Chapter 1 Units and Measurements

NCERT Solutions Class 11 Physics Chapter 1 – Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 1 (Units and Measurements) are the foundation for the entire study of Physics. Every physical quantity—from the speed of a car to the mass of an atom—requires a standard of measurement. This chapter introduces students to the international system of units and the mathematical tools needed to ensure precision and accuracy in scientific calculations.

ALLEN NCERT Solutions for Class 11 Physics Chapter 1 are developed by expert faculty to ensure strong conceptual clarity and numerical accuracy. Solutions are provided in a systematic way according to the exam pattern required for JEE and NEET.

These concepts are very important for JEE and NEET aspirants as Dimensional Analysis and Error Analysis are high scoring topics and asked in almost all competitive exams. These solutions provide a step-by-step guide for solving numerical problems in parallax methods, significant figures and propagation of errors.

1.0Class 11 Physics Chapter 1 : Key Concepts

Class 11 Physics Chapter 1, Units and Measurements, explains the basic principles of physical measurement. It covers SI units, measurement methods, errors, significant figures, dimensions, and dimensional analysis used in Physics.

SI Units: Understand the seven fundamental SI units: metre, kilogram, second, ampere, kelvin, mole, and candela. The chapter also introduces units for measuring plane and solid angles.
Measurement of Large Distances: Learn the parallax method for estimating the distances of stars and other distant objects.
Measurement of Mass and Time: Study different methods used to measure mass and time, including the role of highly accurate atomic clocks.
Accuracy, Precision, and Errors: Understand accuracy and precision along with systematic errors, random errors, absolute error, relative error, and percentage error.
Significant Figures: Learn how to identify significant figures, apply rounding-off rules, and represent measured values with the appropriate precision.
Dimensions of Physical Quantities: Express physical quantities using fundamental dimensions such as [M], [L], [T], and [A].
Dimensional Analysis: Learn how dimensional analysis helps check the correctness of physical equations, derive relationships between physical quantities, and convert units from one system to another.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 1

SOLVED EXAMPLES

Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures? Sol. The number of significant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures.

$Surface area of the cube Volume of the cube = 6 (7.203)^{2} m^{2} = 311.299254 m^{2} = 311.3 m^{2} = (7.203)^{3} m^{3} = 373.714754 m^{3} = 373.7 m^{3}$

5.74 g of a substance occupies $1.2 cm^{3}$ . Express its density by keeping the significant figures in view. Sol. There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.

$Density = \frac{5.74}{1.2} g cm^{- 3} = 4.8 g cm^{- 3} .$

Let us consider an equation $\frac{1}{2} m v^{2} = m g h$ where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct. Sol. The dimensions of LHS are

$[M] [LT^{- 1}]^{2} = [M] [L^{2} T^{- 2}] = [M L^{2} T^{- 2}]$

The dimensions of RHS are

$[M] [LT^{- 2}] [L] = [M] [L^{2} T^{- 2}] = [M L^{2} T^{- 2}]$

The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

The SI unit of energy is $J = kg m^{2} s^{- 2}$ ; that of speed v is $ms^{- 1}$ and of acceleration a is $ms^{- 2}$ . Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body): (a) $K = m^{2} v^{3}$ (b) $K = (1/2) mv^{2}$ (c) $K = ma$ (d) $K = (3/16) m v^{2}$ (e) $K = (1/2) m v^{2} + ma$ Sol. Every correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are $[M^{2} L^{3} T^{- 3}]$ for (a): $[M L^{2} T^{- 2}]$ for (b) and (d); $[ML T^{- 2}]$ for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the dimensions of [ $ML^{2} T^{- 2}$ ], formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy (see chapter 5). The correct formula for kinetic energy is given by (b).
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length $(ℓ)$ . mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions. Sol. The dependence of time period T on the quantities $ℓ, g$ and m as a product may be written as:

$T = k ℓ^{x} g^{y} m^{z}$

where k is dimensionless constant and x, y and z are the exponents. By considering dimensions on both we have

$[L^{o} M^{o} T^{1}] = [L^{1}]^{x} [L^{1} T^{- 2}]^{y} [M^{1}]^{z} = L^{xy} M^{z} T^{- 2 y}$

On equating the dimensions on both sides, we have $x + y = 0; - 2 y = 1$ and $z = 0$ So that $x = \frac{1}{2}, y = - \frac{1}{2}, z = 0$ Then, $T = k ℓ^{1/2} g^{- 1/2}$ or $T = k \frac{ℓ}{g}$ Note that value of constant k can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions. Actually, $k = 2 π$ so that $T = 2 π \frac{ℓ}{g}$

EXERCISE QUESTION WITH SOLUTIONS

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 relatively density of lead is 11.3. Its density is .... $g cm^{- 3}$ or ... $kg m^{- 3}$ . Sol. (a) $1 cm = \frac{1}{100} m$ Volume of the cube $= 1 cm^{3}$ But, $1 cm^{3} = 1 cm \times 1 cm \times 1 cm$

$= (\frac{1}{100}) m \times (\frac{1}{100}) m \times (\frac{1}{100}) m ∴ 1 cm^{3} = 1 0^{- 6} m^{3}$

Hence, the volume of a cube of side 1 cm is equal to $1 0^{- 6} m^{3}$ . (b) The total surface area of a cylinder of radius r and height h is

$S = 2 π r (r + h)$

Give that

$r = 2 cm = 2 \times 1 cm = 2 \times 10 mm = 20 mm h = 10 cm = 10 \times 10 mm = 100 mm ∴ S = 2 \times 3.14 \times 20 (20 + 100) = 15072 = 1.5 \times 1 0^{4} mm^{2}$

$1 km / h = \frac{5}{18} m / s 18 km / h = 18 \times \frac{5}{18} = 5 m / s$

Therefore, distance can be obtained using the relation :

$Distance = Speed \times Time = 5 \times 1 = 5 m$

Hence, the vehicle covers 5m in 1s.

(d) Relative density of a substance is given by the relation, Relative density

$= \frac{Density of substance}{Density of water}$

Density of water $= kg / cm^{3}$ Density of lead = relative density of lead × Density of water

$= 11.3 \times 1 = 11.3 g / cm^{3}$

Again, $l g = \frac{1}{1000} kg$

$1 cm^{3} = 1 0^{- 6} m^{3}$

$1 g . cm^{3} = \frac{1 0 ^{- 3}}{1 0 ^{- 6}} lg / m^{3} = 1 0^{3} kg / m^{3}$

$∴ 11.3 g / cm^{3} = 11.3 \times 1 0^{3} kg / m^{3}$

Fill in the blanks by suitable conversion of units: (a) $1 kgm^{2} s^{- 2} = \dots .. gcm^{2} s^{- 2}$ (b) $1 m = \dots$ . 1 y (c) $3.0 ms^{- 2} = \dots . km h^{- 2}$ (d) $G = 6.67 \times 1 0^{- 11} Nm^{2} (kg)^{- 2} = \dots (cm)^{3} s^{- 2} g^{- 1}$ Sol. (a) $1 kg = 1 0^{3} g$

$1 m^{2} = 1 0^{4} cm^{2}$

$1 kg m^{2} s^{- 2} = 1 kg \times 1 m^{2} \times 1 s^{- 2}$

$= 1 0^{3} g \times 1 0^{4} cm^{2} \times 1 s^{- 2} = 1 0^{7} gcm^{2} s^{- 2}$

Light year is the total distance travelled by light in one year.

$11 y = Speed of light \times One year = (3 \times 1 0^{8} m / s) \times (365 \times 24 \times 60 \times 60 s) = 9.46 \times 1 0^{15} m$

$∴ 1 m = \frac{1}{9.46 \times 1 0 ^{15}} = 1.057 \times 1 0^{- 16} ly$

(b) $1 m = 1 0^{- 3} km$

$Again, 1 s = \frac{1}{3600} h 1 s^{- 1} = 3600 h^{- 1} 1 s^{- 1} = (3600)^{2} h^{- 2} ∴ 3 ms^{- 2} = (3 \times 1 0^{- 3} km) \times (36000)^{2} h^{- 2} = 3.88 \times 1 0^{- 4} kmh^{- 2}$

$∴ ∴ 6.67 \times 1 0^{- 11} Nm^{2} kg^{- 2} = 6.67 \times 1 0^{- 11} \times (1 kg m^{- 2}) (1 m^{2}) (1 s^{- 2}) 6.67 \times 1 0^{- 11} \times (1 kg \times 1 m^{3} \times 1 s^{- 2}) = 6.67 \times 1 0^{- 11} \times (1 0^{- 3} g^{- 1}) \times (1 0^{6} cm^{3}) \times (1 s^{- 2}) = 6.67 \times 1 0^{- 8} cm^{3} s^{- 2} g^{- 1}$

A calorie is a unit of heat or energy and it equals about 4.2 J , where $IJ = 1 kg m$ . 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. Sol. Given that, 1 calorie $= 4.2 (1 kg) (lm^{2}) (1 s^{- 2})$ New unit of mass $α kg$ Hence, in terms of the new unit,

$1 kg = \frac{1}{α} = α^{- 1}$

In terms of the new unit of length,

$lm \frac{1}{β} = β^{- 1} or 1 m^{2} = β^{- 2}$

And, in terms of the new unit of time,

$1 s = \frac{1}{γ} = γ^{- 1} 1 s^{2} = γ^{- 2} 1 s^{- 2} = γ^{2}$

$∴ 1 Calorie = 4.2 (1 α^{- 1}) (1 β^{- 2}) (1 γ^{2}) = 4.2 α^{- 1} β^{- 2} γ^{2}$

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. Sol. The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. (a) An atom is a very small object in comparison to a soccer ball. (b) A jet plane moves with a speed greater than that of a bicycle. (c) Mass of Jupiter is very large as compared to the mass of a cricket ball. (d) The air inside this room contains a large number of molecules as compared to that present in a geometry box. (e) A proton has a mass of $1.67 \times 1 0^{- 27} kg$ which is approximately 1836 times heavier than mass of an electron

$(9.11 \times 1 0^{- 31} kg) .$

(f) Speed of sound is very small than the speed of light. i.e.

$v_{sound} = 332 m / s at 0^{\circ} C v_{light} = 3 \times 1 0^{8} m / s$

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? Sol. Distance between the Sun and the Earth = Speed of light × Time taken by light to cover the distance Given that in the new unit,

$speed of light = 1 unit$

Time taken, $t = 8 min 20 s = 500 s$ ∴ Distance between the Sun and the Earth

$= 1 \times 500 = 500 units$

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?

Sol. A device with minimum count is the most suitable to measure length. (a) Least count of vernier callipers

$= 1 standard division (SD) - 1$

vernier division (VD) $= 1 - \frac{9}{10}$

$= \frac{1}{10} = 0.01 cm$

(b) Least count of screw gauge

$= \frac{pitch}{Number of divisions} = \frac{1}{1000} = 0.001 cm$

$= Wavelength of light = 0.00001 cm$

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

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? Sol. Magnification of the microscope $= 100$ Average width of the hair in the field of view of the microscope $= 3.5 mm$ ∴ Actual thickness of the hair is

$\frac{3.4}{100} = 0.035 mm$

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?

Sol. (a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation.

$Diameter = \frac{Length of thread}{Number of turns}$

(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular seale will increase its accuracy to a certain extent only. (c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

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 $1.55 m^{2}$ is the linear magnification of the projectorscreen arrangement? Sol. Area of the house on the slide $= 1.75 cm^{2}$ Area of the image of the house formed on the screen $= 1.55 m^{2} = 1.55 \times 1 0^{4} cm^{2}$ Arial magnification

$m_{a} = \frac{Area of image}{Area of object} = \frac{1.55}{1.75} \times 1 0^{4}$

$∴$ Linear magnifications,

$m_{ℓ} = m_{a} = \frac{1.55}{1.75} \times 1 0^{4} = 94.11$

State the number of significant figures in the following: (a) $0.007 m^{2}$ (b) $2.64 \times 1 0^{24} kg$ (c) $0.2370 g cm^{- 3}$ (d) 6.320J (e) $6.032 Nm^{- 2}$ (f) $0.0006032 m^{2}$

Sol. (a) The given quantity is $0.007 m^{2}$ "If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant". This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity. (b) The given quantity is $2.64 \times 1 0^{24} kg$ . Here, "the power of 10 is irrelevant for the determination of significant figures". Hence, all digits i.e., 2, 6 and 4 are significant figure. (c) The given quantity is $0.2370 g cm^{- 3}$ "For a number with decimals, the trailing zeroes are significant". Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure. (d) The given quantity is 6.320 J "For a number with decimals, the trailing zeroes are significant". Hence, all four digits appearing in the given quantity are significant figures. (e) The given quantity is $6.032 Nm^{- 2}$ . "All zeroes between two non-zero digits are always significant". (f) The given quantity is $0.0006032 m^{2}$ "If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant". Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence. the remaining four digits are significant figures.

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. Sol. Length of sheet, $ℓ = 4.234 m$ Breadth of sheet $b = 1.005 m$ Thickness of sheet $h = 2.01 cm = 0.0201 m$ The given table lists the respective significant figures:

Quantity	Number	Significant Figure
$ℓ$	4.234	4
b	1.005	4
h	2.01	3

Hence, area and volume both must have least significant figures i.e. 3. Surface area of the sheet

$= 2 (ℓ \times b + b \times h + h \times ℓ) = 2 (4.234 \times 1.005 + 1.005 \times 0.0201 + 0.0201 \times 4.234) = 2 (4.25517 + 0.02020 + 0.08510) = 2 \times 4.360 = 8.72 m^{2}$

Volume of the sheet $= 1 \times b \times h$

$= 4.234 \times 1.005 \times 0.0201 = 0.0855 m^{3}$

This number has only 3 significant figures i.e., 8, 5, and 5.

The mass of a box measured by a grocer's balance is 2.300 kg. Two gold pieces 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? Sol. Mass of grocer's box = 2.300 kg Mass of gold piece $I = 20.15 g = 0.02015 kg$ Mass of gold piece $II = 20.17 g = 0.02017 kg$ Total mass of the

$= 2.3 + 0.02015 + 0.02017 = 2.34032 kg$

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg. Difference in masses $= 20.17 - 20.15 = 0.02 g$ In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

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: $m = \frac{m _{0}}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$

Sol. Given the relation, $m = \frac{m _{0}}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$ Dimension of $m = M^{1} L^{0} T^{0}$ Dimension of $m_{0} = M^{1} L^{0} T^{0}$ Dimension of $v = M^{0} L^{1} T^{- 1}$ Dimension of $v^{2} = M^{0} L^{2} T^{- 2}$ Dimension of $c = M^{0} L^{1} T^{- 1}$ The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, $(1 - v^{2})^{\frac{1}{2}}$ is dimensionless. i.e., $(1 - v^{2})$ is dimensionless. This is only possible if $v^{2}$ is divided by $c^{2}$ . Hence, the correct relation is

$m = \frac{m _{0}}{( 1 - \frac{v ^{2}}{c ^{2}} ) ^{\frac{1}{2}}}$

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A : $1 A = 1 0^{- 10} m$ . The size of a hydrogen atom is about 0.5A what is the total atomic volume in $m^{3}$ of a mole of hydrogen atoms? Sol. Radius of hydrogen atom,

$r = 0.5 A = 0.5 \times 1 0^{- 10} m$

Volume of hydrogen atom $= \frac{4}{3} π r^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times (0.5 \times 1 0^{- 10})^{3} = 0.524 \times 1 0^{- 30} m^{3}$

1 mole of hydrogen contains $6.023 \times 1 0^{23}$ hydrogen atoms. $∴$ Volume of 1 mole of hydrogen atoms

$= 6.023 \times 1 0^{23} \times 0.524 \times 1 0^{- 30} = 3.16 \times 1 0^{- 7} m^{3}$

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 1A). Why is this ratio so large? Sol. Radius of hydrogen atom

$r = 0.5 A = 0.5 \times 1 0^{- 10} m$

Volume of hydrogen atom $= \frac{4}{3} π r^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times (0.5 \times 1 0^{- 10})^{3} = 0.524 \times 1 0^{- 30} m^{3}$

Now, 1 mole of hydrogen contains $6.023 \times 1 0^{23}$ hydrogen atoms. $∴$ Volume of 1 mole of hydrogen atoms

$V_{a} = 6.023 \times 1 0^{23} \times 0.524 \times 1 0^{- 30} = 3.16 \times 1 0^{- 7} m^{3}$

Molar volume of 1 mole of hydrogen atoms at STP

$V_{m} = 22.4 L = 22.4 \times 1 0^{- 3} m^{3} ∴ \frac{V _{m}}{V _{a}} = \frac{22.4 \times 1 0 ^{- 3}}{3.16 \times 1 0 ^{- 7}} = 7.09 \times 1 0^{4}$

Hence, the molar volume is $7.09 \times 1 0^{4}$ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

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).

Sol. Line of sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly. [Parallax] On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $1 0^{7} K$ 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 $= 2.0 \times 1 0^{30} kg$ , radius of the Sun $= 7.0 \times 1 0^{8} m$ Sol. Mass of the Sun, $M = 2.0 \times 1 0^{30} kg$ Radius of the Sun, $R = 7.0 \times 1 0^{8} m$ Volume of the Sun, $V = \frac{4}{3} π R^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times (7.0 \times 1 0^{8})^{3} = \frac{88}{21} \times 343 \times 1 0^{24} = 1437.3 \times 1 0^{24} m^{3}$

Density of the Sun $= \frac{Mass}{Volume}$

$= \frac{2.0 \times 1 0 ^{30}}{1437.3 \times 1 0 ^{24}} \sim 1.4 \times 1 0^{3} kg / m^{3}$

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the out er layer of the Sun.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Explore Chapter-wise NCERT Solutions for Class 11 Physics with detailed explanation for each and every question. Understand textbook questions, learn key concepts and build your Physics fundamentals chapter by chapter.

Chapter Number	Chapterwise NCERT Solutions
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 1

Step-by-Step Error Calculations: Understand absolute error, relative error, percentage error, and error propagation through clear, step-by-step solutions for addition, subtraction, multiplication, and division.
Easy Understanding of Dimensional Analysis: Learn how to check the dimensional consistency of physical equations such as v2=u2+2asv^2 = u^2 + 2as and T=2πl/gT = 2\pi\sqrt{l/g}. The solutions explain how dimensions are used to verify physical relationships.
Accurate Significant Figure Calculations: Rules for significant figures, rounding off, calculations involving measured quantities. Detailed steps help students to avoid mistakes caused by incorrect precision.
Detailed Solutions to Numerical Problems: Get step-by-step solutions for NCERT numerical questions based on density, volume, units, measurements, errors, and significant figures, with proper calculations and units.
Prepared by ALLEN Subject Experts These NCERT Solutions for Class 11 Physics Chapter 1 are prepared by ALLEN subject experts, with emphasis on accuracy, conceptual clarity, and alignment with the latest NCERT syllabus.
Simple and Conceptual Explanation: Important concepts of Units and Measurements are explained in simple language with logical steps, so that students can understand the chapter without much complexity.
Complete NCERT Question Coverage: Get detailed solutions to the NCERT Class 11 Physics Chapter 1 questions and exercises, covering important concepts required for CBSE, JEE, and NEET preparation.

NCERT Solutions Class 11 Physics Chapter 1 – Units and Measurements

1.0Class 11 Physics Chapter 1 : Key Concepts

SI Units: Understand the seven fundamental SI units: metre, kilogram, second, ampere, kelvin, mole, and candela. The chapter also introduces units for measuring plane and solid angles.
Measurement of Large Distances: Learn the parallax method for estimating the distances of stars and other distant objects.
Measurement of Mass and Time: Study different methods used to measure mass and time, including the role of highly accurate atomic clocks.
Accuracy, Precision, and Errors: Understand accuracy and precision along with systematic errors, random errors, absolute error, relative error, and percentage error.
Significant Figures: Learn how to identify significant figures, apply rounding-off rules, and represent measured values with the appropriate precision.
Dimensions of Physical Quantities: Express physical quantities using fundamental dimensions such as [M], [L], [T], and [A].
Dimensional Analysis: Learn how dimensional analysis helps check the correctness of physical equations, derive relationships between physical quantities, and convert units from one system to another.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 1

SOLVED EXAMPLES

Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the cube to appropriate significant figures? Sol. The number of significant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures.

$Surface area of the cube Volume of the cube = 6 (7.203)^{2} m^{2} = 311.299254 m^{2} = 311.3 m^{2} = (7.203)^{3} m^{3} = 373.714754 m^{3} = 373.7 m^{3}$

5.74 g of a substance occupies $1.2 cm^{3}$ . Express its density by keeping the significant figures in view. Sol. There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.

$Density = \frac{5.74}{1.2} g cm^{- 3} = 4.8 g cm^{- 3} .$

Let us consider an equation $\frac{1}{2} m v^{2} = m g h$ where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct. Sol. The dimensions of LHS are

$[M] [LT^{- 1}]^{2} = [M] [L^{2} T^{- 2}] = [M L^{2} T^{- 2}]$

The dimensions of RHS are

$[M] [LT^{- 2}] [L] = [M] [L^{2} T^{- 2}] = [M L^{2} T^{- 2}]$

The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.

The SI unit of energy is $J = kg m^{2} s^{- 2}$ ; that of speed v is $ms^{- 1}$ and of acceleration a is $ms^{- 2}$ . Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body): (a) $K = m^{2} v^{3}$ (b) $K = (1/2) mv^{2}$ (c) $K = ma$ (d) $K = (3/16) m v^{2}$ (e) $K = (1/2) m v^{2} + ma$ Sol. Every correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are $[M^{2} L^{3} T^{- 3}]$ for (a): $[M L^{2} T^{- 2}]$ for (b) and (d); $[ML T^{- 2}]$ for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the dimensions of [ $ML^{2} T^{- 2}$ ], formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy (see chapter 5). The correct formula for kinetic energy is given by (b).
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length $(ℓ)$ . mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions. Sol. The dependence of time period T on the quantities $ℓ, g$ and m as a product may be written as:

$T = k ℓ^{x} g^{y} m^{z}$

where k is dimensionless constant and x, y and z are the exponents. By considering dimensions on both we have

$[L^{o} M^{o} T^{1}] = [L^{1}]^{x} [L^{1} T^{- 2}]^{y} [M^{1}]^{z} = L^{xy} M^{z} T^{- 2 y}$

EXERCISE QUESTION WITH SOLUTIONS

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 relatively density of lead is 11.3. Its density is .... $g cm^{- 3}$ or ... $kg m^{- 3}$ . Sol. (a) $1 cm = \frac{1}{100} m$ Volume of the cube $= 1 cm^{3}$ But, $1 cm^{3} = 1 cm \times 1 cm \times 1 cm$

$= (\frac{1}{100}) m \times (\frac{1}{100}) m \times (\frac{1}{100}) m ∴ 1 cm^{3} = 1 0^{- 6} m^{3}$

Hence, the volume of a cube of side 1 cm is equal to $1 0^{- 6} m^{3}$ . (b) The total surface area of a cylinder of radius r and height h is

$S = 2 π r (r + h)$

Give that

$r = 2 cm = 2 \times 1 cm = 2 \times 10 mm = 20 mm h = 10 cm = 10 \times 10 mm = 100 mm ∴ S = 2 \times 3.14 \times 20 (20 + 100) = 15072 = 1.5 \times 1 0^{4} mm^{2}$

$1 km / h = \frac{5}{18} m / s 18 km / h = 18 \times \frac{5}{18} = 5 m / s$

Therefore, distance can be obtained using the relation :

$Distance = Speed \times Time = 5 \times 1 = 5 m$

Hence, the vehicle covers 5m in 1s.

(d) Relative density of a substance is given by the relation, Relative density

$= \frac{Density of substance}{Density of water}$

Density of water $= kg / cm^{3}$ Density of lead = relative density of lead × Density of water

$= 11.3 \times 1 = 11.3 g / cm^{3}$

Again, $l g = \frac{1}{1000} kg$

$1 cm^{3} = 1 0^{- 6} m^{3}$

$1 g . cm^{3} = \frac{1 0 ^{- 3}}{1 0 ^{- 6}} lg / m^{3} = 1 0^{3} kg / m^{3}$

$∴ 11.3 g / cm^{3} = 11.3 \times 1 0^{3} kg / m^{3}$

Fill in the blanks by suitable conversion of units: (a) $1 kgm^{2} s^{- 2} = \dots .. gcm^{2} s^{- 2}$ (b) $1 m = \dots$ . 1 y (c) $3.0 ms^{- 2} = \dots . km h^{- 2}$ (d) $G = 6.67 \times 1 0^{- 11} Nm^{2} (kg)^{- 2} = \dots (cm)^{3} s^{- 2} g^{- 1}$ Sol. (a) $1 kg = 1 0^{3} g$

$1 m^{2} = 1 0^{4} cm^{2}$

$1 kg m^{2} s^{- 2} = 1 kg \times 1 m^{2} \times 1 s^{- 2}$

$= 1 0^{3} g \times 1 0^{4} cm^{2} \times 1 s^{- 2} = 1 0^{7} gcm^{2} s^{- 2}$

Light year is the total distance travelled by light in one year.

$11 y = Speed of light \times One year = (3 \times 1 0^{8} m / s) \times (365 \times 24 \times 60 \times 60 s) = 9.46 \times 1 0^{15} m$

$∴ 1 m = \frac{1}{9.46 \times 1 0 ^{15}} = 1.057 \times 1 0^{- 16} ly$

(b) $1 m = 1 0^{- 3} km$

$Again, 1 s = \frac{1}{3600} h 1 s^{- 1} = 3600 h^{- 1} 1 s^{- 1} = (3600)^{2} h^{- 2} ∴ 3 ms^{- 2} = (3 \times 1 0^{- 3} km) \times (36000)^{2} h^{- 2} = 3.88 \times 1 0^{- 4} kmh^{- 2}$

A calorie is a unit of heat or energy and it equals about 4.2 J , where $IJ = 1 kg m$ . 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. Sol. Given that, 1 calorie $= 4.2 (1 kg) (lm^{2}) (1 s^{- 2})$ New unit of mass $α kg$ Hence, in terms of the new unit,

$1 kg = \frac{1}{α} = α^{- 1}$

In terms of the new unit of length,

$lm \frac{1}{β} = β^{- 1} or 1 m^{2} = β^{- 2}$

And, in terms of the new unit of time,

$1 s = \frac{1}{γ} = γ^{- 1} 1 s^{2} = γ^{- 2} 1 s^{- 2} = γ^{2}$

$∴ 1 Calorie = 4.2 (1 α^{- 1}) (1 β^{- 2}) (1 γ^{2}) = 4.2 α^{- 1} β^{- 2} γ^{2}$

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. Sol. The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. (a) An atom is a very small object in comparison to a soccer ball. (b) A jet plane moves with a speed greater than that of a bicycle. (c) Mass of Jupiter is very large as compared to the mass of a cricket ball. (d) The air inside this room contains a large number of molecules as compared to that present in a geometry box. (e) A proton has a mass of $1.67 \times 1 0^{- 27} kg$ which is approximately 1836 times heavier than mass of an electron

$(9.11 \times 1 0^{- 31} kg) .$

(f) Speed of sound is very small than the speed of light. i.e.

$v_{sound} = 332 m / s at 0^{\circ} C v_{light} = 3 \times 1 0^{8} m / s$

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? Sol. Distance between the Sun and the Earth = Speed of light × Time taken by light to cover the distance Given that in the new unit,

$speed of light = 1 unit$

Time taken, $t = 8 min 20 s = 500 s$ ∴ Distance between the Sun and the Earth

$= 1 \times 500 = 500 units$

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?

Sol. A device with minimum count is the most suitable to measure length. (a) Least count of vernier callipers

$= 1 standard division (SD) - 1$

vernier division (VD) $= 1 - \frac{9}{10}$

$= \frac{1}{10} = 0.01 cm$

(b) Least count of screw gauge

$= \frac{pitch}{Number of divisions} = \frac{1}{1000} = 0.001 cm$

$= Wavelength of light = 0.00001 cm$

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

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? Sol. Magnification of the microscope $= 100$ Average width of the hair in the field of view of the microscope $= 3.5 mm$ ∴ Actual thickness of the hair is

$\frac{3.4}{100} = 0.035 mm$

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?

$Diameter = \frac{Length of thread}{Number of turns}$

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 $1.55 m^{2}$ is the linear magnification of the projectorscreen arrangement? Sol. Area of the house on the slide $= 1.75 cm^{2}$ Area of the image of the house formed on the screen $= 1.55 m^{2} = 1.55 \times 1 0^{4} cm^{2}$ Arial magnification

$m_{a} = \frac{Area of image}{Area of object} = \frac{1.55}{1.75} \times 1 0^{4}$

$∴$ Linear magnifications,

$m_{ℓ} = m_{a} = \frac{1.55}{1.75} \times 1 0^{4} = 94.11$

State the number of significant figures in the following: (a) $0.007 m^{2}$ (b) $2.64 \times 1 0^{24} kg$ (c) $0.2370 g cm^{- 3}$ (d) 6.320J (e) $6.032 Nm^{- 2}$ (f) $0.0006032 m^{2}$

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. Sol. Length of sheet, $ℓ = 4.234 m$ Breadth of sheet $b = 1.005 m$ Thickness of sheet $h = 2.01 cm = 0.0201 m$ The given table lists the respective significant figures:

Quantity	Number	Significant Figure
$ℓ$	4.234	4
b	1.005	4
h	2.01	3

Hence, area and volume both must have least significant figures i.e. 3. Surface area of the sheet

$= 2 (ℓ \times b + b \times h + h \times ℓ) = 2 (4.234 \times 1.005 + 1.005 \times 0.0201 + 0.0201 \times 4.234) = 2 (4.25517 + 0.02020 + 0.08510) = 2 \times 4.360 = 8.72 m^{2}$

Volume of the sheet $= 1 \times b \times h$

$= 4.234 \times 1.005 \times 0.0201 = 0.0855 m^{3}$

This number has only 3 significant figures i.e., 8, 5, and 5.

The mass of a box measured by a grocer's balance is 2.300 kg. Two gold pieces 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? Sol. Mass of grocer's box = 2.300 kg Mass of gold piece $I = 20.15 g = 0.02015 kg$ Mass of gold piece $II = 20.17 g = 0.02017 kg$ Total mass of the

$= 2.3 + 0.02015 + 0.02017 = 2.34032 kg$

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: $m = \frac{m _{0}}{( 1 - v ^{2} ) ^{\frac{1}{2}}}$

$m = \frac{m _{0}}{( 1 - \frac{v ^{2}}{c ^{2}} ) ^{\frac{1}{2}}}$

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A : $1 A = 1 0^{- 10} m$ . The size of a hydrogen atom is about 0.5A what is the total atomic volume in $m^{3}$ of a mole of hydrogen atoms? Sol. Radius of hydrogen atom,

$r = 0.5 A = 0.5 \times 1 0^{- 10} m$

Volume of hydrogen atom $= \frac{4}{3} π r^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times (0.5 \times 1 0^{- 10})^{3} = 0.524 \times 1 0^{- 30} m^{3}$

1 mole of hydrogen contains $6.023 \times 1 0^{23}$ hydrogen atoms. $∴$ Volume of 1 mole of hydrogen atoms

$= 6.023 \times 1 0^{23} \times 0.524 \times 1 0^{- 30} = 3.16 \times 1 0^{- 7} m^{3}$

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 1A). Why is this ratio so large? Sol. Radius of hydrogen atom

$r = 0.5 A = 0.5 \times 1 0^{- 10} m$

Volume of hydrogen atom $= \frac{4}{3} π r^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times (0.5 \times 1 0^{- 10})^{3} = 0.524 \times 1 0^{- 30} m^{3}$

Now, 1 mole of hydrogen contains $6.023 \times 1 0^{23}$ hydrogen atoms. $∴$ Volume of 1 mole of hydrogen atoms

$V_{a} = 6.023 \times 1 0^{23} \times 0.524 \times 1 0^{- 30} = 3.16 \times 1 0^{- 7} m^{3}$

Molar volume of 1 mole of hydrogen atoms at STP

$V_{m} = 22.4 L = 22.4 \times 1 0^{- 3} m^{3} ∴ \frac{V _{m}}{V _{a}} = \frac{22.4 \times 1 0 ^{- 3}}{3.16 \times 1 0 ^{- 7}} = 7.09 \times 1 0^{4}$

Hence, the molar volume is $7.09 \times 1 0^{4}$ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

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).

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $1 0^{7} K$ 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 $= 2.0 \times 1 0^{30} kg$ , radius of the Sun $= 7.0 \times 1 0^{8} m$ Sol. Mass of the Sun, $M = 2.0 \times 1 0^{30} kg$ Radius of the Sun, $R = 7.0 \times 1 0^{8} m$ Volume of the Sun, $V = \frac{4}{3} π R^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times (7.0 \times 1 0^{8})^{3} = \frac{88}{21} \times 343 \times 1 0^{24} = 1437.3 \times 1 0^{24} m^{3}$

Density of the Sun $= \frac{Mass}{Volume}$

$= \frac{2.0 \times 1 0 ^{30}}{1437.3 \times 1 0 ^{24}} \sim 1.4 \times 1 0^{3} kg / m^{3}$

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the out er layer of the Sun.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Chapter Number	Chapterwise NCERT Solutions
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 1

Step-by-Step Error Calculations: Understand absolute error, relative error, percentage error, and error propagation through clear, step-by-step solutions for addition, subtraction, multiplication, and division.
Easy Understanding of Dimensional Analysis: Learn how to check the dimensional consistency of physical equations such as v2=u2+2asv^2 = u^2 + 2as and T=2πl/gT = 2\pi\sqrt{l/g}. The solutions explain how dimensions are used to verify physical relationships.
Accurate Significant Figure Calculations: Rules for significant figures, rounding off, calculations involving measured quantities. Detailed steps help students to avoid mistakes caused by incorrect precision.
Detailed Solutions to Numerical Problems: Get step-by-step solutions for NCERT numerical questions based on density, volume, units, measurements, errors, and significant figures, with proper calculations and units.
Prepared by ALLEN Subject Experts These NCERT Solutions for Class 11 Physics Chapter 1 are prepared by ALLEN subject experts, with emphasis on accuracy, conceptual clarity, and alignment with the latest NCERT syllabus.
Simple and Conceptual Explanation: Important concepts of Units and Measurements are explained in simple language with logical steps, so that students can understand the chapter without much complexity.
Complete NCERT Question Coverage: Get detailed solutions to the NCERT Class 11 Physics Chapter 1 questions and exercises, covering important concepts required for CBSE, JEE, and NEET preparation.

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NCERT Solutions Class 11 Physics Chapter 1 – Units and Measurements

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3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

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NCERT Solutions Class 11 Physics Chapter 1 – Units and Measurements

1.0Class 11 Physics Chapter 1 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 1

SOLVED EXAMPLES

EXERCISE QUESTION WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 1

Table of Contents