NCERT Solutions for Class 11 Physics Chapter 1: Units and Measurements (NCERT 2026–27)

These Class 11 Physics Chapter 1 solutions cover Units and Measurements with complete, step-by-step answers to every NCERT exercise (1.1–1.17). You will master the SI system of units, scientific notation, significant figures, error and rounding rules, and dimensional analysis. All numericals are solved with units shown at every step and the final answer cross-checked against the NCERT answer key, ready for session 2026–27.

Class: 11 Subject: Physics Chapter: 1 Title: Units and Measurements Exercises: 1.1 – 1.17 Session: 2026–27
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Class 11 Physics Chapter 1 – Overview

Chapter 1, Units and Measurements, lays the quantitative foundation of physics. Every physical quantity is expressed as a number together with a unit. The chapter introduces the seven SI base quantities (length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity) and their units (metre, kilogram, second, ampere, kelvin, mole and candela), from which all derived units are built. It then explains how to report measurements honestly using scientific notation and significant figures, how errors and rounding propagate through calculations, and finally dimensions — the powers of base quantities — which let us check the consistency of equations and even guess relations among physical quantities. These tools are used throughout the rest of the Class 11 and 12 Physics course.

Key Concepts & Definitions

Unit: an internationally accepted reference standard used to compare and express a physical quantity (e.g. metre for length).

Base & derived units: the seven SI base units stand on their own; derived units (m s−1, N, J…) are combinations of base units.

Significant figures: all the reliably known digits in a measurement plus the first uncertain digit. They indicate the precision of a measurement and do not change with a change of units.

Scientific notation: writing a number as a × 10b, where 1 ≤ a < 10. It removes all ambiguity about trailing zeros — every digit in a is significant.

Order of magnitude: the power of ten nearest to a quantity, obtained by rounding the mantissa to 1 (if ≤ 5) or 10 (if > 5).

Dimensions: the powers to which the base quantities [L], [M], [T], [A], [K], [cd], [mol] are raised to represent a physical quantity.

Principle of homogeneity: every term on both sides of a correct physical equation must have the same dimensions; only quantities of identical dimensions can be added or subtracted.

Important Formulas

Significant figures – multiplication/division: the result keeps as many significant figures as the factor with the fewest.

Significant figures – addition/subtraction: the result keeps as many decimal places as the term with the fewest.

Dimensional formulae (mechanics): Area [L2], Volume [L3], Speed/Velocity [L T−1], Acceleration [L T−2], Force [M L T−2], Energy/Work [M L2 T−2], Density [M L−3].

Combination of errors (product/quotient): the percentage (relative) errors add. For Z = AB or A/B, ΔZ/Z = ΔA/A + ΔB/B.

Method of dimensions (example): for a simple pendulum, T ∝ lx gy mz gives T = 2π√(l/g).

NCERT Solutions for Class 11 Physics Chapter 1 – Exercises

Questions are reproduced verbatim from the NCERT textbook (Exercises 1.1–1.17). Note: in stating numerical answers, take care of significant figures.

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

ANSWER (a) 1 cm = 10−2 m, so 1 cm3 = (10−2 m)3 = 10−6 m3. (b) Total surface area of a solid cylinder = 2πr(r + h) = 2π(2.0)(2.0 + 10.0) = 2π(2.0)(12.0) = 48π ≈ 150.8 cm2. Converting: 1 cm2 = 100 mm2, so area = 150.8 × 100 ≈ 1.5 × 104 mm2. (c) 18 km h−1 = 18 × (1000 m / 3600 s) = 5 m s−1. In 1 s it covers 5 m. (d) Relative density 11.3 means density = 11.3 × density of water = 11.3 × 1 g cm−3 = 11.3 g cm−3. Since 1 g cm−3 = 103 kg m−3, density = 11.3 × 103 = 1.13 × 104 kg m−3.

1.2 Fill in the blanks by suitable conversion of units (a) 1 kg m2 s−2 = ….g cm2 s−2 (b) 1 m = ….. ly (c) 3.0 m s−2 = …. km h−2 (d) G = 6.67 × 10−11 N m2 (kg)−2 = …. (cm)3 s−2 g−1.

ANSWER (a) 1 kg = 103 g and 1 m2 = 104 cm2. So 1 kg m2 s−2 = (103)(104) g cm2 s−2 = 107 g cm2 s−2. (b) 1 light year = 9.46 × 1015 m. Therefore 1 m = 1 / (9.46 × 1015) ≈ 1.06 × 10−16 ly (i.e. of the order of 10−16 ly). (c) 3.0 m s−2 = 3.0 × (10−3 km) / (1/3600 h)2 = 3.0 × 10−3 × (3600)2 km h−2 = 3.0 × 10−3 × 1.296 × 107 ≈ 3.9 × 104 km h−2. (d) N = kg m s−2, so N m2 kg−2 = m3 s−2 kg−1. Convert: m3 = 106 cm3 and kg−1 = (103 g)−1 = 10−3 g−1. So G = 6.67 × 10−11 × 106 × 10−3 = 6.67 × 10−8 cm3 s−2 g−1.

1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 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 γ2 in terms of the new units.

ANSWER The dimensional formula of energy is [M L2 T−2]. If n1 is the numerical value in one system and n2 in another, then n2 = n1 (M1/M2)1 (L1/L2)2 (T1/T2)−2. Here n1 = 4.2; M1 = 1 kg, L1 = 1 m, T1 = 1 s; M2 = α kg, L2 = β m, T2 = γ s. n2 = 4.2 (1/α)1 (1/β)2 (1/γ)−2 = 4.2 α−1 β−2 γ2. Hence a calorie has the magnitude 4.2 α−1 β−2 γ2 in the new units. (Proved.)

1.4 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 Explanation: “Large” or “small” are relative terms. A quantity can only be judged large or small compared with a chosen standard. An ant is large compared with a bacterium but small compared with an elephant, so the words mean nothing until the reference is fixed. (a) Atoms are very small objects compared with the tip of a pin (an atom ≈ 10−10 m). (b) A jet plane moves with a speed greater than that of a superfast train (or a car). (c) The mass of Jupiter is very large compared with the mass of the Earth (about 318 times). (d) The air inside this room contains a number of molecules far larger than the number contained in the same volume at high altitude (statement is already comparative once a standard is named). (e) Correct as it stands — the comparison (proton vs electron) is already specified; a proton is about 1836 times more massive than an electron. (f) Correct as it stands — the comparison (sound vs light) is already specified; speed of sound (≈ 340 m s−1) is much smaller than the speed of light (3 × 108 m s−1).

1.5 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 = speed × time. In the new unit, speed of light c = 1 new-unit per second. Time t = 8 min 20 s = (8 × 60) + 20 = 480 + 20 = 500 s. Distance = c × t = 1 × 500 = 500 new units.

1.6 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 = 1 mm / 20 = 0.05 mm = 5 × 10−5 m. (b) Least count = pitch / number of divisions = 1 mm / 100 = 0.01 mm = 1 × 10−5 m. (c) Wavelength of light ≈ 5 × 10−7 m, which is the least count. The smallest least count belongs to (c). Hence the optical instrument (c) is the most precise device.

1.7 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 Observed (magnified) width = 3.5 mm; magnification = 100. Real thickness = observed width / magnification = 3.5 mm / 100 = 0.035 mm (= 3.5 × 10−5 m).

1.8 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) Wind the thread closely (turns touching, no gaps) many times around a pencil so it forms a compact coil. Measure the length L of the coil with the metre scale and count the number of turns n. Then diameter of the thread = L / n. Taking many turns increases accuracy. (b) In principle increasing the number of divisions lowers the least count (least count = pitch / divisions), improving accuracy. But beyond a point the divisions become too fine to read reliably with the eye, and other practical limitations dominate. So accuracy cannot be increased arbitrarily. (c) Random errors are reduced by averaging. The larger the number of independent readings, the more the random (accidental) errors cancel out, so the mean of 100 measurements is more reliable than the mean of only 5.

1.9 The photograph of a house occupies an area of 1.75 cm2 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 m2. What is the linear magnification of the projector-screen arrangement.

ANSWER Areal magnification = area on screen / area on slide. Convert to the same unit: 1.55 m2 = 1.55 × 104 cm2 = 15500 cm2. Areal magnification = 15500 / 1.75 = 8857.1. Linear magnification = √(areal magnification) = √8857.1 ≈ 94.1.

1.10 State the number of significant figures in the following : (a) 0.007 m2   (b) 2.64 × 1024 kg   (c) 0.2370 g cm−3   (d) 6.320 J   (e) 6.032 N m−2   (f) 0.0006032 m2

ANSWER (a) 0.007 → leading zeros are not significant → 1 significant figure. (b) 2.64 × 1024 → in scientific notation only the mantissa counts → 3 significant figures. (c) 0.2370 → digits 2, 3, 7 and the trailing 0 (after decimal) are significant → 4 significant figures. (d) 6.320 → trailing zero after the decimal is significant → 4 significant figures. (e) 6.032 → the zero between non-zero digits is significant → 4 significant figures. (f) 0.0006032 → leading zeros not significant; 6, 0, 3, 2 are → 4 significant figures.

1.11 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 l = 4.234 m (4 s.f.), b = 1.005 m (4 s.f.), t = 2.01 cm = 0.0201 m (3 s.f.). Surface area = 2(lb + bt + tl). Compute: lb = 4.234 × 1.005 = 4.2552; bt = 1.005 × 0.0201 = 0.0202; tl = 0.0201 × 4.234 = 0.0851. Sum = 4.3605; × 2 = 8.7210 m2. Rounded to the least s.f. (3) → 8.72 m2. Volume = l × b × t = 4.234 × 1.005 × 0.0201 = 0.08552 m3. Rounded to 3 s.f. (least, from 2.01) → 0.0855 m3 (= 8.55 × 10−2 m3).

1.12 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 ?

ANSWER Box = 2.30 kg = 2300 g; pieces = 20.15 g and 20.17 g. (a) Total = 2300 + 20.15 + 20.17 = 2340.32 g. In addition, keep the fewest decimal places — the box mass 2.30 kg is known only to 2 decimal places in kg (i.e. nearest 10 g). So the total = 2340.32 g ≈ 2.3 kg (rounded to the precision of the least precise term). (b) Difference = 20.17 − 20.15 = 0.02 g. Both values have 2 decimal places, so the difference is 0.02 g (correct to 2 decimal places).

1.13 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 = m0 / (1 − v2)1/2. Guess where to put the missing c.

ANSWER The quantity inside the bracket must be dimensionless, because we cannot subtract a pure number (1) from a quantity that has dimensions. The term v2 has dimensions [L2 T−2], so it must be divided by another speed-squared to become dimensionless. Dividing v2 by c2 (also [L2 T−2]) makes v2/c2 dimensionless. Hence the correct relation is m = m0 / (1 − v2/c2)1/2.

1.14 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 m3 of a mole of hydrogen atoms ?

ANSWER Radius of one atom r = 0.5 Å = 0.5 × 10−10 m = 5 × 10−11 m. Volume of one atom = (4/3)πr3 = (4/3)(3.14)(5 × 10−11)3 = (4/3)(3.14)(1.25 × 10−31) = 5.24 × 10−31 m3. One mole has NA = 6.022 × 1023 atoms. Total atomic volume = (6.022 × 1023)(5.24 × 10−31) = 3.16 × 10−7 ≈ 3 × 10−7 m3.

1.15 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 ?

ANSWER Molar (gas) volume = 22.4 L = 22.4 × 10−3 m3 = 2.24 × 10−2 m3. Radius of one molecule r = 0.5 Å = 5 × 10−11 m (size 1 Å means diameter 1 Å). Atomic volume of a mole = NA × (4/3)πr3 = 6.022 × 1023 × 5.24 × 10−31 = 3.16 × 10−7 m3 (from Q 1.14). Ratio = molar volume / atomic volume = (2.24 × 10−2) / (3.16 × 10−7) = 7.1 × 104 ≈ 104. Reason: in a gas the molecules are very far apart — the inter-molecular separation is much larger than the size of a molecule — so the volume occupied by the gas is enormously greater than the actual volume of the molecules themselves.

1.16 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. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

ANSWER The apparent (angular) speed of an object depends on the angle it sweeps at the eye in a given time. As the train moves a distance, a nearby object subtends a large angle at the eye, so it appears to shift quickly — and in the direction opposite to the train’s motion. A distant object (hill, Moon, star) subtends a very small angle for the same train displacement, because the angle = (distance moved) / (distance of object) is tiny when the object is far away. Its apparent shift is negligible, so it seems almost stationary — and because you know you are moving, it appears to move along with you.

1.17 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 × 1030 kg, radius of the Sun = 7.0 × 108 m.

ANSWER Volume of the Sun = (4/3)πR3 = (4/3)(3.14)(7.0 × 108)3 = (4/3)(3.14)(3.43 × 1026) = 1.437 × 1027 m3. Density = mass / volume = (2.0 × 1030) / (1.437 × 1027) = 1.4 × 103 kg m−3. This value (≈ 1400 kg m−3) lies in the range of densities of solids and liquids, not gases, even though the Sun is a hot plasma. The high density arises from the enormous inward gravitational pull of the inner layers on the outer layers, which compresses the matter strongly.

Extra Practice Questions

Short Answer Type Questions

Q1. Define one parsec and express it in metres.

ANSWERA parsec is the distance at which an arc of 1 astronomical unit (AU) subtends an angle of 1 second of arc. 1 parsec = 3.08 × 1016 m ≈ 3.26 light years.

Q2. Why are the trailing zeros in 4700 m ambiguous, and how does scientific notation remove the ambiguity?

ANSWERWithout a decimal, the trailing zeros may or may not be significant, so 4700 m could mean 2, 3 or 4 significant figures. Writing it as 4.7 × 103 m (2 s.f.) or 4.700 × 103 m (4 s.f.) fixes the precision unambiguously, since every digit in the mantissa is significant.

Q3. The radius of a sphere is (5.3 ± 0.1) cm. Find the percentage error in its volume.

ANSWERV = (4/3)πr3, so ΔV/V = 3(Δr/r) = 3 × (0.1/5.3) = 3 × 0.0189 = 0.0566 ≈ 5.7%.

Q4. Check whether the equation v = u + at is dimensionally correct.

ANSWER[v] = [L T−1], [u] = [L T−1], [at] = [L T−2][T] = [L T−1]. All terms are [L T−1], so the equation is dimensionally consistent.

Q5. What is meant by the “order of magnitude” of a quantity? Give the order of magnitude of the diameter of the Earth (1.28 × 107 m).

ANSWERThe order of magnitude is the power of ten nearest to the quantity (round the mantissa to 1 if ≤ 5). For 1.28 × 107 m, the mantissa 1.28 rounds to 1, so the order of magnitude is 7 (i.e. of the order of 107 m).

Long Answer Type Questions

Q1. State and explain the principle of homogeneity of dimensions. Discuss its uses and its limitations.

ANSWERThe principle states that in any correct physical equation, every term on both sides must have the same dimensions; only quantities of identical dimensions can be added, subtracted or equated. Uses: (i) to check the dimensional correctness of an equation — if the dimensions of all terms are not equal, the equation is certainly wrong; (ii) to convert a physical quantity from one system of units to another; (iii) to derive a relation among physical quantities when the dependence is known (the method of dimensions). Limitations: it cannot find dimensionless constants (like 2π or ½); it cannot check equations involving trigonometric, logarithmic or exponential functions; it cannot derive relations with more than three unknown factors or those that are sums of several terms; and a dimensionally correct equation is not guaranteed to be physically correct.

Q2. Derive an expression for the time period T of a simple pendulum using the method of dimensions, assuming it depends on length l, mass m and acceleration due to gravity g.

ANSWERAssume T = k lx gy mz, where k is a dimensionless constant. Writing dimensions: [T] = [L]x [L T−2]y [M]z = [Lx+y T−2y Mz]. Comparing powers on both sides: for M: z = 0; for T: −2y = 1 → y = −½; for L: x + y = 0 → x = ½. Therefore T = k l1/2 g−1/2 = k √(l/g). Experiment (or theory) gives k = 2π, so T = 2π√(l/g). Note that the mass of the bob does not appear, which agrees with observation.

Q3. Explain the difference between accuracy and precision, and describe the rules for combining errors in (i) a sum/difference and (ii) a product/quotient.

ANSWERAccuracy is how close a measured value is to the true value; precision is how close repeated measurements are to one another (it reflects the resolution / least count of the instrument). A set of readings can be precise yet inaccurate (a systematic error), or accurate on average but imprecise (large random scatter). (i) Sum or difference: the absolute errors add — if Z = A ± B, then ΔZ = ΔA + ΔB. (ii) Product or quotient: the relative (percentage) errors add — if Z = AB or A/B, then ΔZ/Z = ΔA/A + ΔB/B. For a power Z = An, ΔZ/Z = n(ΔA/A). These rules ensure the final result is reported with a realistic uncertainty.

MCQs & Assertion–Reason

1. The number of base quantities in the SI system is:

(a) 5    (b) 6    (c) 7    (d) 9

2. The number of significant figures in 0.0060230 is:

(a) 3    (b) 4    (c) 5    (d) 7

3. The dimensional formula of force is:

(a) [M L T−1]    (b) [M L T−2]    (c) [M L2 T−2]    (d) [M L−1 T−2]

4. Which of the following pairs has the same dimensions?

(a) work and power    (b) work and energy    (c) force and momentum    (d) pressure and force

5. 1 light year is approximately equal to:

(a) 9.46 × 1012 m    (b) 9.46 × 1015 m    (c) 3.08 × 1016 m    (d) 1.5 × 1011 m

6. In multiplication and division, the result is rounded to the:

(a) most significant figures present    (b) least significant figures present    (c) least decimal places    (d) most decimal places

7. The dimensions of the universal gravitational constant G are:

(a) [M L3 T−2]    (b) [M−1 L3 T−2]    (c) [M−1 L2 T−2]    (d) [M L−1 T−2]

8. Which quantity is dimensionless?

(a) force    (b) strain    (c) velocity    (d) density

9. A dimensionally correct equation:

(a) is always physically correct    (b) may or may not be physically correct    (c) is always wrong    (d) has no meaning

10. The least count of a screw gauge of pitch 1 mm with 100 divisions on the circular scale is:

(a) 0.1 mm    (b) 0.01 mm    (c) 0.001 mm    (d) 1 mm

Answer key: 1-(c), 2-(c), 3-(b), 4-(b), 5-(b), 6-(b), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The number of significant figures in a measurement does not change when its unit is changed.

Reason: Significant figures reflect the precision of measurement, which is fixed by the instrument and not by the choice of unit.

A-R 2. Assertion: A dimensionally correct equation must be physically correct.

Reason: The method of dimensions cannot determine dimensionless constants such as ½ or 2π.

A-R 3. Assertion: In the relation m = m0(1 − v2/c2)−1/2, the term v2/c2 is dimensionless.

Reason: The argument of a function and any quantity subtracted from a pure number must be dimensionless.

A-R 4. Assertion: In addition and subtraction, the result is rounded to the least number of decimal places.

Reason: The uncertainty of a sum or difference is governed by the least precise (fewest decimal places) measurement.

A-R 5. Assertion: Plane angle and solid angle are base quantities in the SI system.

Reason: Radian and steradian are dimensionless derived units.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(D).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Counting leading zeros (as in 0.007) as significant — they are not; only the first non-zero digit onwards counts.
  • Using the multiplication rule (least significant figures) for addition/subtraction — there the rule is least decimal places.
  • Forgetting to convert all quantities to the same unit before combining them (e.g. cm and m in area/volume problems).
  • Assuming a dimensionally correct equation is automatically physically correct — dimensions cannot fix numerical constants.
  • Adding absolute errors in a product/quotient — there you must add relative (percentage) errors.
  • Treating radian/steradian as dimensional or as base quantities — they are dimensionless.

How to score full marks in this chapter

Always carry units through every line of a numerical and box the final answer with its unit. Decide the correct number of significant figures from the data before rounding, and round only at the end. For “check the equation” questions, write the dimensional formula of each term separately and compare. Memorise the dimensional formulae of force, energy, pressure, G and momentum — most dimensional-analysis questions reduce to these. State the rule you are using (e.g. “least decimal places”) to earn method marks.

Frequently Asked Questions

What is Class 11 Physics Chapter 1 Units and Measurements about?

Chapter 1 introduces the SI system of units (seven base and derived units), scientific notation, significant figures, the rules for errors and rounding, and dimensional analysis. It teaches you how to measure physical quantities correctly and how to check or derive equations using dimensions.

How many exercises are there in Class 11 Physics Chapter 1?

The NCERT textbook gives exercises numbered 1.1 to 1.17. All seventeen are solved on this page, with every numerical worked step by step and the final answer verified against the NCERT answer key.

What are the rules for significant figures in addition and multiplication?

In addition or subtraction, the result keeps as many decimal places as the term with the fewest decimal places. In multiplication or division, the result keeps as many significant figures as the factor with the fewest significant figures.

Are these Class 11 Physics Chapter 1 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Physics are free and follow the official NCERT textbook for session 2026–27.

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