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NCERT Class 11
NCERT 11 Physics
Chapter 2: Units And Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

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* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 1.

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Access the answers of NCERT Class 11 Physics Chapter 2 Units and Measurements

2.1 Fill in the blanks.

(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 …(mm) 2 (c) A vehicle moving with a speed of 18 km h –1 covers….m in 1 s

(d) The relative density of lead is 11.3. Its density is ….g cm –3 or ….kg m –3 .

(a) Volume of cube, V = (1 cm) 3 = (10 -2 m) 3 = 10 -6 m 3

(b) Surface area = curved area + area on top /base = 2πrh + 2πr 2 = 2πr (h + r)

r = 2 cm = 20 mm

h = 10 cm = 100 mm

Surface area = 2πr (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm 2

Hence, the answer is 15072 mm 2

1 km = 1000 m

1 hr = 60 x 60 = 3600 s

1 km/hr = 1000 m/3600 s = 5/18 m/s

18 km/h = = (18 x 1000)/3600 = 5 m/s

Distance travelled by the vehicle in 1 s = 5 m

(d) The relative density of lead is 11.3 g cm -3

=> 11.3 x 10 3 kg m -4

2.2 Fill in the blanks by suitable conversion of units.

(a) 1 kg m 2 s –2 = ….g cm 2 s –2

(b) 1 m = ….. ly (c) 3.0 m s –2 = …. km h –2 (d) G = 6.67 × 10 –11 N m 2 (kg) –2 = …. (cm)3s –2 g –1 Answer:

1 kg m 2 s -2 = 1kg x 1m 2 x 1s -2

We know that,

1m = 100cm = 10 2 cm

When the values are put together, we get

1kg x 1m 2 x 1s -2 = 10 3 g x (10 2 cm) 2 x 1s -2 = 10 3 g x 10 4 cm 2 x 1s -2 = 10 7 gcm 2 s -2

=> 1kg m 2 s -2 = 10 7 gcm 2 s -2

(b) 1 m = ….. ly

Using the formula,

Distance = speed x time

Speed of light = 3 x 10 8 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, and we get

One light year distance = (3 x 10 8 m/s) x (365 x 24 x 60 x 60) = 9.46×10 15 m

9.46 x 10 15 m = 1ly

So that, 1m = 1/9.46 x 10 15 ly

=> 1.06 x 10 -16 ly

=> 1 meter = 1.06 x 10 -16 ly

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) -2 h -2 )

= 3 x 10 -3 km x (3600) 2 hr -2 = 3.88 x 10 4 km h -2

=> 3.0 m s -2 = 3.88 x 10 4 km h -2

(d) G = 6.67 × 10 –11 N m 2 (kg) –2 = …. (cm)3s –2 g –1

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

1N = 1kg m s -2

1 kg = 10 3 g

1m = 100cm= 10 2 cm

Put the values together, we get

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

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

=> 6.67 x 10 -11 x (1kg -1 x 1m 3 x 1s -2 )

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

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

=> 6.67 x 10 -8 cm 3 s -2 g -1

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

2.3 A calorie is a unit of heat (energy in transit), which 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 α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α –1 β –2 γ 2 in terms of the new units.

1 calorie = 4.2 J = 4.2 kg m 2 s –2

The standard formula for the conversion is

Here, x = 1, y = 2 and z =- 2

M 1 = 1 kg, L 1 = 1m, T 1 = 1s

and M 2 = α kg, L 2 = β m, T 2 = γ s

Calorie = 4.2 α –1 β –2 γ 2

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.

(a) Atoms are small object

(a) In comparison with a soccer ball, atoms are very small

(b) When compared with a bicycle, a jet plane travels at high speed.

(d) As compared with the air inside a lunch box, the air inside the room has a large number of molecules.

(e) A proton is massive when compared with an electron.

(f) Like comparing the speed of a bicycle and a jet plane, the speed of light is more than the speed of sound.

2.5 A new unit of length is chosen such that the speed of light in a 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 s to cover this distance?

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

Speed of light = 1 unit

Time taken = 8 x 60 + 20 = 480 + 20 = 500s

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

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 to within a wavelength of light

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

= $\begin{array}{l}\frac{1}{10000}\end{array} $ = 0.001 cm

= 0.00001 cm

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

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 the estimate on the thickness of the hair?

Magnification of the microscope = 100 The average width of the hair in the field of view of the microscope = 3.5 mm

Actual thickness of hair =3.5 mm/100 = 0.035 mm

2. 8. Answer the following : (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 arbitrarily increase the accuracy of the screw gauge 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) The thread should be wrapped around a pencil a number of times to form a coil having its turns touching each other closely. Measure the length of this coil with a metre scale. If L be the length of the coil and n be the number of turns of the coil, then the diameter of the thread is given by the relation

Diameter = L/n. (b) Least count of the screw gauge = Pitch/number of divisions on the circular scale

So, theoretically, when the number of divisions on the circular scale is increased, the least count of the screw gauge will decrease. Hence, the accuracy of the screw gauge will increase. However, this is only a theoretical idea. Practically, there will be many other difficulties when the number of turns is increased.

(c) The probability of making random errors can be reduced to a larger extent in 100 observations than in the case of 5 observations.

2.9 . The photograph of a house occupies an area of 1.75 cm 2 on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is 1.55 m 2 . What is the linear magnification of the projector-screen arrangement?

Arial Magnification = Area of the image/Area of the object

= 1.55/1.75 x 10 4

= 8.857x 10 3

Linear Magnification = √Arial magnification

= √8.857x 10 3

2.10 State the number of significant figures in the following: (a) 0.007 m 2 (b) 2.64 × 10 24 kg (c) 0.2370 g cm –3 (d) 6.320 J (e) 6.032 N m –2 (f) 0.0006032 m 2

(a) 0.007 m 2

The given value is 0.007 m 2 .

Only one significant digit. It is 7.

(b) 2.64 × 10 24 kg

The value is 2.64 × 10 24 kg

For the determination of significant values, the power of 10 is irrelevant. The digits 2, 6, and 4 are significant figures. The number of significant digits is 3.

The value is 0.2370 g cm –3

For the given value with decimals, all the numbers 2, 3, 7, and 0 are significant. The 0 before the decimal point is not significant

(d) All the numbers are significant. The number of significant figures here is 4.

(e) 6, 0, 3, and 2 are significant figures. Therefore, the number of significant figures is 4.

(f) 6, 0, 3, and 2 are significant figures. The number of significant figures is 4.

2. 11. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm, respectively. Give the area and volume of the sheet to correct significant figures.

Area of the rectangular sheet = length x breadth

= 4.234 x 1.005 = 4.255 m 2 = 4.3 m 2

The volume of the rectangular sheet = length x breadth x thickness = 4.234 x 1.005 x 2.01 x 10 -2 = 8.55 x 10 -2 m 3 .

2.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is

(a) the total mass of the box,

(b) the difference in the masses of the pieces to correct significant figures?

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g

The mass of the second gold piece = 20.17 g

The total mass = 2.300 + 0.2015 + 0.2017 = 2.7032 kg

Since 1 is the least number of decimal places, the total mass = 2.7 kg.

The mass difference = 20.17 – 20.15 = 0.02 g

Since 2 is the least number of decimal places, the total mass = 0.02 g.

2.13 A physical quantity P is related to four observables a, b, c and d as follows:

P = $\begin{array}{l}\frac{a^{3}b^{2}}{\sqrt{c}d}\end{array} $

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?

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

= 3 x 1 + 2 x 3 + $\begin{array}{l}\frac{1}{2}\end{array} $ x 4 + 2

= 3 + 6 + 2 + 2 = 13 %

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

2.14 A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a) y = a sin ( $\begin{array}{l}\frac{2\pi t}{T}\end{array} $ )

(b) y = a sin vt

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

Dimension of y = M 0 L 1 T 0

The dimension of a = M 0 L 1 T 0

Dimension of sin $\begin{array}{l}\frac{2\pi t}{T}\end{array} $ = M 0 L 0 T 0

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.

Dimension of $\begin{array}{l}\frac{t}{T}\end{array} $ = M 0 L 0 T 0

The formula is dimensionally correct.

2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m o 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 = $\begin{array}{l}\frac{m_{0}}{\sqrt{1 – \nu ^{2}}}\end{array} $

Guess where to put the missing c.

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

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

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

2.16 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10 –10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m 3 of a mole of hydrogen atoms?

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

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

= 0.524 x 10 -30 m 3

1 hydrogen mole contains 6.023 x 10 23 hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10 23 x 0.524 x 10 -30

= 3.16 x 10 -7 m 3 .

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 the hydrogen molecule to be about 1 Å)? Why is this ratio so large?

Radius = 0.5 A = 0.5 x 10 -10 m

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

= 3.16 x 10 -7 m 3

V m = 22.4 L = 22.4 x 10 -3 m 3

The molar volume is 7.1 x 10 4 times more than the atomic volume. Hence, the inter-atomic separation in hydrogen gas is larger than the size of the hydrogen atom.

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 train’s motion, but the distant objects (hilltops, the Moon, the stars etc.) seem to be stationary (In fact, since you are aware that you are moving, these distant objects seem to move with you).

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.

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 × 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?

Diameter of Earth’s orbit = 3 × 10 11 m

Radius of Earth’s orbit r = 1.5 × 10 11 m

Let the distance parallax angle be θ=1″ (s) = 4.847 × 10 –6 rad.

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″

Therefore, D = 1.5 × 10 11 /4.847 × 10 –6 = 0.309 x 10 17

Hence 1 parsec ≈ 3.09 × 10 16 m.

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?

1 light year is the distance travelled by light in a year.

1 light year = 3 x 108 x 365 x 24 x 60 x 60 = 9.46 x 10 15 m

Therefore, distance travelled by light in 4.29 light years = 4.29 x 9.46 x 10 15 = 4.058 x 10 16 m

Parsec is also a unit of distance 1 parsec = 3.08 x 10 16 m Therefore, the distance travelled by light in parsec is given as

4.29 light years =4.508 x 10 16 /3.80 x 10 16 = 1.318 parsec = 1.32 parsec

Using the relation, θ = d / D here, d is the diameter of Earths orbit, d = 3 × 10 11 m D is the distance of the star from the earth, D = 405868.32 × 10 11 m ∴ θ = 3 × 10 11 / 405868.32 × 10 11 = 7.39 × 10 -6 rad But the angle covered in 1 sec = 4.85 × 10 –6 rad ∴ 7.39 × 10 -6 rad = 7.39 × 10 -6 / 4.85 × 10 -6 = 1.52″

2.21 Precise measurements of physical quantities are a need for 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 r adar 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.

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.

2.23 The Sun is a hot plasma (ionised 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: a mass of the Sun = 2.0 × 10 30 kg, the radius of the Sun = 7.0 × 10 8 m.

Mass = 2 x 10 30 kg

Radius = 7 x 10 8 m

= $\begin{array}{l}\frac{4}{3}\end{array} $ x $\begin{array}{l}\frac{22}{7}\end{array} $ x (7 x 10 8 ) 3

= $\begin{array}{l}\frac{88}{21}\end{array} $ x 512 x 10 24 m 3 = 2145.52 x 10 24 m 3

Density = $\begin{array}{l}\frac{Mass}{Volume}\end{array} $ = $\begin{array}{l}\frac{3\times 10^{30}}{2145.52\times 10^{24}}\end{array} $ = 1.39 x 10 3 kg/m 5 .

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.

2.24. When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.

Distance of the planet Jupiter from Earth, D= 824.7 million kilometres = 824.7 x 10 6 km

Angular diameter θ = 35.72 “= 35.72 x 4.85 x 10 -6 rad = 173.242 x 10 -6 rad Diameter of Jupiter d = θ x D= 173.241 x 10 -6 x 824.7 x 10 6 km =142871 = 1.43 x 10 5 km

2.25. A man walking briskly in the rain with speed v must slant his umbrella forward, making an angle θ with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit: as v →0, θ →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.

According to the principle of homogeneity of dimensional equations, Dimensions of L.H.S. = Dimensions of R.H.S.

In relation v = tan θ, tan θ is a trigonometric function, and it is dimensionless. The dimension of v is [L 1 T -1 ]. Therefore, this relation is incorrect. To make the relation correct, the L.H.S. must be divided by the velocity of rain, u.

Therefore, the relation becomes v/u= tan θ

This relation is correct dimensionally

2.26. It is claimed that two caesium 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 caesium clock in measuring a time interval of 1 s?

Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Error in 100 years = 0.02 s Error in 1 second=0.02/100 x 365 x 24 x 60 x 60 =6.34 x 10 -12 s The accuracy of the standard caesium clock in measuring a time-interval of 1 s is 10 -12 s.

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?

The diameter of sodium= 2.5 A = 2.5 x 10 -10 m

Therefore, the radius is 1.25 x 10 -10 m

The volume of a sodium atom, V= (4/3)πr 3

= (4/3) x (22/7) x (1.25 x 10 -10 ) 3 = 8.177 x 10 -30 m 3

Mass of one mole atom of sodium = 23 g = 23 x 10 -3 kg

1 mole of sodium contains 6.023 x 10 23 atoms

Therefore, the mass of one sodium atom, M= 23 x 10 -3 /6.023 x 10 23 = 3.818 x 10 -26 kg

The atomic mass density of sodium, ρ= M/V =3.818 x 10 -26 /8.177 x 10 -30

= 0.46692 x 10 4 = 4669.2 kg m -3 The density of sodium in its solid state is 4669.2 kg m -3, but in the crystalline phase, the density is 970 kg m -3 . Hence, both are in a different order. In the solid phase, atoms are tightly packed, but in the crystalline phase, atoms arrange a sequence which contains a void. So, the density in the solid phase is greater than in the crystalline phase.

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 0 A 1/3 where r is the radius of the nucleus, A its mass number, and r 0 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 the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

The radius of the nucleus

r = r 0 A 1/3

r o = 1.2 f = 1.2 x 10 -15 m

So, the nuclear mass density is much larger than the atomic mass density for a sodium atom we got in 2.27.

2.29. A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of laser light can be exploited to measure long distances. The distance of the Moon from the Earth has already been 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?

Time taken for the laser beam to return to Earth after reflection by the Moon’s surface = 2.56 s

The speed of laser light is c = 3 x 10 8 m/s.

Let d be the distance of the Moon from the Earth,

The time taken by laser signal to reach the Moon, t = 2d/c

Therefore, d = tc/2 = (2.56 x 3 x 10 8 )/2 = 3.84 x 10 8 m

2. 30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between the 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 ).

Speed of sound in water,v = 1450 m s –1

The time between generation and the reception of the echo after reflection, 2t= 77.0 s

Time taken for the sound waves to reach the submarine, t = 77.0/2 = 38. 5 s

Then v = d/t

Distance of enemy submarine, d = tv

Therefore, d=vt=(1450 x 38. 5) =55825 m=55.8 x 10 3 m or 55.8 km.

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?

Time taken by light from the quasar to reach the observer, t = 3.0 billion years = 3.0 x 10 9 years = 3.0 x 10 9 x 365 x 24 x 60 x 60 s

= 94608000 x 10 9 s

= 9.46 x 10 16 m

Speed of light = 3 x 10 8 m/s Distance of quasar from Earth = 3.0 x 10 8 x 9.46 x 10 16 m = 28.38 x 10 24 m

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 the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

From examples 2.3 and 2.4, we get the following data

Distance of the Moon from Earth = 3.84 x 10 8 m

Distance of the Sun from Earth = 1.496 x 10 11 m

Sun’s diameter = 1.39 x 10 9 m

During a total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal.

Therefore, the Angular diameter of the moon, θ = 9.31 x 10 -3 rad The earth-moon distance, S = 3.8452 x 10 8 m

Therefore, the diameter of the moon, D = θ x S = 9.31 x 10 -3 x 3.8452 x 10 8 m = 35.796 x 10 5 m

For students of Class 11 who are looking to give their best for the upcoming annual and competitive exams, it is very important to get accustomed to the solutions to the questions in the textbook. Thus, students are advised to have good practice with different kinds of questions that can be framed from the chapter. Students are suggested to solve the NCERT questions. To clear all the doubts of the students, BYJU’S provides NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements.

Topics Covered in Class 11 Chapter 2 Physics Units and Measurement


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

Scientists gather 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 may be 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 measurements and significant figures. Practising problems from Class 11 Physics NCERT Solutions gives one a good understanding of measurement.

Along with Chapter 2, BYJU’S provides NCERT Solutions for all the subjects of all the classes. BYJU’S also provides notes, study materials, numerical problems, previous years’ question papers, sample papers and competitive exam study materials to help students score good marks in Class 11 examinations.

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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements :


2	Units and Measurements
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

NCERT Solutions Class 11 Physics Physics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 2. 1. Fill in the blanks (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 ……..(mm) 2 . (c) A vehicle moving with a speed of 18 km h -1 covers ………. m in 1 s. (d) The relative density of lead is 11.3. Its density is …….. g cm -3 or ………. kg m -3 . Answer: (a) Volume of cube, V = (1 cm) 3 = (10 -2 m) 3 10 -6 m 3 . Hence, answer is 10 -6 (b) Surface area = 2πrh + 2πr 2 = 2πr (h + r) = 2 x 22/7 x 2 x 10 (10 x 10 + 2 x 10) mm 2 = 1.5 x 10 4 mm 2 Hence, answer is 1.5 x 10 4 . (c) Speed of vehicle = 18 km/h = 18 x 1000/3600 m/s = 5 m/s ; so the vehicle covers 5 m in 1 s. = 11.3 (d) Density= 11.3 g cm -3 =11.3 x 10 3 kg m -3 [1 kg =10 3 g,1m=10 2 cm] =11.3 x 10 3 kg m -4

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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. Answer: Physical quantities are called large or small depending on the unit (standard) of measurement. For example, the distance between two cities on earth is measured in kilometres but the distance between stars or inter —galactic distances are measured in parsec. The later standard parsec is equal to 3.08 x 10 16 m or 3.08 x 10 12 km is certainly larger than metre or kilometre. Therefore, the inter-stellar or intergalactic distances are certainly larger than the distances between two cities on earth. (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 super fast 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 a correct statement. (f) This is a correct statement.

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 s to cover this distance? Answer: Distance between Sun and Earth = Speed of light in vacuum x time taken by light to travel from Sim to Earth = 3 x 10 8 m/ s x 8 min 20 s = 3 x 10 8 m/s x 500 s = 500 x 3 x 10 8 m. In the new system, the speed of light in vacuum is unity. So, the new unit of length is 3 x 10 8 m. .•. distance between Sun and Earth = 500 new units.

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 to within a wavelength of light? Answer: (a) Least count of vernier callipers = 1/20 = 0.05 mm = 5 x 10 -5 m (b) Least count of screw gauge =Pitch/No. of divisions on circular scale = 1 x 10 -3 /100 = 1 x 10 -5 m (c) Least count of optical instrument = 6000 A (average wavelength of visible light as 6000 A) = 6 x 10- 7 m As the least count of optical instrument is least, it is the most precise device out of three instruments given to us.

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 the estimate on the thickness of hair? Answer: As magnification, m =thickness of image of hair/ real thickness of hair = 100 and average width of the image of hair as seen by microscope = 3.5 mm .•. Thickness of hair =3.5 mm/100 = 0.035 mm

Question 2. 8. Answer the following: (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? Answer: (a) Wrap the thread a number of times on a round pencil so as to form a coil having its turns touching each other closely. Measure the length of this coil, mode by the thread, with a metre scale. If n be the number of turns of the coil and l be the length of the coil, then the length occupied by each single turn i.e., the thickness of the thread = 1/n . This is equal to the diameter of the thread. (b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on circular scale is increased, least count is decreased. Hence the accuracy is increased. However, this is only a theoretical idea.Practically speaking, increasing the number of ‘turns would create many difficulties. As an example, the low resolution of the human eye would make observations difficult. The nearest divisions would not clearly be distinguished as separate. Moreover, it would be technically difficult to maintain uniformity of the pitch of the screw throughout its length. (c) Due to random errors, a large number of observation will give a more reliable result than smaller number of observations. This is due to the fact that the probability (chance) of making a positive random error of a given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of observations, positive and negative errors are likely to cancel each other. Hence more reliable result can be obtained.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q9

Question 2. 10. State the number of significant figures in the following: (a) 0.007 m 2 (b) 2.64 x 10 4 kg (c) 0.2370 g cm -3 (d) 6.320 J (e) 6.032 N m -2 (f) 0.0006032 m 2 Answer: (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4.

Question 2 .11. ‘The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. Answer: As Area = (4.234 x 1.005) x 2 = 8.51034 = 8.5 m 2 Volume = (4.234 x 1.005) x (2.01 x 10 -2 ) = 8.55289 x 10 -2 = 0.0855 m 3 .

Question 2. 12. The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box (b) the difference in the masses of the pieces to correct significant figures? Answer: (a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg Since the least number of decimal places is 1, therefore, the total mass of the box = 2.3 kg. (b) Difference of mass = 2.17 – 2.15 = 0.02 g Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is 0.02 g.

Question 2. 16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A: 1 A = 10 -10 m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m 3 of a mole of hydrogen atoms? Answer: Volume of one hydrogen atom = 4/3 πr3 (volume of sphere) = 4/3 x 3.14 x (0.5 x 10 -10 ) m 3 = 5.23 x 10 -31 m 3 According to Avagadro’s hypothesis, one mole of hydrogen contains 6.023 x 10 23 atoms. Atomic volume of 1 mole of hydrogen atoms = 6.023 x 1023 x 5.23 x 10 -31 = 3.15 x 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 A.) Why is this ratio so large? Answer: Volume of one mole of ideal gas, V g = 22.4 litre = 22.4 x 10 -3 m 3 Radius of hydrogen molecule = 1A/2 = 0.5 A = 0.5 x 10 -10 m Volume of hydrogen molecule = 4/3 πr 3 =4/3 x 22/7 (0.5 x 10 -10 ) 3 m 3 = 0.5238 x 10 -30 m 3 One mole contains 6.023 x 10 23 molecules. Volume of one mole of hydrogen, VH = 0.5238 x 10 -30 x 6.023 x 10 23 m 3 = 3.1548 x 10 -7 m 3 Now V g /V H =22.4 x 10 -3 /3.1548 x 10 -7 =7.1 x 10 4 The ratio is very large. This is because the interatomic separation in the gas is very large compared to the size of a hydrogen molecule.

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 train’s 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). Answer: 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 changes its direction rapidly. As a result, the nearby trees and other objects appear to run in a direction opposite to the train’s 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 =3 x 10 n m. However, even the nearest stars are so distant that with such a long baseline, they show parallel 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? Answer: From parallax method we can say θ=b/D,where b=baseline ,D = distance of distant object or star Since, θ=1″ (s) and b=3 x 10 11 m D=b/20=3 x 10 11 /2 x 4.85 x 10 -6 m or D=3 x 10 11 /9.7 x 10 -6 m =30 x 10 16 /9.7 m = 3.09 x 10 16 m = 3 x 10 16 m.

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 Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun? Answer: As we know, 1 light year = 9.46 x 10 15 m .•. 4.29 light years = 4.29 x 9.46 x 10 15 = 4.058 x 10 16 m Also, 1 parsec = 3.08 x 10 16 m .•. 4.29 light years =4.508 x 10 16 /3.80 x 10 16 = 1.318 parsec = 1.32 parsec. As a parsec distance subtends a parallax angle of 1″ for a basis of radius of Earth’s orbit around the Sun (r).In present problem base is the distance between two locations of the Earth six months apart in its orbit around the Sun = diameter of Earth’s orbit (b = 2r). .•. Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec). .•. Parallax angle subtended by the star Alpha Centauri at the given basis θ = 1.32 x 2 = 2.64″.

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 the actual motivation behind the discovery of radar in World War II. Think of different examples in modem science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed. Answer: Extremely precise measurements are needed in modem 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 A° = 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.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q22

Question 2. 23. The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in 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 x 10 30 kg, radius of the Sun = 7.0 x 10 8 m. Answer: Given M = 2 x 10 30 kg, r = 7 x 10 8 m .-. Volume of Sun = 4/3πr 3 x 3.14 x (7 x 10 8 ) 3 = 1.437 x 10 27 m 3 As p = M/V, .’. p = 2 x 10 30 /1.437 x 10 27 = 1391.8 kg m -3 = 1.4 x 10 3 kg m -3 Mass density of Sun is in the range of mass densities of solids/liquids and not gases.

Question 2. 24. When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter. Answer: Given angular diameter θ = 35.72= 35.72 x 4.85 x 10 -6 rad = 173.242 x 10 -6 = 1.73 x 10 -4 rad Diameter of Jupiter D = θ x d = 1.73 x 10 -4 x 824.7 x 10 9 m =1426.731 x 10 3 = 1.43 x 10 8 m

Question 2. 25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v: tanθ = v and checks that the relation has a correct limit: as v—>θ, θ —>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. Answer: According to principle of homogenity of dimensional equations, Dimensions of L.H.S. = Dimensions of R.H.S. Here, v = tan θ i. e., [L 1 T -1 ] = dimensionless, which is incorrect. Correcting the L.H.S., we. get v/u= tan θ, where u is velocity of rain.

Question 2. 26. It is claimed that two cesium clocks, if allowed to run for 180 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? Answer: Total time = 100 years = 100 x 365 x 24 x 60 x 60 s Error in 1 second=0.02/100 x 365 x 24 x 60 x 60 =6.34 x 10 -12 s .•. Accuracy of 1 part in 10 11 to 10 12 .

Question 2. 27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m 3- . Are the two densities of the same order of magnitude? If so, why? Answer: It is given that radius of sodium atom, R = 2.5 A = 2.5 x 10 -10 m Volume of one mole atom of sodium, V = NA .4/3 π R 3 V = 6.023 x 10 23 x –4/3 x 3.14 x (2.5 x 10 -10 ) 3 m 3 and mass of one mole atom of sodium, M = 23 g = 23 x 10 -3 kg .•. Average mass density of sodium atom, p = M/V =(23 x 10 -3 /6.023 x 10 23 x 4/3 x 3.14 x (2.5 x 10 -10 )) = 6.96 x 10 2 kg m -3 = 0.7 x 10 -3 kg m -3 The density of sodium in its crystalline phase = 970 kg m -3 = 0.97 x 10 3 kg m -3 Obviously the two densities are of the same order of magnitude (= 10 3 kg m -3 ). It is on account of the fact that in solid phase atoms are tightly packed and so the atomic mass density is close to the mass density of solid.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q28

Question 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 ). Answer: Here speed of sound in water v = 1450 m s -1 and time of echo t = 77.0 s. If distance of enemy submarine be d, then t = 2d/v .’. d=vt/2 =1450 x 77.0/2 =55825 m=55.8 x 10 3 m or 55.8 km.

Question 2. 31. The farthest objects in our Universe discovered by modem 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? Answer: The time taken by light from the quasar to the observer t = 3.0 billion years = 3.0 x 10 9 years As 1 ly = 9.46 x 10 15 m .•. Distance of quasar from the observer d = 3.0 x 10 9 x 9.46 x 10 15 m = 28.38 x 10 24 m = 2.8 x 10 25 m or 2.8 x 10 22 km.

Question 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. Answer: From examples 2.3 and 2.4, we get θ = 1920″ and S = 3.8452 x 10 8 m. During the total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal. Angular diameter of the moon, θ= Angular diameter of the sun = 1920″ = 1920 x 4.85 x 10 -6 rad [1″ = 4.85 x 10 -6 rad] The earth-moon distance, S = 3.8452 x 10 8 m .’. The diameter of the moon, D = θ x S = 1920 x 4.85 x 10 -6 x 3.8452 x 10 8 m = 35806.5024 x 10 2 m = 3581 x 10 3 m 3581 km.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q33

QUESTIONS BASED ON SUPPLEMENTARY CONTENTS

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Numerical Questions Q1

NCERT Solutions for Class 11 Physics All Chapters

Chapter 1 Physical World
Chapter 2 Units and Measurements
Chapter 3 Motion in a Straight Line
Chapter 4 Motion in a plane
Chapter 5 Laws of motion
Chapter 6 Work Energy and power
Chapter 7 System of particles and Rotational Motion
Chapter 8 Gravitation
Chapter 9 Mechanical Properties Of Solids
Chapter 10 Mechanical Properties Of Fluids
Chapter 11 Thermal Properties of matter
Chapter 12 Thermodynamics
Chapter 13 Kinetic Theory
Chapter 14 Oscillations
Chapter 15 Waves

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Class 11 Physics Chapter 2 Units and Measurements

Topics and Subtopics in Class 11 Physics Chapter 2 Units and Measurements :


2	Units and Measurements
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

Units and Measurement Class 11 Notes Physics Chapter 2

Measurement The process of measurement is basically a comparison process. To measure a physical quantity, we have to find out how many times a standard amount of that physical quantity is present in the quantity being measured. The number thus obtained is known as the magnitude and the standard chosen is called the unit of the physical quantity.
Unit The unit of a physical quantity is an arbitrarily chosen standard which is widely accepted by the society and in terms of which other quantities of similar nature may be measured.
Standard The actual physical embodiment of the unit of a physical quantity is known as a standard of that physical quantity. • To express any measurement made we need the numerical value (n) and the unit (μ). Measurement of physical quantity = Numerical value x Unit For example: Length of a rod = 8 m where 8 is numerical value and m (metre) is unit of length.
Fundamental Physical Quantity/Units It is an elementary physical quantity, which does not require any other physical quantity to express it. It means it cannot be resolved further in terms of any other physical quantity. It is also known as basic physical quantity. The units of fundamental physical quantities are called fundamental units. For example, in M. K. S. system, Mass, Length and Time expressed in kilogram, metre and second respectively are fundamental units.
Derived Physical Quantity/Units All those physical quantities, which can be derived from the combination of two or more fundamental quantities or can be expressed in terms of basic physical quantities, are called derived physical quantities. The units of all other physical quantities, which car. be obtained from fundamental units, are called derived units. For example, units of velocity, density and force are m/s, kg/m3, kg m/s2 respectively and they are examples of derived units.
Systems of Units Earlier three different units systems were used in different countries. These were CGS, FPS and MKS systems. Now-a-days internationally SI system of units is followed. In SI unit system, seven quantities are taken as the base quantities. (i) CGS System. Centimetre, Gram and Second are used to express length, mass and time respectively. (ii) FPS System. Foot, pound and second are used to express length, mass and time respectively. (iii) MKS System. Length is expressed in metre, mass is expressed in kilogram and time is expressed in second. Metre, kilogram and second are used to express length, mass and time respectively. (iv) SI Units. Length, mass, time, electric current, thermodynamic temperature, Amount of substance and luminous intensity are expressed in metre, kilogram, second, ampere, kelvin, mole and candela respectively.
Definitions of Fundamental Units
Supplementary Units Besides the above mentioned seven units,there are two supplementary base units. these are (i) radian (rad) for angle, and (ii) steradian (sr) for solid angle.

(i) It is internationally accepted, (ii) It is a rational unit system, (iii) It is a coherent unit system, (iv) It is a metric system, (v) It is closely related to CGS and MKS systems of units, (vi) Uses decimal system, hence is more user friendly.

Other Important Units of Length For measuring large distances e.g., distances of planets and stars etc., some bigger units of length such as ‘astronomical unit’, ‘light year’, parsec’ etc. are used. • The average separation between the Earth and the sun is called one astronomical unit. 1 AU = 1.496 x 10 11 m. • The distance travelled by light in vacuum in one year is called light year. 1 light year = 9.46 x 10 15 m. • The distance at which an arc of length of one astronomical unit subtends an angle of one second at a point is called parsec. 1 parsec = 3.08 x 10 16 m • Size of a tiny nucleus = 1 fermi = If = 10 -15 m • Size of a tiny atom = 1 angstrom = 1A = 10 -10 m
Parallax Method This method is used to measure the distance of planets and stars from earth. Parallax. Hold a pen in front of your eyes and look at the pen by closing the right eye and ‘ then the left eye. What do you observe? The position of the pen changes with respect to the background. This relative shift in the position of the pen (object) w.r.t. background is called parallax. If a distant object e.g., a planet or a star subtends parallax angle 0 on an arc of radius b (known as basis) on Earth, then distance of that distant object from the basis is given by • To estimate size of atoms we can use electron microscope and tunneling microscopy technique. Rutherford’s a-particle scattering experiment enables us to estimate size of nuclei of different elements. • Pendulum clocks, mechanical watches (in which vibrations of a balance wheel are used) and quartz watches are commonly used to measure time. Cesium atomic clocks can be used to measure time with an accuracy of 1 part in 10 13 (or to a maximum discrepancy of 3 ps in a year). • The SI unit of mass is kilogram. While dealing with atoms/ molecules and subatomic particles we define a unit known as “unified atomic mass unit” (1 u), where 1 u = 1.66 x 10 -27 kg.

After that some lycopodium powder is lightly sprinkled on the surface of water in a large trough and one drop of this solution is put in water. The oleic acid drop spreads into a thin, large and roughly circular film of molecular thickness on water surface. Then, the diameter of the thin film is quickly measured to get its area A. Suppose n drops were put in the water. Initially, the approximate volume of each drop is determined (V cm 3 ). Volume of n drops of solution = nV cm 3 Amount of oleic acid in this solution

The solution of oleic acid spreads very fast on the surface of water and forms a very thin layer of thickness t. If this spreads to form a film of area A cm 2 , then thickness of the film

If we assume that the film has mono-molecular thickness, this becomes the size or diameter of a molecule of oleic acid. The value of this thickness comes out to be of the order of 10 -9 m.

Dimensions The dimensions of a physical quantity are the powers to which the fundamental units of mass, length and time must be raised to represent the given physical quantity.
Dimensional Formula The dimensional formula of a physical quantity is an expression telling us how and which of the fundamental quantities enter into the unit of that quantity. It is customary to express the fundamental quantities by a capital letter, e.g., length (L), mass (AT), time (T), electric current (I), temperature (K) and luminous intensity (C). We write appropriate powers of these capital letters within square brackets to get the dimensional formula of any given physical quantity.
Applications of Dimensions The concept of dimensions and dimensional formulae are put to the following uses: (i) Checking the results obtained (ii) Conversion from one system of units to another (iii) Deriving relationships between physical quantities (iv) Scaling and studying of models. The underlying principle for these uses is the principle of homogeneity of dimensions. According to this principle, the ‘net’ dimensions of the various physical quantities on both sides of a permissible physical relation must be the same; also only dimensionally similar quantities can be added to or subtracted from each other.
Limitations of Dimensional Analysis The method of dimensions has the following limitations: (i) by this method the value of dimensionless constant cannot be calculated. (ii) by this method the equation containing trigonometric, exponential and logarithmic terms cannot be analyzed. (iii) if a physical quantity in mechanics depends on more than three factors, then relation among them cannot be established because we can have only three equations by equalizing the powers of M, L and T. (iv) it doesn’t tell whether the quantity is vector or scalar.
Significant Figures The significant figures are a measure of accuracy of a particular measurement of a physical quantity. Significant figures in a measurement are those digits in a physical quantity that are known reliably plus the first digit which is uncertain.
The Rules for Determining the Number of Significant Figures (i) All non-zero digits are significant. (ii) All zeroes between non-zero digits are significant. (iii) All zeroes to the right of the last non-zero digit are not significant in numbers without decimal point. (iv) All zeroes to the right of a decimal point and to the left of a non-zero digit are not significant. (v) All zeroes to the right of a decimal point and to the right of a non-zero digit are significant. (vi) In addition and subtraction, we should retain the least decimal place among the values operated, in the result. (vii) In multiplication and division, we should express the result with the least number of significant figures as associated with the least precise number in operation. (viii) If scientific notation is not used: (a) For a number greater than 1, without any decimal, the trailing zeroes are not significant. (b) For a number with a decimal, the trailing zeros are significant.
Error The measured value of the physical quantity is usually different from its true value. The result of every measurement by any measuring instrument is an approximate number, which contains some uncertainty. This uncertainty is called error. Every calculated quantity, which is based on measured values, also has an error.
Causes of Errors in Measurement Following are the causes of errors in measurement: Least Count Error. The least count error is the error associated with the resolution of the instrument. Least count may not be sufficiently small. The maximum possible error is equal to the least count. Instrumental Error . This is due to faulty calibration or change in conditions (e.g., thermal expansion of a measuring scale). An instrument may also have a zero error. A correction has to be applied. Random Error. This is also called chance error. It makes to give different results for same measurements taken repeatedly. These errors are assumed to follow the Gaussian law of normal distribution. Accidental Error. This error gives too high or too low results. Measurements involving this error are not included in calculations. Systematic Error. The systematic errors are those errors that tend to be in one direction, either positive or negative. Errors due to air buoyancy in weighing and radiation loss in calorimetry are systematic errors. They can be eliminated by manipulation. Some of the sources of systematic errors are: (i) intrumental error (ii) imperfection in experimental technique or procedure (iii) personal errors
Absolute Error, Relative Error and Percentage Error
Combination of Errors
IMPORTANT TABLES

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NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line - FREE PDF Download

NCERT for Chapter 2 Motion In A Straight Line Class 11 Solutions by Vedantu, introduces the basics of kinematics, which deals with the motion of objects. This chapter explains how objects move along a straight path, providing the essential concepts required for further studies in physics. In this chapter, you will study motion and the different kinds of motion. Also read about concepts like path lengths, displacement, velocity, and speed and their various numerical associated. Through these Class 11 Physics NCERT Solutions , we aim to simplify complex concepts into easy-to-understand explanations, helping you with each topic effectively. With Vedantu's NCERT Solutions, you will find step-by-step explanations of all the questions in your textbook, ensuring that you understand the concepts thoroughly.

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Glance on Physics Chapter 2 Class 11 - Motion in a Straight Line

An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight fine. position to the right of the origin is taken as positive and to the left as negative.

The path to the right of the origin is taken as positive and to the left is taken as negative when motion is in a straight line.

Average Speed - The total path length travelled in the total given time interval is known as average speed.

The path length is defined as the total length of the path traversed by an object.

An object is said to be in uniform motion in a straight line if its displacement is equal in equal intervals of time. Otherwise, the motion is said to be nonuniform.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 2 - Motion In A Straight Line, which you can download as PDFs.

There are 18 fully solved questions in class 11th physics chapter 2 Motion In A Straight Line.

Access NCERT Solutions for Class 11 Physics Chapter 2 – Motion in a Straight Line

1. In which of the following examples of motion, can the body be considered approximately a point object:

(a) a railway carriage moving without jerks between two stations.

Ans: As the size of a carriage is very small as compared to the distance between two stations, the carriage can be treated as a point sized object.

(b) a monkey sitting on top of a man cycling smoothly on a circular track.

Ans: As the size of a monkey is very small as compared to the size of a circular track, the monkey can be considered as a point sized object on the track.

Ans: As the size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground, the cricket ball cannot be considered as a point object.

(d) a tumbling beaker that has slipped off the edge of a table.

Ans: As the size of a beaker is comparable to the height of the table from which it slipped, the beaker cannot be considered as a point object.

2. The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in figure. Choose the correct entries in the brackets below:

The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q.png

a. (A/B) lives closer to the school than (B/A)

Ans: From the graph, it is clear that $OP<OQ$. Therefore, A lives closer to the school than B.

b. (A/B) starts from the school earlier than (B/A)

Ans: From the graph it is clear that for \[x\text{ }=\text{ }0,\text{ }t\text{ }=\text{ }0\] for A and t has some finite value for B. Hence, A starts from the school earlier than B.

c. (A/B) walks faster than (B/A)

Ans: As the velocity is equal to slope of x-t graph, in case of uniform motion and slope of x-t graph for B is greater than that for A. Thus B walks faster than A.

d. A and B reach home at the (same/different) time

Ans: From the graph it is clear that both A and B reach their respective homes at the same time.

e. (A/B) overtakes (B/A) on the road (once/twice).

Ans: As B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.

3. A woman starts from her home at 9.00 am, walks with a speed of \[5\text{ }km/hr\] on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of \[25\text{ }km/hr\]. Choose suitable scales and plot the x-t graph of her motion

Ans: In the above question it is given that:

Speed of the woman \[=\text{ }5\text{ }km/h\].

Distance between her office and home \[=\text{ }2.5\text{ }km\].

\[Time\text{ }taken\text{ }=\text{ }Distance\text{ }/\text{ }Speed\]

\[=\text{ }2.5\text{ }/\text{ }5\text{ }=\text{ }0.5\text{ }h\text{ }=\text{ }30\text{ }min\]

It is given that she covers the same distance in the evening by an auto.

Now, \[speed\text{ }of\text{ }the\text{ }auto\text{ }=\text{ }25\text{ }km/h\].

\[=\text{ }2.5\text{ }/\text{ }25\text{ }=\text{ }1\text{ }/\text{ }10\text{ }=\text{ }0.1\text{ }h\text{ }=\text{ }6\text{ }min\]

The suitable x-t graph of the motion of the woman is shown in the given figure.

he suitable x-t graph of the motion of the woman.png

4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Distance covered with 1 step \[=\text{ }1\text{ }m\]

Time taken \[=\text{ }1\text{ }s\]

Time taken to move first \[5\text{ }m\] forward \[=\text{ }5\text{ }s\]

Time taken to move \[3\text{ }m\]backward \[=\text{ }3\text{ }s\]

\[Net\text{ }distance\text{ }covered\text{ }=\text{ }5\text{ }\text{ }3\text{ }=\text{ }2\text{ }m\]

Net time taken to cover \[2\text{ }m\text{ }=\text{ }8\text{ }s\].

Drunkard covers \[2\text{ }m\]in \[8\text{ }s\].

Drunkard covered \[4\text{ }m\]in \[16\text{ }s\].

Drunkard covered \[6\text{ }m\]in $2\text{4 s}$.

Drunkard covered \[8\text{ }m\]in \[32\text{ }s\].

In the next \[5\text{ }s\], the drunkard will cover a distance of \[5\text{ }m\] and a total distance of \[13\text{ }m\] and then fall into the pit.

Net time taken by the drunkard to cover $km$\[13\text{ }m\text{ }=\text{ }32\text{ }+\text{ }5\text{ }=\text{ }37\text{ }s\]

The x-t graph of the drunkard’s motion can be shown below:

5. A car moving along a straight highway with speed of \[126\text{ }km/hr\] is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Initial velocity of the car is \[u=126\text{ }km/hr=35m/s\].

Final velocity of the car is $v=0km/hr$.

Distance covered by the car before coming to rest is \[200\text{ }m\].

Consider retardation produced in the car $=a$

From third equation of motion, ${{v}^{2}}-{{u}^{2}}=2as$

Therefore, $-{{35}^{2}}=2a\left( 200 \right)$

$a=-3.0625m/{{s}^{2}}$

From first equation of motion, time (t) taken by the car to stop can be obtained as:

$0=35-\left( 3.065 \right)t$

Hence, the retardation of the car (assumed uniform) is $3.0625m/{{s}^{2}}$, and it takes $11.44s$ for the car to stop.

6. A player throws a ball upwards with an initial speed of \[29.4\text{ }m/s\].

a. What is the direction of acceleration during the upward motion of the ball?

Ans: Acceleration of the ball (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth, irrespective of the direction of the motion of the ball.

b. What are the velocity and acceleration of the ball at the highest point of its motion?

Ans: Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e.,n\[g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\]. At maximum height, velocity of the ball becomes zero.

c. Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

Ans: The sign of position is positive, sign of velocity is negative, and sign of acceleration is positive during upward motion. During downward motion, the signs of position, velocity, and acceleration are all positive.

d. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take \[g\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\] and neglect air resistance).

Ans: Initial velocity of the ball, \[u\text{ }=\text{ }29.4\text{ }m/s\].

Final velocity of the ball, \[v\text{ }=\text{ }0\] (At maximum height, the velocity of the ball becomes zero)

Acceleration, \[a=g\text{ }=9.8\text{ }m/{{s}^{2}}\]

From first equation of motion,

$0=-29.4+\left( 9.8 \right)t$

$t=\frac{29.4}{9.8}=3s$

\[Time\text{ }of\text{ }ascent\text{ }=\text{ }Time\text{ }of\text{ }descent\]

Hence, the total time taken by the ball to return to the player’s hands\[=\text{ }3\text{ }+\text{ }3\text{ }=\text{ }6\text{ }s\].

7. Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

a. with zero speed at an instant may have non-zero acceleration at that instant

Ans: The above statement is true. When an object is thrown vertically up in the air, its speed becomes zero at maximum height. It has acceleration equal to the acceleration due to gravity (g) Which acts in the downward direction at that point.

b. with zero speed may have non-zero velocity,

Ans: The above statement is false as speed is the magnitude of velocity. If speed is zero, the magnitude of velocity along with the velocity is zero.

c. with constant speed must have zero acceleration,

Ans: The above statement is true. If a car is moving on a straight highway with constant speed, it will have constant velocity. Acceleration is defined as the rate of change of velocity. Hence, the acceleration of the car is also zero.

d. with a positive value of acceleration must be speeding up.

Ans: The above statement is false. If acceleration is positive and velocity is negative at the instant time is taken as origin. Thus, for all the time before velocity becomes zero, there is slowing down of the particle. This case occurs when a particle is projected upwards. This statement will be true when both velocity and acceleration are positive, at that instant time taken as origin. This case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

8. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between \[\mathbf{t}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{to}\text{ }\mathbf{12}\text{ }\mathbf{s}\].

Ball is dropped from a height is \[s\text{ }=\text{ }90\text{ }m\].

Initial velocity of the ball is \[u\text{ }=\text{ }0\].

Acceleration is $a=g=9.8m/{{s}^{2}}$.

Final velocity of the ball to be v.

Using second equation of motion, time (t) taken by the ball to hit the ground can be obtained

$s=ut+\left( 1/2 \right)a{{t}^{2}}$

$90=0+\left( 1/2 \right)9.8{{t}^{2}}$

$t=\sqrt{18.38}=4.29s$

Using, first equation of motion, final velocity is given as:

$v=0+9.8\left( 4.29 \right)=42.04m/s$

Rebound velocity of the ball is calculated as:

${{u}_{r}}=9v/10=9\left( \frac{42.04}{10} \right)=37.84m/s$

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

$v={{u}_{r}}+at'$

$0=37.84+\left( -9.8 \right)t'$

$t'=37.84/9.8=3.86s$

Total time taken by the ball will be $t+t'=4.29+3.86=8.15s$.

As the time of ascent is equal to the time of descent, the ball takes \[3.86\text{ }s\] to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor will be ${{u}_{r}}=9v/10=9\left( \frac{37.84}{10} \right)=34.05m/s$

Total time taken by the ball for second rebound will be $t+t'=8.15+3.86=12.01s$

The speed-time graph of the ball is represented in the given figure as:

9. Explain clearly, with examples, the distinction between:

a. magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

Ans: The shortest distance (which is a straight line) between the initial and final positions of the particle gives the magnitude of displacement over an interval of time. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle is AC.

the magnitude of displacement of the particle is AC

Whereas, total path length \[=\text{ }AB\text{ }+\text{ }BC\]

We know that the magnitude of displacement can never be greater than the total path length. But, in some cases, both quantities are equal to each other.

b. magnitude of average velocity over an interval of time, and the average speed over the same interval. Average speed of a particle over an interval of time is defined as the total path length divided by the time interval. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? (For simplicity, consider one-dimensional motion only).

Ans: We know that:

\[Magnitude\text{ }of\text{ }average\text{ }velocity\text{ }=\text{ }Magnitude\text{ }of\text{ }displacement\text{ }/\text{ }Time\text{ }interval\]

Hence, for the given particle,

Average velocity $=AC/t$

\[Average\text{ }speed\text{ }=\text{ }Total\text{ }path\text{ }length\text{ }/\text{ }Time\text{ }interval\]

\[=\left( AB+BC \right)/t\]

Since, \[AB+BC>AC\], average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

10. A man walks on a straight road from his home to a market 2.5 km away with a speed of \[5\text{ }km/hr\]. Finding the market closed, he instantly turns and walks back home with a speed of \[7.5\text{ }km/hr\].What is the

(a) magnitude of average velocity, and

(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? (Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero)

Time taken by the man to reach the market from home is ${{t}_{1}}=2.5/5=1/2hr=30\min $.

Time taken by the man to reach home from the market is ${{t}_{2}}=2.5/7.5=1/3hr=20\min $.

Total time taken in the whole journey \[=\text{ }30\text{ }+\text{ }20\text{ }=\text{ }50\text{ }min\].

0 to 30 min

\[Average\text{ }velocity\text{ }=\text{ }Displacement/Time\]

\[Average\text{ }speed\text{ }=\text{ }Distance/Time\]

0 to 50 min

\[Time\text{ }=\text{ }50\text{ }min\text{ }=\text{ }50/60\text{ }=\text{ }5/6\text{ }h\]

\[Net\text{ }displacement\text{ }=\text{ }0\]

\[Total\text{ }distance\text{ }=\text{ }2.5\text{ }+\text{ }2.5\text{ }=\text{ }5\text{ }km\]

\[Average\text{ }velocity\text{ }=\text{ }Displacement\text{ }/\text{ }Time\text{ }=\text{ }0\]

\[Average\text{ }speed\text{ }=\text{ }Distance\text{ }/\text{ }Time\text{ }=\text{ }5/\left( 5/6 \right)\text{ }=\text{ }6\text{ }km/h\]

0 to 40 min

\[Speed\text{ }of\text{ }the\text{ }man\text{ }=\text{ }7.5\text{ }km/h\]

\[Distance\text{ }travelled\text{ }in\text{ }first\text{ }30\text{ }min\text{ }=\text{ }2.5\text{ }km\]

Distance travelled by the man (from market to home) in the next 10 min

\[=\text{ }7.5\text{ }\times \text{ }10/60\text{ }=\text{ }1.25\text{ }km\]

\[Net\text{ }displacement\text{ }=\text{ }2.5\text{ }\text{ }1.25\text{ }=\text{ }1.25\text{ }km\]

\[Total\text{ }distance\text{ }travelled\text{ }=\text{ }2.5\text{ }+\text{ }1.25\text{ }=\text{ }3.75\text{ }km\]

\[Average\text{ }velocity\text{ }=\text{ }Displacement\text{ }/\text{ }Time\text{ }=\text{ }1.25\text{ }/\text{ }\left( 40/60 \right)\text{ }=\text{ }1.875\text{ }km/h\]

\[Average\text{ }speed\text{ }=\text{ }Distance\text{ }/\text{ }Time\text{ }=\text{ }3.75\text{ }/\text{ }\left( 40/60 \right)\text{ }=\text{ }5.625\text{ }km/h\].

11. In Exercises 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Ans: We know that instantaneous velocity is the first derivative of distance with respect to time.

Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. Therefore, both the total path length and magnitude of displacement become equal in this interval of time. Thus, instantaneous speed is always equal to instantaneous velocity.

12. Look at the graphs (a) to (d) (figure) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

a. Consider the x-t graph, given in fig (a). It does not represent the one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.

b. Consider the x-t graph, given in fig (b). It does not represent the one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.

c. Consider the x-t graph, given in fig (c). It does not represent the one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.

d. Consider the x-t graph, given in fig (d). It does not represent the one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

13. Figure shows the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for $t<0$ and on a parabolic path for $t>0$? If not, suggest a suitable physical context for this graph.

Ans: No, this is because the x-t graph does not represent the trajectory of the path followed by a particle. Also from the graph, it is clear that at \[t=0,\text{ }x=0\].

14. A police van moving on a highway with a speed of \[30\text{ }km/hr\] fires a bullet at a thief’s car speeding away in the same direction with a speed of \[192\text{ }km/hr\]. If the muzzle speed of the bullet is \[150\text{ }m/s\], with what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).

Speed of the police van is \[{{v}_{p}}=30\text{ }km/hr=8.33m/s\].

Muzzle speed of the bullet is \[{{v}_{b}}=150\text{ }m/s\].

Speed of the thief’s car is \[{{v}_{t}}=192\text{ }km/hr=53.33m/s\].

As the bullet is fired from a moving van, its resultant speed will be: $150+8.33=158.33m/s$.

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:

${{v}_{bt}}={{v}_{b}}-{{v}_{t}}=158.33-53.33=105m/s$

15. Suggest a suitable physical situation for each of the following graphs (figure):

its velocity increases with time and attains an instantaneous constant value

Ans: Consider fig 3.22 given in the question:

a. From the x-t graph given it is clear that initially a body was at rest. Further, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Further, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately stops after sometime.

b. From the given v-t graph it is clear that the sign of velocity changes and its magnitude decreases with a passage of time. This type of situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.

c. From the given a-t graph it is clear that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This shows that the body again starts moving with the same constant velocity. This type of physical situation arises when a hammer moving with a uniform velocity strikes a nail.

16. Figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion.

(You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at $t=0.3s,1.2s,-1.2s$.

Ans: Negative, Negative, Positive

Positive, Positive, Negative

Negative, Positive, Positive

When a particle executes simple harmonic motion (SHM), acceleration (a) is given by the relation: $a=-{{\omega }^{2}}x$

$\omega $ is the angular frequency ...

i. \[~t\text{ }=\text{ }0.3\text{ }s\]

For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.

ii. \[~t\text{ }=1.2s\]

For this time interval, x is positive. Hence, the slope of the x-t plot will be positive. Thus, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.

iii. \[\text{t }=-1.2s\]

For this time interval, x is negative. Hence, the slope of the x-t plot will be negative. Thus, both x and t are negative, the velocity comes to be positive. From equation (i), it can be interpreted that the acceleration of the particle will be positive.

17. Figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.

Give the sign of average velocity for each interval

Ans: The average speed is greatest in interval 3 and least in interval 2. It is positive in intervals 1 & 2 and negative in interval 3.

The average speed of a particle shown in the x-t graph is given by the slope of the graph in a particular interval of time.

From the graph it is clear that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Thus, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

18. Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

the accelerations at the points A, B, C and D?

Ans: From the graph given in the question,

Average acceleration is greatest in interval \[2\].

Average speed is greatest in intervals of $3$.

v is positive for intervals \[1,\text{ }2\], and $3$.

a is positive for intervals $1$ and $3$ and negative in interval $2$

\[a\text{ }=\text{ }0\] at A, B, C, D

Acceleration is calculated as the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.

As the slope of the given speed-time graph is maximum in interval \[2\], average acceleration will be the greatest in this interval.

From the time-axis, height of the curve gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Thus, average speed of the particle is the greatest in the interval 3.

For interval 1:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2:

As slope of the speed-time graph is negative, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3:

As the slope of the speed-time graph is zero, acceleration is zero in this interval.

However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time-axis. Thus, the slope is zero at these points.

Therefore, at points A, B, C, and D, acceleration of the particle is zero.

Motion in a Straight Line Chapter Summary - Class 11 NCERT Solutions

Path length is defined as the total length of the path traversed by an object.

Displacement is the change in position: $\Delta x=x_{2}-x_{1} $ Path length is greater or equal to the magnitude of the displacement between the same points.

An object is said to be in uniform motion in a straight line if its displacement is equal in equal intervals of time. Otherwise, the motion is said to be nonuniform

Average velocity is the displacement divided by the time Interval in which the displacement occurs: $ v\bar{}=\frac{\Delta x}{\Delta t}

On an x-t graph. The average velocity over a time Interval is the slope of the line connecting the initial and final positions corresponding to that interval.

Average Speed is the ratio of total path length traversed and the corresponding time Interval.

Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval $\Delta t$ becomes infinitesimally small $v=\lim_{\Delta t\rightarrow 0}v\bar{}=\lim_{\Delta t\rightarrow 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}$

The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant.

Average acceleration is the change in velocity divided by the time interval during which the change occurs: $a\vec{}=\frac{\Delta v}{\Delta t}$

Instantaneous acceleration is defined as the limit of the average acceleration as the time interval at goes to zero: $a=\lim_{\Delta \rightarrow t}a\vec{}=\lim_{\Delta t\rightarrow 0}\frac{\Delta t}{\Delta v}=\frac{dv}{dt}$

The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion. acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line parallel to the time axis. For motion with uniform acceleration. x - t graph is a parabola while the v-t graph is a straight line inclined to the time axis.

The area under the velocity-time curve between times t 1 and t 2 is equal to the displacement of the object during that interval of time.

For objects in uniformly accelerated rectilinear motion. the five quantities, displacement x, time taken t. Initial velocity v 0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion: v=v 0 +at x=v 0 t+$\frac{1}{2}$at 2 v 2 =$v_{0}^{2}$+2ax

For Uniform Motion

For Non-Uniform Motion

Overview of Deleted Syllabus for CBSE Class 11 Physics Motion in a Straight Line

Chapter	Dropped Topics
Motion in a Straight Line	3.2 Position, Path Length and Displacement
	3.3 Average Velocity and Average Speed
	3.7 Relative Velocity
	Exercises 3.5, 3.7–3.9 and 3.23–3.28
	Appendix 3.1

NCERT Class 11 Physics Chapter 2 Exercise Solutions on Motion In A Straight Line provided by Vedantu are comprehensive, and helps students understand motion and the different kinds of motion. This chapter explores different types of motion, including uniform motion, non-uniform motion, and uniformly accelerated motion. In Ch 2 physics class 11 ncert solutions we learn how to solve problems involving these concepts using various equations relating to distance, time, speed, velocity, and acceleration. These concepts are essential for mastering the topic and are often tested in exams. From previous year's question papers, typically around 2–3 questions are asked from this chapter. These questions test students' grasp of theoretical concepts as well as their problem-solving skills.

Other Study Material for CBSE Class 11 Physics Chapter 2

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Chapter-Specific NCERT Solutions for Class 11 Physics

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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FAQs on NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line

1. If I'm not a Non-Medical Student, do I Still Need to Learn and Take Physics as One of the Subjects?

The answer to this question is both yes and no, you can opt for Physics even if you are not a non-medical student, but the only other branch which offers Physics as one of its subjects is medical. In that branch, you don't have maths as the main subject, but you can opt for maths as well. On the other hand, if you are from commerce or arts, you don't choose Physics. You can't opt for Physics as it's one of the core subjects and isn't easy. In the medical stream, you can opt for Physics.

2. If an Object has One Dimension Motion with a Positive Acceleration Value and Negative Velocity, is the Object Speeding up?

No, in this case, the object present in the starting point is not speeding up. As the velocity is negative, we can see that the particle is slowing down. This type of scenario happens only when the object is thrown upwards in the air. We can say the object is in acceleration if both the acceleration and the object's velocity have a positive value, or the object is falling from a height.

3. What are the contents of Chapter 3 of the Class 11 Physics textbook?

The theme of the chapter is Kinematics, motion in a straight line. Mechanics is a scientific term that refers to the study of the movement of physical objects. The chapter explains all the relevant terms in a simple and easy-to-understand language. The chapter distinguishes between rest and motion. You will also come across the terms position, distance, and displacement as well as the difference between speed and velocity. The chapter explains acceleration. The chapter also contains the units and mathematical representations.

4. What is the difference between distance and displacement according to Chapter 3 of the Class 11 Physics textbook?

Distance means the actual path traversed by the object whereas displacement refers to the difference between initial and final positions. The former is the scalar quantity while the latter is the vector quantity. The distance covered by an object is always positive; it can never be negative or zero. However, the displacement of an object can be either positive, negative, or even zero. The distance is dependent on the path traversed by the object but displacement is not dependent on the path.

5. What is the difference between speed and velocity Chapter 3 of the Class 11 Physics textbook?

Speed is the total path length covered by the object divided by the total time taken. Velocity is the change in position divided by time intervals. Speed is a scalar quantity. Velocity, however, is a vector quantity. Speed is always positive. Velocity, on the other hand, can be positive, negative, or zero. Speed is the time rate at which an object moves; speed is about how fast an object moves. Velocity is the rate and direction of the movement of an object.

6. What are the types of motion mentioned in Chapter 3 of the Class 11 Physics textbook?

The motion refers to changes in the position or orientation of an object with the change in time. Motion can be of two types- Uniform motion or Non-Uniform motion. Uniform motion exists when the object completes equal distance in equal time intervals. Non-Uniform motion, on the other hand, exists when an object completes unequal distance in equal intervals of time. During a uniform motion, the velocity of the object remains the same. When an object undergoes non-uniform motion, the magnitude of the velocity increases or decreases with time.

7. How can previous years’ questions help in preparation of Motion in a straight line chapter of Class 11 Physics?

Previous years’ questions enable you to understand the pattern of the question and analyse the trend of the questions asked. These papers will give you an idea regarding the topics that you should invest more time and effort on. These question papers should also motivate you to make notes accordingly. The previous years’ question papers will lend a direction to your preparation and you will feel more prepared and confident on the day of the exam. These papers are available on the official website of Vedantu or the Vedantu app. All the resources are free of cost.

8. In Motion In A Straight Line Class 11 NCERT Solutions, which motion travels in a straight line?

In Class 11 Physics Chapter 2, Motion In A Straight Line refers to the movement of an object along a straight path. This type of motion is characterized by having only one spatial dimension. Examples include a car driving on a straight road, a ball rolling down a straight hill, or an object falling freely under gravity (neglecting air resistance).

9. What are the two motions of a straight line mentioned in Physics Class 11 Chapter 2?

The two types of motion in a straight line are:

Uniform Motion : When an object moves with a constant speed in a straight line, covering equal distances in equal intervals of time.

Non-uniform Motion : When an object's speed varies as it moves along a straight line, covering unequal distances in equal intervals of time.

10. What are the important things in Motion In A Straight Line Class 11 Questions?

Important aspects of motion in a straight line include:

Displacement : The change in position of an object, considering direction.

Velocity : The rate of change of displacement with time.

Speed : The rate at which an object covers distance.

Acceleration : The rate of change of velocity with time.

Kinematic Equations : These equations relate displacement, velocity, acceleration, and time.

11. In Class 11 Motion In A Straight Line NCERT Solutions, Is motion always in a straight line?

No, motion is not always in a straight line. While this chapter focuses on straight-line motion, objects can move in various types of paths, such as circular motion, projectile motion, or any arbitrary path in two or three dimensions.

12. What is another name for a straight line motion in Motion In Straight Line Class 11 NCERT Solutions?

Another name for straight-line motion is rectilinear motion. This term is often used in physics to describe motion that occurs along a straight path.

13. What is the other name of motion in a straight line in Class 11 Physics Chapter 2 Exercise Solutions?

The other name for motion in a straight line is linear motion. This term highlights the one-dimensional aspect of the motion, indicating that it occurs along a single straight axis.

NCERT Solutions for Class 11 Physics

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

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NCERT Solutions Class 11 Physics Chapter 2 Updated for 2024-25

Subject specialists have created NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements, according to the updated CBSE syllabus which includes thorough solutions for reference. All of the questions from the NCERT textbook’s exercises are answered here. Students can use these answers to help them prepare for their exams. The NCERT Solutions for Class 11 provide useful solutions for improving conceptual knowledge.

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The solutions are carefully solved using student-friendly terms while still adhering to the norms that must be followed when solving NCERT Solutions for Class 11. Practicing these answers can be incredibly advantageous not only in terms of exams but also in terms of helping Class 11 pupils perform well in upcoming competitive exams.

The approaches for answering have been given special consideration to stay on target while not deviating from the intended answer. Because time is so important in exams, excellent time management when answering questions is essential for getting the best results.

Units and Measurements Class 11 Physics Questions and Answers PDF

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It is critical for students in Class 11 who want to do their best in the next term – I exams and competitive exams to become familiar with the textbook solutions to the questions. As a result, students should practice numerous types of questions that can be framed from the chapter. It is recommended that students answer the NCERT questions. Our specialists provide NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements to clear all of the students’ worries.

NCERT Solutions for Class 11 Physics Chapter 2 Overview

Chapter 2 of the NCERT Class 11 Physics textbook, titled “Units and Measurements,” delves into the fundamental concepts essential for scientific study. This chapter covers the importance of units, the International System of Units (SI), and the methods of measurement for various physical quantities. It also discusses significant figures, dimensional analysis, and error analysis, providing students with the tools needed for precise and accurate measurements in physics experiments. Understanding these concepts is crucial for grasping more complex topics in physics, making this chapter a foundational element of the curriculum.

Important Topics Included in Class 11 Physics Chapter 2 Units and Measurements

Class 11 Physics Chapter 2, “Units and Measurements,” covers several important topics that form the foundation for understanding measurement in physics. Here are the key topics included in this chapter:

Introduction : Overview of the importance of measurement in physics.
The International System of Units (SI) : Detailed discussion on the SI units, which are the standard units used globally for scientific measurements.
Measurement of Length : Techniques and instruments used to measure length accurately.
Measurement of Mass : Methods for measuring mass, including the use of balances.
Measurement of Time : Various ways to measure time, including mechanical and electronic devices.
Accuracy, Precision of Instruments, and Errors in Measurement : Understanding the difference between accuracy and precision, types of errors (systematic and random), and how to minimize them.
Significant Figures : Rules for determining significant figures in measurements and their importance in calculations.
Dimensions of Physical Quantities : Explanation of how physical quantities can be expressed in terms of fundamental dimensions.
Dimensional Formulae and Dimensional Equations : How to derive the dimensional formula for various physical quantities.
Dimensional Analysis and Its Applications : Applications of dimensional analysis in checking the correctness of equations, converting units, and deriving relationships between different physical quantities.

Scientists use their senses, such as their eyes and ears, to gather data and make observations. Some observations are straightforward, such as determining the texture and colour, while others are more complex, necessitating the use of measurements. One of the most fundamental notions in science is measurement. A scientist would be unable to gather data, construct a theory, or conduct experiments without the ability to measure. The units of physical quantities and methods for evaluating them are described in this chapter, whereas the other portion of the chapter deals with measurement mistakes and significant figures. One can gain a strong grasp of measurement by studying problems from Physics NCERT Solutions for Class 11.

Our specialists also provide NCERT Solutions for all subjects in all classes, in addition to Chapter 2. It also includes notes, study materials, numerical problems, previous year question papers, sample papers, and competitive exam preparation tools to help you do well in the first and second terms of Class 11.

Class 11 Physics Chapter 2 Units and Measurements Overview

The overview of Class 11 Physics Chapter 2, “Units and Measurements,” encompasses several fundamental concepts essential for understanding physical quantities and their measurement.

Measurement

Measurement is fundamentally a comparison process where a physical quantity is assessed against a standard unit. The result of a measurement is expressed as a product of a numerical value and a unit, for example, a length of 8 meters indicates that the rod’s length is 8 times the standard meter.

Units Definition and Properties

A unit is a standard reference used to quantify physical quantities. Key properties of a unit include:

Well-defined : The unit must have a clear and precise definition.
Reproducibility : It should be consistent across different locations and conditions.
Stability : The unit should remain unchanged over time and under varying physical conditions.

Types of Units

Fundamental Units : These are the basic units defined for fundamental physical quantities, such as length, mass, and time.
Derived Units : These units are derived from fundamental units. For example, velocity is expressed in meters per second (m/s), and force is measured in newtons (N), which is equivalent to kg·m/s².

Systems of Units

Several systems of units are used globally:

CGS System : Centimeter (cm), gram (g), and second (s).
FPS System : Foot (ft), pound (lb), and second (s).
MKS System : Meter (m), kilogram (kg), and second (s).
Length: meter (m)
Mass: kilogram (kg)
Time: second (s)
Temperature: kelvin (K)
Electric current: ampere (A)
Amount of substance: mole (mol)
Luminous intensity: candela (cd)

Dimensions of Physical Quantities

Each physical quantity can be expressed in terms of its dimensions. For example:

Displacement: m m (M^0L^1T^0)
Velocity: ms − 1 ms−1 (M^0L^1T^{-1})
Force: N N (M^1L^1T^{-2}).

Accuracy and Precision

The chapter also discusses the accuracy and precision of measurements, highlighting the importance of significant figures and the potential errors that can occur during the measurement process. Understanding these concepts is crucial for conducting experiments and interpreting results accurately.

FAQs on NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

What is a unit called.

A unit is a standardized quantity used as a basis for measurement. It is the particular magnitude of a quantity that is used as a standard or reference point.

What is the SI unit of length?

The SI unit of length is the meter (m). The meter is defined as the distance traveled by light in a vacuum in 1/299,792,458 of a second.

What is the SI base unit?

The SI base units are the seven fundamental units defined by the International System of Units (SI). They are: Length: meter (m) Mass: kilogram (kg) Time: second (s) Electric current: ampere (A) Temperature: kelvin (K) Amount of substance: mole (mol) Luminous intensity: candela (cd)

What are the 7 main units of measurement?

The seven main units of measurement are: Length (meter) Mass (kilogram) Time (second) Temperature (kelvin) Amount of substance (mole) Electric current (ampere) Luminous intensity (candela)

What are units and measurement?

Units and measurement refer to the standardized quantities used to quantify and compare physical quantities. A unit is the particular magnitude of a quantity that serves as a standard of measurement. Measurement is the process of quantifying a physical quantity using a unit.

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1st Year Physics Chapter 2 Short Questions Notes PDF

In this post, I am sharing FSC 1st Class Physics Chapter 2 Short Questions with Answers Notes PDF for the students of Intermediate Part 1. This chapter’s name is Vectors and Equilibrium. From this post of ilmihub , students can download the 11th class chapter 2 Vector and Equilibrium Short Questions Notes in PDF format on their laptops or mobiles. These Class 11 Physics Chapter 2 Short Questions with Answers Notes are for all the boards working under the Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Visit this link if you want to download the complete 1st Year Physics Notes in PDF format.

11th Class Physics Chapter 2 Vectors and Equilibrium Short Questions Notes PDF Download

Class 11 physics chapter 2 short questions with answers, can you add a zero to a null vector.

As zero is a scalar and null is a vector. So, they can never be added because physical quantities of different nature cannot be added.

Two vectors have unequal magnitudes. Can their sum be zero?

No, their sum cannot be zero. Their sum will only be zero if both the vectors have the same magnitudes but opposite directions.

A picture is suspended from a wall by two strings. Show by diagram the configuration of the strings for which the tension in the strings will be minimal.

If ‘W’ is the weight of the picture and “T” is the tension in the strings making angle “q” horizontal as the picture is in the equilibrium, then the relation for the tension in the string will be

T sinq + Tsinq = W

2Tsinq = W T = W/2sinq

For minimum tension q = 90 0

T = W/2sin90 = 1/2W

Thus the strings should be vertical to be under minimum tension. The configuration is shown in the figure.

Can a body rotate about its centre of gravity under the action of its weight?

The torque produced in a body given by

i = r x F i = rFsinq

As the weight acts at the centre of gravity. So, the moment arm is zero, i = (0)Fsinq

i = 0

Thus a body cannot rotate about its center of gravity under the action of its weight.

Define equal vector and position vector.

Equal vector:.

Two vectors are said to be equal if they have the same magnitudes and directions, regardless of the position of their initial points. This means that parallel vectors of the same magnitude are equal.

Position vector:

The vector which indicates the position of a body with respect to the origin is called the position vector.

1st Year Physics MCQs with Answers All Chapters

1st year physics short questions, 11th physics short questions notes

1st Year Physics Short Questions Notes of All Chapters PDF

FSC 1st Year Physics Chapter 11 Solved MCQs Notes PDF

FSC 1st Year Physics Chapter 11 MCQs Notes PDF

NCERT Solutions for Class 11 Physics Chapter 2 Free PDF Download

Ncert solutions for class 11 physics chapter 2 – units and measurement.

Class 11 physics is pretty tough and difficult to understand by students. Moreover, there are certain topics that completely bounce off over students head. So to help them know the topic we have prepared NCERT Solutions for Class 11 Physics Chapter 2 that will help them with the topic in which they face difficulty.

Besides, these NCERT Solutions are prepared by our panel of professionals who have developed it after thorough research. In addition, they are in easy language so that students can easily get their meaning.

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CBSE Class 11 Physics Chapter 15 Units and Measurement NCERT Solutions

Unit refers to the international standards that are accepted worldwide. While on the other hand, measurement refers to the use of the unit for measuring the quantity. In this NCERT Solutions for Class 11 Physics Chapter 2, we will discuss all the topic of the chapter in brief and in an easily understandable language so that students can get what the content of the chapter.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 15

2.1 introduction.

This topic discusses what is unit and measurement which we have discussed above.

2.2 The International System of Units

This topic defines that earlier scientist of different countries uses the different unit for measurement. But with time and international usage, they started using SI (Standard Unit).

2.3 Measurement of Length

This topic overview the different unit of measuring the length that is used worldwide.

2.3.1 Measurement of Large Distances- This topic defines the use of parallax method for measuring long distances. Besides, the topic provides various examples of using large distances.

2.3.2 Estimation of very Small Distances: Size of a Molecule- This topic defines the use of the small unit by which we can measure the distance of the molecule.

2.3.3 Range of Lengths- this topic defines the use of such unit that helps to measure the lengths of objects that are spread in this wide universe.

2.4 Measurement of Mass

The weight of any object is its mass and this topic discusses the various unit of measurement of mass.

2.4.1 Range of Masses- It refers to all those objects that are spread over the universe that has a fixed mass and can be measured using these units.

2.5 Measurement of Time

This topic defines the atomic standard of time that is used for measuring the atomic clock and cesium clock.

2.6 Accuracy, Precision of Instruments and Errors in Measurement

This topic firstly defines what are errors and the mistakes we do which causes the errors.

Systematic Errors
Instrumental errors
Imperfection in experimental technique or procedure
Personal errors
Random Errors
Least Count Errors

2.6.1 Absolute Error, Relative Error, and Percentage Error- This topic defines all the three errors and how they affect the results.

2.6.2 Combination of Errors- This topic defines the error that we do while performing several measurements

An error of a sum or a difference
The error of a product or a quotient
Error in case of a measured quantity raised to a power

2.7 Significant Figures

This refers to the first uncertain digit plus the reliable digit. Moreover, this topic several other points related to significant figures.

2.7.1 Rules of Arithmetic Operations with Significant Figures- The topic define the rules related to the arithmetic operations.

2.7.2 Rounding off the Uncertain Digits- This topic defines the rule of rounding off of unclear digits.

2.7.3 Rules of Determining the Uncertainty in the Results of Arithmetic Calculations- This topic defines the rules that help to govern the doubt of the results of mathematics calculations.

2.8 Dimensions of Physical Quantities

This topic defines the scope of physical quantities.

2.9 Dimensional Formulae and Dimensional Equations

This topic describes the dime national equations and formula that we use to equitize a physical quantity.

2.10 Dimensional Analysis and its Applications

This topic defines the various applications by which we can analyze the physical quantities of a dimension.

2.10.1 Checking the Dimensional Consistency of Equations- This topic checks the consistency of dimensional equations.

2.10.2 Deducing Relation among the physical Quantities- This topic defines how we can use the relation of the physical quantities for reasoning.

You can download NCERT Solutions for Class 11 Physics Chapter 2 PDF by clicking on the button below.

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NCERT Solutions for Class 11

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RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

June 18, 2021 by Veerendra

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement 1

Question 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. Answer:

Atoms are very small objects when compared to a cricket ball.
A jet plane moves with great speed when compared to a car.
The mass of Jupiter is much larger than that of Earth.
The air inside this room contains a large number of molecules when compared to the number of objects in the room.
No change necessary.

Question 5. A new unit of length is chosen such that the speed of light in a 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 s to cover this distance? Answer: New unit of length = 3 × 10 8 ms -1 Distance between the Earth and the sun = $(8 \min 20 \mathrm{s}) \times 3 \times 10^{8} \mathrm{ms}^{-1}$ = 500 × 3 × 10 8 ms -1 ∴Distance between the Earth and the Sun in terms of the new units =$\frac{500 \times 3 \times 10^{8}}{3 \times 10^{8}}$ = 500 new units.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement 4

Question 10. State the number of significant figures in the following:

2.64 x 10 4 kg
0.2370 gem -3
6.032 N nr -2
0.0006032 m 2

Question 11. The length, breadth, and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. Answer: Length l = 4.234 m, Breadth b = 1.005 m, Thickness t = 2.01 cm = 0.0201 m Area = 2 × (lb + bt + It) = 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 4.234 × 0.0201) = 8.72 m 2 (Rounding off to 3 significant figures) Volume = lbt = 4.234 × 1.005 × 0.0201 = 8.55 × 10 -2 m 3

Question 12. The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box (b) the difference in the masses of the pieces to correct significant figures ? Answer: (a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg Since the least number of significant figure is 2, therefore, the total mass of the box = 2.3 kg. (b) Difference of mass = 2.17 – 2.15 = 0.020 g Since there are two significant figures so the difference in masses to the correct significant figures is 0.020 g.

Question 16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A : 1 A = 10 – 10 m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m 3 of a mole of hydrogen atoms? Answer: Volume of one hydrogen atom = $ \cfrac {4}{ 3 } $ $ \cfrac {4}{ 3 } $ X 3.14 x (0.5 x l(T -10 )m 3 = 5-23 x 10 -31 m 3 . According to Avogadro’s hypothesis, one mole of hydrogen contains 6 023 x 10 23 atoms. ∴ The atomic volume of 1 mole of hydrogen atoms = 6.023 x 10 23 x 5.23 x 10 – 31 = 3.15 x 10 – 7 m 3 .

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement 15

This high ratio is because of intermolecular spaces in gas being much larger than the size of the molecules.

Question 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 train’s motion, but the distant objects (hilltops, the Moon, the stars, etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you). Answer: Objects nearer to the eye subtend a greater angle in the eye than distant objects. When we move, the change in this angle is less for distant objects than for near objects. So the distant objects seem stationary but nearer objects seem to move in the opposite direction.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement 16

Length: In, sports, sending satellites
Mass: To add proper proportions of different salts for preparing medicines, the mass of the satellite should be accurately measured.
Time: For the study of various chemical reactions, the study of various activities in the universe.

Question 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 (d) The number of strands of hair on your head. (e) The number of air molecules in your classroom. Answer: (a) Firstly to calculate the total rain in India, we can get an estimate of it and then knowing the weight of water we can estimate the weight of clouds. (b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x 1 Therefore, volume of water displaced by the boat, V 1 – AX 1 Move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into the water. Let it be x 2 . ∴ Volume of water displaced by boat and elephant V 2 = Ax 2 ∴ Volume of water displaced by the elephant V = V 2 – V 1 = A(x 2 – X 1 ) If p is the density of water, then the mass of elephant = mass of water displaced by it. = V ρ = A(x 2 – x 1 )ρ (c) The pressure generated by the wind can give us an estimation of its speed. (d) Estimating no. of hairs per cm 2 area of the head, we can estimate the total no. of hairs on our head, ( ∴ we can estimate the area of our head) (e) We can get the density of air and hence an estimation of no. of molecules in 1 cm 3 can be made by which we can estimate no. of molecules in our room.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement 18

Question 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. Answer: Given angular diameter θ = 35.72 = 35.72 x 4.85 x 10 -6 rad = 173.242 x 10 -6 = 1.73 x 10 -6 rad ∴Diameter of Jupiter, D = θ x d = 1.73 x 10 -4 x 824.7 x 10 9 m = 1426.731 x 10 5 = 1.43 x 10 8 m

Question 25. A man walking briskly in rain with speed v> must slant his umbrella forward making an angle with the vertical. A student derives the following relations between θ and v: tan θ = υ and checks that the relation has a correct limit: as υ→0, θ→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. Answer: According to the principle of homogeneity of dimensional equations, Dimensions of L.H.S. = Dimensions of R.H.S. Here υ = tan θ i.e. [L 1 T -1 ] = dimensionless, which is incorrect. Correcting die L.H.S, we get υ/u = tan θ, where u is the velocity of rain.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement 19

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102 comments:

Very excillent information thanks brother

Assalamu alaikum

When will unit 10 full syllabus will be ready

How to get full note in hard form

sir can you tell me the password that you given to pdf files for locking the editing

Important long question for 2024

So you not solve the MCQs

stop complaining

Its solved sister

Tusa saryan ni mao ch lan rakhan

hey are you mad ??????? STOP CURSING MAN

why you lots abusing bruvvv grow the hell up average 17 year olds cant even respect anyone these days.

Please upload HSSC 1 physics chapters full notes from chapter 4 to 10 ... Plz

Yes upload full sallybus

Yes😩😩we can't survive

Sir, Can u upload ch 4 to ch 10 full syllabus.

it is a good job from your side sir please upload the full sylabu.

please update the chapters

Numercels in short form pkz upload

please upload full syllabus

Please upload complete notes please

This notes are not show in pdf it want some password to your notes why?

Sir plz upload full syllabus conceptual as soon as possible

Sir plz upload full syllabus conceptual questions as soon as possible

Excellent notes Sir it's a humble request that our exams will held in june so plz fastly upload notes of all subjects (full syllabus)

Fastly isn't a word Einstein .....

Please upload full syllabus notes

Are these notes for smart syllabus or Full syllabus?

PLZ UPLOAD FULL NOTES PLZ PLZ PLZ PLZ PLZ

kindly upload full syllabus

Plzz upload important topics for 11 class physics

sir upload full syllabus notes

sir upload 11th physics important topics fbise

sir plzzzz upload chap 4 😭😭😭😭

chap 4 😭😭😭😭😭

Please upload full notes

Sir plzzz chapter 1and 2 ka notes full syllabus k according h

Complete 1st year course sir notes and lectures

Plz upload solutions of mcqs

Impressive Note sir plz plz upload the other chapter. Waiting 😶😶

sir plzz no response no result plzz sir give me a important topics and important question

khud nots bana lo bachy mery pass time nahi hy

beta sab yaad karo

Please upload full notes plz

sir plz upload full syllabus notes of class 11

Comprehensive question nai ha in notes ma

Sir u r notes r outstanding in how much time more notes will be prepare

sir I m a student of KPK board I am preparation for MDCAT TEST in these notes .are you make these notes from fadral board?

Bacha mera aap is english ky sath exam kesy pass karo gy , yes federal board is consistent with it

English doesn't matters

Plz upload full syllabus plzzzz

sir try to upload full syllabus notes soon extremely waiting for it

SIR NOTES OF COMPREHENSIVE QUESTIONS

Very good notes but not complete please complete the notes

Sir please upload full syllabus notes. Your notes are too good.

Lajawab ha sir

exams are coming sir , it would be alot more easy to prepare if u share the full syllabus. we all would be obliged. thank u

Kindly upload full syllabus notes from chapter 4 to 10

Sir these notes are great and easy to understand kindly upload full syllabus notes of remaining chapters it Wil be a lot easier for us to prepare for tests and exams

AOA sir your notes are exellent and usefull but kindly upload full syllabus

Chapter 10 Conceptual Questions And Chapter 9 Conceptual Questions Are Full Syllabus Not Reduced

please post full notes as soon as possible and DONT forget to include answers of quizes and and pont to ponder boxes ,, Waiting , JazakAllah

sir ap bilkul tension na lain, arram say notes bnayn, mai board ky papers ky baad inshaAllah tafseel say prhun ga

aoa sir please complete fast full syllabus becuase exams is near

Plz make complete notes

Board 2023 se pehle mil sakte hain sare?!!

sir can you please make full notes..it would be extremely helpful..

Pehla pta hota to parh leta ab akhri din muh marna Aya Hun🥷

where are numerical question of c.h.p 4

Notes are very good

Very Good Notes..............

Chapter 4 numerical

Chapter 4 numericals

aoa plz upload the comprehension questions answers of chapter no 1 measurement

numericals of chapter no 4 please

thanks for your notes

Awesome notes

Chapter 4 numericals please

Very informative notes Great

How to download ?

IT WAS A GREAT HELP

Very very very nice Sir MashaAllah .THANKU

Jazak Allah Sir

Where are the MCQs potions?

Great work. No doubt you are serving the nation.

جزاك اللهُ بہت مہربانی

I write your conceptual answers in sendup exams and I got fail😢

Mashallah very helpful in my board exams. May Allah bless you

sir can u mark the imp questions for fbise ,

or list down the important numerical as well as theoretical qs

Excellent solutions for FSC student

How to download these pdf notes plz tell

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NCERT Solutions for Class 11 Physics Chapter 2

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Measurement is a fundamental concept in science. A scientist would be unable to gather information, form a theory, or conduct experiments without the ability to measure. Chapter 2 Physics Class 11 discusses physical quantity units and methods of evaluating them, while the other section of the chapter discusses measurement errors and significant figures.

The chapter covers a variety of physical units that are used quantitatively in Physics. Students will go over the international unit system and how it is used. Measuring lengths over short and long distances is also included in the chapter. Students are also introduced to mass measurement and its range. They will also learn how to measure and calculate significant numbers. The chapter concludes with a discussion of physical quantity dimensions and formulae.

After students have referred to NCERT books for Class 11 Physics, they must solve the questions listed at the end of every chapter. To help students, Extramarks offers NCERT Solutions Class 11 Physics Chapter 2 . The solutions include accurate answers to all the questions that are discussed in the practise exercise of Chapter 2.

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

All of the important topics covered in the chapter on Units and Measurements are discussed and summarised in NCERT Solutions for Class 11 Physics Chapter 2.

Students can look through the Class 11 Physics Chapter 2 NCERT Solutions for a complete list of topics. It is important for the students to fully understand the topics and plan ahead to set up their own schedule. Let’s take a look at the topics and subtopics covered in this chapter before diving into the detailed NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements:


2	Units and Measurements
2.1	Introduction
2.2	The International System of Units
2.3	Measurement of Length
2.3.1	Measurement of Long Distances
2.3.2	Estimation of Very Small Distances: Size of a Molecule
2.3.3	Range of Lengths
2.4	Measurement of Mass
2.4.1	Range of Masses
2.5	Measurement of Time
2.6	Accuracy, Precision of Instruments and Errors in Measurement
2.6.1	Absolute Error, Relative Error, and Percentage Error
2.6..2	Error Combination
2.7	Significant Figures
2.7.1	Rules of Arithmetic Operations with Significant Figures
2.7.2	Rounding off Uncertain Digits
2.7.3	Rules of Determining the Uncertainty in the Results of Arithmetic Calculations
2.8	Dimensions of Physical Quantities
2.9	Dimensional Formulae and Dimensional Equations
2.10	Dimensional Analysis and its Applications
2.10.1	Checking the Dimensional Consistency of Equations
2.10.2	Deducing Relationships Among Physical Quantities

2.1 Introduction

This topic expands the discussion of unit and measurement.

2.2 The International System of Units (SI)

This topic defines how earlier scientists from various countries used different measurement units. However, as time passed and international usage increased, they began to use SI (Standard Unit).

2.3 Length Measurement

This topic provides a description of the different length measurement units used around the world.

2.3.1 Measurement of Long Distances

This topic describes how to use the parallax method to measure long distances. Furthermore, the topic provides a variety of examples of how to use long distances.

2.3.2 Estimation of Very Small Distances: Size of a Molecule

This topic explains how to use a small unit to quantify a molecule’s distance.

2.3.3 Range of Lengths

This topic defines the use of such a unit that aids in the measurement of the lengths of objects spread throughout the universe.

2.4 Mass Measurement

The mass of an object is its weight, and this topic explores the various units of mass measurement.

2.4.1 Range of Masses

This term refers to all objects in the universe that have a fixed mass and can be measured using these units.

2.5 Measurement of Time

This topic discusses the atomic time standard, which is used to calculate the atomic clock and the caesium clock.

2.6 Accuracy, Precision of Instruments and Errors in Measurement

This topic first defines errors and the mistakes we make that cause errors.

Instrumental errors
Imperfection in experimental technique or procedure
Personal errors
Random Errors
Least Count Errors

2.6.1 Absolute Error, Relative Error, and Percentage Error

This section defines all three errors and their effects on the results.

2.6.2 Error Combination

This topic defines the error that we make when performing multiple measurements.

A summing or difference error
A product’s or a quotient’s error
Error when a measured quantity is raised to a power

2.7 Significant Figures

This is the combination of the first uncertain digit and the reliable digit. Furthermore, this topic contains a number of other points concerning significant figures.

2.7.1 Rules of Arithmetic Operations with Significant Figures

The section discusses arithmetic operations rules.

2.7.2 Rounding off Uncertain Digits

This section defines how uncertain digits need to be rounded off.

2.7.3 Rules of Determining the Uncertainty in the Results of Arithmetic Calculations

This topic defines the rules that help to govern the uncertainty of the results of mathematical calculations.

2.8 Physical Quantity Dimensions

The scope of physical quantities is defined in this topic.

2.9 Dimensional Formulae and Dimensional Equations

This topic explains how to equitize a physical quantity using dime national equations and formulas.

2.10 Dimensional Analysis and Its Applications

This topic defines the various applications that can be used to analyse physical dimensions.

2.10.1 Checking the Dimensional Consistency of Equations

This topic evaluates the dimensional consistency of equations.

2.10.2 Deducing Relationships Among Physical Quantities

This topic defines how we can use physical quantity relationships to reason.

As explained in Chapter 2 Physics Class 11 , Physics is a quantitative science based on the measurement of physical quantities. As base or fundamental quantities, certain physical quantities have been chosen (such as the amount of substance, electric current, length, mass, thermodynamic temperature, time, and luminous intensity). Students can access the solutions for this chapter by clicking the download button given below.

Access NCERT Solutions for Physics Chapter 2 – Units and Measurement

The Chapter 2 Physics Class 11 will reinforce your fundamental understanding of unit, base units, derived units, system of units, and other topics. Students will understand the need for an international system of units and how they are decided. The complexities of units and measurement will be thoroughly addressed in this chapter. Units are used to measure various dimensions such as length, width, mass, and so on. Through this chapter, you will learn about the parallax method, the parallax angle, and other methods for measuring length. Alternative methods for measuring length and estimating very small distances, such as the size of a molecule, will also be covered.

In the same way, this chapter will teach you how to measure time, accuracy, and precision of instruments. Various instruments are used to measure various dimensions. Students will become adept at using a pendulum clock, a Vernier calliper, a scale, and other instruments to measure physical quantities at the micro and macro levels. This chapter also discusses small and large ranges of length and mass measurement. Similarly, identifying errors during measurements, such as absolute errors, zero errors, and so on will also be taught.

Benefits of the NCERT Solutions for Class 11 Physics Chapter 2

NCERT Solutions Class 11 Physics Chapter 2 (Units and Measurements) include all questions with detailed explanations and solutions. This will help students clear up any questions they may have and improve their application skills as they prepare for board exams. The detailed, step-by-step solutions will assist students in understanding the concepts and solve any confusions they may have.

The following are the advantages of using the NCERT Solutions for Class 11 Physics Chapter 2 :

Fully-solved answers have been provided to all of the questions in the NCERT textbook.
Simple, easy-to-understand language is used to make learning enjoyable for students.
After conducting extensive research on each concept, professionals prepare the solutions.
Apart from their Term – I exams, students can also prepare for various competitive exams such as JEE, NEET and so on.
Solutions are available in chapter and exercise format to assist students in learning the concepts.
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Q.1 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 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?

One such quantity consisting of some fundamental constants and having dimensions of time is given as: t = e 2 4πε 0 2 × 1 m p m e 2 c 3 G Here, ε 0 represents absolute permittivity e representing charge of electrons = 1.6 × 10 -19 C m e representing mass of electrons = 9 .1×10 -31 kg m p representing mass of protons = 1 .67×10 -27 kg 1 4πε 0 = 9×10 9 Nm 2 C -2 c representing speed of light = 3× 10 8 ms -1 and G = 6 .67×10 -11 Nm 2 kg -2 Putting the above values in the given relation, we obtain: t = 1 .6×10 -19 4 × 9×10 9 2 9 .1×10 -31 2 ×1 .67×10 -27 × 3× 10 8 3 ×6 .67×10 -11 = 2 .18×10 16 s

Q.2 It is a well known fact that during a 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.

The given figure shows the positions of Sun, Moon and Earth at the time of a lunar eclipse. Here, distance between Moon and Earth, UE = 3 .84×10 8 m Distance between Sun and Earth, SE = 1 .496×10 11 m Diameter of Sun, AB = 1 .39×10 9 m ∵ ΔEMN and ΔEAB are similar ∴ AB MN = SE UE 1 .39×10 9 m MN = 1 .496×10 11 m 3 .84×10 8 m ∴MN = 1.39×3.84 1.496 ×10 6 m = 3 .57×10 6 m ∴Diameter of Moon = 3 .57×10 6 m

Q.3 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 yet not been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

Given, time taken by quasar light to reach Earth, t = 3 billion years t = 3×10 9 ×365×24×60×60 s Velocity of light in vacuum, c = 3×10 8 ms -1 As, distance = velocity×time ∴Distance between Earth and quasar, d = ct d = 3×10 8 ms -1 ×3×10 9 ×365×24×60×60 s = 2 .84×10 22 km

Q.4 A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 ms -1 ).

Let distance between the ship and the enemy submarine = s Speed of sound in water, v = 1450 ms -1 Time delay between transmission and reception of SONAR waves, t = 77 s From the relation: s = v×t 2 ∴Distance between ship and enemy submarines is given as: s = 1450 ms -1 ×77 s 2 = 55825 m

Q.5 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?

Given, time taken by the LASER beam to return earth , t = 2.56 s Velocity of light in vacuum, c = 3×10 8 ms -1 Radius of lunar orbit = Separation between Earth and Moon Let radius of lunar orbit be x From the relation: x = c×t 2 x = 3×10 8 ms -1 ×2.56 s 2 = 3 .84×10 8 m

Q.6 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 0 A 1/3 where r is the radius of the nucleus, A its mass number, and r 0 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.

Given, radius of nucleus, r = r 0 A 1/3 ∴Volume of nucleus, V = 4 3 π r 3 = 4 3 π r 0 A 1/3 3 = 4 3 π r 0 3 A ∵ the mass of the nucleus = Mass number ∴ M = A amu = A × 1 .66×10 -27 kg ∴Density of sodium nucleus, ρ = 3×1 .66×10 -27 kg 4×3.14× 1 .2×10 15 m 3 ρ = 4.98 21.71 ×10 18 kgm -3 = 2 .29×10 17 kgm -3 Density of sodium atom, ρ’ = 4 .67×10 3 kgm -3 ∴ ρ ρ’ = 2 .29×10 17 kgm -3 4 .67×10 3 kgm -3 = 4 .9×10 13

Q.7 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 density of sodium in its crystalline phase: 970 kgm -3 . Are the two densities of the same order of magnitude? If so, why?

Given, size of sodium atom = Diameter of sodium atom = 2.5 Å ∴Radius of sodium atom, r = 1 2 ×2.5 Å = 1.25 Å = 1.25 × 10 -10 m Volume of sodium atom, V = 4 3 πr 3 = 4 3 × 22 7 × 1 .25×10 -10 m 3 As per Avogadro hypothesis, Number of atoms in one mole of sodium = 6 .023×10 23 Mass of one mole of sodium = 23 g = 23×10 -3 kg ∴Mass of one atom of sodium = 23×10 -3 6 .023×10 23 kg ∴Density of sodium atom = mass volume Density of sodium atom = 23×10 -3 6 .023×10 23 kg 4 3 × 22 7 × 1 .25×10 -10 m 3 = 4 .67×10 3 kgm -3 Due to the inter-atomic separation in the crystalline phase, the two densities are not of the same order.

Q.8 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?

Given, error in time of cesium clocks = 0.02 s Time taken = 100 years = 100×365 1 4 ×24×60×60 s Error in time of cesium clocks in 100 years = 0.02 s ∴ Error in time of cesium clocks in 1 s, Δt = 1 s× 0.02 s 100×365 1 4 ×24×60×60 s Δt = 6 .337×10 -12 s ≈ 10 -11 s ∴Accuracy of a standard cesium clock for measuring the time interval of 1 s = 10 -11 s

Q.9 A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit: as v → 0, θ → 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 given relation is incorrect dimensionally. One way to make the given relation correct is by dividing the R.H.S of the given relation by the speed of rainfall v’. The given relation becomes: tanθ = v v’ This relation is dimensionally correct.

Q.10 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?

Given, distance between Jupiter and Earth, r = 824 .7×10 6 km Angular diameter of Jupiter, θ = 35.72″ θ = 35.72 60×60 × π 180 rad Let diameter of Jupiter be l. From the relation: θ = l r ∴l = θr l = 35.72 60×60 × π 180 rad×824 .7×10 6 km = 1 .429×10 5 km

Q.11 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 ×10 30 kg, radius of the Sun = 7.0 × 10 8 m.

Given,Mass of the Sun, M = 2 .0×10 30 kg Radius of the Sun, R = 7 .0×10 8 m ∴ Volume of the Sun, V = 4 3 πR 3 ∴V = 4 3 × 22 7 × 7 .0×10 8 m = 1 .4377×10 27 m 3 ∴Density of Sun = Mass Volume Density of Sun = 2 .0×10 30 kg 1 .4377×10 27 m 3 = 1 .3911×10 3 kgm -3 The density of sun is of order of the density of solids and liquids. The enormous gravitational attraction of the inner layers on the outerlayer of Sun is responsible for the high density of Sun.

Q.12 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 (d) the number of strands of hair on your head (e) the number of air molecules in your classroom.

(a) During the monsoon season, a metrologist records nearly 215 cm of rainfall in India. ∴Height of water column, h = 215 cm = 2.15 m Area of the Country, A = 3.3 × 10 12 m 2 ∴Volume of rain water, V = A × h = 7.09 × 10 12 m 3 ∵ density of rain water, ρ = 1 × 10 3 kgm -3 ∴Mass of rain water = ρV = 7.09 × 10 15 kg

(b) Let us assume a boat of known base area A. Let depth of boat in sea be d 1 Volume of water displaced by boat, V 1 = Ad 1 Now move the elephant on the boat Let depth of boat when elephant is moved on the boat be d 2 ∴Volume of water displaced by boat and elephant, V 2 = Ad 2 ∴Volume of water displaced by elephant, V = V 2 – V 1 = A d 2 – d 1 Let density of water be ρ Mass of elephant = Mass of water displaced by elephant = Vρ = A d 2 -d 1 ρ

(c) With the help of an anemometer we can measure the wind speed. The wind speed is given by the rotation made by the anemometer in one second.

(d) Let area of head carrying hair be A. The radius of a hair can be measured with help of a screw gauge. Let radius of a hair be r ∴Area of one hair = π r 2 Considering that the uniform distribution of hair over the head, Number of strands of hair = Total surface area Area of one hair = A π r 2

(e) Volume occupied by one mole of air at NTP = 22.4 l = 22 .4×10 -3 m 3 Number of molecules in one mole of air = 6 .023×10 23 Let volume of the room be V ∴Number of air molecules in room of volume V = 6 .023×10 23 22 .4×10 -3 m 3 ×V = V×1 .35×10 28 m -3

Q.13 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.

It is true that precise measurements of physical quantities are essential for the development of laws of Physics or any other Science. For example:

(a) In various physical and chemical processes, ultra-shot LASER pulses are used to determine small time intervals.

(b) Mass spectrometer can be used to measure the mass of atoms precisely.

Q.14 The nearest star to our solar system is 4.29 light years away. How much is the distance in terms of parsec? 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?

Distance between the star and the solar system=4.29 ly 1 light year = speed of light × 1 year 1 light year = 3×10 8 ms -1 ×365×24×60×60 s = 94608×10 11 m ∴4.29 ly = 4 .29×94608×10 11 m = 405868 .32×10 11 m ∵1 parsec = 3 .08×10 16 m ∴4.29 ly = 405868 .32×10 11 3 .08×10 16 = 1.32 parsec Here, diameter of Earth’s orbit, d = 3×10 11 m Distance of Star from the Earth, D = 405868.32 × 10 11 m From the relation: θ = d D ∴θ = 3×10 11 405868 .32×10 11 = 7 .39×10 -6 rad ∵ 1 sec = 4 .85×10 -6 rad ∴7 .39×10 -6 rad = 7 .39×10 -6 4 .85×10 -6 = 1.52″

Q.15 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 × 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?

Given, dameter of Earth’s orbit = 3×10 11 m ∴Radius of Earth’s orbit,r = 1 .5×10 11 m Given that, parallax angle, θ = 1″ θ = 1′ 60 = 1° 60×60 = π 180 × 1 60×60 rad = 4 .847×10 -6 rad Let the distance of star be D From the relation: θ = r D ∴D = r θ D = 1 .5×10 11 4 .847×10 -6 = 3 .09×10 16 m ∴ 1 parsec≈3 .09×10 16 m

Q.16 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 train’s motion, but the distant objects (hill tops, the moon, the stars etc.) seem to be stationary.

The imaginary line joining an object to the eye is called the line of sight. When a train is moving rapidly, the line of sight of a nearby house changes its direction very rapidly. Therefore, the house seems to move rapidly in a direction opposite to the train’s motion.

Since distant objects such as the hill top, the moon, the stars, etc. are extremely large distances apart, therefore, the line of sight does not change its direction rapidly and they appear to be stationary.

Q.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 Å). Why is this ratio so large?

Given, size of hydrogen molecule = 1 Å ∴Atomic radius, r = 0 .5×10 -10 m Volume of each H atom, V = 4 3 πR 3 V = 4 3 ×3.14× 0 .5×10 -10 m 3 = 5 .236×10 -31 m 3 Since number of hydrogen atoms in 1 g mole of hydrogen, Avogadro’s Number, N 0 = 6 .023×10 23 ∴Atomic volume of 1 g mole of H atom = N 0 ×V = 5 .236×10 -31 m 3 ×6 .023×10 23 = 3 .154×10 -7 m 3 Molar volume = 22.4 L = 22 .4×10 -3 m 3 Molar volume Atomic volume = 22 .4×10 -3 m 3 3 .154×10 -7 m 3 = 7 .1×10 4 This ratio is large because of large intermolecular separations.

Q.18 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :m = m 0 1 – v 2 1/2 Guess where to put the missing c.

∵ As per the principle of Homogeneity of dimensions, dimensions of M, L, and T on one side of dimensional physical relation should be equal to their respective dimensions on the other side of the relation. ∴On RHS, the denominator 1 – v 2 1/2 must be dimensionless. ∴In place of 1 – v 2 1/2 , there should be 1 – v 2 c 2 1/2 ∴The correct relation is: m = m 0 1 – v 2 c 2 1/2 Sr19 (1702789) Instruction Question

Q.19 A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion:

(i) y = a sin 2πt T

(ii) y = a sin vt

(iii) y = a T sin t a

(iv) y = a 2 [ sin 2πt T + cos 2πt T ]

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

Angle is a dimensionless physical quantity. (i) 2πt T = T T = 1 = M 0 L 0 T 0 The above formula is dimensionless. (ii) vt = LT -1 T = L = M 0 L 1 T 0 The above formula is not dimensionless. (iii) t a = T L = L -1 T 1 (iv) 2πt T = T T = 1 = M 0 L 0 T 0 The above formula is not dimensionless. ∴Formulae (ii) and (iii) are wrong. Sr20 (1702761) Instruction Question

Q.20 A physical quantity P is related to four observables a, b, c and d as follows:P = a 3 b 2 ( c d ) The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error is 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?

Given, P = a 3 b 2 c d ∵ Fractional error in a quantity raised to power n is n times the fractional error in the individual quantity, ∴Maximum fractional error in P = ΔP P ΔP P = 3 Δa a + 2 Δb b + 1 2 Δc c + Δd d ΔP P = 3 1 100 + 2 3 100 + 1 2 4 100 + 2 100 = 13 100 = 0.13 ∴Percentage error in P = ΔP P ×100 = 0.13×100 = 13% Since the result has two significant figures, ∴ The result, P = 3.763 would be rounded off to 3.8. Sr21 (1702746) Instruction Question

Q.21 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Given, mass of grocer’s box, m = 2.3 kg Mass of first gold piece, m 1 = 20.15 g = 0.02015 kg Mass of the second gold piece, m 2 = 20.17 g = 0.02017 kg (a) Total mass of box, m T = m + m 1 + m 2 m T = 2.3 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg Since the result is accurate only up to one place of decimal, ∴On rounding off, we obtain: Total mass of box = 2.3 kg (b) Difference in masses of the gold pieces = m 2 – m 1 = 20.17 g – 20.15 g = 0.02 g

Q.22 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10 –10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m 3 of a mole of hydrogen atoms?

Given, radius of each hydrogen atom, r = 0.5 Å = 0 .5×10 -10 m ∴Volume of each hydrogen atom, V = 4 3 πR 3 V = 4 3 ×3.14 0 .5×10 -10 3 = 5 .236×10 -31 m 3 ∵ Number of H atoms in 1 g mole of hydrogen = 6 .023×10 23 ∴Atomic volume of 1 g mole of H atom = 6 .023×10 23 ×V = 6 .023×10 23 ×5 .236×10 -31 m 3 = 3 .154×10 -7 m 3

Q.23 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Given, length of the metal sheet, l = 4.234 m Breadth of the metal sheet, b = 1.005 m Thickness of the metal sheet, t = 2.01 cm = 2 .01×10 -2 m Area of the metal sheet, A = 2 l×b + b×t + t×l A = 2 4.234 m×1.005 m + 1.005 m×2 .01×10 -2 m + 2 .01×10 -2 m×4.234 m A = 2 4 .3604739 m 2 = 8.7209478 m 2 Since area and volume can contain a maximum of three significant figures, ∴ On rounding off, we obtain Area = 8.72 m 2 Volume of metal sheet, V = l×b×t V = 4.234 m×1.005 m×0.0201 m = 0.0855289 = 0.0855 m 3 Sr24 (1702663) Instruction Question

Q.24 State the number of significant figures in the following: (a) 0.007 m 2 (b) 2.64 × 10 24 kg (c) 0.2370 gcm -3 (d) 6.320 J (e) 6.032 Nm -2 (f) 0.0006032 m 2

As per the rules of finding significant figures, the number of significant figures is given below:

Q.25 The photograph of a house occupies an area of 1.75 cm 2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m 2 . What is the linear magnification of the projector-screen arrangement?

Given, area of the object = 1.75 cm 2 area of the image = 1.55 m 2 = 1 .55×10 4 cm 2 Areal magnification is given as: m A = area of image area of object = 1 .55×10 4 cm 2 1 .75 cm 2 = 8857 Linear magnification of projector screen arrangement = m A = 8857 = 94.1

Q.26 Answer the following : (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) Since the diameter of a thread is very small, therefore, it cannot be measured by using a metre scale. Wind a number of turns of the thread on the metre such that the turns are very closely touching each other. Now measure the length (L) of the windings of thread on the scale having n number of turns.

∴ diameter of thread= Length of thread Number of turns = l n

(b) Least count of screw gauge = Pitch of screw gauge Number of divisions on the circular scale

Since on increasing the number of divisions on the circular scale, the least count decreases, therefore, accuracy increases.

However, due to the low resolution of the human eye, it may not be possible to take the reading precisely.

(c) Since the probability of making a positive random error of definite magnitude is equal to that of making a negative random error of the same magnitude, therefore, when the number of observations is large, random errors are probable to cancel and the result may become more accurate. Therefore, a large number of observations (100) will give more accurate results than a smaller number of observations.

Q.27 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 the estimate on the thickness of hair?

Here, magnification of microscope = 100

Observed width of hair, y = 3.5 mm Let real width of hair be x Magnification of microscope is given as:

m = observed width real width ∴m = y x ∴Real width, x = y m x = 3.5 mm 100 = 0.035 mm Sr28 (1702657) Instruction Question

Which of the following is the most precise device for measuring length : (a) a vernier calipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light?

(a) Least count of the given vernier callipers = Value of one main scale division Total number of divisions on the sliding scale = 1 mm 20 = 0.05 mm = 0.005 cm

(b) Least count of screw gauge = Pitch of screw gauge Total no. of divisions on circular scale = 1 100 mm = 0.01 mm = 0.001 cm

(c) Here, wavelength of light, λ≈ 10 -5 cm = 0.00001 cm The most precise device is that whose least count is minimum. ∴An optical instrument is the most precise device.

Q.28 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 of light takes 8 min and 20 s to cover this distance?

Given, velocity of light in vacuum, c = 1 new unit of length s -1 Time taken by sun light to reach the Earth, t = 8 min 20 s t = 8×60 s + 20 s = 500 s ∴ Distance between Sun and Earth is given as: x = c×t = 1 new unit of length s -1 ×500 s ∴Distance between Sun and Earth = 500 new units of length Sr30 (1702654) Instruction Question

Q.29 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.

(i) Since a dimensionless quantity can be large or small in comparison to some standard reference, therefore, the given statement is true. For example, the angle is a dimensionless quantity.

∠θ = 60° is greater than ∠θ = 30° , but smaller than ∠θ = 90°

(ii) (a) An atom is smaller as compared to the sharp tip of a pin.

(b) A jet plane moves with a speed greater than a superfast train.

(d) The air inside this room contains a large number of molecules as compared to that present in one mole of air.

(e) The given statement is already correct.

(f) The given statement is already correct.

Q.30 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kgm 2 s –2 . Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α –1 β –2 γ 2 in terms of the new units.

Given, 1 calorie = 4.2 J = 4.2 kg m 2 s -2 Since new unit of mass = α kg ∴1 kg = 1 α new unit of mass = α -1 new unit of mass In the same way, 1 m = β -1 new unit of length 1 s = γ -1 new unit of time Substituting these values in (i), we obtain: 1 calorie = 4.2 α -1 new unit of mass β -1 new unit of length 2 γ -1 new unit of time -2 1 calorie = 4.2 α -1 β -2 γ 2 new unit of energy

Q.31 Fill in the blanks by suitable conversion of units (a) 1 kgm 2 s –2 = ….gcm 2 s –2 (b) 1 m = ….. ly (c) 3.0 ms –2 = …. kmh –2 (d) G = 6.67 × 10 –11 Nm 2 (kg) –2 = …. (cm) 3 s –2 g –1

(a) 1 kgm 2 s -2 = 1×1000 g 100 cm 2 s -2 = 10 7 gcm 2 s -2

(b) Light year is the distance travelled by light in one year. 1 light year = 9 .46×10 15 m ∴ 1 m = 1 9 .46×10 15 light year = 1 .053×10 -16 light year

(c) ∵1 m = 10 -3 km 1 s = 1 3600 h ∴1 s -1 = 3600 h -1 ∴3 ms -2 = 3×10 -3 km 3600 h -2 3 ms -2 = 3×10 -3 ×3600×3600 kmh -2 = 3 .888×10 4 kmh -2

(d) G = 6 .67×10 -11 Nm 2 kg -2 = 6 .67×10 -11 kgms -2 m 2 kg -2 G = 6 .67×10 -11 m 3 s -2 kg -1 = 6 .67×10 -8 cm 3 s -2 g -1

Q.32 Fill in the blanks (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 …(mm) 2 . (c) A vehicle moving with a speed of 18 kmh –1 covers….m in 1 s. (d) The relative density of lead is 11.3. Its density is ….gcm –3 or ….kgm –3 .

(a) Given, length of side of cube, L = 1 cm = 10 -2 m ∴ Volume of cube = L 3 = (10 -2 m) 3 = 10 -6 m 3

(b) Given, radius of cylinder, r = 2.0 cm = 20 mm Height of cylinder, h = 10.0 cm = 100 mm Surface area of solid cylinder is given as: A = (2πr)h = 2× 22 7 ×20×100 mm 2 Surface area of solid cylinder = 1 .26×10 4 mm 2

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Faqs (frequently asked questions), 1. is ncert sufficient for class 11 physics.

For Class 11 Physics exam, the NCERT textbook and past years’ papers are sufficient. However, for exams such as the JEE, you will need to consult some advanced books as well to understand the underlying principle of various concepts.

2. What are CGS,MKS and FPS units?

The fundamental units in the CGS system are centimetre, gram and second. While the MKS system is an abbreviation for the Metre, kilogramme, and second, which is also the system’s fundamental unit. FPS stands for Foot, Pound, and Second, which are the fundamental units of the FPS system.

3. Do I have to go over all of the questions in Class 11 Physics Chapter 2?

When preparing for your Class 11 Physics exams, the questions provided in Class 11 Physics Chapter 2 NCERT Solutions are critical. The exam question papers always include questions from the Physics NCERT for Class 11 textbook. Practising all of the questions will only help you improve your understanding of all of the concepts taught in Chapter 2 as well as your chances of scoring well on the chapter’s questions.

4. What is the distribution of marks for Class 11 Physics Chapter 2?

Unit – I in Class 11 Physics includes Chapter 2 – Units and Measurement as well as Chapter 1 – Physical World. According to the CBSE marks distribution for Class 11 Physics, Unit – I, which includes both chapters, carries a total of 23 marks. As a result, preparation from both chapters should be given equal priority in order to avoid losing any marks in exam questions framed from them.

5. Which topics are important in Class 11 Physics Chapter 2?

The International System of Units, Measurement of length, mass, and time, Application of Significant Figures, and other topics are covered in NCERT Solutions for Class 11 Physics Chapter 2 – Units and Measurements. The chapter’s most important topics include SI Units, Absolute Errors, Dimensional Analysis, and Significant Figures. Short-answer and numerical-based questions about evaluating errors in quantity measurement can also be asked.

6. What is the best way for me to fully understand Chapter 2 of Class 11 Physics?

This chapter discusses in detail the various units used to determine the measurement of various physical quantities, the instruments used for such measurements and their accuracy, and so on. Students can easily understand this chapter if they read it on a regular basis and answer the questions provided in the NCERT. Students can also find Class 11 Physics Chapter 2 NCERT Solutions on the Extramarks website for additional assistance.

7. What is the best book for Class 11 Physics?

The best book for Class 11 CBSE Physics is Ans. Concepts of Physics by HC Verma.

8. What exactly is dimensional analysis?

Dimensional Analysis is used to quantify the size and shape of things. It allows us to study the nature of objects mathematically. It includes lengths and angles, as well as geometrical properties like flatness and straightness. We can only add and subtract quantities that have the same dimensions, according to the basic concept of dimension. Likewise, if two physical quantities have the same dimensions, they are equal.

9. What are the benefits of downloading the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements?

The NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements include diagrams and answers to all of the textbook’s questions. Every question is answered keeping the students’ comprehension abilities in mind. The solutions developed strictly adhere to the latest term-wise CBSE Syllabus and exam pattern, allowing students to face the first term exams with confidence. It also improves their time management skills, which are important for exams.

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Solutions for chapter 2: units and measurements.

Below listed, you can find solutions for Chapter 2 of CBSE NCERT Exemplar for Physics Class 11.

NCERT Exemplar solutions for Physics Class 11 Chapter 2 Units and Measurements Exercises [Pages 5 - 12]

The number of significant figures in 0.06900 is ______.

The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is ______.

The mass and volume of a body are 4.237 g and 2.5 cm 3 , respectively. The density of the material of the body in correct significant figures is ______.

1.6048 g cm –3

1.69 g cm –3

1.7 g cm –3

1.695 g cm –3

The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give ______.

2.75 and 2.74

2.74 and 2.73

2.75 and 2.73

2.74 and 2.74

The length and breadth of a rectangular sheet are 16.2 cm and 10.1cm, respectively. The area of the sheet in appropriate significant figures and error is ______.

164 ± 3 cm 2

163.62 ± 2.6 cm 2

163.6 ± 2.6 cm 2

163.62 ± 3 cm 2

Which of the following pairs of physical quantities does not have same dimensional formula?

Work and Torque.

Angular momentum and Planck’s constant.

Tension and Surface tension.

Impulse and Linear momentum.

Measure of two quantities along with the precision of respective measuring instrument is A = 2.5 ms –1 ± 0.5 ms –1 B = 0.10 s ± 0.01 s. The value of A B will be ______.

(0.25 ± 0.08) m

(0.25 ± 0.5) m

(0.25 ± 0.05) m

(0.25 ± 0.135) m

You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m. We should report correct value for `sqrt(AB)` as ______.

1.4 m ± 0.4 m

1.41 m ± 0.15 m

1.4 m ± 0.3 m

1.4 m ± 0.2 m

Which of the following measurements is most precise?

The mean length of an object is 5 cm. Which of the following measurements is most accurate?

Young’s modulus of steel is 1.9 × 10 11 N/m 2 . When expressed in CGS units of dynes/cm 2 , it will be equal to (1N = 10 5 dyne, 1m 2 = 10 4 cm 2 )

1.9 × 10 10

1.9 × 10 11

1.9 × 10 12

1.9 × 10 13

If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula ______.

(P 1 A –1 T 1 )

(P 2 A 1 T 1 )

(P 1 A –1/2 T 1 )

(P 1 A 1/2 T – 1 )

On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct ______.

y = `a sin (2πt)/T`
y = `a sin vt`
y = `a/T sin (t/a)`
y = `asqrt(2) (sin (2pit)/T - cos (2pit)/T)`

If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?

(P – Q)/R
PQ – R
(PR – Q 2 )/R

Photon is quantum of radiation with energy E = h ν where ν is frequency and h is Planck’s constant. The dimensions of h are the same as that of ______.

Linear impulse
Angular impulse
Linear momentum
Angular momentum

If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?

Mass of electron (m e )
Universal gravitational constant (G)
Charge of electron (e)
Mass of proton (m p )

Which of the following ratios express pressure?

Energy/Volume
Energy/Area
Force/ Volume

Which of the following is not a unit of time?

Why do we have different units for the same physical quantity?

The radius of atom is of the order of 1 Å and the radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Name the device used for measuring the mass of atoms and molecules.

Express unified atomic mass unit in kg.

A function f(θ) is defined as: `f(θ) = 1 - θ + θ^2/(2!) - θ^3/(3!) + θ^4/(4!)` Why is it necessary for q to be a dimensionless quantity?

Why length, mass and time are chosen as base quantities in mechanics?

The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?
Moon is seen to be of (½)°diameter from the earth. What must be the relative size compared to the earth?
From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Which of the following time measuring devices is most precise?

A wall clock.

A stop watch.

A digital watch.

An atomic clock.

The distance of a galaxy is of the order of 10 25 m. Calculate the order of magnitude of time taken by light to reach us from the galaxy.

The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

Give an example of a physical quantity which has a unit but no dimensions.

Give an example of a physical quantity which has neither unit nor dimensions.

Give an example of a constant which has a unit.

Give an example of a constant which has no unit.

Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of `pi/6` at the centre.

Calculate the solid angle subtended by the periphery of an area of 1 cm 2 at a point situated symmetrically at a distance of 5 cm from the area.

The displacement of a progressive wave is represented by y = A sin(ωt – kx), where x is distance and t is time. Write the dimensional formula of (i) ω and (ii) k.

Time for 20 oscillations of a pendulum is measured as t 1 = 39.6 s; t 2 = 39.9 s; t 3 = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement?

A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s. How much will 5 J measure in this new system?

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as `v = π/8 (pr^4)/(ηl)` where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula ML –1 T –1 . Check whether the equation is dimensionally correct.

A physical quantity X is related to four measurable quantities a, b, c and d as follows: X = a 2 b 3 c 5/2 d –2 . The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity X? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

In the expression P = E l 2 m –5 G –2 , E, m, l and G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

If velocity of light c, Planck’s constant h and gravitational contant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that `T = k/R sqrt(r^3/g)`. where k is a dimensionless constant and g is acceleration due to gravity.

In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions:

Why do we dissolve oleic acid in alcohol?
What is the role of lycopodium powder?
What would be the volume of oleic acid in each mL of solution prepared?
How will you calculate the volume of n drops of this solution of oleic acid?
What will be the volume of oleic acid in one drop of this solution?

How many astronomical units (A.U.) make 1 parsec?

Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.

Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.

Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m ) to energy (E ) as E = mc 2 , where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV= 1.6 × 10 –13 J; the masses are measured in unified atomic mass unit (u) where 1u = 1.67 × 10 –27 kg.

Show that the energy equivalent of 1 u is 931.5 MeV.
A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

NCERT Exemplar solutions for Physics Class 11 chapter 2 - Units and Measurements

Shaalaa.com has the CBSE Mathematics Physics Class 11 CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT Exemplar solutions for Mathematics Physics Class 11 CBSE 2 (Units and Measurements) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT Exemplar textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Physics Class 11 chapter 2 Units and Measurements are International System of Units, Measurement of Length, Measurement of Mass, Measurement of Time, Accuracy, Precision and Least Count of Measuring Instruments, Significant Figures, Dimensions of Physical Quantities, Dimensional Formulae and Dimensional Equations, Dimensional Analysis and Its Applications, Introduction of Units and Measurements, Errors in Measurements, Need for Measurement, Units of Measurement, Fundamental and Derived Units, Length, Mass and Time Measurements.

Using NCERT Exemplar Physics Class 11 solutions Units and Measurements exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Exemplar Solutions are essential questions that can be asked in the final exam. Maximum CBSE Physics Class 11 students prefer NCERT Exemplar Textbook Solutions to score more in exams.

Get the free view of Chapter 2, Units and Measurements Physics Class 11 additional questions for Mathematics Physics Class 11 CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

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NCERT Solutions for Class 11th: Ch 2 Units and Measurements Physics

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11th Class Physics Chapter 2
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Class 11 Physics
Instructor: Ms. Maimoona Altaf (MPhil Physics)Dear Students, in this video you will learn about how to solve assignment 2.2 of the physics grade 11 new book...
Assignment 2.2 || Class 11 Physics Chapter 2 || Ilmi Stars Academy
This video lecture by ilmi stars academy is on the topic of Assignment 2.2 chapter 2 vectors and equilibrium 11th physics federal board
Physics class 11th assignment 2.2 & Example 2.2 Kpk & Federal board
This lecture is about fsc part 1 physics chapter 2 Example 2.2 And assignment 2.2 For Federal and kpk board students... FSC part 1 physics chapter 3 numerica...
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All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Physics Class 11. Students can click on the links below and download all Pdf Assignments for Physics class 11 for free. All latest Kendriya Vidyalaya Class 11 Physics Assignments with Answers and test papers are given below.
NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
Access the answers of NCERT Class 11 Physics Chapter 2 Units and Measurements. 2.1 Fill in the blanks. (a) The volume of a cube of side 1 cm is equal to …..m3. (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 18 km h-1 covers….m in 1 s.
Class 11 Physics Ch.2 Vectors and Forces
Download Class 11 Physics Ch.2 Vectors and Equilibrium PDF Notes ⇒ Class 11 Physics Ch.2 Vectors and Equilibrium Theory Notes ⇒ Class 11 Physics Ch.2 Vectors and Equilibrium Assignments ⇒ Class 11 Physics Ch.2 Vectors and Equilibrium Conceptual Questions ⇒ Class 11 Physics Ch.2 Vectors and Equilibrium Numerical Questions
NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements. ... =11.3 x 10 3 kg m-4. Question 2. 2. Fill in the blanks by suitable conversion of units (a) 1 kg m2 s-2 = …. g cm 2 s-2 (b) 1 m ...
CBSE Class 11 Physics Chapter 2 Units and Measurements Study Materials
Units and Measurement Class 11 Notes Physics Chapter 2. The process of measurement is basically a comparison process. To measure a physical quantity, we have to find out how many times a standard amount of that physical quantity is present in the quantity being measured. The number thus obtained is known as the magnitude and the standard chosen ...
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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
Chapter 2 of the NCERT Class 11 Physics textbook, titled "Units and Measurements," delves into the fundamental concepts essential for scientific study. This chapter covers the importance of units, the International System of Units (SI), and the methods of measurement for various physical quantities. It also discusses significant figures ...
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In this post, I am sharing FSC 1st Class Physics Chapter 2 Short Questions with Answers Notes PDF for the students of Intermediate Part 1. This chapter's name is Vectors and Equilibrium. From this post of ilmihub, students can download the 11th class chapter 2 Vector and Equilibrium Short Questions Notes in PDF format on their laptops or mobiles.
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CBSE Class 11 Physics Chapter 15 Units and Measurement NCERT Solutions Unit refers to the international standards that are accepted worldwide. While on the other hand, measurement refers to the use of the unit for measuring the quantity.
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Class 11 Physics Notes (Unit # 2) for FBISE (Federal Board) Islamabad (F.Sc, ICS , 1st Year): Very Easy, Updated & Comprehensive Notes for all types of students i.e. Average / Extra Ordinary. Notes, Guides & Key Books (Main Page) About Our Notes: All academic material which includes notes, key books, guides, handouts, assignments, solutions ...
NCERT Solutions for Class 11 Physics Chapter 2
Solution: These type of question can be simply solved by putting the respective conversion units in the place. (a) 1kgm2s − 2 = 1 × 103g(102cm)2s − 2 = 107gcm2s − 2. (b) Now we know that distance traveled by light in 1 year is called 1 light year. Speed of light = 3 × 108 m/s and 1 year = 365 × 24 × 60 × 60 s.
NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement
1; 3; 4; 4; 4; 4; Question 11. The length, breadth, and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Class 11 Physics Notes FBISE & KPK Board
11th Class Physics Chapter 10 Thermodynamics. Chapter 10 Theory (complete) Chapter 10 Conceptual Questions (complete) Chapter 10 Assignments (selected) Chapter 10 Numerical Questions (selected) December 17, 2021 at 1:12 AM. Very excillent information thanks brother. Reply. Replies.
NCERT Solutions for Class 11 Physics Chapter 2
The International System of Units, Measurement of length, mass, and time, Application of Significant Figures, and other topics are covered in NCERT Solutions for Class 11 Physics Chapter 2 - Units and Measurements. The chapter's most important topics include SI Units, Absolute Errors, Dimensional Analysis, and Significant Figures.
Class 11 Physics
Instructor: Ms. Maimoona Altaf (MPhil Physics)Dear Students, in this video you will learn about how to solve assignment 2.1 and activity on page # 42 of the ...
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Using NCERT Exemplar Physics Class 11 solutions Units and Measurements exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Exemplar Solutions are essential questions that can be asked in the final exam.
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NCERT Solutions for Class 11th: Ch 1 Units and Measurements Physics Science. NCERT Solutions for Class 11th: Ch 1 Units and Measurements Physics Science. NCERT Solutions; ... (1 - v 2 /c 2) 1/2 2.16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10-10 m. The size of a hydrogen atom is ...
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