# NCERT Solutions for Class 11 Physics Chapter 2

## NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurement

In this page we have Units and measurements class 11 ncert solutions . These will help you in doing your homework and can serve as guidelines for solving further problems in yhis chapter. Do not forget to look for worksheets and assignments of this chapter.

Question 2.1
Fill in the blanks
(a) The volume of a cube of side 1cm is equal to__________$m^3$.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to_________$mm^2$.
(c) A vehicle moving with a speed of 18km $h^{-1}$ cover_______$m$ in $1s$.
(d) The relative density of lead is 11.3. Its density is_______$gm cm^{-3}$ or ________ $kgm^{-3}$.
Solution :
(a) As we know that $1 cm = \frac {1}{100} m = 10^{-2}m$ $Length,L=1cm=10^{-2}m,\\ \therefore \, Volume, \, L^3=\left( 10^{-2} \right) ^3=10^{-6}m^3$ (b)
As we know that $1 cm = 10 mm$ $r=2cm=20mm, \, h=10cm=100mm\\ \text{surface area of solid cylinder}=\left( 2\pi r \right) h +2 \pi r^2\\ =2\times \frac{22}{7}\times 20\times 100 + 2 \times \frac{22}{7} \times 20 \times 20=1.5\times 10^{4}mm^2$ (c)
As we know that $1 km = 1000 m$ and $1 hour = 3600 s$ $Speed, \, v=18kmh^{-1}=\frac{18 \times 1000}{60 \times 60} = 5 m/s\\ \therefore \text{Distance covered in 1s}=5m\\$ (d)
Relative Density = 11.3
Now density of water = 1 gm/cm3
So, Density of Lead will be = 11.2 gm/m3
Now we know that $1g = \frac {1}{1000} kg = 10^{-3} kg$ and $1 cm = \frac {1}{100} m =10^{-2} m$
So density in kg/m3 can be expressed as
$density=11.3\, g/cm^3=\frac{11.3 \times 10^{-3}kg}{\left( 10^{-2}m \right) ^3}\\ =11.3\times 10^3kgm^{-3}.$

Question 2.2
Fill in the blank by suitable conversion of units:
(a) $1kgm^2s^{-2}=\_\_\_\_\_\_gcm^2s^{-2}$
(b) $1m=\_\_\_\_\_\_\_light\,\,year$
(c) $3ms^{-2}=\_\_\_\_\_\_\_kmh^{-2}$
(d) $G=6.67\times 10^{-11}Nm^2kg^{-2}=\_\_\_\_\_cm^3s^{-2}g^{-1}\,$
Solution: These type of question can be simply solved by putting the respective conversion units in the place
(a)
$1kgm^2s^{-2}=1\times 10^3g\left( 10^2cm \right) ^2s^{-2}=10^7gcm^2s^{-2}$
(b) Now we know that distance traveled by light in 1 year is called 1 light year
Speed of light = $3 \times 10^8$ m/s and $1\,\, year =365 \times 24 \times 60 \times 60$ s
So 1 light year = $3 \times 10^8 \times 365 \times 24 \times 60 \times 60$ m
$1\,\,light\,\,year=9.46\times 10^{15}m$
$\therefore 1m=\frac{1}{9.46\times 10^{15}}=1.053 \times 10^{-16}light\,\,year$
(c)
\begin{align*} 3ms^{-2} &=3 \times 10^{-3}km\left[ \frac{1}{60 \times 60}h \right] ^{-2}\\ & =3 \times 10^{-3} \times 3600 \times 3600\,\,kmh^{-2}\\ & =3.888 \times 10^{4}kmh^{-2} \end{align*} (d)
\begin{align*} G & =6.67 \times 10^{-11}Nm^2\,kg^{-2}\\ & =6.67 \times 10^{-11}\left( kgms^{-2} \right) m^2kg^{-2}\\ & =6.67 \times 10^{-11}m^3s^{-2}kg^{-1}\\ & =6.67 \times 10^{-11} \times 10^6cm^3s^{-2}10^{-1}g^{-1}\\ & =6.67 \times 10^{-8}cm^3s^{-2}g^{-1} \end{align*}

Question 2.3
A calorie is a unit or energy and it equals about $4.2J$, where $1J=1kgm^2s^{-2}$. Suppose, we employ a system of units in which the unit of mass equals $\alpha\, kg$, the unit of length equals $\beta \, m$ and the unit of time is $\gamma \, s$. Show that a calorie has a magnitude of 4.2 $\alpha ^{-1}\beta ^{-2}\gamma ^2$ in terms of new units.
Solution:
Given,
$1 \, calorie = 4.2J = 4.2 \,kg m^2s^{-2}$
If $\alpha$ kg = new unit of mass
Then,
$1kg=\frac{1}{\alpha}$ is the new unit of mass
Similarly,
$1m=\beta ^{-1}$ is the new unit of length
$1s=\gamma ^{-1}$ new unit of time
Now, $1 \, calorie =4.2 \alpha ^{-1} (\beta ^{-1})^{-2}(\gamma ^{-1})^{-2}$ unit of energy
or
$1 \; calorie = 4.2 \alpha ^{-1} \beta ^{-2} \gamma ^{2}$ unit of energy
where,
$\alpha ^{-1}$ is new unit of mass
$\beta ^{-1}$ new unit of length
and $\gamma ^{-1}$ new unit of time

Question 2.4.
Explain this statement clearly :
"To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison." In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with 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
Physical quantities are called large or small depending on the unit (standard) of measurement. For example the unit of parsec is equal to $3.08 \times 10^{16} m$ which is certainly very bigger than km or m.So we used these type of unit when we have to measure distance between the stars
(a) The size of an atom is much smaller than even the sharp tip of a pin.
(b) A jet plane moves with a speed greater than that of a superfast train.
(c) The mass of Jupiter is very large compared to that of the earth.
(d) The air inside this room contains more number of molecules than in one mole of air.
(e) This is already a correct statement.We don't require any change
(f) This is already a correct statement.We don't require any change

Question 2.5
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 light takes 8 min and 20 sec to cover this distance?
Solution
Velocity of light = c = 1 new unit of length $s^{-1}$
Time taken by light of sun to reach the earth
is $t = 8\, min\, 20 s = (8\times 60)+20=500 s$
Distance between the sun and earth,
$x = c \times t = 1 \text{ new unit of length}=500\text{ new units of length}$ .

Question 2.6
Which of the following is the most precise device for measuring length ?
(a) a Vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length within a wavelength of light ?
Solution.
(a) We need to first calculate the least count of each of these device.
Least count of Vernier callipers = 1 SD - 1 VD
\begin{align*} \text{least count } &= 1SD - \frac{19}{20}SD = \frac{1}{20}SD =\frac{1}{20}mm\\ &=0.005cm \end{align*} (b) Least count of screw gauge
$=\frac{pitch}{\text{no. of division on circular scale}}\\ \frac{1}{100}mm=0.001 cm$ (c) Wavelength of light is of the order of,
$\lambda =10^{-5}cm=0.00001cm$
Since most precise device should have minimum least count, optical instrument is the most precise one.

Question 2.7
A student measures the thickness of a 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 his estimate on the thickness of hair?
Solution:
\begin{align*} Magnification(m)=\frac{\text{observed width}\left( y \right)}{\text{real width}\left( x \right)}\\ \therefore x=\frac{y}{m}=\frac{3.5}{100}=0.035mm \end{align*}

Question 2.8
(a) You are given a thread and a meter scale. How will you estimate the diameter of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by Vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Solution.
(a) Meter scale can not measure small diameter of thread. But it can be found by wrapping the thread number of times on on round pencil so as to form a coil having it turns touching each other . Now diameter can be found be measuring the length of the coil using meter scale and counting the turns as below
Let $l$ is measured length of coil on the scale which contains $n$ number of turns.
$\therefore \text{Diameter of thread }=\frac{l}{n}$ (b) $Least \, count =\frac{pitch}{\text{number of division in circular scale}}$ i.e. least count decreased when no. of division on the circular scale increases. Thereby accuracy would increase. But practically speaking , it is impossible to take precise reading due to low resolution of human eye.
(c) The random errors are those errors, which occur irregularly and hence are random with respect to sign and size. Large no. of observations (say 100) gives more reliable result, because probability of making random error in positive side of a physical quantity would be same that of in negative side. Therefore, when no. of observations is large random errors would cancel each other and hence result would be reliable.

Question 2.9
The photograph of a house occupies an area of $1.75cm^2$ on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55m^2$ What is the linear magnification of the projector screen arrangement?
Solution:
Area of object $=1.75cm^2$ Area of image $=1.55m^2=1.55 \times 10^4cm^2$
$\therefore \text{Area magnification}=\frac{\text{area of image}}{\text{area of object}} =\frac{1.55x10^4}{1.75}=8875\\ \text{Linear magnification}=\sqrt{8857}=94.1$

Question 2.10
State the number of significant figures in the following :
$\left( i \right) 0.007m^2\\ \left( ii \right) 2.64\times 10^{24}kg\\ \left( iii \right) 0.2370gcm^{-3}\\ \left( iv \right) 6.320J\\ \left( v \right) 6.032Nm^{-2}\\ \left( vi \right) 0.0006032m^2\\$ Solution:
No. of significant figures :
(i) one
(ii) three
(iii) four
(iv) four
(v) four
(vi) four

Question 2.11
The length breadth and thickness of a metal sheet are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct number of significant figure.
Solution:
Given, length, $l=4.234m$, Breadth, $b=1.005m$,
Thickness, $t=2.01cm=2.01\times 10^{-2}m$
Here the least significant figure is 3,so we will be calculating area and volume in 3 significant figures
Area of sheet $=2(lb+bt+tl)=2$
$=[4.234\times 1.005+1.005\times 0.0201+0.0201\times 4.234]\\ =8.7209478m^2$ So, rounding off to 3 significant digits
$Area = 8.72 m^2$
$Volume=ibt=4.234\times 1.005\times 0.0201 \\ =0.0855289 m^3\approx \text{0.0855m}^3$ (rounding off to 3 significant digits)

Question 2.12
The mass of a box measured by a grocer's balance is 2.30kg. 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 gold pieces to correct significant figures.
Solution:
Mass of the box, m = 2.30 kg
Mass of one gold piece, $m_1=20.17g=0.02017kg$
Mass of second gold piece, $m_2=20.15g=0.02015kg$
(a) The total mass of the box $= 2.3+0.0215+0.0217kg = 2.3432$
Now In addition , the final result should retain as many decimal places as are there in the number with the least decimal places., therefore after rounding off, total mass = 2.3kg (b) Difference in masses of the gold pieces
$=m_2-m_1=20.17-20.15=0.02g$
(correct up to 2 places of decimal)

Question 2.13
A physical quantity P is related to four observation $a$, $b$, $c$ and $d$ as follows : $P=\frac{a^3b^2}{\left( \sqrt{cd} \right)}$ 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?
Solution:
$P=\frac{a^3b^2}{\left( \sqrt{cd} \right)}$ Maximum error in $P$
\begin{align*} \frac{\Delta P}{\Delta P}& =3\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{\Delta d}{d}+\frac{1}{2}\frac{\Delta c}{c}\\ & =3\left[ \frac{1}{100} \right] +2\left[ \frac{3}{100} \right] +\frac{2}{100}+\frac{1}{2}\left[ \frac{4}{100} \right]\\ & =\frac{13}{100}=0.13 \end{align*} Percentage error in $P$ is
$P=\frac{\Delta P}{P}\times 100=0.13\times 100=\text{13\%}$ $\therefore \, 13%$ has two significant figures.
Therefore , if $P\approx 3.763$, $P$ would be rounded off to two significant figures hence $P=3.8$.

Question 2.14
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
\begin{align*} \left( a \right) y&=a\sin \frac{2\pi t}{T}\\ \left( b \right) y&=a\sin vt\\ \left( c \right) y&=\left( \frac{a}{T} \right) \sin \frac{t}{a}\\ \left( d \right) y&=\left( a\sqrt{2} \right) \left( \sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T} \right) \end{align*} ($a$ = maximum displacement of the particle, $v$ = speed of the particle, $T$ = time-period of motion). Rule out the wrong formulas on dimensional grounds.
Solution:
According to dimensional analysis an equation must be dimensionally homogeneous and also trigonometric function should be dimensionless. We can easily rule out the incorrect formula based on that
(a) $y=a\sin \frac{2\pi t}{T}$
Here,
$\left[ L.H.S. \right] =\left[ y \right] =\left[ L \right]$
and
$\left[ R.H.S. \right] =\left[ a\sin \frac{2\pi t}{T} \right] =\left[ L\sin \frac{T}{T} \right] =\left[ L \right]$
So, it is correct.
(b) $y = a sin vt$
Here, LHS = $[y] = [L]$
And RHS = $\left [a sin vt \right ] = \left[ L\sin \left( LT^{-1}.T \right) \right] =\left[ L\sin L \right]$
Here the trigonometric function is not dimensionless, So the equation is incorrect.
(c) $y= \left ( \frac{a}{T}\right )\sin\frac{t}{a}$
Here, $[y]=[l]$
and,
$\left[ \left( \frac{a}{T} \right) \sin \frac{t}{a} \right] =\left[ \frac{L}{T}\sin \frac{T}{L} \right] =\left[ LT^{-1}\sin TL^{-1} \right]$ Here the trigonometric function is not dimensionless, So the equation is incorrect.
(d) \begin{align*} y=\left[ a\sqrt{2} \right] \left[ \sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T} \right]\\ \end{align*} Here,
$\left[ y \right] =\left[ L \right] ,\left[ a\sqrt{2} \right] =\left[ L \right]$ and
$\left[ \sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T} \right] =\left[ \sin \frac{T}{T}+\cos \frac{T}{T}\, \right]\\ \therefore \left[ LHS \right] =\left[ RHS \right]=\text{dimensionless}$ So,the equation is correct.

Question 2.15
A famous relation in physics related 'moving mass' $m$ to the 'rest mass' $m_0$ 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 : $m=\frac{m_0}{\left( 1-v^2 \right) ^{1/2}}$ Guess where to put the missing c.
Solution:
The given equation can be written as
$\frac{m_0}{m}=\sqrt{1-v^2}$
We can easily see that Left hand side(LHS) is dimensionless
$\frac {m_0}{m} = \frac {M^1}{M^1} = 1$ ( a dimension less quantity) Therefore, right hand side should also be dimensionless.
It is possible only when $\sqrt{1-v^2}$ should be dimension less. Which all means $(1- v^2)$ is dimensionless
Now $(1- v^2)$ can be dimensionless only when this becomes $(1- \frac {v^2}{c^2})$
Thus, the correct formula is
$m=m_0\left[ 1-\frac{v^2}{c^2} \right] ^{-1/2}$

Question 2.16
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by $\unicode[.8,0]{x212B} : 1\unicode[.8,0]{x212B} = 10^{-10}$ The size of a hydrogen atom is about $0.5 A^o$. What is the total atomic volume in of a mole of hydrogen atoms?
Solution:
Given, $r=0.5 \unicode[.8,0]{x212B} = 0.5\times 10^{-10}m$
Volume of each atom of hydrogen $=\frac{4}{3}r\pi ^3 =\frac{4}{3}\times 3.14\left( 0.5\times 10^{-10} \right) ^3=5.236\times 10^{-31}m^3$
No. of atoms in 1 mole of$H_2=6.023\times 10^{23} =3.154\times 10^{-7}m^3$ (Avogadro Number)
Therefore, Atomic Volume $= 5.236 \times 10^{-31}\times 6.023 \times 10^{23} = 3.154 \times 10^{-7}m^3$

Question 2.17
One mole of an ideal gas at standard temperature and pressure occupies $22.4$ L (molar volume). 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 $1\unicode[.8,0]{x212B})$. Why is this ratio so large?
Solution:
Volume of one mole of ideal gas, $V_g =\text{22.4 }litre=22.4\times 10^{-3}m^3$ Radius of hydrogen molecule $=\frac{1\unicode[.8,0]{x212B}}{2}\\ =0.5\unicode[.8,0]{x212B}=0.5\times 10^{-10}m$ Volume of hydrogen molecule (Sphere Volume) $=\frac{4}{3}r\pi ^3\\ =\frac{4}{3}\times \frac{22}{7}\left( 0.5\times 10^{-10} \right) ^3m^3\\ =0.5238\times 10^{-30}m^3$ Now One mole contains $6.023\times 10^{23}$ molecules (Avogadro number)
Therefore, Volume of one mole of hydrogen,
$V_H=0.5238\times 10^{-30}\times 6.023\times 10^{23}m^3\\ =3.1548\times 10^{-7}m^3$ Now, $\frac{V_g}{V_H}=\frac{22.4\times 10^{-3}}{3.1548\times 10^{-7}}=7.1\times 10^4$ We can see that this ratio is very large. The reason is the inter-atomic separation in the gas which is very large compared to the size of a hydrogen molecules.

Question 2.18
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 rapidly in a direction opposite to the trains motion, but the distant objects (hill tops, 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).
Solution:
This happens because of the line of sight. The line joining a given object to our eye is known as the line of sight. When a train moves rapidly, the line of sight of a passenger sitting in the train for nearby trees,cars, objects changes its direction rapidly. As a result, the nearby trees ,cars and other objects appear to run in a direction opposite to the trains motion. However, the line of sight of distant and large size objects e.g., hill tops the Moon, the stars etc., almost remains unchanged (or changes by an extremely small angle). As a result, the distant object seems to be stationary.

Question 2.19
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 $\approx 3\times 10^{11}m.$ However, even the nearest starts 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 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?
Solution:

\begin{align*} parallax \,\,angle=\theta =1''&=\frac{1}{60}=\frac{1^0}{60\times 60}\\ &=\frac{\pi}{180}\times \frac{1}{60x60}radian\\ Now,\\ \theta =\frac{arc}{radius}=\frac{d}{D}\\ \therefore D&=\frac{d}{\theta}=\frac{1.5\times 10^{11}}{\frac{\pi}{180}\times \frac{1}{60\times 60}}\\ &=3.1\times 10^{16}m\\ \therefore 1\,\,parsec =3.1\times 10^{16}m. \end{align*}

Question 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 centaur) show when viewed from two locations of the Earth six months apart in its orbit around the sun?
Solution:
Lets first calculate the light year
Now we know that distance travelled by light in 1 year is called 1 light year
Speed of light = $3 \times 10^8$ m/s and $1 year =365 \times 24 \times 60 \times 60$ s
So 1 light year = $3 \times 10^8 \times 365 \times 24 \times 60 \times 60$ m
$1light\,\,year=9.46\times 10^{15}m$
\begin{align*} x=4.29\,light\,\,year &=4.29\times 9.46\times 10^{15}m\\ &=\frac{4.29\times 9.46\times 10^{15}}{3.08\times 10^{16}}par\sec =1.323 parsec \end{align*} In an orbit around the sun, the distance between six months apart locations is diameter of the Earth orbit itself which is $3 \times 10^{11}m$.
Now, using parallax method

\begin{align*} \therefore \theta =\frac{d}{D}&=\frac{3 \times 10^{11}}{x}=\frac{3 \times 10^{11}}{4.29\times 9.46\times 10^{15}}rad\\ &=1.512\,sec \end{align*}

Question 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 at closely separated instants of time. This was the actual motivation behind the discovery of radar 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.
Solution
Extremely precise measurements are needed in modern science. As an example, while launching a satellite using a space launch rocket system we must measure time to a precision of 1 micro second. Again working with lasers we require length measurements to an angstrom unit ( 1 Å = $10^{-10}m$ ) or even a fraction of it. For estimating nuclear sizes we require a precision of $10^{-15}m$ To measure atomic masses using mass spectrograph we require a precision of $10^{-30}Kg$ and so on.

Question 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.
Solution
(a) The average rainfall of nearly 100 cm or 1 m is recorded by meteorologists, during monsoon, in India. The area of India is \begin{align*} A&=\text{3.3 }million\,\,sq.km=3.3\times 10^6\left( km \right) ^2\\ &=3.3\times 10^6\times 10^6m^2=3.3\times 10^{12}m^2 \end{align*} Therefore, mass of rain-bearing clouds \begin{align*} &=volume\,\,of\,\,rain\,\,water\,\, \times \,density\,\,of\,\,water\\ &= Area \, of \, India \times height \, of \, the \, rainfall \, recorded \times density \, of \, the \,water\\ &=3.3\times 10^{12}m^2\times 1m\,\,\times \,\,\text{1000 }kg/m^3\\ &=\text{3.3 }\times 10^{15}\,\,kg.\\ \end{align*} (b) We can measure the mass of the elephant using the below technique
Measure the depth of an empty boat in water. Let it be . If $A$ be the base area the boat, then volume of water displaced by boat,
$V_1=Ad_{1}$
Let $d_2$ be the depth of boat in water when the elephant is moved into the boat. Volume of water displaced by (boat + elephant), $V_2=Ad_2$ volume of water displaced by elephant, $V= V_2 - V_1 = A\left( d_2-d_1 \right)$ if $\rho$ be the density of water, then $\text{mass of elephant = mass of water displaced by it}\\ \text{mass of elephant}=A\left( d_2-d_1 \right) \rho .$ (c) Wind speed during storms can be estimated by floating a gas-filled balloon in air at a known height $h$. When there is no wind, the balloon is at $A$. Suppose the wind starts blowing to the right such that the balloon reaches position B in 1 second.
Now,
$AB = d = h$
The value of $d$ directly gives the wind speed.
(d) Let us assume that the hair on our head are uniformly distributed. Thickness of a human hair is $5\times 10^{-5}m.$
Number of hair on the head $=\frac{\text{Area of the head}}{\text{Area of cross}-\sec\text{tion of a hair}}$
Taking head to be a circle of radius 8 cm, number of hair on head $=\frac{\pi \left( 0.08 \right) ^2}{\pi \left( 5\times 10^{-5} \right) ^2}=\frac{64\times 10^{-4}}{25\times 10^{-10}}=2.56\times 10^6$ The number of hair on the human head is of the order of one million.
(e) Consider a class room of size $10\, m \times 8 \,m \times 4\, m$. volume of this room is $320m^3$ We know that $22.4\,L$ or $22.4\times 10^{-3} m^3$ of air has $6.02\times 10^{23}$ molecules ( equal to Avogadro's number) . Therefore, Number of molecules of air in the class room $=\frac{6.02\times 10^{23}}{22.4\times 10^{-3}} \times 320=8.6\times 10^{27}$

Question 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 such 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, liquid or gases? Check if your guess is correct from the following data:
$\text{mass of Sun }= 2\times 10^{30} Kg$, $\text{radius of the sun}=7\times 10^8\text{m}$
Solution
Given that, $M=2\times 10^{30}kg\,\, R=7\times 10^8m$ Lets calculate the density based on the given data
\begin{align*} Density, \,\, \rho \,\,&=\frac{mass}{volume}=\frac{M}{\frac{4}{3}\pi R^3}\\ &=\frac{3M}{4\pi R^3}\\ &=\frac{3\times 2\times 10^{30}}{4\times 3.14\times \left( 7\times 10^8 \right) ^3}\\ &=\text{1.392 }\times 10^3kg/m^3 \end{align*} Which is the order of density of solids and liquids not gases.High density of sun is because of inward gravitational attractions or the outer layers of the sun.

Question 2.24
When the planet Jupiter is at a distance of 824.7 million km from earth, its angular diameter is measured to be 35.72 sec of arc. Calculate the diameter of Jupiter?
Solution
Using Parallax Method

\begin{align*} D&=824.7 \, million \, km = 824.7\times 10^6km\\ \theta &=35.72^{''}=\frac{35.72}{60\times 60}\times \frac{\pi}{180}radian\\ Now \theta =\frac {d}{D}\\ \because d&=D\theta =824.7\times 10^6\times \frac{35.72}{60\times 60}\times \frac{\pi}{180}km.\\ &=1.429\times 10^5km. \end{align*}

Question 2.25
A man walking briskly in rain with speed v must slant his umbrella making forward an angle $\theta$ with the vertical. A student derives the following relation between and v: $\tan \theta =v$ and checks that the relation has a correct limit: as $v\rightarrow \text{0, }\theta \rightarrow \text{0,}$ 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.
Solution.
According to principle of homogeneity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.
Here, $v=\tan \theta$
i.e. $L^1 T^{-1} =$ Dimensionless, which is incorrect.
For the equation to be correct, L.H.S and RHS should both be dimensionless, So We get
$\frac{v}{u} = \tan\theta$
where $u$ is velocity of rain.

Question 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 ?
Solution:
Given,
Error in 100 years = 0.02 s
Now $100 \, years = 100 \times 365 \times 24 \times 60 \times 60 = 3.15 \times 10^{9} sec$
Now .02 sec error in $3.15 \times 10^{9} sec$ time
so error in 1 sec will be given by
$= \frac {.02}{3.15 \times 10^{9}}$
so, accuracy of the clock will be given by
$= \frac {3.15 \times 10^{9}}{.02}=1.5 \times 10^{11}$

Question 2.27
Estimate the average atomic mass density of a sodium atom, assuming its size to be $2.5 A^o$ compare it with the density of sodium in its crystalline phase$\text{(970 kgm}^{-3}\text{)}.$ Are the two densities of the same order of magnitude? If so, why ?
Solution.
Given,
Radius of Sodium atom(R) = $2.5 A^o$
Now volume of sphere = $\frac {4}{3} \pi R^3$
So, \begin{align*} \text{Atomic volume} &= \frac{4}{3}\pi \text{R}^3\\ &= \frac{4}{3}\times \frac{22}{7} \left( 2.25\times 10^{-10} \right) ^3\\ &= 65.25\times 10^{-30} m^3 \end{align*} Since, \begin{align*} \text{Mass of one sodium atom}= \frac{\text{Atomic mass}}{\text{Avogardo's No.}} \\ = \frac{23}{6.023\times 10^{23}}g\\ = 3.82\times 10^{-26}kg \end{align*} Now,
$\text{Average mass density} = \frac{mass}{volume} = \frac{3.82\times 10^{-26}}{65.25\times 10^{-30}} = \text{0.58 }\times 10^3 kg/m^3$ The two densities are of same order $\left( \sim 10^3 \right)$ ,which is due to close packing of atoms in the crystalline phase.

Question 2.28
The unit of length convention on the nuclear scale is a Fermi: $1f=10^{-15}\,\,m.$ Nuclear sizes obey roughly the following empirical relation, $r=r_0A^{\frac{1}{3}},$ Where $r$ is the radius of the nucleus, $A$ its mass number, and $r_0$ is a constant equal to $1.2 f$. 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 average mass density of sodium atom.
Solution
Let $m =$ average mass of proton or nucleon
Therefore Nuclear mass, $M = mA$ and radius of nucleus, $r = r_0 A^{1/3}$ Nuclear density, \begin{align*} \rho &= \frac{\text{mass of nucleus}}{\text{volume of nucleus}}\\ &=\frac{M}{\frac{4}{3}\pi r^3}=\frac{mA}{\frac{4}{3}\pi \left( r_0A^{1/3} \right) ^3}\\ &=\frac{3m}{4\pi r_0^3} \end{align*} As $r_0$ and m are constant, hence nuclear density would be constant for all nuclei.
Now, $m = 1.60\times 10^{-27}kg$ and $r_0 = 1.2\,\,f = 1.2\times 10^{-15} m$ \begin{align*} \therefore \rho &= \frac{3m}{4\pi r_0^3} = \frac{3\times 1.66\times 10^{-27}}{4\times 3.14\times \left( 1.2\times 10^{-15} \right) ^3} \\ &= 2.29\times 10^{17} kg m^{-3} \end{align*} From last question, density of sodium atom $=\text{0.58 }\times 10^3 kg/m^3$ \begin{align*} \therefore \frac{\text{Nuclear mass density}}{\text{Atomic mass density}}&=\frac{2.3\times 10^{17}}{0.58\times 10^3}\\ &=\text{3.96 }\times 10^{14} \end{align*} Therefore Nuclear density is around $10^{15}$ times the atomic density of matter.

Question 2.29
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?
Solution :
Here $t = 256 s$, $c = 3\times 10^8 ms^{-1}$
Radius of the lunar orbit around the earth $r_l$=Distance of the moon from the earth.
So, $r_l= \frac{c\times t}{2} = \frac{3\times 10^8 \times 2.56}{2} = 3.84\times 10^8 m.$

Question 2.30
A sonar (sound Navigation and Ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with 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 s. what the distance of the enemy submarine? (Speed of sound in water $= 1450\, ms^{-1}$ ).
Solution: Here,
$t=77s\,\,v=1450ms^{-1}$
Distance of enemy submarine $=\frac{v\times t}{2}=\frac{1450\times 77}{2}=55825m.$

Question 2.31
The farthest objects(known as quasars) in our universe are so distant that light emitted by them takes billion of years to reach the earth. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?
Solution:
Here t = 3.0 billion years $=3.0 \times 10^9\times 365.25 \times 24 \times 60\times 60 s$
speed of light, $c = 3 \times 10^5Km/s$
Distance of quasar $=ct=3 \times 10^5\times 3.0 \times 10^9\times 365.25 \times 24\times 60\times 60 =2.84 \times 10^{22} km.$

Question 2.32
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 and from the information that sun's angular distance $\theta$ is measured to be 1920", determine the approximate diameter of the moon. Given earth-moon distance = $3.8452 \times 10^8\,m$
Solution
During total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameters of both the sun and the moon must be equal.

Therefore Angular diameter of the moon,
$\theta =$ Angular diameter of the sun
$\theta= 1920''= 1920 \times 4.85 \times 10^{-6} rad$
Since $1'' = 4.85 \times 10^{-6} rad$
Earth-moon distance, $s = 3.8452 \times 10^8 m$
Using parallax method

Diameter of the moon, \begin{align*} d &= \theta \times D\\ &=1920 \times 4.85 \times 10^{-6} \times 3.8452 \times 10^8\\ &=3.581 \times 10^6 m = 3581 km. \end{align*}

Question 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?
Solution
Using basic constants such as speed of light (c), charge on electron (e), mass of electron $m_e$ , mass of proton $m_p$ and gravitational constant (G), we can construct the quantity, $\tau =\left( \frac{e^2}{4\pi \varepsilon _0} \right) \times \frac{1}{m_p m_{e}^{2}c^3G}$ Now \begin{align*} \left[ \frac{e^2}{4\pi \varepsilon _0} \right] &= \left[ \frac{1}{4\pi \varepsilon _0} \frac{e^2}{r^2}r^2 \right] = \left[ Fr^2 \right]\\ &= \left[ MLT^{-2}. L^2 \right] = \left[ ML^3T^{-2} \right]\\ \therefore \left[ \tau \right] &=\frac{\left[ ML^3T^{-2} \right] ^2}{\left[ M \right] \left[ M \right] ^2 \left[ LT^{-1} \right] ^3 \left[ M^{-1}L^3T^{-2} \right]} =\left[ T \right] \end{align*} Clearly, the quantity $\tau$ has the dimensions of time.
put $G = 6.67\times 10^{-11}Nm^2 kg^{-2}$ ,$c = 3\times 10^8 m/s$, $e = \text{1.6 }\times 10^{-19} C$ ,$m_e= 9.1\times 10^{-31} kg,$ , $m = 1.67x10^{-27} kg$
and
$\frac{1}{4\pi \varepsilon _0}=9\times 10^9Nm^2C^2$ therefore, \begin{align*} \tau &= \frac{\left[ 9\times 10^9\times \left( 1.6\times 10^{-19} \right) ^2 \right] ^2}{1.67\times 10^{-27}\times \left( 9.1\times 10^{-31} \right) ^2\times \left( 3\times 10^8 \right) ^3\times 6.67\times 10^{-11}}\\ &= 2.13\times 10^{16} s\\ &= \frac{2.13\times 10^{16}}{3.15\times 10^7} years = 0.667\times 10^9 years.\\ &= 0.667 \text{billion years}.\\ \end{align*}