NCERT Solutions  Class 11 Physics Chapter 2  Units and Measurement
CBSE Solutions  Physics Class 11
Check the below NCERT Solutions for Class 11 Physics PartI Chapter 2 Units and Measurement Pdf free download. NCERT Solutions Class 11 Physics were prepared based on the latest exam pattern. We have Provided Units and Measurement Class 11 Physics NCERT Solutions to help students understand the concept very well.
NCERT  Class 11 Physics PartI
Book:  National Council of Educational Research and Training (NCERT) 

Board:  Central Board of Secondary Education (CBSE) 
Class:  11th 
Subject:  Physics PartI 
Chapter:  2 
Chapters Name:  Units and Measurement 
Medium:  English 
Units and Measurement  Class 11 Physics  NCERT Books Solutions
You can refer to MCQ Questions for Class 11 Physics Chapter 2 Units and Measurement to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.
Question 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 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:
Question 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 m s^{2} = ………. km h^{–}^{2}
(d) G = 6.67 x 10^{11} N m^{2} (kg)^{–}^{2} =……………. (cm)^{3} s^{–}^{2} g^{1}.
Answer:
Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 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, 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.
Answer:
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:
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.
Question 6.
Which of the following is the most precise device for measuring length :
(a) vernier calipers 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 the length to within a wavelength
Answer:
(c) Least count of optical instrument w 6000 A
(average wavelength of visible light as 6000 A) = 6 x 10^{7} m
∴ (c) is the most precise instrument.
Question 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 of the thickness of the hair?
Answer:
Question 8.
Answer the following :
(a) You are given a thread and a meter scale. How will you estimate the diameter of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier calipers. 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) It if; done by winding a known number of turns over a pencil, turns touching each other closely. Then the length occupied by every single turn will be equal to the diameter of the thread.
(b) Yes, because the least count of the screw gauge is given by
i.e. least count of screw gauge is inversely proportional to the number of divisions on a circular scale. So with the increase in a number of the divisions on the circular scale, the least count will improve. Thus the accuracy of the screw gauge will increase.
(c) Increasing the number of observations, increases the reliability as the mean error is also reduced. The best possible value is
Question 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 projectorscreen arrangement?
Answer:
Given Area of the house in photograph = 1.75 cm^{2}
Area of house on screen = 1.55 m^{2} = 1.55 x 10^{4} cm^{2}
Question 10.
State the number of significant figures in the following:
Answer:
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 13.
A physical quantity P is related to four observables a, b, c and d as follows :
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 ?
Answer:
Question 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 2π \( \frac { t}{ T } \)
(b) y = a sin υt
(c) y = \( \frac { a}{ T } \) sin \( \frac { t}{ a } \)
(d) y = (a√2) (sin 2π \( \frac { t}{ T } \) + cos 2π \( \frac { t}{ T } \) )
(a = maximum displacement of the particle, υ= speed of the particle, T = time period of motion). Rule out the wrong formulas on dimensional grounds.
Answer:
The argument of a trigonometric function i. e., angle is dimensionless
Question 15.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m_{0} of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein).A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :
Guess where to put the missing c.
Answer:
On rearranging, we have
Since the lefthand side is dimensionless, so the righthand side should be dimensionless. This will be so ……………
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} = 523 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}.
Question 17.
One gram 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 A). Why is this ratio so large?
Answer:
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 fastmoving 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.
Question 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^{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 meters ?
Answer:
Question 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
Question 21.
Precise measurements of physical quantities are a need of modern times. For example, to ascertain the speed of an enemy fighter plane, one must have an accurate method to find its positions at closely separated instants of time. Only then we can hope to shell it with an antiaircraft gun. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precision measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Answer:
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 rainbearing 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.
Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10^{7} K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be? In the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: a mass of the Sun = 2.0 x 10^{30} kg. radius of the Sun = 7.0 x 10^{8} m.
Answer:
Mass density of Sun is in the range of mass densities of solid/liquids and not gases.
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 3572 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.
Question 26.
It is claimed that two cesium clocks if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time interval of 1 s?
Answer:
Total time = 100 years = 100 x 365 x 24 x 60 x 60 s
Question 27.
Estimate the average mass density of a sodium atom assuming its size to 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 kg m^{3}. Are the two densities of the same order of magnitude? If so, why?
Answer:
The volume of 1 sodium atom,
Both the densities are of the order of 10^{3} i.e. the atoms are tightly packed. They belong to the solid phase.
Question 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 2.20.
Answer:
Question 29.
A LASER is a source of very intense, monochromatic, and a 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 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?
Answer:
Question 30.
A SONAR (sound navigation and ranging) used 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}).
Answer:
Question 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?
Answer:
Given t = 3 x 10^{9} years = 3 x 10^{9} x 365.25 x 24 x 60 x 60 s
c = 3 x 10^{5} km s^{1} (velocity of e.m. waves)
∴ d = c x t = 3 x 10^{5} x 3 x 10^{9} x 365.25 x 24 x 60 x 60
= 2840184 x 10^{16} km = 2.84 x 10^{22} km.
Question 32.
It is a wellknown fact that during a total solar eclipsethe disk of the moon almost completely covers the disk of the Sun. From this fact and the information, you can gather from examples 1 and 2, determine the approximate diameter of the moon.
Answer:
Here CD = diameter of Sun
AB = diameter of moon
and E = Position of earth
As Δ CDE and ABE are similar,
Question 33.
A great physicist of this century (PA.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, the mass of the electron, the mass of a proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (≈15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Answer:
Trying out with basic constants of atomic physics (speed of light c, the charge on electron e, the mass of electron m_{e}, the mass of proton m_{p}) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is
NCERT Class 11 Physics
Class 11 Physics Chapters  Physics Class 11 Chapter 2
NCERT Solutions For Class 11 physics
NCERT Solutions  Class 11 Physics PartI

NCERT Solutions For Class 11 Physics Chapter 1 Physical World
NCERT Solutions For Class 11 Physics Chapter 2 Units and Measurement
NCERT Solutions For Class 11 Physics Chapter 3 Motion in a Straight Line
NCERT Solutions For Class 11 Physics Chapter 4 Motion in a Plane
NCERT Solutions For Class 11 Physics Chapter 5 Laws of Motion
NCERT Solutions For Class 11 Physics Chapter 6 Work, Energy and Power
NCERT Solutions For Class 11 Physics Chapter 7 System of Particles and Rotational Motion
NCERT Solutions For Class 11 Physics Chapter 8 Gravitation
NCERT Solutions  Class 11 Physics PartII

NCERT Solutions For Class 11 Physics Chapter 9 Mechanical Properties of Solids
NCERT Solutions For Class 11 Physics Chapter 10 Mechanical Properties of Fluids
NCERT Solutions For Class 11 Physics Chapter 11 Thermal Properties of Matter
NCERT Solutions For Class 11 Physics Chapter 12 Thermodynamics
NCERT Solutions For Class 11 Physics Chapter 13 Kinetic Theory
NCERT Solutions For Class 11 Physics Chapter 14 Oscillations
NCERT Solutions For Class 11 Physics Chapter 15 Waves
NCERT Solutions for Class 12 All Subjects  NCERT Solutions for Class 10 All Subjects 
NCERT Solutions for Class 11 All Subjects  NCERT Solutions for Class 9 All Subjects 
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