Question 2. [T] = [M L-1 T-2]a [M L-3]b [M L2 T-2]c Equating powers of M, L and T on both sides, Answer: Volume of the block is given by \(1\mathrm{g} / \mathrm{cm}^ 3=\frac{10^{-3}}{10^{-6}} \mathrm{kg} / \mathrm{m}^{3}=10^{3} \mathrm{kg} / \mathrm{m}^{3}\) NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements are part of NCERT Solutions for Class 11 Physics. NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 11 Entrepreneurship, NCERT Solutions Class 11 Indian Economic Development, NCERT Solutions Class 11 Computer Science, Chapter 7 System of particles and Rotational Motion, Chapter 9 Mechanical Properties Of Solids, Chapter 10 Mechanical Properties Of Fluids, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10, Accuracy, precision of instruments and errors in measurement, Dimensional formulae and dimensional equations, Dimensional analysis and its applications. Find the Young’s modulus of the material of the wire from this data. Suresh again got confused because he never used these units in India. The solution drops spread into a thin, large and roughly circular film of molecular thickness on water surface. Why is this ratio so large? Answer: We know that speed of laser light, c = 3 x 108m/ s Time of echo, t = 8.2 minutes = 8.2 x 60 seconds Value-Based Questions Answer: Power P = V x I, Question 3. One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). This led him to an interesting observation. 1 cm³ = 10⁻⁶ m³ Check by the method of dimensional analysis whether the following relations are correct. So, the new unit of length is } 3 \times 10^{8} \mathrm{m} \text { . }} Name two pairs of physical quantities whose dimensions are same. 12. = A/(4/3πr03A) = 3/4πr03 = 1920″ = 1920 x 4.85 x 10-6 rad [1″ = 4.85 x 10-6 rad] 1. NCERT Solutions for Class 7 Science in PDF file format free to download. 6. 13. Answer: No, all physical quantities do not possess dimensions. (i) Acceleration (ii) Angle (iii) Density Volume of cube = 1 cm³ Question 5. (a) 1 kg m2 s-2 = …. Applying principle of homogeneity of dimensional equation, we find that (c) (αa – βb — γc)% (d) none of the above Now Vg/VH=22.4 x 10-3/3.1548 x 10-7 =7.1 x 104 Ch 13 Physics Class 11 … Answer: No, it has dimensions. = 1.5 x 1011 m. Calculate the diameter of Jupiter. Answer: [M L2 T-2]. Find the dimensions of the following quantities Question 25. The number of particles crossing per unit area perpendicular to X-axis in unit time is Pages. If A be the base area of the boat, then volume of water displaced by boat, V1 = Ad2 Question 1. If x = a + bt + ct2 where x is in metre and t in second, then what is the unit of e? Are all constants dimensionless? The percentage errors in m, D and l are 1%, 1.5% and 0.5%. 29. Answer: The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. Ans : 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. Volume of nucleus As an example, while launching a satellite using a space launch rocket system we must measure time to a precision of 1 micro second. g cm2 s-2 Answer: 1 a.m.u. For a glass prism of refracting angle 60°, the minimum angle of deviation Dm is found to be 36° with a maximum error of 1.05°. Answer: Here n1 = 60 W. Obviously, the physical quantity is power whose dimensional formula is [M1 L2 T3-]. (e) the number of air molecules in your classroom. t = 3.0 billion years = 3.0 x 109 years As 1 ly = 9.46 x 1015 m What is the dimensional formula for torque? What does LASER stand for? And, in terms of the new unit of time, Cover your all chapters of physics like Physical World , Units and Measurements , Oscillations , Waves etc. Order of magnitude of nuclear radius 1.5 x 10-14 m is -14. .•. Study Material and Notes of Ch 2 Units and Measurements Class 11th Physics. Question 1. Suppose we employ a system of units in which the unit of mass equals a kg, the unit of length equals j8 m, the. 32. In fact its dimensional formula is [mol-1]. State the number of significant figures in the following: 24. }}\end{array}\) 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. (b) a jet plane moves with great speed .•. (d) G = 6.67 x 10-11 N m2 (kg)-2 = …. (In fact, since you are aware that you are moving, these distant objects seem to move with you). Name at least six physical quantities whose dimensions are ML2 T-2. \(\begin{array}{l}{\text { (b) The total surface area of a cylinder of radius } r \text { and height } h \text { is }} \\ {S=2 \pi r(r+h)} \\ {\text { Given that, }} \\ {r=2 \mathrm{cm}=2 \times 1 \mathrm{cm}=2 \times 10 \mathrm{mm}=20 \mathrm{mm}} \\ {h=10 \mathrm{cm}=10 \times 10 \mathrm{mm}=100 \mathrm{mm}} \\ {\therefore S=2 \times 3.14 \times 20 \times(20+100)=15072=1.5 \times 104 \mathrm{mm} 2}\end{array}\) Answer: The time taken by light from the quasar to the observer Chapter 6: Tissues Chapter 7: Diversity in Living Organisms Chapter 8: Motion Chapter 9: Force and Laws of Motion Chapter 10: Gravitation Chapter 11: Work and Energy Chapter 12: Sound Chapter 13: Why Do We Fall Ill Chapter 14: Natural Resources Chapter 15: Improvement in Food Resources NCERT Solutions for Class 9 Science. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the Moon. ‘The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. (a) 0.007 m2 (b) 2.64 x 104 kg 21. Define Atomic mass unit (a.m.u.). How much is the radius of the lunar orbit around the Earth? The thickness of a hair can be measured by using a screw gauge. Answer: The line joining a given object to our eye is known as the line of sight. Consider a class room of size 10 m x 8 m x 4 m. Volume of this room is 320 m3. Answer: (i) Work (ii) Torque (iii) Moment of force (iv) Couple (v) Potential energy (vi) Kinetic energy. The radius of the Earth is 6.37 x106 m and its average density is 5.517 x 103 kg m-3. So, the nuclear mass density is independent of mass number. Question 4. A physical quantity P is related to four observables a, b, c and d as follows: =11.3 x 103 kg m-3 [1 kg =103 g,1m=102 cm] .-. Answer: Question 12. \(\begin{array}{l}{\text { Therefore, distance can be obtained using the relation: }} \\ {\text { Distance }=\text { Speed } \times \text { Time }=5 \times 1=5 \mathrm{m}} \\ {\text { Hence, the vehicle covers } 5 \mathrm{m} \text { in } 1 \mathrm{s} \text { . Which is a bigger unit-light year or parsec? Chapter 3 Motion In A Straight Line Download in pdf . Home; ... CBSE Rankers is a Free Platform for all CBSE Students to get access to Free sample papers and Study Material from our Wide Range of Study Materials. θ=D/d Explain why? [Energy] = [M L2 T-2], hence 1 joule = 1 kg x 1 m2 x 1 s-2 = 1 kg m2 s-2. The speed of light in air is 3.00 x 108 ms 1. A body travels uniformly a distance of (13.8 ± 0.2) m in a time (4.0 ± 0.3) s. What is the velocity of the body within error limits? (a) The size of an atom is much smaller than even the sharp tip of a pin. The voltage across a lamp is (6.0 ± 0.1) volt and the current passing through it is (4.0 ± 0.2) ampere. (cm)3 s-2 g-1. Hence, mass of the earth = 5.97 x 1024 kg. 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. Volume of cube = 1 cm³ NR is a function of the density of the liquid r, its average speed is v and the coefficient of viscosity of the liquid is h. If NR is given directly proportional to d (the diameter of the pipe), Study Materials for Class 1 to 12 Maths & Science (Physics, Chemistry & Biology), English, Hindi & All Subjects – Free PDF Download. .•. Find the area of the rectangular block. A new unit of length is chosen such that the speed of light in vacuum is unity. (a) the total mass of rain-bearing clouds over India during the Monsoon To express these in terms of F, A and V, we must express, M, L and T in terms of these new ‘fundamental’ quantities. (e) a proton is much more massive than an electron \(\begin{array}{l}{1 \mathrm{s}=\frac{1}{\gamma}=\gamma^{-1}} \\ {1 \mathrm{s}^{2}=\gamma^{-2}} \\ {1 \mathrm{s}^{-2}=\gamma^{2}}\end{array}\) (b) A Jet plane moves with a speed greater than that of a super fast train. Find the Earth’s average density to appropriate significant figures. = (1.37 x 4.11 x 2.56) cm3 = 14.41 cm3 Also, 1 parsec = 3.08 x 1016 m \(\begin{array}{l}{ 1 \mathrm{cm} 3=10-6 \mathrm{m} 3} \\ {\text { Hence, the volume of a cube of side } 1 \mathrm{cm} \text { is equal to } 10-6 \mathrm{m} 3 \text { . Question 4. Volume = (4.234 x 1.005) x (2.01 x 10-2) = 8.55289 x 10-2 = 0.0855 m3. 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. Since the least number of decimal places is 1, therefore, the total mass of the box = 2.3 kg. Question 2. =(23 x 10-3/6.023 x 1023 x 4/3 x 3.14 x (2.5 x 10-10)) The parallactic angle subtended by a distant star is 0.76 on the earth’s orbital diameter (1.5 x 1011 m). They ensure a smooth and easy knowledge of advanced concepts. \(1\mathrm{g} / \mathrm{cm}^ 3=\frac{10^{-3}}{10^{-6}} \mathrm{kg} / \mathrm{m}^{3}=10^{3} \mathrm{kg} / \mathrm{m}^{3}\) Answer: We first note that the dimension of I are [ML2]. Answer: Question 10. .•. Question 2. Since, in this case significant figures in one quantity (3.00 x 108 ms–1) are 3 and the significant figures in the other quantity (3.154 x 107 s) are 4, therefore, the final result should have 3 significant figures. Explain this statement clearly: Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (-15 billion years). =55.4 cm2, Question 2. The value of d directly gives the wind speed. Answer: Let the distance covered is S, (a) atoms are very small objects \\ {\text { In the new system, the speed of light in vacuum is unity. = 1.45 x 109m. Show that Distance of quasar from the observer d = 3.0 x 109 x 9.46 x 1015 m “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. You can also watch the video solutions of NCERT Class11 Physics chapter 2 Units And Measurement here. (c) the mass of Jupiter is very large Ncert Solutions For Class 11 Physics Download Pdf, ncert solutions for Class 11 physics download pdf . Solving these equations, we get and average width of the image of hair as seen by microscope = 3.5 mm Download Class 11 Physics NCERT Solutions in pdf free. (b) A Jet plane moves with a speed greater than that of a super fast train. Question 27. NCERT Revision Notes for Class 11 (Hindi and English Medium) – All Subjects and Chapters NCERT Revision Notes for class 11 includes chapter-wise important points of the exercises, which equip students with the key to unlock their problem-solving skills. How will you estimate the diameter of the thread? Answer: (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4. These properties of a laser light can be exploited to measure long distances. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of fundamental constants of nature. = 4/3 x 3.14 x (0.5 x 10-10) m3 = 5.23 x 10-31 m3 What does SONAR stand for? .•. (a) You are given a thread and a metre scale. Obtain Dirac’s number, given that the desired number is proportional to mp-1 and me-2. Question 2. IV. NCERT Solutions Class 11 can be extremely useful in understanding the methods of framing answers and writing them in a way that exhibits the student’s analysis of phenomena supported by facts. The SI units of magnetic field is Chapter 1 Physical World Download in pdf. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements. 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. Answer: Physical quantities are called large or small depending on the unit (standard) of measurement. This solution contains questions, answers, images, explanations of the complete chapter 2 titled Of Units And Measurement taught in Class 11. Question 11. f= x2, then what is the relative error in f? Ncert Solutions For Class 11 Chemistry Chapter 5 States Of by schools.aglasem.com. The ratio is very large. Volume of one mole of hydrogen, VH Question 2. A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm and length 6 ± 0.06 cm. It is claimed that two cesium clocks, if allowed to run for 180 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s? The distance travelled by light in one year (i.e., 365 days = 3.154 x 107 s) is known as light year. In terms of the new unit length, 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. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: (Given: r = r0 A1/3) 14. The least count of Vernier Caliper is ± 0.01 cm Uncertain values can be written as Its density is ______g a-n 30r ______kg m⁻⁵. Answer: No, since density = Mass/Volume. (c) Speed of vehicle = 18 km/h = 18 x 1000/3600 m/s .•. 1 nano metre (1 nm) = 10-9 m 6. = 22.4 litre = 22.4 x 10-3 m3 What is the ratio of molar volume to the atomic volume of a mole of hydrogen? Question 16. To find the value of ‘g by using a simple pendulum, the following observations were made : Length of thread l = (100 ± 0.1) cm Short Answer Type Questions Question 2. Answer: To determine the molecular diameter of oleic acid, we first of all dissolve 1 mL of oleic acid in 20 mL of alcohol. Class 11th is a critical phase in the education of a person. = 9.7 x 10-3 rad d = 1.496 x 1011 m Calculate the distance of the star from the earth. 33. V = I b t Ans : \(\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 { . }} Answer: According to the principle of homogeneity of dimensions. And selecting the correct study material and study tools will have a huge impact on the student’s result and academic performance. NCERT Solutions for Class 11. Question 15. One mole contains 6.023 x 1023 molecules. Question 2. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. If so, why? (Take the size of hydrogen molecule to be about 1 A.) Answer: Question 3. = 4/3 x 3.142 x (6.37 x 106)3m3 II. In Searle’s experiment, the diameter of the wire as measured by a screw guage of least count 0.001 cm, is 0.500 cm. The diameter of the moon, D = θ x S When a train moves rapidly, the line of sight of a passenger sitting in the train for nearby trees changes its direction rapidly. If the radius of earth is 6.4 x 106 m, determine the distance of the heavenly body from the centre of earth. Its density is …….. g cm-3 or ………. —> Work and Energy. \(1 \mathrm{cal}=4.2 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}\) But we choose the higher of these two values as the uncertainty i.e. Question 19. Estimate the mass density of sodium nucleus. If you have any problem in finding the correct answers of Physics Part I Textbook then you can find here. The radius of the Earth is 6.37 x 106 m and its mass is 5.975 x 1024 kg. 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. Answer: (a) The average rainfall of nearly 100 cm or 1 m is recorded by meteorologists, during Monsoon, in India. 16. NCERT Solutions 2020-2021 and Offline Apps are updated according to NCERT Books 2020-21 following the latest CBSE Syllabus 2020-2021. The length, measured by a scale of least count 0.1 cm is 110.0 cm. 1 A.U. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only? = 28.38 x 1024 m = 2.8 x 1025 m or 2.8 x 1022 km. Answer: If v be the velocity of sound, E the elasticity of the medium and p the density of the medium, then (e) This is a correct statement. Answer: We know that the usual dimensions of Y are [MLT–2]/[L2] i.e.,[M L-2T-2] }}\end{array}\) .’. Experimentally, molecular diameter of oleic acid is found to be of the order of 10-9 m. Question 5. Study Rankers Free Ncert Solutions And Free Download by study-rankers.soft112.com. Answer: Volume of one hydrogen atom = 4/3 πr3 (volume of sphere) Similarly the upper limit can also be calculated as follows. Answer: Question 2. \\ {\text { In the new system, the speed of light in vacuum is unity. Error in 1 second=0.02/100 x 365 x 24 x 60 x 60 Question 17. He could not understand pound and how it is converted into rupees. NCERT Solutions for Class 11 Chemistry: Chapter 8 (Redox Reactions) are provided on this page for the perusal of Class 11 Chemistry students studying under the syllabus prescribed by CBSE.Detailed, student-friendly answers to each and every intext and exercise question provided in Chapter 8 of the NCERT Class 11 Chemistry textbook can be found here. By using the method of dimension, check the accuracy of the following formula: T =rhρg/2cos θ , where T is the surface tension, h is the height of the liquid in a capillary tube, p is the density of the liquid, g is the acceleration due to gravity, 6 is the angle of contact, and r is the radius of the capillary tube. If d be the distance of Moon from the earth, the time taken by laser signal to return after reflection at the Moon’s surface. Since r0 is a constant therefore the right hand side is a constant. Thus, the correct value of one light year = 9.46 x 1015 m. Question 9. Average mass density of sodium atom, p = M/V Answer: 6.67 x 10-8 dyne cm2 g-2 = 6.67 x 1011 Nm2 /kg2 . Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec). 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. =11.3 x 103 kg m-4. If instead of mass, length and time as fundamental quantities, we choose velocity, acceleration and force as fundamental quantities and express their dimensions by V, A and F respectively, show that the dimensions of Young’s modulus can be expressed as [FA2 V-4]. If n be the number of turns of the coil and l be the length of the coil, then the length occupied by each single turn i.e., the thickness of the thread = 1/n . It means, distance is measured in foot, mass in pound and time in seconds whereas in India it is MKS system. Question 2. (c) Wind speed can be estimated by floating a gas-filled balloon in air at a known height h. When there is no wind, the balloon is at A. Answer: Question 4. As an example, the low resolution of the human eye would make observations difficult. What does RADAR stand for? NCERT Book Solutions for Class 11 for the Humanities subjects are also available here. (d) Let us assume that the man is not partially bald. If volume of each drop of solution be V, then volume of n drops = n V – 2c = – 1 …(Hi) 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? Average density (D)=Mass/Volume=M/V= 0.005517 x 106 kg m-3 4.29 light years =4.508 x 1016/3.80 x 1016 = 1.318 parsec = 1.32 parsec. Click Here (e) We can determine the volume of the class-room by measuring its length, breadth and height. Study every day: A student should study NCERT text book one hour per day for seven days. NCERT Solutions for Class 11 Physics will provide you all topics of CBSE class 11 Physics syllabus we will explain all the solution in detail. V. Question On High Order Thinking Skills (HOTS) Calculate the radius of the lunar orbit around the eazth. He explained his brother that here F.P.S. \(\begin{array}{l}{\text { (b) The total surface area of a cylinder of radius } r \text { and height } h \text { is }} \\ {S=2 \pi r(r+h)} \\ {\text { Given that, }} \\ {r=2 \mathrm{cm}=2 \times 1 \mathrm{cm}=2 \times 10 \mathrm{mm}=20 \mathrm{mm}} \\ {h=10 \mathrm{cm}=10 \times 10 \mathrm{mm}=100 \mathrm{mm}} \\ {\therefore S=2 \times 3.14 \times 20 \times(20+100)=15072=1.5 \times 104 \mathrm{mm} 2}\end{array}\) All the solutions of Motion in a Straight Line - Physics explained in detail by experts to … What is the linear magnification of the projector-screen arrangement? Calculate the mass of earth to correct significant figures. NCERT Solutions for Class 11 Physics Physics Chapter 2 Units and Measurements includes all the important topics with detailed explanation that aims to help students to understand the concepts better. NCERT Solutions for Class 11 Humanities Subjects. (d) The air inside this room contains more number of molecules than in one mole of air. A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm² s⁻². Toppr provides free study materials, 1000+ hours of video lectures, last 10 years of question papers for free. Also, wherever you can, give a quantitative idea of the precision needed. Answer: Question 5. (d) the number of strands of hair on your head Mass of nucleus = A (b) Difference of mass = 2.17 – 2.15 = 0.02 g What is the distance in km of a quasar from which light takes 3.0 billion years to reach us? Answer: In order to find out the accuracy of the given equation we shall compare the dimensions of T and rh ρg/2cos θ. III. (14.41 ± 0.20) cm3, Question 12. If the percentage errors in measurements of a, b and c are ± 1%, ±2% and ± 1.5% respectively, then calculate the maximum percentage error in value of x obtained.
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