Applied Physics-I Suggestions (NatiTute) PDF

 Contents Sl No Chapters No. of Question PDF Page No. 1 Unit and Measurement 50 3 2 Force and Motion 44 10 3 Work, Power and E...

 Contents Sl No Chapters No. of Question PDF Page No. 1 Unit and Measurement 50 3 2 Force and Motion 44 10 3 Work, Power and Energy 36 16 4 Rotational Motion 26 21 5 Properties of Matter 87 25 6 Heat and Thermometry 60 37 Total Number of Question= 303 Click Here to watch full playlist of Applied Physics-I on NatiTute YouTube Channel (If your mobile can not open the above link, please install Adobe pdf Reader App) © Copyright Disclaimer Any physical or digital format of this PDF are not allowed to share or re-selling without proper permission from NatiTute YouTube Channel. এটা একটা প঩ইড pdf ebook যারা ফইটি সংগ্রহ করছো তাছের কাছে অনুছরাধ পতামরা এটা পলয়ার কছরা না, কারণ সফার কাছে pdf পাই঱ টা প঩ৌছে পেছ঱ পতামরা তাছের পথছক এগেয়া থাকছত ঩ারছফ না এফং এই ফই পকনার পকাছনা মযয াো থাকছফ না.  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic Polytechnic 1st Semester 5. The physical quantity which has dimensionless unit Applied Physics-I in SI is? (a) Mass Unit 1: Physical World, Unit & Measurement (b) Temperature (Most Important 50 Questions) (c) Plane angle (d) Electric current 1. Which of the following is not a fundamental quantity? Explanation: Plane angle is a dimension less (a) Temperature physical quantity because it is the ratio of two (b) Luminous intensity same quantity (length) but it has a SI unit i.e. (c) Electric current radian. (d) Quality of substance Answer: (c) Explanation: Quantity of Substance is fundamental 6. The physical quantity which has zero dimension in quantity not quality of substance. SI is Answer: (d) (a) Specific heat (b) Specific gravity 2. Dimension formula of surface tension? (c) Specific volume (a) 𝑴𝑳𝑻 (d) None of these (b) 𝑴𝑳−𝟐 (c) 𝑴𝑻−𝟐 Explanation: The ratio of density of a substance to (d) None of these the density of water is known as specific gravity. As it is the ratio of two same physical quantities so, Explanation: As force per unit length is called its dimension is zero. surface tension. So, dimension of surface tension, Answer: (b) 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝐹𝑜𝑟𝑐𝑒 𝑀𝐿𝑇 −2 = = = 𝑀𝑇 −2 𝐷𝑖𝑚𝑒𝑛𝑡𝑖𝑜𝑛 𝑜𝑓 𝐿𝑒𝑛𝑔𝑡 𝑕 𝐿 7. The dimensional formula of work is same as that of Answer: (c) (a) Impulse (b) Moment 3. Which of the following is derived quantity? (c) Force (a) Electric current (d) Momentum (b) Temperature (c) Luminous intensity Explanation: (d) Energy Dimension of work= 𝑀𝐿𝑇 −2 × 𝐿 = 𝑀𝐿2 𝑇 −2 Dimension of moment Explanation: Energy is derived physical quantity = 𝑀𝐿𝑇 −2 × 𝐿 = 𝑀𝐿2 𝑇 −2 because energy depends on two or more physical Answer: (b) quantity like force, displacement, velocity, mass etc. 8. Which of pairs have the same dimension? Answer: (d) (a) Torque, work (b) Stress, energy 4. Which of the following is the unit of a basic physical (c) Force, stress quantity? (d) Force, work (a) Pascal (b) Newton Explanation: Dimension of both of torque and work (c) Candela is = 𝑀𝐿𝑇 −2 × 𝐿 = 𝑀𝐿2 𝑇 −2 (d) Joule Answer: (a) Explanation: Candela is a basic or fundamental physical quantity because it does not depend on any other physical quantity. Answer: (c)  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic 9. Dimension of angular velocity is- 13. Watt is the SI unit of – a) Zero a) energy b) 𝑳𝟏 b) power c) 𝑳𝟏 𝑻𝟏 c) force d) 𝑻−𝟏 d) electricity Explanation: When 1 Joule work done in 1 second Explanation: 𝜃 then it is called 1 Watt. So, Watt = Jule/Second Angular velocity 𝜔 = 𝑡 Answer: (b) 𝜃 = Angular displacement, 𝑡 =Time taken ∴ Dimension of angular velocity 14. Temperature can be expressed as derived quantity 1 = [∵ 𝜃 is a dimensionless quantity] in terms of any of the following – 𝑇 = 𝑇 −1 (a) length and mass Answer: (d) (b) mass and time (c) length, mass and time 10. The dimensional formula of stress is the same as (d) in terms of none that of – Explanation: As temperature is a fundamental a) energy physical quantity so it can’t be expressed as b) pressure derived quantity in terms of any other quantity. c) force Answer: (d) d) surface tension 15. The dimension of strain is- Explanation: Dimensional formula of both of stress a) 𝑴𝟏 𝑳𝟏 𝑻𝟏 𝑀𝐿𝑇 −2 b) 𝑴𝟏 𝑳𝟎 𝑻𝟎 and pressure is = = 𝑀𝐿−1 𝑇 −2 𝐿2 c) 𝑴𝟎 𝑳𝟎 𝑻𝟏 Answer: (b) d) 𝑴𝟎 𝑳𝟎 𝑻𝟎 11. The SI unit of stress is- Change in length (dl ) (a) 𝑵 Explanation: Strain = Initial length (l) (b) 𝑵𝒎𝟐 𝑀 0 𝐿1 𝑇 0 So, Dimension of strain = 𝑀 0 𝐿1 𝑇 0 = 𝑀0 𝐿0 𝑇 0 (c) 𝑵𝒎−𝟐 Answer: (d) (d) 𝑵−𝟏 𝒎−𝟐 Explanation: Force 16. The number of significant figures in 0.06900 is- Stress = Area a) 3 unit of force 𝑁 So, unit of stress = unit of area = 𝑚 2 = 𝑁𝑚−2 b) 4 Answer: (c) c) 5 d) 2 12. The SI unit of Poisson’s ratio is – a) 𝑵 Explanation: In a number of less than one, all zeros b) 𝑵𝒎𝟐 to the right of decimal point and to the left of a non c) 𝑵𝒎−𝟐 zero digit are not significant. d) 𝑵𝒐𝒏𝒆 𝒐𝒇 𝒕𝒉𝒆𝒔𝒆 Answer: (b) Explanation: The ratio of lateral strain to longitudinal strain is called Poisson’s ratio. As it is the ratio of two similar quantities, s, it have not unit. Answer: (d)  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic 17. The number of significant figures in 60.023 is- 20. If the error in the measurement of the volume of a a) 1 sphere is 6%, then the error in the measurements of b) 4 its surface area will be c) 5 (a) 2% d) 2 (b) 3% (c) 4% Explanation: All non-zero digits are significant and (d) 7.5% all zeros occurring between two non-zero digit are Explanation: significant. 4 Volume of sphere 𝑉 = 3 𝜋𝑟 3 Answer: (c) 4 ⇒ log 𝑉 = log 3 𝜋𝑟 3 4 18. The measured value of a capacitor is 205.3 𝝁𝑭 ⇒ log 𝑉 = log 3 𝜋 + log 𝑟 3 whereas the true value is 201.4 𝝁𝑭. The percentage 4 ⇒ log 𝑉 = log 3 𝜋 + 3 log 𝑟 error is ∆𝑉 Δ𝑟 Δ𝑟 6% (a) 1.94% ⇒ = 3 ,⇒ = = 2% 𝑉 𝑟 𝑟 3 (b) 2.94% Surface area 𝑆 = 4𝜋𝑟 2 (c) 0.94% ⇒ log 𝑆 = log 4𝜋 + 2 log 𝑟 (d) 0.094% ∆𝑆 Δ𝑟 ⇒ 𝑆 =2 𝑟 = 2 × 2% = 4% Explanation: Percentage Error Absolute error Answer: (c) = True value × 100% 205.3−201.4 = 201.4 × 100% 21. Dimensional formula of power is- = 0.01936 × 100% a) 𝑴𝑳𝟐 𝑻−𝟑 = 1.94% b) 𝑴𝑳𝟐 𝑻𝟑 Answer: (a) c) 𝑴𝑳𝟑 𝑻−𝟐 d) 𝑴𝑳𝟑 𝑻−𝟑 19. The error in measurement of diameter of a circular disk is 1%. The percentage error in the Work Explanation: Power = measurement of its area is Time (a) 1% So, Dimension of power Dimension of wok ML 2 T −2 (b) 2% = = = ML2 T −3 Dimension of time T (c) 3% Answer: (a) (d) 4% 𝜋 22. The relation between force 𝑭 and time 𝒕 is Explanation: Area of circle 𝐴 = 4 𝐷 2 𝑭 = 𝒂𝒕 + 𝒃𝒕𝟐. The dimensional formula of 𝒂 and 𝒃 𝜋 ⇒ log 𝐴 = log 4 𝐷 2 are respectively- 𝜋 ⇒ log 𝐴 = log 4 + log 𝐷 2 (a) 𝑴𝑳𝑻−𝟏 and 𝑴𝑳𝑻𝟎 𝜋 (b) 𝑴𝑳𝑻−𝟑 and 𝑴𝑳𝟐 𝑻𝟒 ⇒ log 𝐴 = log 4 + 2 log 𝐷 𝑑𝐴 𝑑𝐷 (c) 𝑴𝑳𝑻−𝟒 and 𝑴𝑳𝑻 ⇒ =0+2 𝐴 𝐷 (d) 𝑴𝑳𝑻−𝟑 and 𝑴𝑳𝑻−𝟒 ∴ Error in area=2×Error in Dia. Explanation: According to the principle of =2×1%=2% dimensional homogeneity Answer: (b) 𝑎𝑡 = 𝐹 𝐹 𝑀𝐿𝑇 −2 ⇒ 𝑎 = = = 𝑀𝐿𝑇 −3 𝑡 𝑇 and, 𝑏𝑡 2 = 𝐹 𝐹 𝑀𝐿𝑇 −2 ⇒ 𝑏 = = = 𝑀𝐿𝑇 −4 𝑡2 𝑇2 Answer: (d)  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic 23. The dimension of thermal conductivity is- 27. What do you mean by “the dimension of physical (a) 𝑴𝑳𝑻−𝟐 𝑲−𝟏 quantity”? (b) 𝑴𝑳𝑻−𝟑 Answer: (c) 𝑴𝑳𝑻−𝟑 𝑲−𝟏 Dimension of physical quantity: The dimension of a physical quantity are the powers to which the (d) 𝑴𝑳𝑻−𝟏 fundamental unit of mass (M), length (L), time (T), Explanation: Δ𝑄 temperature (K), electric current (A) have to be Δ𝑄.Δ𝑠 We know, thermal conductivity 𝑘 = Δ𝑡 Δ𝜃 = raised in order to completely represent that 𝐴 𝐴.Δ𝜃.Δ𝑡 Δ𝑠 physical quantity. 𝑀𝐿2 𝑇 −2.𝐿 ∴ 𝑘 = = 𝑀𝐿𝑇 −3 𝐾 Example: Dimension of velocity = 𝐿𝑇 −1 𝐿2 𝐾𝑇 Answer: (c) Dimension of density = 𝑀𝐿−3 24. What is fundamental or basic physical quantity? 28. Write SI unit and dimension of velocity, force, Answer: energy, power, pressure, density and surface Fundamental Quantities: The unit of those physical tension. quantities which do not dependent on others and Answer: by which all other quantities may be obtained is No Quantity SI unit Dimension −1 called fundamental quantities. 1 Velocity 𝑚𝑠 𝐿𝑇 −1 2 Force N 𝑀𝐿𝑇 −2 3 Work/Energy J 𝑀𝐿2 𝑇 −2 25. Give names of the basic physical quantities and 4 Power W 𝑀𝐿2 𝑇 −3 their SI unit. 5 Pressure Pa 𝑀𝐿−1 𝑇 −2 Answer: 6 Density Kgm -2 𝑀𝐿−3 Fundamental quantities and their SI units: 7 Surface tension 𝑁𝑚 −1 𝑀𝑇 −2 Quantity SI unit Symbol 29. What are dimensionless physical quantities? 1. Length Meter m Answer: 2. Mass Kilogram kg Dimensionless physical quantities: A physical 3. Time Second s quantity which is not related to any physical 4. Temperature Kelvin K dimension. A number representing the property of 5. Current Ampere A physical system but not measured on a scale of 6. Illuminating Intensity Candela Cd physical unit. 7. Quantity of Matter Mol mol Example: Plane angle, Specific gravity 30. Can a quantity unit be dimensionless? Give one 26. What is derived physical quantity? Give example. example. Answer: Answer: Yes, a quantity can have dimensionless Derived physical quantities: The unit of those unit. As for example angular displacement is a physical quantities which can be derived from one dimensionless quantity and its SI unit is Radian. or more fundamental units these physical quantities are called derived physical quantities. 31. State “the principle of dimensional homogeneity”. Example: Velocity is a derived physical quantity Answer: because it is expressed in meter/second. Principal of dimensional homogeneity: The dimension of each the terms of an equation will be same. If it does not occur then it can be said that the equation is wrong. Example: Suppose in the case of the equation 𝑣 = 𝑢 + 𝑓𝑡. If we compare the both sides we can found that dimensions of both sides are same 𝐿𝑇 −1.  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic 32. What is the importance of dimensional analysis? 36. What do you mean by percentage error? Answer: Answer: Importance of dimensional analysis: Percentage error: when the relative error or a) Easily we can check the correctness of an proportional error is expressed in percentage we equation. call it percentage error. b) We can deduce the relation among different 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟 i.e. percentage error = 𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒 × 100% physical quantities. c) It is very useful for conversion of unit. 37. What do you mean by systematic error? d) The most important to determine the unit Answer: constant in an equation. Systematic Error: Systematic error is consistent, repeatable error associated with faulty equipment 33. What are the limitations of dimensional analysis? or a flawed experiment design. These errors are Answer: usually caused by measuring instruments that are Limitations of dimensional analysis: incorrectly calibrated or are used incorrectly. a. It gives no information about dimensionless constant of proportionality involved in the 38. What do you mean by random error? equation. Answer: b. It can’t be applied if a quantity is related to Random Error: Random errors are not fixed on other quantities as sum or difference of them. general perimeters and depend on measurements c. This method fails if a quantity depends on to measurements. They are named random errors trigonometric functions or exponential as they are random in nature. Random errors functions. occur due to- d. The method fails if a physical quantity depends a. Sudden and unexpected shifts in on more than three quantities. experimental condition of the environment. e. If the unknown relation contains two more b. Personal bias errors which even the student quantities with the same dimensions this is unaware of. method fails to determine the relation. Example: A spring balance will give different f. We can’t check the relation is absolutely readings if the temperature of the environment is correct or wrong. As an example: the relation not constant. 𝑣 2 = 𝑢2 + 4𝑓𝑠 is dimensionally correct but it is wrong. 39. What is least count in measurement? Answer: 34. What do you mean by absolute error? Least Count: The least count of an instrument is the Answer: size of the smallest scale division on the Absolute error: The difference between the true instrument. The smaller the least count the precise value and the individual measured value of a the instrument is said to be. physical quantity is called absolute error. It may be For example a metric ruler which has positive of negative. millimeter marks has a least count of 1mm or 0.1 Absolute error = True value – Measured value cm. 35. What do you mean by proportional error? Answer: Proportional error: The ratio of the absolute error to the true value is known as proportional error. 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑒𝑟𝑟𝑜𝑟 i.e. proportional error = 𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic 40. What is the difference between accuracy and 44. Length of a rod as measured in an experiment was precision? found to be 2.48 m, 2.46 m, 2.49 m, 2.50 m and 2.48 Answer: m. Find the- Difference between accuracy and precision i) Average length Accuracy Precision ii) Mean absolute error Accuracy means how Precision means how iii) Relative error close the results to the close the results are iv) Percentage error actual value. with each other. Answer: Accuracy is the Precision measures how i) Average length, measure of quantity well measurements can 2.48+2.46+2.49+2.50+2,48 to reality. be reproduced. 𝑙𝑚 = 5 = 2.482 𝑚 Accuracy takes the Precision does take the ii) Mean absolute error ∆𝑙 , accepted value. accepted value. Now, 2.48 − 2.482 = 0.002 It is concerned with a It is concerned with 2.46 − 2.482 = 0.022 systematic error. random error. 2.49 − 2.482 = 0.008 Accuracy has a single Precision has multiple 2.50 − 2.482 = 0.018 factor. factors. 2.48 − 2.482 = 0.002 0.002+0.022+0.008+0.018+0.002 41. If force, velocity and time are taken as fundamental ∆𝑙 = 5 = 0.0104 𝑚 quantities, find the dimension of work. ∆𝑙 0.0104 iii) Relative error = 𝑙 = 2.482 = 0.004 𝑚 Answer: 𝑚 ∆𝑙 Let, dimension of force, velocity and time are iv) Percentage error = 𝑙 × 100% = 0.4% 𝑚 𝐹 , 𝑉 and 𝑇 respectively. As displacement = velocity × time ∴ 𝐿 = 𝑉𝑇 45. A wire has mass 𝟎. 𝟑 ± 𝟎. 𝟎𝟎𝟑 𝒈 , radius Now, work = force × displacement 𝟎. 𝟓 ± 𝟎. 𝟎𝟎𝟓 𝒎, and length 𝟔 ± 𝟎. 𝟎𝟔 𝒄𝒎. What = force × velocity × time will be the maximum percentage error in the So, dimension of work = 𝐹 × 𝑉 × 𝑇 measurement of density of the wire? = 𝐹𝑉𝑇 (Ans.) Answer: 𝑀 𝑀 We know density 𝜌 = 𝑉 = 𝜋𝑟 2 𝑙 𝟐 42. If 𝒙 = 𝒂 + 𝒃𝒕 + 𝒄𝒕 where 𝒙 is in meter and 𝒕 is in 𝑀 second, then what is the unit of 𝒄? Taking log in both side, log 𝜌 = log 𝜋𝑟 2 𝑙 Answer: Given, 𝑥 = 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 ⇒ log 𝜌 = log 𝑀 − log 𝜋 − log 𝑟 2 − log 𝑙 By Principle of dimensional homogeneity we get, ⇒ log 𝜌 = log 𝑀 − log 𝜋 − 2 log 𝑟 − log 𝑙 unit of 𝑐𝑡 2 = unit of 𝑥 Differentiating both side we get, ∆𝜌 ∆𝑀 ∆𝑟 ∆𝑙 ⇒ 𝑐 × 𝑠𝑒𝑐𝑜𝑛𝑑2 = 𝑚𝑒𝑡𝑒𝑟 = −0−2 − 𝑚𝑒𝑡𝑒𝑟 𝜌 𝑀 𝑟 𝑙 ⇒ 𝑐 = 𝑠𝑒𝑐𝑜𝑛 𝑑 2 = 𝑚/𝑠 2 = 𝑚𝑠 −2 For maximum fractional error of 𝜌, −2 Hence, unit of 𝑐 is 𝑚𝑠 (Ans.) all sign will be (+ve). ∆𝜌 ∆𝑀 ∆𝑟 ∆𝑙 𝜌 = 𝑀 +2 𝑟 + 𝑙 𝟏 43. Check whether the equation 𝑺 = 𝒖𝒕 + 𝒂𝒕𝟐 is 0.003 0.005 0.06 𝟐 = +2× + 0.3 0.5 6 dimensionally correct. (Symbols have their usual = 0.01 + 2 × 0.01 + 0.01 = 0.04 meaning) ∴ Maximum percentage error in density Answer: 1 = 0.04 × 100% = 4% (Ans.) We have, 𝑆 = 𝑢𝑡 + 2 𝑎𝑡 2 Checking the dimensions of each terms, Dimension of 𝑆 = 𝐿 Dimension of 𝑢𝑡 = 𝐿𝑇 −1 × 𝑇 = 𝐿𝑇 −1 × 𝑇 = 𝐿 1 Dimension of 2 𝑎𝑡 2 = 𝐿𝑇 −2 × 𝑇 2 = 𝐿 As dimensions of each terms are same i.e. 𝐿 Hence, the formula is dimensionally correct.  ⦿ Unit and Measurements ⦿ Applied Physics-I Suggestions ⦿ 1st Semester ⦿ WB Polytechnic 46. The centripetal force 𝑭 acting on a particle moving Now, from the principle of dimensional uniformly in a circle may depend upon mass 𝒎 , homogeneity we get, velocity 𝒗 and radius 𝒓 of the circle. derive the 𝐿𝑇 −1 = 𝑀 𝑎 𝐿 𝑏 𝐿𝑇 −2 𝑐 formula for 𝑭 using the method of dimensions. ⇒ 𝑀 0 𝐿 1 𝑇 −1 = 𝑀 𝑎 𝐿 𝑏+𝑐 𝑇 −2𝑐 Answer: Comparing both sides we get, Let, 𝐹 ∝ 𝑚𝑎 , 𝐹 ∝ 𝑣 𝑏 and 𝐹 ∝ 𝑟 𝑐 𝑎 = 0 , −2𝑐 = −1 , ⇒ 𝑐 = 2 1 We can say, 𝐹 ∝ 𝑚𝑎 𝑣 𝑏 𝑟 𝑐 1 1 𝑏 + 𝑐 = 1, ⇒ 𝑏 + = 1, ⇒ 𝑏 = or, 𝐹 = 𝑘𝑚𝑎 𝑣 𝑏 𝑟 𝑐 (𝑘 is a dimension less constant) 2 2 1 1 Now writing the dimensions of individual terms So, required relation is 𝑣 = 𝑘𝑚0 𝑟 2 𝑔2 we get, ⇒ 𝑣 = 𝑘 𝑟𝑔 𝑀𝐿𝑇 −2 = 𝑘 𝑀 𝑎 𝐿𝑇 −1 𝑏 𝐿 𝑐 ⇒ 𝑀𝐿𝑇 −2 = 𝑘 𝑀𝑎 𝐿(𝑏+𝑐) 𝑇 −𝑏 𝜸𝑷 Now comparing both side we get, 49. Velocity of sound in a gas is given by 𝑽 = 𝑫 𝑎 = 1, 𝑏 = 2 and 𝑏 + 𝑐 = 1 ⇒ 2 + 𝑐 = 1 ⇒ 𝑐 = −1 where 𝑷 = Pressure, 𝑫 = Density and 𝜸 is a ∴ Now, 𝐹 = 𝑘𝑚𝑎 𝑣 𝑏 𝑟 𝑐 dimensionless constant. Show that the formula ⇒ 𝐹 = 𝑘𝑚𝑣 2 𝑟 −1 obeys the principle of dimensional homogeneity. 𝑚 𝑣2 Answer: ⇒𝐹=𝑘 𝑟 𝛾𝑃 It is given that, 𝑉 = 𝐷 47. The time period 𝑻 of a simple pendulum may Dimension of L.H.S. = 𝐿𝑇 −1 depend on mass of the bob 𝒎 , effective length of 𝑀𝐿−1 𝑇 −2 the pendulum 𝒍 and acceleration due to gravity Dimension of R.H.S. = 𝑀𝐿 −3 𝒈. By the principle of dimensional homogeneity, 𝒍 = 𝐿2 𝑇 −2 prove that 𝑻 = 𝒌 𝒈 , where 𝒌 is dimensionless = 𝐿𝑇 −1 constant. As dimension of both LHS and RHS are same Answer: So, the formula obeys the principle of dimensional Let, 𝑇 ∝ 𝑙 𝑎 𝑚𝑏 𝑔𝑐 homogeneity. ⇒ 𝑇 = 𝑘𝑙 𝑎 𝑚𝑏 𝑔𝑐 Now, from the principle of dimensional 50. Convert 10 Joule into erg with the help of homogeneity we get, dimensional analysis. 𝐿 𝑎 𝑀 𝑏 𝐿𝑇 −2 𝑐 = 𝑇 Answer: ⇒ 𝐿 𝑎+𝑐 𝑀 𝑏 𝑇 −2𝑐 = 𝐿 0 𝑀 0 𝑇 Joule is the SI unit of work and the dimensional Comparing both sides we get, formula of work = 𝑀𝐿2 𝑇 −2 1 In SI system 𝑏 = 0, −2𝑐 = 1, ⇒ 𝑐 = − 2 𝑀1 = 1𝑘𝑔; 𝐿1 = 1𝑚; 𝑇1 = 1𝑠; 𝑛1 = 10 1 1 𝑎 + 𝑐 = 0, ⇒ 𝑎 − 2 = 0, ⇒ 𝑎 = 2 In CGS system 1 1 So, required relation is 𝑇 = 𝑘𝑙 2 𝑚0 𝑔−2 𝑀2 = 1𝑔; 𝐿2 = 1𝑐𝑚; 𝑇2 = 1𝑠; 𝑛2 =? Since the quantity is same in both the system. 𝑙 ⇒𝑇=𝑘 𝑔 (Proved) ∴ 𝑛1 𝑀1 𝐿21 𝑇1−2 = 𝑛2 𝑀2 𝐿22 𝑇2−2 𝑀1 𝐿1 2 𝑇1 −2 ⇒ 𝑛2 = 𝑛1 𝑀2 𝐿2 𝑇2 48. The orbital velocity of an artificial satellite may 1𝑘𝑔 1𝑚 2 1𝑠 −2 depend on its mass 𝒎 , radius of orbit 𝒓 and on = 10 1𝑔 1𝑐𝑚 1𝑠 the acceleration due to gravity 𝒈. Using the 1000 𝑔 100𝑐𝑚 2 = 10 = 108 method of dimensions obtain an expression for the 1𝑔 1𝑐𝑚 8 orbital velocity. ∴ 10 Joule= 10 erg (Ans.) Answer: Let, orbital velocity of the satellite is 𝑣. and 𝑣 ∝ 𝑚𝑎 𝑟 𝑏 𝑔𝑐 ⇒ 𝑣 = 𝑘𝑚𝑎 𝑟 𝑏 𝑔𝑐 [𝑘 is a dimensionless constant]  ⦿ Force and Motion ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic Polytechnic 1st Semester 5. Impulse of a force acting on a body is equal to – Applied Physics-I a) change in momentum of the body b) rate of change of momentum of the body Unit 2: Force and Motion c) change in kinetic energy of the body (Most Important Questions) d) none of these Explanation: 1. Which of the following remains constant for a Impulse = Force × Time particle executing uniform circular motion? =𝐹×𝑡 a) linear speed = 𝑚𝑎 × 𝑡 b) linear velocity =𝑚 𝑣−𝑢 ×𝑡 𝑡 c) linear acceleration = 𝑚𝑣 − 𝑚𝑢 d) none of these = Change in momentum Answer: (a) Explanation: When a particle executing uniform circular motion it always changes its direction but 6. In velocity-time graph, area enclosed with the time magnitude may constant. axis represents- Answer: (a) a) displacement b) velocity 2. The recoil of a gun after firing shell is due to the c) force principle of conservation of – d) distance a) mass b) linear momentum Displacement c) angular momentum Explanation: Velocity = Time d) kinetic energy Therefore Velocity × Time = Displacement Answer: (a) Explanation: The total momentum of the system remains constant after firing so, it obey the 7. A 10N force applied on a body to produce in it an principle of conservation of linear momentum. acceleration of 1m/𝐬 𝟐. The mass of the body is- Answer: (b) a) 15 kg b) 20 kg 3. The rocket propulsion is based on the principle of c) 10 kg conservation of- d) 5 kg (a) mass Explanation: (b) linear momentum We know, Force = Mass × Acceleration (c) angular momentum ⇒ 𝐹 = 𝑚𝑎 (d) kinetic energy ⇒ 10 = 𝑚 × 1, ⇒ 𝑚 = 10 kg Explanation: When the rocket is ready to take off Answer: (c) the momentum of small mass of gas ejected at high speed is compensated by increased momentum of the rocket with the remaining fuel. 8. The net force required to accelerate a 1000 kg car at Answer: (b) 𝟒 𝒎/𝒔𝟐 is- a) 250 N 4. Rate of change of linear momentum is- b) 1000 N a) work c) 4000 N b) acceleration d) 0.004 N c) force d) torque Explanation: We know, Force = Mass × Acceleration Explanation: Rate of change of linear momentum ⇒ 𝐹 = 𝑚𝑎 𝑚𝑣 −𝑚𝑢 𝑣−𝑢 ⇒ 𝐹 = 1000 × 4, ⇒ 𝐹 = 4000 𝑁 = =𝑚 = 𝑚𝑎 = 𝐹 = Force 𝑡 𝑡 Answer: (c) Answer: (c)  ⦿ Force and Motion ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 9. 1 N is equal to- 12. A car moving with a speed of 50 km/hr can be a) 100 dyne stopped by brakes after at least 6m. If the same car b) 1000 dyne is moving at a speed of 100 km/hr, the minimum c) 10000 dyne stopping distance is- d) 100000 dyne (a) 12 m (b) 18 m Explanation: (c) 24 m 1𝑘𝑔 ×1𝑚 1000 𝑔×100𝑐𝑚 (d) 6 m 1𝑁 = = = 100000 dyne 1𝑠 1𝑠 Explanation: We know, 𝑣 2 = 𝑢2 ± 2𝑎𝑠 Answer: (d) Here, 𝑣 = 0 and 𝑎 is constant ∴ 𝑠 ∝ 𝑢2 10. A body of mass 4kg is hanged at lower end of a So, when speed increased by 2 times spring balance whose upper end is fixed at the ceiling of a lift. If the lift moves up with an then, distance increased by 22 = 4 times ∴ Minimum stopping distance is 4×6 = 24 m acceleration of 6 𝒎/𝒔𝟐 , the reading of the spring Answer: (c) balance is (Take𝒈 = 𝟗. 𝟖 𝒎/𝒔𝟐 ). (a) 15.2 N 13. What do you mean by force? (b) 24 N Answer: (c) 39.2 N (d) 63.2 N Force 𝑭 : Force is an external effort in the form of push or pull. Explanation: 1) It produces or tries to produce motion in a Here, mass, 𝑚 = 4 𝑘𝑔, body at rest. acceleration, 𝑎 = 6 + 9.8 = 15.8 𝑚/𝑠 2 2) It stops or tries to stop a moving body. ∴ Force 𝐹 = 𝑚𝑎 = 4 × 15.8 = 63.2 N 3) It changes or tries to change the direction of Answer: (d) motion of the body. Mathematically, 𝐹 = 𝑚𝑎 11. A large vertical cylinder of height 𝟐𝟎 𝒎 is filled with SI unit of force is Newton (N) or 𝑘𝑔𝑚𝑠 −2 water. Find the velocity of efflux of water through a Dimension of force: 𝑀𝐿𝑇 −2 very small hole made at the bottom of its vertical wall. (take 𝒈 = 𝟏𝟎 𝒎/𝒔𝟐 ). 14. Define conservative force with example. (a) 𝟏𝟎 𝒎/𝒔 Answer: (b) 𝟐𝟎 𝒎/𝒔 Conservative Force: A conservative force is a force (c) 𝟒𝟎 𝒎/𝒔 done in moving a particle from one point to (d) 𝟖𝟎 𝒎/𝒔 another, such that the force is independent of the Explanation: path taken by the particle. It depends only on the The velocity of efflux of water initial and final position of the particle. 𝑣= 2𝑔𝑕 = 2 × 10 × 20 Example: Gravitational force and elastic spring forces are two examples of conservative forces. = 400 = 20 𝑚/𝑠 Answer: (b) 15. Define non-conservative force with example. Answer: Non-Conservative Force: A non-conservative force is a force for which the work done depends on the path taken. A force is said to be a non-conservative force if it results in the change of mechanical energy, which is nothing but the sum of potential and kinetic energy. Example: Air drag and friction are two example of non-conservative force.  ⦿ Force and Motion ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 16. What is linear momentum? 21. Deduce an expression of the recoil velocity of gun. Answer: Answer: Linear Momentum (𝒑) : It is defined as the total Recoil of gun: When a bullet is fired from a gun, the quantity of motion contained in a body and is gun recoils i.e., moves in a direction opposite to measured as the product of the mass of the body direction of the bullet. The recoil velocity of the and its velocity. gun can be calculated from the principle of Mathematically, 𝑝 = 𝑚𝑣 conservation of linear momentum. SI unit: 𝑘𝑔𝑚𝑠 −1 Let, 𝑀 = mass of the gun Dimension: 𝑀𝐿𝑇 −1 𝑉 = velocity of the gun 17. Define Impulse of a force. 𝑚 = mass of the bullet Answer: 𝑣 = velocity of the bullet Impulse (𝑰): It is the product of force and interval According to the principle of conservation linear of time for which the force act on body is called momentum, total linear momentum after firing is impulse. 𝐼 = 𝐹. 𝑡 equal to total momentum before firing. 𝒎𝒗 SI unit: Ns, Dimension: 𝑀𝐿𝑇 −1 i.e. 𝑴𝑽 + 𝒎𝒗 = 𝟎 ⇒ 𝑽 = − 𝑴 The negative sign shows that direction of recoil 18. How Impulse related to momentum? velocity is opposite to the direction of velocity of Answer: bullet. Relation between Impulse and Linear Momentum: We know, Impulse = Force × Time of action 22. State Newton’s Law of Motion. ⇒𝐼 =𝐹×𝑡 Answer: ⇒ 𝐼 = 𝑚𝑎 × 𝑡 ∵ 𝐹 = 𝑚𝑎 Newton’s Law of Motion: 𝑣−𝑢 𝑣−𝑢 ⇒𝐼=𝑚 ×𝑡 ∵𝑎 = 1) 1st Law: It states that every body continues in 𝑡 𝑡 ⇒ 𝐼 = 𝑚𝑣 − 𝑚𝑢 its state of rest or of uniform motion in a ⇒ 𝐼 = Change in linear momentum straight line, unless it is compelled by some i.e. Impulse = Change in linear momentum external force to change that state. 2) 2nd Law: The rate of change of momentum of a 19. Define Impulsive force. body is directly proportional to the force Answer: applied on it and the change in momentum is Impulsive Force: A large amount of force acting on along the direction in which force act. a body for very small interval of time is called 3) 3rd Law: To every action there is an equal and impulsive force. opposite reaction. Example: 1) Striking cricket ball by bat. 23. How to measure force from Newton’s 2nd law of 2) Kicking to football motion. 3) Hammering on a nail. Answer: SI unit: Newton (N), Dimension: 𝑀𝐿𝑇 −2 Measurement of force from Newton’s 2nd Law: From the 2nd law of motion it is state that the rate of 20. State the principle of conservation of linear change of momentum of a body is directly momentum. proportional to the impressed force and takes Answer: place in a direction of the force. Principle of conservation of linear momentum: ∴ 𝐹 ∝ Rate of change of momentum 𝑚𝑣 −𝑚𝑢 It states that if no external force acts on a system, ⇒ 𝐹∝ 𝑡 the linear momentum of the system remains 𝑣−𝑢 ⇒ 𝐹∝𝑚 𝑡 constant. ⇒ 𝐹 ∝ 𝑚𝑎 , ⇒ 𝐹 = 𝑘𝑚𝑎 If 𝑚1 and 𝑚2 be the masses of two particle and If a unit of force is chosen to the force which before impact 𝑢1 and 𝑢2 be their velocities and produces a unit acceleration on a unit mass then after impact if 𝑣1 and 𝑣2 be their respective 𝑘 = 1. ∴ 𝑭 = 𝒎𝒂 i.e. Force=Mass×Acceleration. velocities, then we have, 𝒎𝟏 𝒖𝟏 + 𝒎𝟐 𝒖𝟐 = 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐  ⦿ Force and Motion ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 24. Prove that Newton’s 1st law of motion is special case 27. Define angular displacement. of 2nd law. Answer: Answer: Angular displacement: The angular displacement of Newton’s 1st law from 2nd law: From Newton’s 2nd a particle undergoing circular motion, during a law we know, if 𝑚 be the mass of a particle, when certain interval of time, is defined as the angle applied force 𝐹 on a body then its acceleration is 𝑎. swept out by the radius during that interval of So, 𝐹 = 𝑚𝑎 time. 𝑣−𝑢 𝑎𝑟𝑐 𝑠 ⇒𝐹=𝑚 [symbols used in usual meaning] Angular displacement, 𝜃 = 𝑟𝑎𝑑𝑖𝑢𝑠 = 𝑟 𝑡 ⇒ 𝐹𝑡 = 𝑚𝑣 − 𝑚𝑢 The SI unit of angular displacement is radian. Now, when 𝐹 = 0, then 𝑣 = 𝑢. That is, in the absence of force, the object continues to move with 28. Define angular velocity. same velocity throughout. Answer: And, when F = 0 and u = 0, then v = 0. That is, Angular Velocity: The rate of angular displacement an object at rest will remain at rest if no force is undergone by a particle performing uniform acting on it. circular motion is called its angular velocity. 𝜃 Thus, Newton's first law is derived from Angular Velocity, 𝜔 = 𝑡 second law. The SI unit of angular velocity is radian/second. 25. Establish Newton’s 3rd law from 2nd law. 29. The angular velocity of second hand of a clock is- Answer: (a) 𝟐𝝅 𝒓𝒂𝒅/𝒔 Newton’s 3rd law from 2nd law: If no external force 𝝅 (b) 𝟔𝟎 𝒓𝒂𝒅/𝒔 is applied on two colliding body 𝐴 and 𝐵 and only 𝝅 (c) 𝒓𝒂𝒅/𝒔 the action and reaction 𝐹𝐴𝐵 and 𝐹𝐵𝐴 present in the 𝟑𝟎 𝝅 system. If 𝑝1 and 𝑝2 be the momentum of two (d) 𝟓 𝒓𝒂𝒅/𝒔 bodies, then we can write from 2nd law Explanation: 𝑑𝑝 1 𝑑𝑝 2 𝑑 𝐹𝐴𝐵 + 𝐹𝐵𝐴 = + = 𝑝1 + 𝑝2 = 0 We know, second’s hand completes 2𝜋 in 60 s. 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝜃 2𝜋 𝜋 Since external force is zero. So total momentum ∴ Angular Velocity, 𝜔 = 𝑡 = 60 = 30 𝑟𝑎𝑑/𝑠 will be conserved. So derivative of constant Answer: (c) momentum will be zero. ∴ 𝐹𝐴𝐵 + 𝐹𝐵𝐴 = 0 , ⇒ 𝐹𝐴𝐵 = −𝐹𝐵𝐴 30. Define angular acceleration and sate SI unit. (-ve) sign denotes action is opposite to reaction. Answer: Angular Acceleration is defined as the time rate of change of angular velocity. It is usually 26. What do you mean by ‘Inertia’? expressed in radians per second per second. Answer: The SI unit of angular acceleration is 𝑟𝑎𝑑/𝑠 2. Inertia: Inertia is the property of a body by virtue Angular acceleration 𝛼 of which it is unable to change its state of rest or of = Linear acceleration 𝑎 × Radius 𝑟 uniform motion in a straight line without the help ∴ 𝛼 = 𝑎𝑟 of an external force. More massive bodies have greater inertia. Hence mass is defined as a 31. What is the direction of velocity vector of a particle quantitative measure of inertia. in circular motion? Inertia may be two kinds, Answer: The particle is moving tangent to the 1) Inertia of rest circle. Since the direction of velocity vector is the 2) Inertia of motion same as the direction of the particle motion, the velocity vector is directed tangent to the circle as well.  ⦿ Force and Motion ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 32. The angle 𝜽 covered by a body in rotational motion 37. Why does a cyclist bend inwards while riding along is given by the equation 𝜽 = 𝟔𝒕 + 𝟓𝒕𝟐 + 𝟐𝒕𝟑. a curved road? Determine the angular velocity and angular Answer: A cyclist bends inwards because then the acceleration at time 𝒕 = 𝟐𝒔. horizontal component of the normal reaction of Answer: the ground provides the necessary centripetal Given, 𝜃 = 6𝑡 + 5𝑡 2 + 2𝑡 3 force for going along the curved road. 𝑑𝜃 ∴ Angular Velocity 𝜔 = 𝑑𝑡 = 6 + 10𝑡 + 6𝑡 2 38. What are the differences between centripetal and At 𝑡 = 2𝑠, centrifugal force? 𝜔 = 6 + 10 × 2 + 6 × 22 = 50 rad/s (Ans.) 𝑑𝜔 Answer: Again angular acceleration 𝛼 = = 10 + 12𝑡 Differences between centripetal and centrifugal 𝑑𝑡 At 𝑡 = 2𝑠, 𝛼 = 10 + 12 × 2 = 34 𝑟𝑎𝑑/𝑠 2 (Ans.) force Centripetal Force Centrifugal Force 33. Define centripetal force. It is real force which is It is a pseudo force of Answer: exerted on the body by fictitious force which Centripetal Force: Centripetal force is defined as the external agencies cannot arise from that constant external force which acting like gravitational force, gravitational force, tension tension in string, by the string, normal force continuously at right angle to the motion of a normal force etc. etc. particle causes it to move in a circle with constant It acts both in inertial It acts only rotating speed. and non inertial frame accelerated non inertial 𝑚 𝑣2 of reference. frame of reference. Centripetal force, 𝐹𝑐𝑝 = 𝑟 = 𝑚𝜔2 𝑟 Origin of centripetal Origin of centrifugal force force is interaction is inertia of the body. 34. Why centripetal force is a no work force? between the objects. Answer: It is a real force and has If is a pseudo force but has Centripetal force is a no work force: Since the real effects. real effects. centripetal force acts at right angles to the It acts towards the It acts radially outward direction of motion. So there is no displacement in centre of the circular from the centre of the the direction of the centripetal force; the path. circular path. 2 𝑚𝑣 𝑚𝑣 2 centripetal force therefore does not perform any 2 𝐹𝑐𝑝 = −𝑚𝜔 𝑟 = − 2 𝐹𝑐𝑓 = 𝑚𝜔 𝑟 = 𝑟 𝑟 work. Centripetal force is thus a no work force. 35. Define centrifugal force. 39. Calculate the centripetal force acting on a small Answer: mass of 0.25 kg rotating 1800 RPM on a radius 200 Centrifugal Force: Centrifugal force is defined as an mm. outward radial fictitious force that is experienced Answer: by an object moving in a non inertial frame of Here, Frequency, 𝑓 = 1800 𝑅𝑃𝑀 reference in circular path directed away from the 1800 = 𝑅𝑃𝑆 centre of rotation. It is equal in magnitude with 60 centripetal force. = 30 𝑅𝑃𝑆 𝑚 𝑣2 ∴ Angular speed, 𝜔 = 30 × 2𝜋 = 60𝜋 𝑟𝑎𝑑/𝑠 ∴ 𝐹𝑐𝑓 = = 𝑚𝜔2 𝑟 𝑟 Geven, Radius 𝑟 = 200 𝑚𝑚 = 0.2 𝑚 Mass 𝑚 = 0.25 𝑘𝑔 36. Why centrifugal force is called a pseudo force? ∴ Centripetal Force 𝐹𝐶𝑃 = 𝑚𝑟𝜔2 Answer: The centrifugal force in the non-inertial = 0.25 × 0.2 × 60𝜋 2 frame of reference of a particle in circular motion = 1776.5 𝑁 (Ans.) is the effect of the acceleration of the frame of reference with respect to an inertial frame of reference. Therefore, it is called a pseudo or fictitious force.  ⦿ Force and Motion ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 40. A ball of mass 𝟓 𝒌𝒈 strikes a wall with a speed of 43. If the velocity of a car moving along a straight path 𝟏𝟎 𝒎/𝒔 and rebounds with the same speed. What is be doubled, the distance by which it can be stopped the change in momentum of the ball? becomes four times. Explain. Ans: Given, Mass of the ball, 𝑚 = 5 𝑘𝑔 Answer: Speed of the ball, 𝑣 = 10 𝑚/𝑠 Let, the initial and final velocity of the car is 𝑢 and ∴ Initial momentum 𝑚𝑣 = 5 × 10 = 50 𝑘𝑔. 𝑚/𝑠 𝑣 respectively and acceleration 𝑎. and Terminal momentum= −𝑚𝑣 = −50 𝑘𝑔. 𝑚/𝑠 ∴ 𝑣 2 = 𝑢2 ± 2𝑎𝑠 [𝑠 =displacement] So, Change in momentum of the ball, As the is stopped so 𝑣 = 0 = 50 − −50 = 100 𝑘𝑔. 𝑚/𝑠 (Ans.) ∴ 0 = 𝑢2 − 2𝑎𝑠 𝑢2 ⇒ 2𝑎𝑠 = 𝑢2 , ⇒ 𝑠 = 2𝑎 41. A 𝟑𝟎 𝒈 bullet leaves a rifle with a velocity of If doubled the initial velocity the 𝑢𝑛𝑒𝑤 = 2𝑢 𝟑𝟎𝟎 𝒎/𝒔 and the rifle recoils with a velocity of 4𝑢 2 𝑢2 𝟔𝟎 𝒄𝒎/𝒔. Find the mass of the rifle. ∴ 𝑠𝑛𝑒𝑤 = =4 = 4𝑠 2𝑎 2𝑎 Answer: Let, Mass of rife = 𝑀 Given, mass of bullet, 𝑚 = 30 𝑔 = 0.03 𝑘𝑔 44. A bullet moving with a speed 𝟐𝟎𝟎 𝒎𝒔−𝟏 can Recoil velocity of rifle, 𝑉 = 60 𝑐𝑚/𝑠 = 0.6 𝑚/𝑠 penetrate 𝟒 𝒄𝒎 thick wooden board. What will be Velocity of bullet, 𝑣 = 300 𝑚/𝑠 the suitable speed of the bullet to penetrate 𝟏𝟔 𝒄𝒎 By law of conservation of linear momentum, thick same wooden block? 𝑀𝑉 + 𝑚𝑣 = 0 Answer: 𝑚𝑣 0.03×300 ⇒𝑀=− =− = −15 Given initial speed 𝑢 = 200 𝑚𝑠 −1 𝑉 0.6 ∴ Mass of rifle is 15 𝑘𝑔 Displacement 𝑠 = 4 𝑐𝑚 = 0.04 𝑚 −𝑣𝑒 sign shows that recoil velocity of rifle is Final speed 𝑣 = 0 opposite to velocity of bullet. We know, 𝑣 2 = 𝑢2 − 2𝑎𝑠 ⇒ 0 = 2002 − 2𝑎 × 0.04 200×200 42. Find the angle of banking of a road of radius of ⇒𝑎= 2×0.04 = 500000 𝑚𝑠 −2 curvature of 𝟔𝟎𝟎 𝒎 so that a car can move safely at 2nd case, a maximum speed of 𝟏𝟖𝟎 𝒌𝒎/𝒉. Initial speed 𝑢 =? Answer: Displacement 𝑠 = 16 𝑐𝑚 = 0.16 𝑚 𝑘𝑚 5 Given, speed of car 𝑣 = 180 = 180 × 𝑚/𝑠 Final speed 𝑣 = 0 𝑕 18 = 50 𝑚/𝑠 Acceleration 𝑎 = 500000 𝑚𝑠 −2 Radius of curvature of road 𝑟 = 600 𝑚 0 = 𝑢2 − 2 × 500000 × 0.16 Let, Angle of banking of road = 𝜃 ⇒ 𝑢2 = 160000 Mass of car = 𝑀 ⇒ 𝑢 = 160000 = 400 𝑚𝑠 −1 Normal reaction of car = 𝑅 ∴ Initial speed of bullet is 400 𝑚𝑠 −1 to penetrate Now, Vertical component of 𝑅 =Weight of car 16 𝑐𝑚 wooden block. ⇒ 𝑅 cos 𝜃 = 𝑀𝑔 Horizontal component of 𝑅 = Centripetal force 𝑀𝑣 2 ⇒ 𝑅 sin 𝜃 = 𝑟 𝑀 𝑣2 𝑅 sin 𝜃 𝑣2 ∴ 𝑅 cos 𝜃 = 𝑀𝑔 𝑟 ⇒ tan 𝜃 = 𝑟𝑔 50 2 ⇒ tan 𝜃 = 600×9.81 = 0.42 −1 ⇒ 𝜃 = tan 0.42 ≈ 23° (Ans.)  ⦿ Work, Power & Energy ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic Polytechnic 1st Semester 4. A body of mass 𝒎 has momentum 𝒑, its kinetic Applied Physics-I energy will be- 𝒑𝟐 Unit 3: Work, Power and Energy a) 𝟐𝒎 𝒑 (Most Important Questions) b) 𝟐𝒎𝟐 𝒑𝟐 c) 𝟐𝒎𝟐 1. Mechanical energy of a body falling under gravity- 𝒑𝟐 a) will always be conserved d) 𝟐𝒎 b) will actually never be conserved Explanation: c) will be conserved if the body falls through 1 Kinetic energy 𝐸𝑘 = 2 𝑚𝑣 2 vacuum 1 𝑚 2𝑣2 d) will be conserved if the body falls through air = 2 𝑚 𝑝2 = 2𝑚 (𝑝 =momentum= 𝑚𝑣) Explanation: Answer: (a) Mechanical energy = kinetic energy + potential energy 1 5. A body of mass 𝒎 has kinetic energy 𝑬 , its = 2 𝑚𝑣 2 + 𝑚𝑔𝑕 = Constant momentum will be- Answer: (a) (a) 𝟐𝒎 𝑬 (b) 𝟐𝒎𝑬 2. Kinetic energy with any reference, must be- 𝑬𝟐 (a) zero (c) 𝟐𝒎 (b) negative (d) none of these (c) positive (d) none of these Explanation: 1 1 𝑚 2𝑣2 𝑝2 𝐸 = 𝑚𝑣 2 = = (𝑝 =momentum) Explanation: 2 2 𝑚 2𝑚 2 The kinetic energy of a body of mass 𝑚 which is ⇒ 𝑝 = 2𝑚𝐸 1 moving with velocity 𝑣 is = 2 𝑚𝑣 2. ⇒ 𝑝 = 2𝑚𝐸 Answer: (b) Mass is always positive and if the velocity is either positive or negative, the kinetic energy is 6. The ratio of masses of two bodies is 3:2. If kinetic always positive due to the square of velocity. energies are equal, the ratio of their linear Answer: (c) momentum is- (a) 3:2 3. BOT is the unit of – (b) 2:3 a) power (c) 𝟑: 𝟐 b) energy c) current (d) 𝟐: 𝟑 d) speed Explanation: Linear momentum 𝑝 = 2𝑚𝐸 Explanation: One B.O.T. is 1 kWhr. BOT stands ∴ The ratio of momentum for Board Of Trade unit of electric energy. A = 𝑝1 : 𝑝2 kilowatt-hour is a composite unit of energy = 2𝑚1 𝐸: 2𝑚2 𝐸 (∵ Kinetic energies are equal) equivalent to one kilowatt (1 kW) of power = 𝑚1 : 𝑚2 sustained for one hour. = 3: 2 Answer: (b) Answer: (c)  ⦿ Work, Power & Energy ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 7. The ratio of kinetic energy of two body of mass 𝒎 10. Work is a _____ quantity. and 𝟒𝒎 is 2:1, then the ratio of their linear a) vector momentum will be- b) scalar (a) 𝟏: 𝟐 c) tensor (b) 1:2 d) none of these (c) 1:4 (d) 1:6 Explanation: Work is scalar quantity because it has magnitude only no direction is needed. Explanation: Momentum 𝑝 = 2𝑚𝐸 Answer: (b) So, the ratio of momentum of two bodies = 𝑝1 : 𝑝2 = 2𝑚1 𝐸1 : 2𝑚2 𝐸2 11. Define Work. Ans: = 𝑚 × 2 : 4𝑚 × 1 Work: Work is defined as the product of = 2: 4 displacement and the component of the force along = 2 : 2 = 1: 2 the displacement. Work is a scalar quantity. The SI Answer: (a) unit of work is Joule(J). Mathematically, Work 𝑊 = 𝐹. 𝑆 = 𝐹𝑆 cos 𝜃 8. A body at rest may have- 12. What is no work? a) energy Ans: b) momentum No-Work: When a force acts in a direction at right c) speed angles to the direction of displacement, then it d) velocity called no work or zero work. Here angle between 𝐹 and 𝑆 is 𝜃 = 90° Explanation: When a body is at rest its velocity or So, work done, 𝑊 = 𝐹𝑆 cos 90° = 0 kinetic energy must be zero but it may have potential energy for its shape or position. 13. Give an example of no work. Answer: (a) Answer: Example of no work: A person carrying a box over 9. A light body and a heavy body have equal linear his head is walking on a horizontal road. The momentum. Which has greater kinetic energy? gravitational force does not work since the force a) light body and displacement are always perpendicular to b) heavy body each other. c) both have equal kinetic energy d) none of these 14. “A body executing uniform circular motion in a circular path does no work” – Explain. Explanation: Let, mass of light and heavy body be Answer: 𝑚 and 𝑀 respectively and 𝑝 be their momentum. When a body executing uniform circular motion in So their kinetic energy is 𝑝2 𝑝2 a circular path, the centripetal force acting on it 2𝑚 and 2𝑀. along the radius and its displacement along the 𝑝2 𝑝2 tangent are always mutually perpendicular to each As 𝑀 > 𝑚 , ∴ 2𝑚 > 2𝑀 Answer: (a) other. Here angle between 𝐹 and 𝑆 is 𝜃 = 90° So, work done, 𝑊 = 𝐹𝑆 cos 90° = 0 Therefore centripetal force does no work.  ⦿ Work, Power & Energy ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 15. What is negative work? 21. What is energy? Write SI unit of Energy. Answer: Ans: Negative work: Negative work is done when the Energy: It is the capacity of a physical system to displacement is in the opposite direction of the perform work. Energy exists in several forms such force delivered. as heat energy, mechanical energy, light energy, Here angle between 𝐹 and 𝑆 is 𝜃 = 180° sound energy or other forms. So, work done, 𝑊 = 𝐹𝑆 cos 180° = −𝐹𝑆 The SI unit of energy is Joule. 16. Give an example of negative work. 22. State work-energy principle. Ans. Ans: Example of negative work: When a body being Work-Energy Principle or Work-Energy Theorem: pulled vertically upwards. The work done by the Work done by a force in displacing a body is gravitational force is negative. measured by the change in kinetic energy of the body. So, according to work energy principle, work 17. Define the term Power. Is it a scalar or vector and kinetic energy are equivalent. quantity? Give its dimension and unit in SI system. Proof: Let us suppose, Answer: 𝑚 = Mass of a body Power: The rate of work i.e. the amount of work 𝑢 = Initial velocity of the body done in unit time is called power. 𝑣 = Final velocity of the body Power= 𝑊𝑜𝑟𝑘 i.e. 𝑃 = 𝑊 𝐹 = Force applied on the body 𝑇𝑖𝑚𝑒 𝑡 𝑆 = Displacement of the body Power has magnitude only and no direction so 𝑊 = Work done by applied force it is a scalar quantity. Here, 𝑊 = 𝐹𝑆 [𝐹 and 𝑆 are in same direction] The SI unit of power is Joule/Second or Watt ⇒ 𝑊 = 𝑚𝑎𝑆 [𝑎 =acceleration] Dimension of Power = 𝑀𝐿2 𝑇 −3 𝑣 2 −𝑢 2 =𝑚 2𝑆 ×𝑆 [From 𝑣 2 = 𝑢2 + 2𝑎𝑠] 2 2 18. Establish relation between Joule and erg. = 𝑚 𝑣 −𝑚 𝑢 2 Ans: 1 1 1 Joule = 𝑚𝑣 2 − 𝑚𝑢2 =Final K.E. – Initial K.E. 2 2 = 1 𝑁 × 1 𝑚 [∵ 𝑊 = 𝐹𝑆] So, Work done = Change in kinetic energy. = 1 𝑘𝑔 × 1 𝑚 × 1 𝑠 −2 × 1 𝑚 [∵ 𝐹 = 𝑀𝑎] = 1000 𝑔 × 100 𝑐𝑚 × 1 𝑠 −2 × 100 𝑐𝑚 23. State law of conservation of energy. = 107 𝑔. 𝑐𝑚2. 𝑠 −2 = 107 𝑒𝑟𝑔 Answer: Law of Conservation of Energy: Energy can neither 19. Establish relation between B.O.T. and Joule. be created nor destroyed. It can be converted from Ans: one form to another. The sum total of energy, in 1 B.O.T. this universe remains constant. = 1𝑘𝑊 × 1 hour = 1000 𝑊 × 60 minute 24. State the principle of conservation of mechanical = 1000 𝐽𝑜𝑢𝑙𝑒 × 60 × 60 Second energy. 𝑠𝑒𝑐𝑜𝑛𝑑 Answer: 5 = 36 × 10 Joule Principle of conservation of mechanical energy: In = 3.6 × 106 Joule absence of any dissipating force such as friction, sum total of potential energy and kinetic energy i.e. 20. Prove that 𝟏 𝑯𝑷 = 𝟕𝟒𝟔 𝑾. total mechanical energy is constant. Answer: Example: When a body falling freely under gravity 1 𝐻𝑃 = 550 foot-pound/Second the total mechanical energy always conserved. Now, 1 foot = 0.3048 𝑚, 1 pound = 0.4536 𝑘𝑔 ∴ 1𝐻𝑃 = 550 × 0.3048 × 0.45356 𝑘𝑔. 𝑚𝑠 −1 = 550 × 0.3048 × 0.45356 × 9.81 J/s = 745.90 𝑊 ≈ 746 𝑊 (Proved)  ⦿ Work, Power & Energy ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 25. Prove that the total mechanical energy of freely 28. The power of a water pump is 𝟐 𝒌𝑾. How much falling body under gravity is conserved. time will it take to lift a mass of 𝟐𝟎𝟎 𝒌𝒈 of water to Answer: a height of 𝟒𝟎 𝒎? Suppose a body of mass 𝑚 falling freely height 𝑕 in Solution: a straight way 𝐴𝐵𝐶. Given, mass of water 𝑚 = 200 𝑘𝑔 At point 𝑨, height lifted 𝑕 = 40 𝑚 𝑃𝐸 = 𝑚𝑔𝑕 Work done = Potential energy 𝐾𝐸 = 0 = 𝑚𝑔𝑕 Total energy = 𝑚𝑔𝑕 + 0 = 𝒎𝒈𝒉 = 200 × 9.8 × 40 At point 𝑩, = 78400 J 𝑃𝐸 = 𝑚𝑔 𝑕 − 𝑥 = 𝑚𝑔𝑕 − 𝑚𝑔𝑥 Given, power of pump = 2 𝑘𝑊 = 2000 𝑊 1 𝑊 𝑊 78400 𝐾𝐸 = 2 𝑚𝑣𝐵2 Now, Power 𝑃 = ,⇒ 𝑡 = = = 39.2 𝑠 𝑡 𝑃 2000 1 = 2 𝑚 2𝑔𝑥 [∵ 𝑣𝐵2 = 𝑣𝐴2 + 2𝑔𝑥, 𝑣𝐴 = 0] ∴ Time required to lift water is 39.2 second. = 𝑚𝑔𝑥 Total energy = 𝑚𝑔𝑕 − 𝑚𝑔𝑥 + 𝑚𝑔𝑥 = 𝒎𝒈𝒉 29. What is friction? Answer: At point 𝑪, Friction: Whenever a body moves or tends to 𝑃𝐸 = 0 1 moves over another body in contact, forces 𝐾𝐸 = 𝑚𝑣𝐶2 2 opposing the motion come into play, which are 1 = 𝑚 2𝑔𝑕 [∵ 𝑣𝐶2 = 𝑣𝐴2 + 2𝑔𝑕, 𝑣𝐴 = 0] tangential to the surface in contact. These 2 = 𝑚𝑔𝑕 opposing forces are called friction between the Total energy = 0 + 𝑚𝑔𝑕 = 𝒎𝒈𝒉 two surfaces in contact. ∴ Total energy at points 𝐴, 𝐵 and 𝐶 are equal. So, the mechanical energy of falling body is 30. State the laws of limiting friction. constant. Answer: Laws of Limiting Friction: 26. An apple of mass 𝟎. 𝟏 𝒌𝒈 is dropped from 𝟏𝟎 𝒎 1) The magnitude of the force of limiting friction height. Find out the value of the kinetic energy 𝐹 between any two bodies in contact is when it hits the ground. directly proportional to the normal reaction Solution: 𝑅 between them i.e.,𝐹 ∝ 𝑅 , ⇒ 𝐹 = 𝜇𝑅. By the principle of conservation of energy, the (𝜇 is constant of proportionality) potential energy of the body at height 𝑕 is 2) The direction of the force of limiting friction is converted into kinetic energy before it strikes the always opposite to the direction in which one ground. Thus, 𝐾𝐸 = 𝑚𝑔𝑕 body is at the verge of moving over the other. = 0.1 × 9.8 × 10 3) The force of limiting friction is independent of = 9.8 Joule the apparent area of contact as long as normal ∴ The value of kinetic energy is 9.8 Joule. (Ans.) reaction between two bodies in contact remain the same. 27. A man weighing 𝟔𝟎 𝒌𝒈 climbs up a staircase 4) The limiting friction depends upon the nature carrying a load of 𝟑𝟎 𝒌𝒈 on his head. The staircase of materials (smoothness) of the surfaces in has 20 steps each of height 𝟎. 𝟑 𝒎. If he takes 𝟏𝟎 𝒔 contact and acts tangentially to the interface to climb, find his power. between them. Solution: Here, Total mass 𝑚 = 60 + 30 = 90 𝑘𝑔 Total height lifted 𝑕 = 20 × 0.3 = 6 𝑚 ∴ Work done = Potential energy gained = 𝑚𝑔𝑕 = 90 × 9.8 × 6 = 5292 Joule Time taken 𝑡 = 10 𝑠 𝑊 5292 ∴ Power = 𝑡 = 10 = 529.2 𝑊 (Ans.)  ⦿ Work, Power & Energy ⦿ Applied Physics-I Suggestion ⦿ 1st Semester ⦿ WB Polytechnic 31. Discuss some ways to reduce friction. 34. A body is pulled along a horizontal plane through a Answer: distance of 𝟓 𝒎 by applying a constant force of 𝟒 𝑵. Methods of Reducing Friction: Find the work done by the force when- 1) By polishing: Surfaces in contact become i) the force is applied along the plane smoother on polishing, thereby reducing ii) the force is applied at an angle of 𝟔𝟎° with the friction. plane. 2) By Lubrication: Lubricants like oil, grease etc. Solution: Given, force applied 𝐹 = 4𝑁, fill up the roughness of the surfaces and Displacement 𝑠 = 5 𝑚 thereby reducing friction. When the force applied along the plane, the angle 3) By using bearing: With use of ball bearings or between force and displacement 𝜃 = 0 roller bearings, sliding friction converted into ∴ Work done = 𝐹𝑆 cos 𝜃 rolling friction, thereby reducing friction. = 4 × 5 × cos 0 = 20 J (Ans.) 4) By Streamlining: Friction due to air in fast When the force applied at 𝜃 = 60° moving car, jets etc. is reduced by giving them ∴ Work done = 𝐹𝑆 cos 𝜃 proper shape called streamlining. = 4 × 5 × cos 60° 1 = 20 × 2 = 10 J (Ans.) 32. “Static friction is a self adjusting force” - Explain. Answer: When a force is applied to a body within 35. Rotational kinetic energy of two bodies of inertia the limiting friction, the frictional force will be just 𝟗 𝒌𝒈𝒎𝟐 and 𝟏 𝒌𝒈𝒎𝟐 are same. What will be the equal to the force applied to remain rest. ratio of their angular momentum? Even if a small amount of force is increased or Answer: decreased within limiting friction, the force As rotational kinetic energy of two bodies are provided by the friction will still be the same as the 1 1 same. We have 2 𝐼1 𝜔12 = 2 𝐼2 𝜔22 present applied force only, so as to remain in rest. 𝐼1 𝜔 1 2 𝐼2 𝜔 2 2 So it is called self adjusting force. ⇒ = 𝐼1 𝐼2 𝐿1 2 𝐿2 2 33. An ice skater moving at 𝟏𝟎 𝒎𝒔−𝟏 comes to a halt in ⇒ 𝐼1 = 𝐼 [∵ angular momentum 𝐿 = 𝐼𝜔] 2 𝟏𝟎𝟎 𝒎 on an ice surface. Calculate the coefficient of 𝐿1 2 𝐼1 ⇒ 𝐿2 2 =𝐼 friction between the ice and skater. 2 𝐿 𝐼1 9 3 Solution: ⇒ 𝐿1 = = =1