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

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement: As far as CBSE board is considered a good percentage of questions are asked directly from NCERT concepts and exercise. This is where a tool like  CBSE NCERT solutions for class 11 physics chapter 2 units and measurement is important. NCERT solutions for class 11 physics chapter 2 units and measurement give an explanation to all the exercise questions. Measurements are always specified with a unit. Without units, measurement is meaningless. For example, a man tells the distance between my home and the nearest city is 5. This statement by the man is meaningless. Whether it is 5m, 5cm or 5Km makes it meaningful. The solutions of NCERT class 11 physics chapter 2 units and measurement start with questions related to the conversion of units. The concept of accuracy, precision and error can be studied well with the help of CBSE NCERT solutions for class 11 physics chapter 2 units and measurement. NCERT solutions help in self-evaluation of the concept studied in the chapter. The important topics of chapter 2 units and measurement are the concepts of errors and measurement of errors, dimensional analysis and significant figures. Questions from these topics are explained in NCERT solutions for class 11 physics chapter 2 units and measurement.

## NCERT solutions for class 11 physics chapter 2 units and measurement exercises

(a) The volume of a cube of side 1 cm 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 ...$\small (mm)^2$

(c) A vehicle moving with a speed of $\small 18\ km h^{-1}$ covers....m in 1 s

(d) The relative density of lead is 11.3. Its density is .... $\small g cm^{-3}$ or .... $\small kg m^{-3}$.

(a) We know, $\small 1 cm = .01 m$    (Tip: Divide by 100 to convert cm to m)

The volume of a cube of side a  = $\small a^3\ m^3$

Volume of cube of side 1 cm (i.e, .01 m ) = $\small (.01)^3 = 10^{-6}\ m^3$

(b) We know, $1 cm = 10 mm$   (Tip: Multiply by 10 to convert cm to mm)

The surface area of a solid cylinder of radius r and height h = $2(\pi r^2) + 2 \pi rh = 2\pi r(r + h)$

Required area = $2\pi (20)(\ 20 + 100) mm^3 = 40 \pi (120) mm^3 = 4800 \pi\ mm^3 = 1.5 \times 10^{4 }\ mm^3$

(c)  (Tip: multiply by 5/18 to convert  $km h^{-1}\ to\ m s^{-1}$ )

Ditance covered = $Speed \times time = (18 \times 5/18)ms^{-1} \times (1s) = 5 m$

(d)  Density = Relative Density $\times$ Density of water

(Density of water = $1 \ gcm^{-3} = 1000\ kgm^{-3}$)

$(11.3)\times1\ gcm^{-3} = 11.3\ gcm^{-3} \\ (11.3)\times1000\ kgm^{-3} = 11300\ kgm^{-3} = 1.13 \times 10^4\ kgm^{-3}$

(a) $1 kg m^2s^-^2 = \ \ \ \ \ \ \ \ gcm^2s^-^2$

b) $1 m = \ \ \ \ ly$

(c) $3.0 \ m s^-^2 = \ \ \ \ km h^-^2$

(d)$G = 6.67 \times 10^{-11} Nm^2(kg)^-^2 = \ \ \ \ \ (cm)^3 s^{-2}g^{-1}$.

(a) $(1 kg \rightarrow 1000g ;\ 1m\rightarrow 100 cm)$

$1 kgm^2s^{-2} = (1000g)(100cm)^2s^{-2} = (10^3 \times10^{4})gcm^2s^{-2} = 10^{7}\ gcm^2s^{-2}$

(b) $1 m = 1.057\times10^{-16} ly$   (1 ly = Distance travelled by light in 1 year )

(c) $(1 m \rightarrow 10^{-3}km ;\ (60\times60 =) 3600 s \rightarrow 1hr)$

$\\3.0 ms^{-2} = 3.0 (10^{-3}km)(1/3600 hr)^{-2} = 3.0 \times (3600)^2 / 1000\ kmhr^{-2} \\ = 3.9 \times 10^4 \ kmhr^{-2}$

(d)    $(N \rightarrow J.s = kgms^{-2};\ 1 m \rightarrow 100cm;\ 1 kg\rightarrow 1000g)$

$G = 6.67 \times 10^{-11} (kgms^{-2})m^2kg^-^2 = (6.67 \times 10^{-11})(m)^3 s^{-2}kg^{-1} = 6.67 \times 10^{-8}cm^3 s^{-2}g^{-1}$

## Q 2.3: A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where $1J = 1 kg m^2 s^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha \ kg$, the unit of length equals $\beta \ m$, the unit of time is $\gamma \ s$. Show that a calorie has a magnitude 4.2 $\alpha^{-1}\beta ^{-2}\gamma^{2}$ in terms of the new units.

Given,

$1 Cal = 4.2 (kg)(m)^{2}(s)^{-2}$

Given new unit of mass = $\alpha kg$     (In old unit 1kg corresponded to a unit mass, but in new unit $\alpha kg$ corresponds to a unit mass)

$\therefore$ In terms of the new unit, $1 kg = 1/\alpha = \alpha ^{-1}$

Similarly in terms of new units$1 m = 1/\beta = \beta^{-1}$ and  $1 s = 1/\gamma = \gamma ^{-1}$

$\therefore$ $1 Cal = 4.2 (kg)(m)^{2}(s)^{-2} = 4.2 (\alpha ^{-1})(\beta ^{-1})^{2}(\gamma ^{-1})^{-2} = 4.2 \alpha ^{-1}\beta ^{-2}\gamma ^{2}$

(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

The given statement is true. A dimensional quantity may be small with respect to one reference and maybe large with respect to another reference. Hence, we require a standard reference to judge for comparison.

(a) An atom is a very small object with respect to a tennis ball. (but larger than an electron!)

(b) A jet plane moves with great speed with respect to a train.

(c) The mass of Jupiter is very large as compared to an apple.

(d) The air inside this room contains a large number of molecules as compared to in your lungs.

(e) A proton is much more massive than an electron

(f) The speed of sound is less than the speed of light

Distance between Sun and Earth = Speed of light x Time taken by light to cover the distance

Speed of light = 1 unit

Time taken by light to reach earth is 8  minute 20 seconds

Time taken =  $(8\times60) + 20 = 500 s$

The distance between Sun and Earth = 1 x 500 = 500 units.

(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 to within a wavelength of light?

To judge which tool is more precise, we have to find out their least count. Least count defines the margin of error and hence the precision. Hence the instrument with lower least count will be more precise.

(a) Least count = $1MSD - 1VSD = 1 - (19/20)= 1/20 = 0.05\ cm$   (Taking 1 MSD as 1 mm)

(b) Least count = pitch/ number of divisions

$1/10000 = 0.001 cm$

(c) least count = wavelength of light = 400nm to 700nm, that is in the range of $10^{-7}$ m

Therefore, the optical instrument is the most precise device used to measure length.

Given,

Magnification of Microscope = 100

The average width of hair under the microscope = 3.5 mm

(20 observations were made to calculate the average i.e. 3.5 mm as an experimental procedure. No need in our calculations.)

(Note: When magnified, the width is 3.5 mm. Hence actual width will be less by a factor of magnification value)

The average thickness of hair = $\frac{3.5 mm}{100} = 0.035 \ mm.$

(a)You are given a thread and a metre 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?

(a) Take the thread and wrap it around the metre scale. Make sure the coils are packed closely without any space in between. If the diameter of the thread is d and number of turns obtained are n, then (n x d) corresponds to the marking on the metre scale, l.

Therefore, the diameter of the thread would be, d = l/n

(b) Theoretically, by increasing the number of divisions on the circular divisions, the value of least count decreases and hence accuracy increases.(Lower the value of least count, better will be the reading)

But practically, the number of divisions can be increased only up to a certain limit. (Also two adjacent divisions cannot be separated by a distance less than the human eye resolution!)

(c) With an increase in the number of observations, the accuracy of the experiment increases as the error is now distributed over a large range. Hence, a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements.

Given,

Area of the house in the photo = $1.75\ cm^{2}$

Area of the house on the screen = $1.55 m^{2} = 1.55 \times10^{4} cm^2$

$\therefore$ Arial magnification, $m_{a}$ = Area on the screen / area on photo = $\frac{1.55\times 10^{4}}{1.75} = 0.886 \times 10^4$

Linear magnification of the projector- screen arrangement  $m_{l} = \sqrt{m_{a}} = \sqrt{.886 \times 10^{4}}= 94.1$

(b) $2.64 \times 10^{24} kg$

(c) $0.2370 \ g cm^{-3}$

(d) $6.320 \ J$

(e) $6.032 \ Nm^{-2}$

(f) $0.0006032 \ m^{2}$

(a) The given value is $0.007 \ m^2$.

Since, the number is less than 1, the zeros on the right to the decimal before the first non-zero integer is insignificant. So, the number 7 is the only significant digit.

$\therefore$ It has 1 significant digit.

(b) The value is $2.64 \times 10^{24} kg$

For the determination of significant values, we do not consider the power of 10 (Number is not less than 1). The digits 2, 6, and 4 are significant figures.

$\therefore$ It has 3 significant digits.

(c)The value is $0.2370 \ g cm^{-3}$.

For the given value with decimals, all the numbers 2, 3, 7 and 0 are significant.

$\therefore$ It has 4 significant digits.

(d) The value is $6.320 \ J$.

It has 4 significant digits.

(e)  The value is $6.032 \ Nm^{-2}$.

All the four digits are significant as the zeros in between two non-zero values are also significant.

$\therefore$ It has 4 significant digits.

(f) The value is $0.0006032 \ m^{2}$

Same as (a), first three zeroes after the decimal is insignificant. Only 6, 0, 3, 2 are significant.

$\therefore$ It has 4 significant digits.

Given,

Length, l = 4.234 m ; Breadth, b = 1.005 m; Height, h = 2.01 cm = 0.0201 m

The length has 4 significant figures

The breadth has 4 significant figures

The height has 3 significant figures (Since the number is less than 1, hence zeroes after decimal before the first non-zero integer is insignificant)

We know,

Surface area of a cuboid = 2(l x b + b x h + h x l)

= $2(4.234 \times 1.005 + 1.005 \times 0.0201 + 0.0201 \times 4.234)$

=$2(4.255 + 0.020 + 0.085)$

=  $8.72 \ m^{2}$                (Note: For addition/subtraction, the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal.)

Volume = l x b x h

= $4.234 \times 1.005 \times 0.0201$ = $0.0855 \ m^3$   ($\therefore$ 3 significant digits)

(Note: For Multiplication The LEAST number of significant digits in any number determines the number of significant figures in the answer; i.e decided by the number of significant digits)

The area has three significant values 2, 7 and 8.

The volume has three significant values 5, 5 and 8.

Given,

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g = 0.02015 kg

The mass of the second gold piece = 20.17 g = 0.02017 kg

(a)  The total mass = 2.30 + 0.02015 + 0.02017 = 2.34032 kg

Since one is the least number of decimal places, the total mass = 2.3 kg.

(Note: For addition/subtraction, the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal.)

Given,     $P = a^3b^2/(\sqrt{c}d)$

(a) $\frac{\Delta P}{P} = 3\frac{\Delta a}{a} +2\frac{\Delta b}{b} +\frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}$

$\frac{\Delta P}{P} \times 100 \%= (3\frac{\Delta a}{a} +2\frac{\Delta b}{b} +\frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}) \times 100 \%$

$\ \ \ \ \ \ \ = (3\times1 + 2\times3 + .5\times4 + 1\times2)\% = 13\%$    $(\Delta a/a = 1\% = 1/100)$

$\therefore$  The percentage error in the quantity P = 13 %

(b) Rounding off the value of P = 3.8

(a) $y = a sin 2\pi t/T$

(b) $y = a sin vt$

(c) $y = (a/T) sin\ t/a$

(d) $y = ( a\sqrt{2}) (sin 2\pi t / T+ cos 2\pi t / T )$

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Ground rules:

$\\sin\Theta , cos\Theta \ are\ DIMENSIONLESS.\ \\ and\ \Theta\ must\ be\ dimensionless$

[y] = L  ($M^0 L^1 T^0$)

[a] = L

[v] = $\dpi{80} LT^{-1}$

[t/T] is Dimenionless.

(a) The dimensions on both sides are equal, the formula is dimensionally correct.

(b) $\dpi{80} \because$ [vt] = ($\dpi{80} LT^{-1}$)(T) = L ($\dpi{100} \therefore \Theta$ is not dimensionless)

The formula is dimensionally incorrect

(c) [a/T] = (L)/(T)

It is dimensionally incorrect, as the dimensions on both sides are not equal.

(d) The dimensions on both sides are equal, the formula is dimensionally correct. (Don't get confused by summation of trigonometric functions !)

Guess where to put the missing c.

The relation given is $m = \frac{m_{o}}{(1-v^2)^{1/2}}$

Divide both sides by $m_{o}$;  $\therefore$ L.H.S becomes $m / m_{o}$ which is dimensionless.

Hence, R.H.S must be dimensionless too. (After Dividing by $m_{o}$ !)

$\frac{1}{(1-v^2)^{1/2}}$ can be dimensionless only when $v \rightarrow (v/c)$

Therefore, the dimensional equation is $m = \frac{m_{o}}{(1-(\frac{v}{c})^2)^{1/2}}$

Radius of an Hydrogen atom = 0.5 $\AA$= 0.5 x 10-10 m

Volume = $\frac{4}{3} \pi r^3$

$4/3 \times 22/7 \times (0.5 \times 10^{-10})^3$

= $0.524 \times 10^{-30} m^3$

1 hydrogen mole contains $6.023 \times 10^{23}$ hydrogen atoms.

The volume of 1 mole of hydrogen atom = $6.023 \times 10^{23} \times 0.524 \times 10^{-30}$

= $3.16 \times 10^{-7} m^3 \approx 3 \times 10^{-7} m^3$.

Radius of hydrogen atom = 0.5 $\AA$ = 0.5 x 10-10 m  (Size here refers to Diameter!)

Volume occupied by the hydrogen atom= $\frac{4}{3} \pi r^3$

$4/3 \times 22/7 \times ( 0.5 \times 10^{-10})^3$

= $0.524 \times 10^{-30} m^3$

1 mole of hydrogen contains 6.023 x 1023 hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 1023 x 0.524 x 10-30

= 3.16 x 10-7 m3

$V_{m} = 22.4 L = 22.4 \times 10^{-3} m^3$

$\frac{V_{m}}{V_{a}} = \frac{22.4\times10^{-3}}{3.16\times10^{-7}} = 7.09 \times 10^4$

The molar volume is $7.09 \times 10^4$ times greater than the atomic volume.

Hence, intermolecular separation in gas is much larger than the size of a molecule.

Our eyes detect angular velocity, not absolute velocity. An object far away makes a lesser angle than an object which is close. That's why the moon (which is so far away!) does not seem to move at all angularly and thus seems to follow you while driving.

In other words, while in a moving train, or for that matter in any moving vehicle, a nearby object moves in the opposite direction while the distant object moves in the same direction. !

The diameter of Earth’s orbit = $3 \times 10^{11} m$
$\therefore$ The 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 $= 3.09 \times 10^{16}\ m$.

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 $1\ sec = 4.85 \times10^{-6} rad$

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

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 the half-life of radioactive material, a very precise value of the mass of nuclear particles is required. Similarly, in Spectroscopy precise value of the length in Angstroms is required.

(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.

(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 a density less than water.

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

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

The 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 the approximate mass of the elephant.

(c) A rotating device can be used to determine the speed of the 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 the hair, the area of the hair strand, $a = \pi r^2$

So, the 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.

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

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

Given,

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

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

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

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

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!)

Given,

The 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 2.25: 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.

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 the speed of rain)

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

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

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  $10^{11}\ to\ 10^{12}$

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 the same order. In the solid phase, atoms are tightly packed and thus interatomic space is very small.

## 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 it's 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.

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

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

The volume of the nucleus using the 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.

∴ The density of the 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 the atomic density of matter!

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 = $3\times10^8\ ms^{-1}\times1.28\ s = 3.84\times 10^8 \ m$

Given,

77.0 s is the total time between the 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 the total time = $\frac{1}{2}\times 77 s = 38.5 s$

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

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

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

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

= $2.8\times 10^{22}\ m$

From Examples 2.3 and 2.4 , we have,

The 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, $D_{sun}$ = $1.496\times10^{11} m$

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

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

Now, During Solar eclipse, the 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, the diameter of the moon = $3.56\times10^3 km$

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

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.!

## Chapterwise NCERT solutions for class 11 physics

 Chapter 1 NCERT solutions for class 11 physics chapter 1 Physical world Chapter 2 NCERT Solutions for class 11 physics chapter 2 Units and Measurement Chapter 3 CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line Chapter 4 NCERT solutions for class 11 physics chapter 4 Motion in a Plane Chapter 5 Solutions of NCERT for class 11 physics chapter 5 Laws of Motion Chapter 6 CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power Chapter 7 NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion Chapter 8 Solutions of NCERT for class 11 physics chapter 8 Gravitation Chapter 9 CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids Chapter 10 NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids Chapter 11 Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter Chapter 12 CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics Chapter 13 NCERT solutions for class 11 physics chapter 13 Kinetic Theory Chapter 14 Solutions of NCERT for class 11 physics chapter 14 Oscillations Chapter 15 CBSE NCERT solutions for class 11 physics chapter 15 Waves

## Subject wise NCERT solutions for class 11

 NCERT solutions for class 11 biology NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry NCERT solutions for class 11 physics

## Use of NCERT solutions for class 11 physics chapter 2 units and measurement is important:

• This is an easy and important chapter in the annual exam as well as competitive exam point of view.
• For competitive exams like NEET, JEE Mains, JEE advance, NEET and KVPY etc the CBSE NCERT solutions for class 11 physics chapter 2 units and measurement is important.
• One question can be expected from this chapter in NEET and JEE Mains too.
• The solutions of NCERT class 11 physics chapter 2 units and measurement helps to understand the chapter well and also to score well in class 11 exams.