Volume of earth = 4/3π R3 Free PDF download of NCERT Solutions for Class 11 Physics Chapter 3 - Motion in a Straight Line solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. Since the least number of decimal places is 1, therefore, the total mass of the box = 2.3 kg. 29. Question 2.7. Volume = (4.234 x 1.005) x (2.01 x 10-2) = 8.55289 x 10-2 = 0.0855 m3. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals j8 m, the. The voltage across a lamp is (6.0 ± 0.1) volt and the current passing through it is (4.0 ± 0.2) ampere. Question 19. .•. (ii) How many unit system are there? }}\end{array}\) (d) The air inside this room contains more number of molecules than in one mole of air. Question 25. 1. Answer:  Total time = 100 years = 100 x 365 x 24 x 60 x 60 s unit of time is ys. Since, θ=1″ (s) and b=3 x 1011 m Mass density of Sun is in the range of mass densities of solids/liquids and not gases. Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec). Given 1 A.U. Answer:  Let the measured values be : Volume of the Earth (V) = 4/3 πR3m3 = 4/3 x (3.142) x (6.37 x 106)3 m3 Thus, nuclear mass density is constant for different nuclei. Answer: Here n1 = 60 W. Obviously, the physical quantity is power whose dimensional formula is [M1 L2 T3-]. So, the new unit of length is } 3 \times 10^{8} \mathrm{m} \text { . }} Answer: (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4. .’. Are the two densities of the same order of magnitude? In view of this, reframe the following statements wherever necessary: The value of d directly gives the wind speed. Answer: From examples 2.3 and 2.4, we get θ = 1920″ and S = 3.8452 x 108 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. 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. Answer: Question 2. V(max) = (1.37 + 0.01) x (4.11 + 0.01) x (2.56 + 0.01) cm3 = (1.38 x 4.12 x 2.57) cm3 = 14.61 cm3 And, in terms of the new unit of time, He could not understand pound and how it is converted into rupees. (e) 6.032 N m-2 (f) 0.0006032 m2 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? This solution contains questions, answers, images, explanations of the complete chapter 2 titled Of Units And Measurement taught in Class 11. What is the total atomic volume in m3 of a mole of hydrogen atoms? Dimensions of L.H.S. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Physics Chapter 2 Units and Measurements. Compare it with the density of sodium in its crystalline phase: 970 kg m3-. Question 4. where n1 and n2 are number of particles per unit volume for the value of x1 and x2 Question 2. Download Class 11 Physics NCERT Solutions in pdf free. Calculate the radius of the lunar orbit around the eazth. Obtain the dimensions of relative density. What is the estimate on the thickness of hair? The heat dissipated in a resistance can be obtained by the measurement of resistance, the current and time. Download NCERT Solutions class 11 Physics chapter wise. Physics NCERT Solutions for Class 11 – Chapter wise Class 11 Physics – Physics Part I. In the new system, the speed of light in vacuum is unity. 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? Question 3. The distance travelled by light in one year (i.e., 365 days = 3.154 x 107 s) is known as light year. ‘The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. The solution drops spread into a thin, large and roughly circular film of molecular thickness on water surface. Do A and A.U. Average density (D)=Mass/Volume=M/V= 0.005517 x 106 kg  m-3 Also, 1 parsec = 3.08 x 1016 m 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 calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm² s⁻². 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.) The speed of light in air is 3.00 x 108 ms 1. 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. 1 A.U. (c) modulus of elasticity (d) all the above The least count of Vernier Caliper is ± 0.01 cm Uncertain values can be written as (e) This is a correct statement. (a) You are given a thread and a metre scale. Name at least six physical quantities whose dimensions are ML2 T-2. Answer: Given angular diameter θ = 35.72= 35.72 x 4.85 x 10-6 rad (f) This is a correct statement. Answer: Briefly explain how you will estimate the molecular diameter of oleic acid. The first system, in which unit of power is 1 watt, is SI system in which M1 = 1 kg,L1 = 1 m and T1= ls in second system, M2 = 100 g, L2 = 20 cm and T2 = 1 min = 60 s. Question 24. \(\mathrm{But}, 1 \mathrm{cm} 3=1 \mathrm{cm} \times 1 \mathrm{cm} \times 1 \mathrm{cm}=\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m}\) Density of water = 1 g/cm³ Since r0 is a constant therefore the right hand side is a constant. =4/3 x 22/7 (0.5 x 10-10)3 m3 = 1.496 x 1011 m. Question 7. [Energy] = [M L2 T-2], hence 1 joule = 1 kg x 1 m2 x 1 s-2 = 1 kg m2 s-2. Given that the value of G in the CGS system as 6.67 x 10-8dyne cm2 g-2 , find the value in MKS system. Question 2. When a beam of parallel light is incident on the prism, find the range of experimental value of refractive index ‘μ’. As an example, the low resolution of the human eye would make observations difficult. A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm2 s-2. Answer: LASER stands for ‘Light Amplification by Stimulated Emission of Radiation’. Question 4. IV. Answer: Question 15. = 7.4 x 1010 m. Question 4. NCERT Solutions for Class 11 Physics Chapter 13 come with elaborate explanations and precise answers on crucial concepts that students need to have a solid understanding of, in order to clear their entrance exams with an excellent score. Question 1. (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale. a=0,b=+(1/2) and c=+(1/2). = Speed of light in vacuum x time taken by light to travel from Sim to Earth = 3 x 108  m/ s x 8 min 20 s = 3 x 108 m/s x 500 s = 500 x 3 x 108 m. As the error lies in first decimal place, the answer should be rounded off to first decimal place. Answer: Young’s modulus of the material of the wire is given as. He explained his brother that here F.P.S. If the maximum error in the measurement of these quantities is 1 %, 2 % and 1 % respectively, what is the maximum error in determination of the dissipated heat? =>t=nV/400S }}\end{array}\) (a) 0.007 m2 (b) 2.64 x 104 kg Explain. What is the order of precision of an atomic clock? Chapter 3 Motion In A Straight Line Download in pdf . (c) the wind speed during a storm (e) a proton is much more massive than an electron 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: One light year = speed x time = 9.462 x 1015 m. When two physical quantities are multiplied, the significant figures retained in the final result should not be greater than the least number of significant figures in any of the two quantities. \(\begin{aligned} \text { Density of lead } &=\text { Relative density of lead } \times \text { Density of water } \\ &=11.3 \times 1=11.3 \mathrm{g} / \mathrm{cm}^{3} \end{aligned}\) You can also download here the NCERT Solutions Class 11 Physics chapter 2 Units And Measurement in PDF format. D=θ d In fact its dimensional formula is [mol-1]. The earth-moon distance, S = 3.8452 x 108 m .’. With these expertly curated NCERT Solutions for Class 11 Physics, all the student’s doubts and queries will be cleared. Angular diameter of the moon, θ= Angular diameter of the sun 11.3 g/cm³ = 11.3 x 10. kg/m³. Fill in the blanks by suitable conversion of units = 28.38 x 1024 m = 2.8 x 1025 m or 2.8 x 1022 km. The number of particles crossing per unit area perpendicular to X-axis in unit time is P.A.M. Dirac, a great physicist of 20th century found that from the following basic constants, a number having dimensions of time can be constructed: .•. Nuclear mass density = Mass of nucleus/Volume of nucleus If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 2 Units And Measurement After you have studied lesson, you must be looking for answers of its questions. Toppr provides free study materials, 1000+ hours of video lectures, last 10 years of question papers for free. Question 2. What do you understand by fundamental physical quantities? 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. A lot of youngsters wish to make a career in the field of physics. (a) The volume of a cube of side 1 cm is equal to ______m³ The later standard parsec is equal to 3.08 x 1016 m or 3.08 x 1012 km is certainly larger than metre or kilometre. Find the power consumed by the lamp. 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? A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm and length 6 ± 0.06 cm. Quickly measure the diameter of thin circular film and calculate its surface area S. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Question 2. Find the value of the gas constant R. 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). = [FA-1] [V2 A-1]-1[VA-1]-2=FA2V-4 Thus the ‘new’ dimensions of Young’s modulus are [FV-4 A2]. radius of sodium nucleus, r = 2.1 and Ar = ± 0.5 What is the distance in km of a quasar from which light takes 3.0 billion years to reach us? Study Rankers - Free download of NCERT Solutions for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance solved by Expert Teachers as per NCERT (CBSE) textbook guidelines. Answer: No, all physical quantities do not possess dimensions. To determine acceleration due to gravity, the time of 20 oscillations of a simple pendulum of length 100 cm was observed to be 40 s. Calculate the value of g and maximum percentage error in the measured value of g. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 … Parallax angle subtended by the star Alpha Centauri at the given basis θ = 1.32 x 2 = 2.64″. 1 cm³ = 10⁻⁶ m³ The mass of a box measured by a grocer’s balance is 2.3 kg. Question 2. In an experiment on determining the density of a ectangular Mock, the dimensions of the Mock are measured with a Venier Caliper (with a least count of 0.01 cm) and its mass is measured with a beam balance of least count of 0.1 gm. The SI units of magnetic field is (a = maximum displacement of the particle, v = speed of the particle, T = time-period of motion)Rule out the wrong formulas on dimensional grounds. As the film is extremely thin, this thickness t may be considered to be the size of one molecule of oleic acid i.e., t is the molecular diameter of oleic acid. In view of this, reframe the following statements wherever necessary. If so, why? \(1 \mathrm{m} \quad=\frac{1}{9.46 \times 10^{5}} \mathrm{ly} \approx \frac{1}{10^{16}} \mathrm{ly}=10^{-16} \mathrm{ly}\) The two specific heat capacities of a gas are measured as Cp = (12.28 ± 0.2) units and Cv = (3.97 ± 0.3) units. Volume of nucleus Assuming that T = k Pa db Ec and substituting dimensions of all the quantities involved, we have where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to about,1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. You can also check out NCERT Solutions of other classes here. Volume of hydrogen molecule = 4/3 πr3 To find the value of ‘g by using a simple pendulum, the following observations were made : Length of thread l = (100 ± 0.1) cm How will you estimate the diameter of the thread? Question 2. (Given: r = r0 A1/3) Answer: Question 11. 1 cm³ = 10⁻⁶ m³ 4.29 light years = 4.29 x 9.46 x 1015 = 4.058 x 1016 m You can view them online or download PDF file for future use. Do specific heat and latent heat have the same dimensions? Now, AB = d = hθ. Answer: (c) 0.2370 g cm-3 (d) 6.320 J 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). Answer the following: = A/(4/3πr03A) = 3/4πr03 Applying principle of homogeneity of dimensional equation, we find that Compute the error in measurement of radius of curvature. (d) the number of strands of hair on your head Can a physical quantity have dimensions but still have no units? .•. -2 a-2 c = 1 28. .•. The value of l and h are 4.0 cm and 0.065 cm respectively where l is measured by a metre scale and h by the spherometer. We have to try to make permutations and combinations of the universal constants and see if there can be any such combination whose dimensions come out to be the dimensions of time. \(\mathrm{I} \mathrm{m}=\frac{1}{\beta}=\beta^{-1} \text { or } \mathrm{Im}^{2}=\beta^{-2}\) 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. Explain this statement clearly: PDF Download Free. Areal magnification =Area on screen/Area on slide = 1.55 m2 / 1.75 x 10-4 m2 = 8.857 x 103 (i) What values are displayed by Suresh? Answer: Parsec is bigger unit than light year (1 parsec = 3.26 light year). = 1.45 x 109m. Answer: The order of magnitude of a numerical quantity (N) is the nearest power of 10 to which its value can be written.For example. (a) the total mass of rain-bearing clouds over India during the Monsoon .•. Question 13. As we approach the higher class like class 11, students need more help along with the NCERT textbooks. Study Rankers Free Ncert Solutions And Free Download by study-rankers.soft112.com. Answer: We known that speed of laser light = c = 3 x 108 m/s. Answer: N m-1 s2 is nothing but SI unit of mass i.e., the kilogram. Answer: It t be the thickness of oleic acid film formed over water surface then the volume of oleic acid film = St We have everything! We can estimate the area of the head. II. breadth (b) = 2.56 cm Suresh found there f the currency is quite different from his country. So, the new unit of length is 3 x 108 m. b + c = 1 …(H) What is (a) the total mass of the box (b) the difference in the masses of the pieces to correct significant figures? St =nV/400 The number of particles crossing per unit area perpendicular to x-axis in unit time N is given by N= -D(n2-n1/x2-x1), where n1 and n2 are the number of particles per unit volume at x1 and x2 respectively. N=-D n2-n1/x2-x1 = 4/3 πr3 = 4/3 π [r0 A1/3]3 = 4/3 πr03A (14.41 ± 0.20) cm3, Question 12. Reynold’s number NR (a dimensionless quantity) determines the condition of laminar flow of a viscous liquid through a pipe. NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements are part of NCERT Solutions for Class 11 Physics. (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. 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? Check if your guess is correct from the following data: mass of the Sun = 2.0 x 1030 kg, radius of the Sun = 7.0 x 108 m. 5. (d) Relative density of a substance is given by the relation, (c) the mass of Jupiter is very large We hope the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements help you. =(6.02 x 1023 /22.4 x 10-3 ) x 320 =8.6 x 1027. It follow that M = FA-1, T = VA~X, L = V2 A-1 Answer: Power P = V x I, Question 3. Fill in the blanks The number of hair on the human head is of the order of one million. Question 15. 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 intergalactic distances are measured in parsec The later standard parsec is equal to \(3.08 \times 10^{16} \mathrm{m} \text { or } 3.08 \times 10^{12} \) 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. Linear magnification. \\ {\therefore \text { distance between Sun and Earth }=500 \text { new units. =2000/3600 x π /180 rad V = l x w x h E, m, 1 and G denote energy, mass, angular momentum and gravitational constant respectively.Determine the dimensions of El2/m5G2. = 25.2 x 16.8 Hence more reliable result can be obtained. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. (c) newton per ampere per metre (d) all the above Calorie = \(4.2(1 \mathrm{a}-1)(1 \beta-2)(1 \mathrm{y} 2)=4.2 \mathrm{a}-1 \beta-2 \mathrm{y}^2\). Question 4. .-. Check by the method of dimensional analysis whether the following relations are correct. Answer:  —> Stress and Young’s modulus. 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. i. e., [L1 T-1] = dimensionless, which is incorrect. So, the new unit of length is } 3 \times 10^{8} \mathrm{m} \text { . }} Question 2. Question 12. V ) = (1.37 – 0.01) x (4.11 – 0.01) x (2.56 – 0.01) cm3 = (1.36 x 4.10 x 2.55) cm3 = 14.22 cm3 What does SONAR stand for? = 0.5 A = 0.5 x 10-10 m NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance exercises are given below to use it online or download in PDF form for offline. If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants? \(\begin{array}{l}{\text { Distance between Sun and Earth }} \\ {=\text { Speed of light in vacuum } x \text { time taken by light to travel from Sim to Earth }=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 8} \\ {\min 20 \mathrm{s}=3 \times 10^{8} \mathrm{m} / \mathrm{s} \times 500 \mathrm{s}=500 \times 3 \times 10^{8} \mathrm{m} \text { . }} Same dimensions I ) what values are found to be measured by callipers! Different examples in modem science where precise Measurements of length, time, mass etc., are in... 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