$ and checks that the relation has a correct limit: as $v \rightarrow 0, \Theta \rightarrow 0$, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation.

S safeer

The derived formula $tan\Theta = v$ is dimensionally incorrect.

We know, Trigonometric functions are dimensionless.

Hence , [$tan\Theta$ ] = $M^0L^0T^0$

and [v] = $M^0L^1T^{-1} = LT^{-1}$.

$\therefore$ To make it DIMENSIONALLY correct , we can divide v by $v_{r}$ (where $v_{r}$ is speed of rain)

Thus, L.H.S and R.H.S are both dimensionless and hence dimensionally satisfied.

The new formula is : $tan\Theta = v / v_{r}$

2.19 The principle of ‘parallax’ in section 2.3.1 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 ≈ $3 \times 10^{11} m$. 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 opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

S safeer

Diameter of Earth’s orbit = $3 \times 10^{11} m$
$\therefore$ Radius of Earth’s orbit, r = $1.5 \times 10^{11} m$
Let the distance parallax angle be = $4.847 \times 10^{-6} rad$.
Let the distance between earth and star be R.
(Parsec is the distance at which average radius of earth’s orbit
subtends an angle of $1{}''$.)
We have     $\Theta = r/R$    (Analogous to a circle, R here is the radius, r is the arc length and $\Theta$ is the angle covered ! )

$R = \frac{r}{\Theta } = \frac{1.5 \times 10^{11}}{4.847 \times 10^{-6}} = 0.309 \times 10^{17}$

$= 3.09 \times 10^{16}\ m$

Hence, 1 parsec $= 3.09 \times 10^{16}\ m$.

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