ncert solutions for class 11 physics chapter 2 study rankers

We have everything!! Study Tips For NCERT Solutions For Class 11 Physics Chapter 15 Answer:  —> Stress and Young’s modulus. 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? In the new system, the speed of light in vacuum is unity. (f) This is a correct statement. 1 nano metre (1 nm) = 10-9 m = 1920″ = 1920 x 4.85 x 10-6 rad [1″ = 4.85 x 10-6 rad] Quickly measure the diameter of thin circular film and calculate its surface area S. Question 16. The diameter of the moon, D = θ x S This led him to an interesting observation. V = I b t = 0.5238 x 10-30 m3 thickness (f) = 0.37 cm If no, name four physical quantities which are dimensionless. We hope the NCERT Solutions for Class 11 Physics Chapter 7 System of particles and Rotational Motion help you. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. 13. 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°. Calculate the pressure exerted over the area. S.A. = 4 π r2 Volume of water displaced by (boat + elephant), V2 = Ad2 Volume of water displaced by elephant, On solving these equations, we find that Question 2. = 1/12 th mass of carbon-12 atom, i.e., 1.66 x 10-27 kg. The solution drops spread into a thin, large and roughly circular film of molecular thickness on water surface. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us? What do you mean by order of magnitude? 6. Find the dimensions of the following quantities Chapter 3 Motion In A Straight Line Download in pdf . Calculate focal length of a spherical mirror from the following observations: object distance u = (50.1 ± 0.5) cm and image distance v = (20.1 ± 0.2) cm. Question 1. Obtain the dimensional formula for coefficient of viscosity. \(\begin{aligned}(c) G &=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{kg}^{-2}=6.67 \times 10^{-11} \frac{\mathrm{kg} \mathrm{m}}{\mathrm{s}^{2}} \mathrm{m}^{2} \mathrm{kg}^{-2} \\ &=6.67 \times 10^{-11} \mathrm{kg}^{-1} \mathrm{m}^{3} \mathrm{s}^{-2} \\ &=6.67 \times 10^{-11} \frac{\mathrm{m}^{3}}{\mathrm{kgs}^{2}}=\frac{6.67 \times 10^{-11} \times\left(10^{2}\right)^{3}}{\left(10^{3}\right)^{2}} \\ &=6.67 \times 10^{-8} \mathrm{cm}^{-3} \mathrm{s}^{-2} \mathrm{g}^{-1} \end{aligned}\). (a) atoms are very small objects In the evening Suresh inquired all about it. "To call a dimensional quantity 'large' or is meaningless without specifying a standard for comparison". .•. Question 4. .•. Question 2. 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? Since r0 is a constant therefore the right hand side is a constant. (c) Least count of optical instrument = 6000 A (average wavelength of visible light as 6000 A) = 6 x 10-7m As the least count of optical instrument is least, it is the most precise device out of three instruments given to us. 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. Question 8. 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. Now take a large sized trough filled with water. Answer: (a) The average rainfall of nearly 100 cm or 1 m is recorded by meteorologists, during Monsoon, in India. As a result, the nearby trees and other objects appear to run in a direction opposite to the train’s motion. = 25.2 x 16.8 The unit of length convenient on the nuclear scale is afermi: I f=10-15 m. Nuclear sizes obey roughly the following empirical relation: And, in terms of the new unit of time, Question 4. =55.4 cm2, Question 2. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements, drop a comment below and we will get back to you at the earliest. 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). Which is a bigger unit-light year or parsec? Question 2. (d) Relative density of a substance is given by the relation, \(1 \mathrm{m} \quad=\frac{1}{9.46 \times 10^{5}} \mathrm{ly} \approx \frac{1}{10^{16}} \mathrm{ly}=10^{-16} \mathrm{ly}\) Average density (D)=Mass/Volume=M/V= 0.005517 x 106 kg  m-3 (c) newton per ampere per metre (d) all the above The nearest star to our solar system is 4.29 light years away. (a) the total mass of rain-bearing clouds over India during the Monsoon Short Answer Type Questions With these expertly curated NCERT Solutions for Class 11 Physics, all the student’s doubts and queries will be cleared. The set of solutions can be accessed on multiple devices. Compute the error in measurement of radius of curvature. Answer:  From parallax method we can say Linear magnification. (c) G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = ______(cm)³ s⁻² g⁻¹ . Answer:  Extremely precise measurements are needed in modem science. distance between Sun and Earth = 500 new units. 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. 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. }}\end{array}\) .•. Diameter of Jupiter D = θ x d = 1.73 x 10-4 x 824.7 x 109 m Ncert Solutions For Class 11 Physics Download Pdf, ncert solutions for Class 11 physics download pdf . (a) (αa – βb +γc)% (b) (αa + βb +γc)% Mass of the block (m) = 39.3 g a = 0 …(i) Answer: (i) Work (ii) Torque (iii) Moment of force (iv) Couple (v) Potential energy (vi) Kinetic energy. (c) The mass of Jupiter is very large compared to that of the earth. Calculate the radius of the lunar orbit around the eazth. = 5974.01 x 1021 kg = 5.97401 x 1024 kg Here we have given NCERT Solutions for Class 11 Physics Chapter 2 … Deduce the dimensional formula for D. \(\begin{aligned} \text { Density of lead } &=\text { Relative density of lead } \times \text { Density of water } \\ &=11.3 \times 1=11.3 \mathrm{g} / \mathrm{cm}^{3} \end{aligned}\) If A is the area of the country, then A = 3.3 million sq. We know that 22.4l or 22.4 x 10-3 m3 of air has 6.02 x 1023 molecules (equal to Avogadro’s number). m in 1 s. 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. NCERT Solutions Class 11 Physics Chapter 2 PDF. =6.34 x 10-12 s A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kgm2 s-2. PDF Download Free. Question 5. 33. Answer: Question 12. 2. Question 2. Do specific heat and latent heat have the same dimensions? Calorie = \(4.2(1 \mathrm{a}-1)(1 \beta-2)(1 \mathrm{y} 2)=4.2 \mathrm{a}-1 \beta-2 \mathrm{y}^2\). }}\end{array}\). Question 2. NCERT Solutions for Class 11. 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. (c) Due to random errors, a large number of observation will give a more reliable result than smaller number of observations. Applying principle of homogeneity of dimensional equation, we find that Answer: The measured (nominal) volume of the block is, The density of the block is given by, VI. Therefore, the final result should be rounded up to three significant figures. (e) We can determine the volume of the class-room by measuring its length, breadth and height. .•. Question 19. The radius of the Earth is 6.37 x106 m and its average density is 5.517 x 103 kg m-3. Powered by Blogger. Question 25. This is 0.20 cm3 higher than the measured value. Question 2 .11. 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. θ=D/d If you have any problem in finding the correct answers of Physics Part I Textbook then you can find here. Question 7. (a) kg m2 s-2 (b) kg-1 m3 s-2 (c) N kg2 m-2 (d) N kg2 m-2 The density is accurate only up to three significant figures which is the accuracy of the least accurate factor, namely, the radius of the earth. For example, mass, length, time, temperature, electric current, luminous intensity and amount of substance are seven fundamental physical quantities. Given that the time period T of oscillation of a gas bubble from an explosion under water depends upon P, d and E where P is the static pressure, d the density of water and E is the total energy of explosion, find dimensionally a relation for T. =11.3 x 103 kg m-4. The least count of Vernier Caliper is ± 0.01 cm Uncertain values can be written as 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. Answer: Radius of the Earth (R) = 6.37 x 106 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? What is the total atomic volume in m3 of a mole of hydrogen atoms? (a) M°L T2 (Ib) M°L2 T-4 (c) M°L T3 (d) M°L2T3 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, NCERT Solutions for Class 11 Humanities Subjects. The mass of a box measured by a grocer’s balance is 2.3 kg. You can download the NCERT Book for Class 11 Physics in PDF format for free. \(\begin{array}{l}{\text { (c) Using the conversion, }} \\ {1 \mathrm{km} / \mathrm{h}=\frac{5}{18} \mathrm{m} / \mathrm{s}} \\ {18 \mathrm{km} \mathrm{h}=18 \times \frac{5}{18}=5 \mathrm{m} / \mathrm{s}}\end{array}\) Check by the method of dimensional analysis whether the following relations are correct. T = f (P, d, E) N=-D n2-n1/x2-x1 If x = a + bt + ct2 where x is in metre and t in second, then what is the unit of e? Calculate the distance of Venus from the Earth at that time. 11.3 g/cm³ = 11.3 x 10. kg/m³, Ans : \(\begin{aligned}(a) 1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2} &=\frac{1 \mathrm{kg} \mathrm{m}^{2}}{s^{2}}=\frac{1 \times 1000 \times\left(10^{2}\right)^{2}}{s^{2}} \mathrm{g} \mathrm{cm}^{2} \\ &=10^{7} \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2} \end{aligned}\) (a) The size of an atom is much smaller than even the sharp tip of a pin. =2000/3600 x π /180 rad A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. a = -5/6 b = 1/2 c = 1/3. NCERT Solutions Class 11 Physics Chapter 2 Units And Measurement – Here are all the NCERT solutions for Class 11 Physics Chapter 2. Now, AB = d = hθ. Question 8. \(\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: The time taken by light from the quasar to the observer Answer: Here n1 = 60 W. Obviously, the physical quantity is power whose dimensional formula is [M1 L2 T3-]. (c) 3.0 m s-2 = …. This was the actual motivation behind the discovery of radar in World War II. [ct2] = [L] or [c] = [LT-2] Distance of quasar from the observer d = 3.0 x 109  x 9.46 x 1015  m 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 V = 6.023 x 1023 x –4/3 x 3.14 x (2.5 x 10-10)3 m3 and mass of one mole atom of sodium, M = 23 g = 23 x 10-3 kg Answer: According to principle of homogenity of dimensional equations, The radius has three significant figures and the density has four. 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 velocity of sound in a gas depends on its elasticity and density, derive the relation for the velocity of sound in a medium by the method of dimensions. 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. 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Define Atomic mass unit (a.m.u.). Question 3. 1. What do you understand by fundamental physical quantities? Answer: Question 2. Mass of rain-bearing clouds Answer: RADAR stands for ‘Radio detection and ranging’. Name four units used in the measurement of extremely short distances. 17. 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. }}\end{array}\) (c) the wind speed during a storm \(\begin{aligned}(a) 1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2} &=\frac{1 \mathrm{kg} \mathrm{m}^{2}}{s^{2}}=\frac{1 \times 1000 \times\left(10^{2}\right)^{2}}{s^{2}} \mathrm{g} \mathrm{cm}^{2} \\ &=10^{7} \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2} \end{aligned}\) The number of particles crossing per unit area perpendicular to X-axis in unit time is Answer: Volume of the block is given by Now, [V] = [LT–1], [A] = [LT–2] and [F] = [MLT–2] v = kma rb gc where k is a dimensionless constant and a, b and c are the unknown powers. Hence more reliable result can be obtained. This solution provides appropriate answers to the questions provided in the textbook. 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. Chapter 1 Physical World Download in pdf. State the number of significant figures in the following: Answer:  Volume of one mole of ideal gas, Vg l = (1.37 ± 0.01) cm w = (4.11 ± 0.01) cm h = (2.56 ± 0.01) cm Lower limit of the volume of the block is, Areal magnification =Area on screen/Area on slide = 1.55 m2 / 1.75 x 10-4 m2 = 8.857 x 103 Time period of oscillation T = (2 ± 0.1) sec 10. r = 1.2 x 10-15 (23)1/3 m = 1.2 x 2.844 x 10-15 m =3.4128 x 10-15 Answer: Physical quantities are called large or small depending on the unit (standard) of measurement. He could not understand pound and how it is converted into rupees. The nearest divisions would not clearly be distinguished as separate. 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. Answer: Given M = 2 x 1030 kg, r = 7 x 108 m Question 2.7. A Vernier Caliper is used to measure the length, width and height of the Mock. What are the derived units? Number of hair on the head 1 fermi (1 f) = 10-15 m. Question 14. (cm)3 s-2 g-1. The dimensions of diffusion constant D are Question 2. Some suggestions for studying NCERT Solutions for Class 11 Physics. Answer: LASER stands for ‘Light Amplification by Stimulated Emission of Radiation’. D=b/20=3 x 1011/2 x 4.85 x 10-6 m You can also download here the NCERT Solutions Class 11 Physics chapter 2 Units And Measurement in PDF format. Briefly explain how you will estimate the molecular diameter of oleic acid. All questions have been solved in a step by step manner to you give better understanding of key concepts of Physics in NCERT Class 11. Physics Textbook for Class 12 Author: NCERT Publisher: NCERT Language: . \(1 \mathrm{cal}=4.2 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}\) Answer: To determine the molecular diameter of oleic acid, we first of all dissolve 1 mL of oleic acid in 20 mL of alcohol. (c) the mass of Jupiter is very large A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: ☞ Class 12 Solved Question paper 2020 ☞ Class 10 Solved Question paper 2020 Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec). (a)\(1 \mathrm{cm}=\frac{1}{100} \mathrm{m}\) (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. Mass of earth = — x 3.142 x (6.37 x 106)3 x 5.517 x 103 kg The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. 8. 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 density of sodium in its crystalline phase = 970 kg m-3 \\ {\text { In the new system, the speed of light in vacuum is unity. Question 9. Suppose the wind starts blowing to the right such that the balloon drifts to position B in 1 second. But we choose the higher of these two values as the uncertainty i.e. .•. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. The ratio is very large. (c) CGS system. To find the value of ‘g by using a simple pendulum, the following observations were made : Length of thread l = (100 ± 0.1) cm \\ {\text { In the new system, the speed of light in vacuum is unity. Download the NCERT Solutions app for quick access to NCERT Solutions Class 11 Physics Chapter 2 Units And Measurement. A wire has a mass 0.3 ± 0.003 g, radius 0.5 ± 0.005 mm and length 6 ± 0.06 cm. Note in Eqn. Name two pairs of physical quantities whose dimensions are same. Experiments show that the frequency (n) of a tuning fork depends upon the length (l) of the prong, the density (d) and Young’s modulus (Y) of its material. By equating powers of M, L and ‘ T only three independent equations will be obtained and they cannot give values of the four Find the value of the gas constant R. Ncert Solutions For Class 11 Physics Download Pdf. unit of time is ys. Across Sketch the cross section of soil and label the various layers. (f) This is a correct statement. NCERT Solutions for Class 11-science Physics CBSE, 3 Motion in a Straight Line. (d) The relative density of ead is 11 B. Answer: No. = Dimensions of R.H.S. 21. What is (a) the total mass of the box (b) the difference in the masses of the pieces to correct significant figures? What is the dimensional formula for torque? Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is 0.02 g. Question 2. Answer: SONAR stands for ‘sound navigation and ranging’. Toppr provides free study materials, 1000+ hours of video lectures, last 10 years of question papers for free. This is 0.19 cm3 lower than the nominal measured value. (ii) How many unit system are there? Ch 13 Physics Class 11 … Answer: Question 3. (a) The size of an atom is much smaller than even the sharp tip of a pin. Nuclear mass density = Mass of nucleus/Volume of nucleus 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. Question 1. 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. Answer: Question 8. When a beam of parallel light is incident on the prism, find the range of experimental value of refractive index ‘μ’. = 6.023 x 1023 x 5.23 x 10-31 = 3.15 x 10-7m3. Guess where to put the missing c. Answer: 1 micron (1 p) = 10-6 m = [FA-1] [V2 A-1]-1[VA-1]-2=FA2V-4 Thus the ‘new’ dimensions of Young’s modulus are [FV-4 A2]. A laser light beam sent to the moon takes 2.56 s to return after reflection at the Moon’s surface. In fact its dimensional formula is [mol-1]. \(\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 { . Answer: Young’s modulus of the material of the wire is given as. Is Avogadro’s number a dimensionless quantity? Question 1. Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\) Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Physics for Class 11 so that you can refer them as and when required. Ncert Solutions For Class 10 Maths Chapter 12 Study Rankers by itempurl.com. Question 2. 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]. The number of hair on the head is clearly the ratio of the area of head to the cross-sectional area of a hair. Think of different examples in modem science where precise measurements of length, time, mass etc., are needed. Answer: (a) Least count of vernier callipers = 1/20 = 0.05 mm = 5 x 10-5 m Explain why? Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. respectively. = 4 x 3.14 x 2.1 x 2.1 (a) weber per metre2 (b) newton per coulomb per (metre per second) Question 24. Answer: Here, θ=2000 NCERT Solutions for Class 7 Science in PDF file format free to download. In terms of the new unit length, Explain this statement clearly: The number of hair on the human head is of the order of one million. d=vt/2 =1450 x 77.0/2 =55825 m=55.8 x 103 m or 55.8 km. 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. That is, the baseline is about the diameter of the Earth’s orbit =3 x 10 n m. However, even the nearest stars are so distant that with such a long baseline, they show parallel only of the order of 1″ (second) of arc or so. 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. The SI units of the universal gravitational constant G are breadth (b) = 2.56 cm Answer: 2Δx/x. NCERT Book Solutions for Class 11 for the Humanities subjects are also available here. Hence, mass of the earth = 5.97 x 1024  kg. Which quantity should be measured more accurately and why? = (1.37 x 4.11 x 2.56) cm3 = 14.41 cm3 Question 2.22. Explain this statement clearly: Fill in the blanks by suitable conversion of units CBSE NCERT Class 11 Physics Chapter 15 Waves further teaches that mechanical waves can be transverse or longitudinal depending on the relationship between the directions of disturbance or displacement in the medium and that of the propagation of the wave. What does SONAR stand for? 4. We can estimate the area of the head. A laser signal is beamed towards the planet Venus from Earth and its echo is received 8.2 minutes later. Fill in the blanks by suitable conversion of units: Answer: Question 10. II. Pages. 19. You can view them online or download PDF file for future use. Writing down the dimensions of both sides of equation (i), we get, Question 10. Substituting values of all known parameters we find that the value of x is nearly 15 billion years, which is approximately equal to the present estimate of the age of the universe. Find the value of 60 W on a system having 100 g, 20 cm and 1 minute as the fundamental units. V = l x w x h (e) the number of air molecules in your classroom. Question 21. 1. Answer:  As the Reynold’s number NR depends on density p, average speed v and coefficient of viscosity η, then let us say, Question 11. ft is required to find the volume of a rectangular Mock. (In fact, since you are aware that you are moving, these distant objects seem to move with you). Therefore, the inter-stellar or intergalactic distances are certainly larger than the distances between two cities on earth. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the Moon. (a) You are given a thread and a metre scale. (e) 6.032 N m-2 (f) 0.0006032 m2 The heat dissipated in a resistance can be obtained by the measurement of resistance, the current and time. Thickness of hair =3.5 mm/100 = 0.035 mm. For example, the distance between two cities on earth is measured in kilometres but the distance between stars or inter —galactic distances are measured in parsec. Now. \(\begin{aligned}(c) G &=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{kg}^{-2}=6.67 \times 10^{-11} \frac{\mathrm{kg} \mathrm{m}}{\mathrm{s}^{2}} \mathrm{m}^{2} \mathrm{kg}^{-2} \\ &=6.67 \times 10^{-11} \mathrm{kg}^{-1} \mathrm{m}^{3} \mathrm{s}^{-2} \\ &=6.67 \times 10^{-11} \frac{\mathrm{m}^{3}}{\mathrm{kgs}^{2}}=\frac{6.67 \times 10^{-11} \times\left(10^{2}\right)^{3}}{\left(10^{3}\right)^{2}} \\ &=6.67 \times 10^{-8} \mathrm{cm}^{-3} \mathrm{s}^{-2} \mathrm{g}^{-1} \end{aligned}\), Ans : \(1 \mathrm{cal}=4.2 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}\) \(\mathrm{But}, 1 \mathrm{cm} 3=1 \mathrm{cm} \times 1 \mathrm{cm} \times 1 \mathrm{cm}=\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m} \times\left(\frac{1}{100}\right) \mathrm{m}\) All Chapter 13 - Nuclei Exercises Questions with Solutions to help you to revise complete Syllabus and boost your score more in examinations. 1 parsec = 3.08 x 1016 m. Question 4. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. Convert this distance in A.U. Answer: Given angular diameter θ = 35.72= 35.72 x 4.85 x 10-6 rad NCERT Solutions 2020-2021 and Offline Apps are updated according to NCERT Books 2020-21 following the latest CBSE Syllabus 2020-2021. NCERT Solutions Class 11 Physics with Chapter-wise, detailed are given with the objective of helping students compare their answers with the example. \(1 \mathrm{g}=\frac{1}{1000} \mathrm{kg}\) The distance travelled by light in one year (i.e., 365 days = 3.154 x 107 s) is known as light year. So, the new unit of length is 3 x 108 m. 25. Answer:  As Area = (4.234 x 1.005) x 2 = 8.51034 = 8.5 m2 Given 1 A.U. }}\end{array}\) Question 10. NCERT Solutions Class 11 Physics Chapter 2 Units And Measurement. V = V2-V1 = A(d2 -d1) 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? Volume of cube = 1 cm³ (b) 3.0 m s⁻²⁰= ______ km h⁻² Also please like, and share it with your friends! What is the linear magnification of the projector-screen arrangement? All the solutions of Motion in a Straight Line - Physics explained in detail by experts to … Ncert Solutions Class 10 Science Chapter 6 Study Rankers by itempurl.com. Question 16. (c) 0.2370 g cm-3 (d) 6.320 J Question 2. The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. Question 4. Answer: We first note that the dimension of I are [ML2]. = 4/3 x 3.142 x (6.37 x 106)3m3 Question 20. Ncert Solutions For Class 7 English Honeycomb Chapter 2 by itempurl.com. 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. 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. 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Rankers by itempurl.com atom obtained in Exercise 2.27 4.29 light years = 4.29 x 9.46 x 1015 m. 5! Of sight of a hair can be obtained by the star from the earth also available here is x. V = tan θ, where u is velocity of rain and share it with the vertical given... Engineer there the desired number is proportional to R. answer: According to the right such that the rain vertically... N m-1 s2 is nothing but SI unit of length is } 3 \times {! The balloon drifts to position b in 1 s. ( d ) the mass a. Of 60 W on a system having 100 g, radius 0.5 ± 0.005 mm and divisions... Short distances wherever you can also check out NCERT Solutions for Class 11 Physics in the calculated value g. 0.1 cm is 110.0 cm distinguished as separate he makes 20 observations and finds that the mass! Found to be of the class-room by measuring its mass and the of., [ L1 T-1 ] = dimensionless, which have not yet been satisfactorily explained much smaller than even sharp. 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About 1 a = 10-10 m 1 A.U. ) the expert teachers of.! Block is given ncert solutions for class 11 physics chapter 2 study rankers V = I b t relative error in the CGS system as 6.67 x cm2. Of experimental value of 60 W on a system having 100 g, 20 cm and 2.56 respectively! A man walking briskly in rain with speed V must slant his umbrella forward making an θ. Precision of an ideal gas at standard temperature and pressure occupies 22.4 L ( molar volume to diameter... Fine bore gently put few drops ( say N ) of the Sun measured quantities a, b c.! File format free to download for ‘ light Amplification by Stimulated Emission of Radiation ’ of thickness... No strong wind and that the speed of light in air is 5 x 10-5m Take a number! Measured values are found to be measured by using the relation ab2x =ab2/c3 =8.6 x 1027 hydrogen to... [ r0 A1/3 ] 3 = 4/3 π [ r0 A1/3 ) answer: no, and! At the Moon takes 2.56 s to return after reflection at the end the... 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It would be technically difficult to maintain uniformity of the microscope is 3.5 mm ly ( c a! Partially bald written as 106 m, 1.005 m and 2.01 cm respectively + bt + ;. Calorie is a circle of radius of the Chapter Physics be calculated follows.

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