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

## NCERT Solutions For Class 11 Physics Chapter 2 PDF Free Download

NCERT solutions for class 11 physics chapter 2 Units and measurement is the best study resource you can get to understand the topics and to score good grades in your class 11 final examination. This solution provides appropriate answers to the questions provided in the textbook. Along with the textbook question, this solution has exemplary problems, worksheets, questions from previous question papers, numerical problems, MCQ’s, short answer questions, tips and tricks.

For students of class 11 who are looking to give their best for the upcoming class 11 final exams and competitive exams, it is very important to get accustomed with the solution to the questions given in the textbook. Thus students are advised to have a good practice of different questions that can be framed from the chapter. Students are suggested to solve the NCERT question. To avoid any questions and to clear all the doubts of the students; BYJU’S provide Solution to NCERT Class 11 Physics Chapter 2 Units and Measurement.

## Topics covered in Class 11 Chapter 2 Physics Units and Measurement

 Section Number Topic 2.1 Introduction 2.2 The International System Of Units 2.3 Measurement Of Length 2.4 Measurement Of Mass 2.5 Measurement Of Time 2.6 Accuracy, Precision Of Instruments And Errors In Measurement 2.7 Significant Figures 2.8 Dimensions Of Physical Quantities 2.9 Dimensional Formulae And Dimensional Equations 2.10 Dimensional Analysis And Its Applications

### Chapter 2 Units and Measurement class 11 physics NCERT Solutions

Question 1:  Complete the following:

(1) A cube whose sides measure 2 cm each has a volume equal to ________

Ans:

2 cm = $\frac{1}{200}$ m

Volume of the cube = $a^{3}$

= $\frac{1}{200}$ m$\times$$\frac{1}{200}$ m$\times$$\frac{1}{200}$ m

= 8 x 106

Therefore, the volume = 8 x 106

(2) A solid cylinder having a radius of 3 cm and a height of 9 cm has a surface area of _________

Ans:

Surface area of a cylinder = 2$\pi$r(r + h)

r = 3 x 10 mm = 30 mm

h = 9 x 10 mm = 90 mm

Surface area = 2 x 3.14 x 30 x (30 + 90) = 22608 = 2.2 x 104 mm2.

(3) A car covers _________ meters in 1 second when moving with a speed of 18 km/hr.

Ans:

The speed of the car is said to be 18 km/hr = 18 x 1000/3600 meter/second

=> 5 m/s, this means that the car covers 5 meters in one second = 11.3

(4) The density of lead is said to be _______ g cm-3 or ­­_________ kg m-3 when its relative density is 11.3.

Ans:

The Relative density of lead is 11.3 g cm-3

=> 11.3 x 103 kg m-3 [1 kilogram = 103g, 1 meter = 102 cm]

=> 11.3 x 103 kg m-4

Question 2 : Complete the sentences with its suitable conversion of unit.

(1) 1 kilogram m2 s-2 = _________ g cm2 s-2

(2) 1 meter = ________ light year

(3) 3.0 m s-2 = _________ km h­-2

(4) G = 6.67 x 10-11 Nm2(kg)-2 = ________ (cm)3 s-2 g -1

Ans:

(1) 1kg m2 s-2 = ______ g cm2 s-2

1 kg m2 s-2 = 1kg x 1m2 x 1s -2

We know that,

1kg = 103

1m = 100cm = 102cm

When the values are put together, we get:

1kg x 1m2 x 1s-2 = 103g x (102cm)2 x 1s-2  = 103g x 104 cm2 x 1s-2  = 107 gcm2s-2

=>1kg m2 s-2 = 107 gcm2s-2

(2) 1 light year is the distance travelled by sunlight in 1yr.

Using the formula,

Distance = speed x time

Speed of light = 3 x 108 m/s

Time = 1 yr = 365 days = 365 x 24 hr = 365 x 24 x 60 x 60 sec

Put these values in the formula mentioned above, we get:

One light year distance = (3 x 108 m/s) x (365 x 24 x 60 x 60) = 9.46×1015m

9.46 x 1015 m = 1ly

So that, 1m = 1/9.46 x 1015ly

=> 1.06 x 10-16ly

=>1 meter = 1.06 x 10-16ly

(3) 1 hour = 3600 seconds so that 1 second = 1/3600 hour

1 km = 1000m so that 1m = 1/1000 km

3.0 m s-2 = 3.0 (1/1000 km) (1/3600 hour) -2 = 3.0 x 10-3 km x ((1/3600)-2h-2)

= 3 x 10-3km x (3600)2 hr-2 = 3.88 x 104 km h-2

=> 3.0 m s-2 = 3.88 x 104 km h­-2

(4) Given that,

G = 6.67 x 10-11 N m2 (kg)-2

We know that,

1N = 1kg m s-2

1 kg = 103 g

1m = 100cm= 102 cm

Put the values together, we get:

=> 6.67 x 10-11 Nm2 kg-2 = 6.67 x 10-11 x (1kg m s -2) (1m2) (1kg-2)

Solve the following and cancelling out the units, we get:

=> 6.67 x 10-11 x (1kg -1 x 1m3 x 1s-2)

Put the above values together to convert kg to g and m to cm

=> 6.67 x 10-11 x (103g)-1 x (102 cm)3 x (1s-2)

=> 6.67 x 10-8 cm3 s-2 g -1

=>G = 6.67 x 10-11 Nm2(kg)-2 = 6.67 x 10-8  (cm)3 s-2 g -1

Question 3: The unit of heat or energy is calorie and is equal to about 5.2 J where 1J = 1 kg m2s-2. A system which has a unit mass of $\alpha$ kg, unit length of $\beta$ m and unit time of y s is employed, then in terms of new units show that the magnitude is 5.2 $\alpha$ $\beta^{-2}$ y2.

Ans: 1 calorie = 5.2 (1kg) (1m2) (1 s-2)

New unit of mass = $\alpha$ kg

Then, 1 kg = $\frac{1}{\alpha }$ = $\alpha ^{-1}$

New unit of length = $\frac{1}{\beta }$ = $\beta ^{-1}$ or 1 m2 = $\beta ^{-2}$

New unit of time = $\frac{1}{\gamma }$ = $\gamma ^{-1}$

1 s2 = $\gamma ^{-2}$

1 s2 = $\gamma ^{2}$

Therefore, 1 calorie = 5.2$\alpha$$\beta^{-2}$ y2.

Question 4: Explain the following:

The dimensional quantity must be specified in comparison with a standard reference.

(a) Atoms are small object

Ans: In comparison with a soccer ball, atoms are very small

(b) Jet plane travel at high speed

Ans: When compared with a bicycle, jet plane travels at high speed.

(c) A large number of molecules are present in the air inside the room.

Ans: As compared with the air inside a lunch box, the air inside the room has a large number of molecules.

(d) Jupiter’s mass is very large.

Ans: When compared with the mass of a cricket ball, the mass of Jupiter is very large.

(e) Speed of light is much more than the speed of sound.

Ans: Like comparing the speed of a bicycle and a jet plane, the speed of light is more than the speed of sound.

(f) An electron is much less massive than a proton.

Ans: A proton is massive when compared with an electron.

Question 5: A unit of length is taken so as to make the speed of light in vacuum to unity. If the light takes 9 min and 30 sec to cover the distance between the Sun and the Earth, then what is the distance between them?

Ans:

Distance between them = Speed of light x Time taken by light to cover the distance

Speed of light = 1 unit

Time taken = 570 s

The distance between Sun and Earth = 1 x 570 = 570 units.

Question 6: Which device of the following measures the length precisely?

(a)  Vernier caliper which has 20 divisions on the sliding scale

(b) A screw gauge of circular scale with pitch 1 mm and 100 divisions

(c) An optical instrument which can even measure wavelength of light

Ans:

(a) Least count = 1- $\frac{9}{10}$= $\frac{1}{10}$ = 0.01cm

(b) Least count = $\frac{pitch}{number of divisions}$

= $\frac{1}{10000}$ = 0.001 cm

(c) least count = wavelength of light = 10-5 cm

= 0.00001 cm

We can come to the conclusion that the optical instrument is the most precise device used to measure length.

Question 7: A human hair thickness is measured through a microscope of magnification of 200. 30 observations are made and the average width of hair under the microscope is 4.5 mm. Estimate the thickness of the hair.

Ans:

Microscope magnification = 200

The average width of hair under the microscope = 4.5 mm

Average thickness of hair = $\frac{4.5}{200}$ = 0.0225 mm.

Question 8:

(i)Estimate the diameter of the thread if a scale is given to you.

Ans:

Take the thread and wrap it around a rod in such a way that the coil turns are very close to each other. Measure the whole length of the thread.

Diameter = $\frac{Length\;of\;thread}{number\;of\;turns}$

(ii) A screw gauge of circular scale has a pitch of 1 mm and 200 division. By increasing the number of divisions on the circular scale can the accuracy be increased?

Ans:

Even if the number of divisions on the circular scale is increased, the accuracy will not be increased. The accuracy will increase only to a certain extent if the number of divisions is increased.

(iii) Vernier calliper is used to measure the mean diameter of the iron rod. A more reliable estimate is got from a set of 200 measurements when compared with a set of 10 measurements. Why is that?

Ans:
A reliable estimate is got from a set of 200 measurements as the random errors involved are less when compared with the set of 10 measurements.

Question 9: A house in a photograph occupies an area of 2.75 cm2 on a 45 mm slide. It is projected on the screen, its area is 2.45 m2. Find the linear magnification.

Ans:

Area of the house in the photo = 2.75 cm2

Area of the house on the screen = 2.45 m2 = 2.45 x 104 cm2

Arial magnification, ma = $\frac{Area\;of\;image}{area\;of\;object}$ = $\frac{2.45}{2.75}$ x 104

Linear magnification m1 = $\sqrt{m_{a}}$ = $\sqrt{\frac{2.45}{2.75}\times 10^{4}}$ =  94.38

Question 10: How many significant figures are there in the given:

(i) 0.008 m2

Ans:

The given value is 0.008 m2.

The given number is below one so the zeros on the right to the decimal are insignificant. So, the number 7 is the only significant figure in the following.

(ii) 3.45 x1032 kg

Ans:

The value is 3.45 x 1032 kg

For the determination of significant values, the power of 10 is irrelevant. The digits 3, 4, and 5 are significant figures.

(iii) 0.8745 g cm-3

Ans:

The value is 0.8745 g cm-3.

For the given value with decimals, all the numbers 8, 7, 4 and 5 are significant.

(iv) 8.4510 J

Ans:

For the given values, the trailing zeros are also significant. Therefore, all the five digits including the zeros are significant.

(v) 7.0215 N m-2

Ans:

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

(vi) 0.000051201 m2.

Ans:

Since the given value is below one, the zeros on the right side of the decimals are insignificant. The other five digits 5, 1, 2, 0 and 1 are significant values as the zeros in-between two non-zeros are also significant.

Question 11: The rectangular sheet has a length of 4.325 m, breadth of 1.402 m and thickness of 3.12 cm. Find the area and volume and find out how many significant values are there.

Ans:

Length = 4.325 m

Height = 3.12 m = 0.0312 m

The length has 4 significant figures

The breadth has 4 significant figures

The height has 3 significant figures

Surface area formulae = 2(l x b + b x h + h x l)

= 2(4.325 x 1.402 + 1.402 x 0.0312 + 0.0312 x 4.325)

=2(6.06365 + 0.0437424 + 0.13494)

= 2 x 6.2423

=12.484 m2

Volume = l x b x h

= 4.325 x 1.402 x 0.0312

= 0.189 m3

The volume has three significant values 1, 8 and 9.

The area has five significant values 1, 2, 4, 8 and 4.

Question 12: A box has a mass of 2,400 kg. To the box, two gold pieces whose masses are 18.12 g and 18.43 g are added. Find the total mass of the box and also find the difference of mass between the two pieces to correct significant figures.

Ans:

The mass of the box = 2.400 kg

and the mass of the first gold piece = 18.12 g

The mass of the second gold piece = 18.43 g

The total mass = 2.400 + 0.01812 + 0.01843 = 2.43655 kg

Since 1 is the least number of decimal places, the total mass = 2.4 kg.

The mass difference = 18.43 – 18.12 = 0.31 g

Since 2 is the least number of decimal places, the total mass = 0.31g.

Question 13: P which is a physical quantity is related to a, b, c and d.

P = $\frac{a^{3}b^{2}}{\sqrt{c}d}$

If 1%, 3%, 4% and 2% are the percentage error, then what percentage of error does P have? If the calculated value of P is 4.235, then to what value do we round off the result?

Ans:

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

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

= 3 x 1 + 2 x 3 + $\frac{1}{2}$ x 4 + 2

= 3 + 6 + 2 + 2 = 13 %

P = 4.235

$\Delta P$ = 13 % of P

$\frac{13P}{100}$

= $\frac{13\times 4.235}{100}$

= 0.55

The error lies in the first decimal point, so the value of p = 4.3

Question 14: In a book which is full of errors, the displacement y of a particle is given in four different formulas. Find the correct one.

(a) y = a sin ($\frac{2\pi t}{T}$)

(b) y = a sin vt

(c) y = $\frac{a}{T}$ sin $\frac{t}{a}$

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

Ans:

(a)  y = a sin $\frac{2\pi t}{T}$

Dimension of y = M0 L1 T0

The dimension of a = M0 L1 T0

Dimension of sin $\frac{2\pi t}{T}$ = M0 L0 T0

Since the dimensions on both sides are equal, the formula is dimensionally correct.

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

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

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

Dimension of y = M0 L1 T0

The dimension of a = M0 L1 T0

Dimension of $\frac{t}{T}$ = M0 L0 T0

The formula is dimensionally correct.

Question 15: In physics, a relation relates the moving mass m to the rest mass m0 which his represented in terms of speed v and the speed of light c. A girl writes the relation for the above but she doesn’t know where to put the constant c.

m = $\frac{m_{0}}{\sqrt{1 – \nu ^{2}}}$

Ans:

The relation given is $\frac{m_{0}}{\sqrt{1 – \nu ^{2}}}$

We can get, $\frac{m_{0}}{m}$ = $\sqrt{1-\nu ^{2}}$ $\frac{m_{0}}{m}$ is dimensionless. Therefore, the right hand side should also be dimensionless.

To satisfy this, $\sqrt{1-\nu ^{2}}$ should become $\sqrt{1-\frac{\nu ^{2}}{c^{2}}}$.

Thus, m = $m_{0}\sqrt{1-\frac{\nu ^{2}}{c^{2}}}$.

Question 16: Angstrom is the unit of length which is convenient on the atomic scale and it is represented by A: 1 A = 10-10 m. The hydrogen atom size is 0.5 A. Give the atomic volume of a mole of a hydrogen atom in m3.

Ans:

hydrogen atom radius = 0.5 A = 0.5 x 10-10 m

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

= $\frac{4}{3}$ x $\frac{22}{7}$ x (0.5 x 10-10)3

= 0.524 x 10-30 m3

1 hydrogen mole 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.

Question 17: At standard temperature and pressure, one mole of an ideal gas occupies 24.2 L. Find the ratio of molar volume to the volume of 1 mole of hydrogen. Why is this ratio too large?

Ans:

Radius = 0.5 A = 0.5 x 10-10 m

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

= $\frac{4}{3}$ x $\frac{22}{7}$ x ( 0.5 x 10-10)3

= 0.524 x 10-30 m3

1 hydrogen mole 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

Vm = 24.2 L = 22.4 x 10=3 m3

$\frac{V_{m}}{V_{a}}$ = $\frac{24.2\times 10^{3}}{3.16 \times 10^{-7}}$ = 7.65 x 104

The molar volume is 7.65 x 104 times more than the atomic volume. Hence, the inter-atomic separation in hydrogen gas is larger than the size of the hydrogen atom.

Question 18: Explain the following: If you are in a moving train and look outside, the nearby objects moves rapidly in the opposite direction whereas the distant object like the hill seems to be stationary.

Ans:

An imaginary line which joins the object and the observer’s eye is called the line of sight. When we observe the nearby objects, they move fast in the opposite direction as the line of sight changes constantly. Whereas, the distant objects seem to be stationary as the line of sight does not change rapidly.

Question 19: The nearest star is 5.32 light-years away from our solar system. In terms of parsecs how much is the distance? When viewed from two locations, how much parallax would the start show when viewed six months apart?

Ans:

The distance of the solar system from the star = 5.32 ly

1 light year = light speed x 1 year

= 3 x 108 x 365 x 24 x 60 x 60

= 94608 x 1011 m

5.32 ly = 503314.56 x 1011 m

1 parsec = 3.08 x 1016 m

Therefore, 5.32 ly = $\frac{503314.56\times 10^{11}}{3.08\times 10^{16}}$ = 1.63 parsec

$\theta$ = $\frac{d}{D}$

We know that, d = 3 x 1011 m

D = 503314.56 x 1011 m

$\theta$ = $\frac{3\times 10^{11}}{503314.56\times 10^{11}}$ = 5.96 x 10-6 rad

1 sec = 4.85 x 10-6

5.96 x 10-6 rad = $\frac{5.96\times 10^{-6}}{4.85\times 10^{-6}}$  = $1.22^{\circ}$.

Question 20: Measurement of physical quantity precisely is the need for science. To measure the speed of an aircraft, they must know the exact position at the separate instant of time. This lead to the discovery of radar in World War II. Measurement of mass, time, length etc. are also required.

Ans:

Precise measurement is essential for the development of science. The ultra-short laser pulse is used for measurement of time intervals. X-ray spectroscopy is used to find the interatomic separation. To measure the mass of atoms, the mass spectrometer is developed.

Question 21: As the precise measurement is required in science, rough estimates are also needed in common observations. Think of a way to estimate the given:

(a) During the monsoon, the total mass of rain clouds over India.

Ans. The metrologist records 325 cm of rain, which is the height of the water column. h = 325 cm = 3.25 m

Area = 3.3 x 1012 m2

Volume of water = A x h = 3.25 x 3.3 x 1012 = 10.725 x 1012 m3

Density of water = 1 x 103 kg m-3

Therefore, the mass of rain water = $\rho$ x V = 1 x 103 x 10.725 x 1012 = 10.725 x 1015 kg

Thus, the total mass of rain bearing clouds is 10.725 x 1015 kg

(b) An elephant’s mass

Ans. Let a known base area be floating in the sea. Let the depth of sea be d1.

Volume of water displaced = A d1

Now measure the depth of the ship with an elephant onboard.

Volume of water displaced = A d2

From the above equations, the volume of water displaced by the elephant = A d1 – A d2

Water density = D

Elephant’s mass = AD(d2 – d1)

(c) The speed of the wind in a storm.

Ans. Anemometer is used to measure the speed of the wind. As the wind blows, it rotates the anemometer and the number of rotations per second gives the wind speed.

Ans. The surface area of the head = A

Area of one hair = $\pi$r2

Number of strands of hair = $\frac{Total\;surface\;area}{area\;of\;one\;hair}$ = $\frac{A}{r^{2}}$

(e) Air molecules in a room

Ans. Let V be the volume of the room

In mole, the number of molecules = 6.023 x 1023

One mole of air = 22.4 x 10-3 m3 volume

Number of molecules in room = $\frac{6.023\times 10^{23}}{22.4\times 10^{-3}}$ = 134.915 x 1026 V

Question 22: The inner core of the sun has 107 k and the outer surface of the sun has a temperature of 6000 k. No substance remains in a solid or liquid phase at this high temperature. What is the mass density of the Sun in terms of density of mass and liquid? The mass is 3 x 1030 kg and the radius is 8 x 108 m.

Ans:

Mass = 3 x 1030 kg

Radius = 8 x 108 m

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

= $\frac{4}{3}$ x $\frac{22}{7}$ x (8 x 108)3

= $\frac{88}{21}$ x 512 x 1024 m3 = 2145.52 x 1024 m3

Density = $\frac{Mass}{Volume}$ = $\frac{3\times 10^{30}}{2145.52\times 10^{24}}$ = 1.39 x 103 kg/m5.

The density is in the range of solids and liquids. Its density is due to the high gravitational attraction on the outer layer by the inner layer of the sun.

Question 23: From the Earth, Jupiter is 954.3 million kilometres far away and its angular diameter is ${38.23}”$of arc. Find Jupiter’s diameter.

Ans:

Distance D = 954.3 x 106 km = 954.3 x 109 m

Angular diameter = ${38.23}”$ = 38.23 x 4.874 x 10-6 rad

Jupiter’s diameter = d

We know that, $\theta = \frac{d}{D}$

d = $\theta D$ = 954.3 x 109 x 38.23 x 4.874 x 10-6

= 177817.60 x 103

=1.778 x 105 Km.

Question 24: In rain, a man walks at a speed v and he must slant his umbrella at an angle $\theta$ with the vertical. The relation $tan \theta = v$ is derived and check whether the limits c $\rightarrow$ 0, $\theta$ $\rightarrow$ 0 are right. There is no strong wind and the raindrops fall vertically. Check whether the relation is correct, if not explain.

Ans:

In the relation $tan\theta = v$,

Dimension of v = M0 L1 T-1

Dimension of $tan\theta$  = M0 L0 T0

Both the dimensions are not equal, hence the relation is not correct.

To make it correct, v should be dimensionless. For that, the relation should be altered as

$tan\theta = \frac{\nu }{{\nu }’}$.

This is the correct relation.

Question 25: If two caesium clocks are allowed to run for 200 years without any disturbance, then it differs by 0.03 s. Then for measuring a time interval of 1 s, what is the accuracy of standard caesium clock?

Ans. Difference of clock = 0.03 s

The time period for the difference = 200 years

= 200 x 365 x 24 x 60 x 60 = 6.30 x 109 s

Time difference in 1 s = $\frac{0.03}{6.30\times 10^{9}}$ s.

The accuracy for time interval of 1 s = $\frac{6.30\times 10^{9}}{0.03}$ = 210 x 109 s = 2.1 x 1011 s.

Question 26: A sodium atom has a size of 3.5 Å. Find the mass density for it. The crystalline form of sodium has a density of 860 kg m-3. Compare both the densities and check whether they are of the same order.

Ans:

Diameter of sodium = 3.5 Å

Radius r = $\frac{1}{2}$ x 3.5 Å = 1.75 Å = 1.75 x 10-10 m

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

= $\frac{4}{3}$ x 3.14 x ( 1.75 x 10-10 )3

One mole of sodium has 6.023 x 1023 atoms and has a mass of 23 x 10-3 kg

One atom’s mass = $\frac{23 \times 10^{-3}}{6.023\times 10^{23}}$kg

Density $\rho$ = $\frac{\frac{23\times 10^{-3}}{6.023\times 10^{23}}}{\frac{4}{3}\times 3.14\times (1.75\times 10^{-10})^{3}}$

= 1.693 x 103 kg m-3

Therefore, it is clear that the density of sodium in its crystalline form is not the same as the density of sodium atom. This is because the atoms are closely packed in its crystalline form and the inter-atomic space is very small.

Question 27: On a nuclear scale, the convenient unit of length is Fermi. I f = 10-15 m. The relation for the nuclear size is r = r0 $A^{\frac{1}{3}}$. R is the radius. The mass number is A and r0 is constant which is equal to 1.2f. Show that the nuclear mass is almost constant to different nuclei and also find the mass density of the sodium nucleus.

Ans:

radius r = r0$A^{\frac{1}{3}}$

R0 = 1.2 f = 1.2 x 10-15 m

Volume of nucleus, V = $\frac{4}{3}\pi r^{3}$

= $\frac{4}{3}\pi (r_{0}A^{\frac{1}{3}})^{3}$ = $\frac{4}{3}\pi r_{o}^{3}A$

Nuclei mass M is equal to mass number

M = A x 1.66 x 10-27 kg

Density $\rho$ = $\frac{nucleus\;mass}{nucleus\;volume}$

= $\frac{A\times 1.66\times10^{-27}}{\frac{4}{3}\pi r_{0}^{3}A}$

= $\frac{3\times 1.66\times10^{-27}}{4\times\pi r_{0}^{3}}$ kg/m3

The above relation shows that r0 constant depends on nucleus mass. Thus, the mass density of all nuclei are almost same.

$\rho _{sodium}$ = $\frac{3\times 1.66\times 10^{-27}}{4\times 3.14\times (1.2\times 10^{-15})^{3}}$

= $\frac{4.98\times 10^{18}}{21.71}$ = 2.29 x 1017 kg m-3.

Question 28: A LASER is a monochromatic and intense unidirectional beam of light. This property of LASER is used to measure the long distance. Using laser the distance between the Moon and the Earth can be found. A laser beam takes 3.56 s to reach Earth after reflecting from the surface of the Moon. Find the lunar orbit radius.

Ans:

Time taken for laser after reflection = 3.56 s

Speed of light = 3 x 108 m/s

Time taken for laser to hit Moon = $\frac{1}{2}$ x 3.56 = 1.78 s

Lunar orbit radius = 1.78 x 3 x 108 = 5.34 x 108  m = 5.34 x 105 km.

Question 29: Ultrasonic waves are used by SONAR to detect and locate the objects which are submerged in water. SONAR is equipped in submarines to detect the enemy submarines and the time delay between transmission and reception is 84 s. Where is the enemy submarine located?

Ans:

Speed of sound in water = 1450 m/s

Time delay = 84 s

Time taken for the wave to make contact with the submarine = $\frac{1}{2}$ x 84 = 42 s

Hence, the distance between the submarines = 1450 x 42 = 60900 m = 60.9 km

Question 30: The modern astronomers discovered the farthest object which takes billions of years for the emitted light to reach Earth. These quasars have puzzling features. What is the distance of the quasar from Earth for which it takes 4 billion years for light to reach us?

Ans:

Time taken = 4 billion years

= 4 x 109 years

= 4 x 109 x 365 x 24 x 60 x 60 s

Speed of light = 3 x 108 m/s

Distance between quasar and Earth = 3 x 108 x 4 x 109 x 365 x 24 x 60 x 60

= 378432 x 1020 m

=3.7 x 1022 km

Question 31: During a solar eclipse, the disk of Sun completely covers the disk of Sun. From the given diagram, find the diameter of the Moon. Ans:

The diagram tells the positions.

Distance between the Moon and the Earth = 3.84 x 108+

Distance between the Sun and the Earth = 1.496 x 1011 m

Diameter of the Sun = 1.39 x 109 m

It can be observed that

$\frac{PQ}{RS}$ = $\frac{VT}{UT}$ $\frac{1.39\times 10^9}{RS}$ = $\frac{1.496 \times 10^11}{3.84\times 10^8}$

RS = $\frac{1.39\times 3.84}{1.496}$ x 106 = 3.57 x 106 m

Diameter of the Moon = 3.57 x 106 m.

 Also Access NCERT Exemplar for class 11 Physics Chapter 2 CBSE Notes for class 11 Physics Chapter 2

Scientist gathers information with their senses like eyes, ears, etc. and make observations. Some observations are simple like figuring out the texture and colour while other observations maybe complex for which they may need to take measurements. Measurement is one of the fundamental concepts in science. Without the ability to measure, a scientist wouldn’t be able to gather information and form a theory or conduct experiments. In this chapter, the units of physical quantities and methods of evaluating them are discussed. While the other section of the chapter deals with the errors that can occur while taking measurement and significant figures. By practising problems from NCERT Solutions Class 11 Physics Units and Measurement one get a good understanding of measurement.

Along with chapter 2 BYJU’s provide NCERT solution for all the subjects of all the classes. BYJU also provides notes, study materials, numerical problems, previous year question papers, sample papers and competitive exam study materials to help you score good marks in class 11 examination and competitive examinations. Related Links Solution Of Class 10 NCERT Maths Class 9 Science NCERT Solutions Pdf NCERT Class 10 Chemistry Solutions NCERT Solution For Class 12 Chemistry NCERT Solutions Class 7 NCERT Solutions Class 8 All Subjects NCERT Questions For Class 9 NCERT Class 10 NCERT Solution Of Maths Of Class 7 NCERT Solutions For Class 8 Maths Free Download NCERT Class 9 Maths NCERT Solutions for Class 5 Science