1. Engineering Mechanics Full Material

["162. Which of the following statements of Following equation obtained from Newton's second law D\u2019Alembert\u2019s principle are correct? 1. The net external force F actually acting on Fnet = R - mg R - mg = ma the body and the inertia force Ft together keep the body in a state of fictitious R = mg + ma = m(g + a) R = 70 (9.81 + 2) = 826.7 N equilibrium 2. The equation of motion may be written as 165. Two blocks of 50 N and 100 N are connected by F + (\u2013 ma) = 0 and the fictitious force (\u2013 a light cord passing over a smooth frictionless ma) is called an inertia force pulley. The acceleration in blocks and tension 3. It tends to give solution of a static problem in rope are respectively given by: [g = an appearance akin to that of a dynamic acceleration due to gravity] (a) g\/2, 40 N (b) g\/3, 66.67 N problem. (a) 1 and 3 only (b) 1 and 2 only (c) g, 200 N (d) g\/5, 40 N (c) 2 and 3 only (d) 1, 2 and 3 (e) g\/4, 50 N ESE 2020 CGPSC AE 2014- I Ans. (d) : 1, 2 and 3 Ans. (b) : Using Newton's second law of motion 163. A 2000 kg of automobile is driven down a 5 degree inclined plane at a 100 km\/h, when the brakes are applied causing a constant total braking force (applied by the road on the tires) of 7 kN. Determine the distance traveled by the automobile as it comes to a stop. (a) 146 m (b) 152 m (c) 135.86 m (d) 122.44 m (e) 149.5 m CGPSC 26th April 1st Shift Ans. (a) : Given, In equilibrium ...........(i) mass (m) = 2000 kg Fnet = T-50 = m1a inclination (\u03b8) = 5\u00ba 100-T = m2a ..........(ii) velocity (v1) = 100 km\/hr = 100 \u00d7 5 = 27.78 m\/s from equation (i) 18 braking force (fb) = 7 kN = 7000 N T-50= 50 \u00d7 a g T = 50 \u00d7 a + 50 .............(iii) g from equation (ii) 100 - T = 100 a g Apply work-energy principle T = 100 - 100 a .............(iv) g W1-2 = (K.E)2 \u2212 (K.E)1 equating equation (iii) & (iv) (2000 g sin 5\u00ba)x - 7000 x = 0\u2212 1 mv12 100 a + 50 = 100 \u2212 100 a 2 gg - 5290 x = - 771728.4 150 a = 50 x = 145.88 \u2243 146 m g 164. Determine the apparent weight of a 70 kg man a=g in a elevator when the acceleration of elevator 3 is 2 m\/s2 upwards. (a) 700 N (b) 686.7 N (c) 826.7 N (d) 560 N summation of equation (iii) & (iv) (e) 4.67 N 2T = 150 \u2212 50a = 150 \u2212 50 g 3 CGPSC 26th April 1st Shift Ans. (c) : As the elevator is moving upwards i.e. in the T = 66.67 direction of reaction force, therefore, the net force 166. A car of mass 150 kg is traveling on a would be R-mg upwards. horizontal track at 36 Km\/hr. The time needed to stop the car is ______ (Take \u00b5 = 0.45). (a) t = 2.26 sec (b) t = 3.20 sec (c) e = 3.8 sec (d) e = 4.2 sec TNPSC AE 2014 414","Ans. (a) : Given, 168. A car starting from rest attains a maximum m = 150 kg, V = 36 km\/hr = 10 m\/s, \u00b5 = 0.45 speed of 100 kmph in 20 seconds. What will be We know that, its acceleration assuming it is uniform? Friction force = \u00b5 \u00d7 mg (a) 1.0 m\/s2 (b) 1.4 m\/s2 (c) 1.8 m\/s2 (d) 2.0 m\/s2 F = 0.45 \u00d7 150 \u00d7 9.81 F = 662.175 N CIL MT 2017 2017 IInd shift Let's suppose that the car stops after traveling a Ans. (b) : Initial velocity (u) = 0 Final velocity (v) = 100 kmph distance d. During this time the entire kinetic energy will be exhausted into the work done by the friction = 100 \u00d7 5 = 500 m \/ s 18 18 force F \u00d7 d = 1 mv2 2 Time (t) = 20 sec. 662.175 \u00d7 d = 1 \u00d7 150 \u00d7 (10)2 v = u + at 2 500 d = 11.326 m \u21d2 a = v \u2212 u = 18 = 500 = 1.38 m \/ s t 20 18\u00d7 20 Then v2 = u2 \u2013 2 a d v=0 169. A body exerts a force of 1 kN on the floor of the u2 = 2 a d (10)2 = 2 a \u00d7 11.362 lift which moves upward with a retardation of 1.5 m\/s2. What is the mass of the body, in kg, a = 4.414 m \/ s 2 carried in the lift? (a) 120.33 (b) 101.94 Then v = u \u2013 at (c) 88.42 (d) 77.32 10 = 4.414 \u00d7 t APPSC-AE-2019 t = 2.2652 sec. Ans. (a) : FBD of mass in the lift 167. The tension in the cable supporting a lift moving upwards is twice the tension when the lift moves downwards. The acceleration of the lift is equal to (b) g\/2 (a) g (c) g\/3 (d) g\/4 TNPSC AE 2014 N = normal reaction between lift and mass Vizag Steel (MT) 2017 Ans. (c) : Case - I \u2211 Fnet = 0 (N - mg) = m(-a) (-a = retardation) 1\u00d7103 = m(g \u2212 a) 1\u00d7103 = m(9.81 \u22121.5) m = 1\u00d7103 = 120.33 kg (9.81 \u2212 1.5) Then, T1 \u2013 mg = ma ...(1) 170. During elastic impact, the relative velocity of Case - II the two bodies after impact is _____ the relative velocity of the two bodies before impact (a) equal to (b) equal and opposite to (c) less than (d) greater than TSPSC AEE 2015 Ans. (b) : During elastic impact, the relative velocity of mg \u2013 T2 = ma ...(2) the two bodies after impact is equal and opposite to the Given that relative velocity of the two bodies before impact. T1 = 2 T2 From equation (1) and (2) 171. A spring scale indicates a tension 'T' in the right hand cable of the pulley system shown in mg \u2013 T2 = 2T2 \u2013 mg the figure. Neglecting the mass of the pulleys 3 T2 = 2 mg and ignoring friction between the cable and pulley the mass 'm' is : 2 T2 = 3 mg Putting the value of T2 in equation (2) mg \u2212 2 mg = ma a=g 3 3 415","(a) 2T (b) T(1 + e4\u03c0 ) 174. A circular disc rolls down without slipping on g g an inclined plane. The ratio of its rotational (c) 4T g kinetic energy to the total kinetic energy is. Ans. (c) : (d) None of the above (a) 1 (b) 1 4 2 OPSC Civil Services Pre. 2011 UPRVUNL AE 2014 (c) 1 (d) 2 3 3 UPPSC AE 12.04.2016 Paper-I Ans : (c) Total kinetic Energy = 1 I\u03c92 + 1 mv2 22 Rotational kinetic Energy = 1 I\u03c92 2 By resolving vertical forces, \u03a3 Fv=0 Moment of Inertia of Circular disc = 1 mR 2 \u21d2 T + 2T + T = mg 2 \u21d2 4T = mg Radius of gyration of circular disc = R 2 \u21d2 m = 4T Total kinetic Energy = 1 \u00d7 1 mR 2 \u00d7 \u03c92 + 1 m (R.\u03c9)2 g 22 2 172. A car moving with uniform acceleration covers Total kinetic Energy = 1 mR 2\u03c92 + 1 mR 2\u03c92 42 450 meter in first 5 second interval and covers Total kinetic Energy = 3 mR 2\u03c92 700 meters in next 5 second interval. The 4 acceleration of the car is\u2013 (a) 7 m\/sec2 (b) 50 m\/sec2 (c) 25 m\/sec2 (d) 10 m\/sec2 Rotational kinetic Energy = 1 \uf8eb mR 2 \uf8f6 \u03c92 \uf8ec \uf8f7 2 \uf8ed 2 \uf8f8 OPSC Civil Services Pre. 2011 Ans. (d) : Let, the car travels with the initial velocity u Rotational kinetic Energy = 1 m2R2\u03c92 4 and acceleration a. Ratio of rotational kinetic Energy of the Total Kinetic considering case-1 When displacement (s) = 450 m and time taken Energy = 1 m2R 2\u03c92 =1 t = 5 seconds 4 3 The displacement 1 mR2\u03c92 + 1 mR 2\u03c92 42 s = ut + 1 at2 2 175. The wheels of a moving car posses 450 = u \u00d7 5 + 1 at2.............(i) (a) potential energy only 2 (b) kinetic energy of translation only considering case-2 When displacement (S) = 450+700 = 1150m (c) kinetic energy of rotation only and time taken (t) = 10 seconds (d) kinetic energy of translation and rotation both 1150 = u\u00d710+1\/2a\u00d7102....(ii) TNPSC 2019 By solving equation (i) and (ii) we get, Ans. (d) : The wheels of a moving car posses kinetic energy of translation and rotation both. a = 10 m \/ sec2 176. A boy walks up a stalled escalator in 90 seconds. When the same escalator moves, he is 9. Work, Power and Energy carried up in 60 seconds. How much time would it take him to walk up the moving 173. The units of energy in SI units escalator? (b) 36 seconds (a) Joule (b) Watt (a) 48 seconds (c) Joule\/sec. (d) Watt\/sec. (c) 30 seconds (d) 24 seconds UJVNL AE 2016 ESE 2017 Ans : (a) The units of energy in SI unit Joule Ans. (b) : Let '\u2113' be length of escalator, Energy:- Velocity of boy vb = \u2113 It may be defined as the capacity to do work. The 90 \u2113 energy exists in many forms e.g. mechanical, electrical, Velocity of escalator ve = 60 chemical, heat, light, etc. But we are mainly concerned with mechanical energy. 416","If both start moving, 182. A block of 500N is to be moved upward for a distance of 1.6 m on an inclined plane of 45\u00ba Time = \u2113 = 36 sec. \uf8eb \u2113 \uf8f6 + \uf8eb \u2113 \uf8f6 with horizontal. Work done will be ( = 0.25): \uf8ed\uf8ec 90 \uf8f7\uf8f8 \uf8ed\uf8ec 60 \uf8f7\uf8f8 (a) 1000 Nm (b) 100 Nm 177. If a particle is in static equilibrium, then the (c) 500 2 Nm (d) 500 Nm work done by the system of force acting on that UPRVUNL AE 2016 particle is: Ans. (c) : Total force acting in upward direction (a) Negative (b) Infinity F = 500 sin 45 + \u00b5R (c) Zero (d) Positive CIL MT 2017 2017 IInd shift Ans. (c) : Static equilibrium is a state in which the net force and net torque acted upon the system is zero. In other words, both linear momentum and angular momentum of the system are conserved. 178. Which conversion is incorrect? UPPSC AE 12.04.2016 Paper-I F = 500\u00d7 1 + 0.25\u00d7 500\u00d7 1 (a) 1 kWh = 3.6\u00d7106 Nm 22 (b) 1 Nm = 0.238\u00d710-3 kcal (c) 1 HP hr = 0.746 kWh (d) 1kcal = 4.1868 Nm F = 500 \u00d7 1 \u00d71.25 Ans : (d) (i) 1 kWh = 3.6\u00d7106 Nm 2 (ii) 1 Nm = 0.238\u00d710-3 kcal Then to be moved upward for a distance 1.6 m on an (iii) 1 HP hr = 0.746 kWh inclined plane. (iv) 1cal = 4.1868 Nm W = F \u00d7 d = 500\u00d7 1 \u00d71.25\u00d71.6 179. A body of mass 20 kg is lifted up through a 2 height of 4 m. How much work is done? (take g W = 500 2 N \u2212 m = 9.81 m\/s2) (a) 648 J (b) 684 J 183. If m is the mass of the body and g is the (c) 748 J (d) 784 J acceleration due to gravity then the (e) 848 J gravitational force is given by: (CGPCS Polytechnic Lecturer 2017) (a) m \u00d7 g3 (b) m \u00d7 g2 Ans. (d) : Work done to left a mass of 20 kg upto 4 m (c) m\/g (d) m \u00d7 g W = Force \u00d7 distance CIL MT 2017 2017 IInd shift =m\u00d7g\u00d74 Ans. (d) : Gravitational force = mass \u00d7 Acceleration = 20 \u00d7 9.81 \u00d7 4 due to gravity = m \u00d7 g = 784.80 N-m 184. The energy possessed by a body, for doing = 784 J work by virtue of its position, is called- 180. A man weighing 900 N climbs a staircase of 15 (a) Potential energy (b) kinetic energy m height in 30 seconds. How much power is (c) electrical energy (d) chemical energy consumed? (b) 250 watt TNPSC AE 2018 (a) 150 watt (c) 350 watt (d) 450 watt Ans. (a) : The energy possessed by a body, for doing (e) 2500 watt work by virtue of its position, is called Potential energy. P. E. = mgh (CGPCS Polytechnic Lecturer 2017) K. E. = 1 mv 2 Ans. (d) : W = 900 N, h = 15 m, t = 30 sec. E. E. = i22Rt Power (P) = ? 185. A block A is dropped down along a smooth We know that, inclined plane, while another block B is P = Work done time released for free fall from the same height P = 900 \u00d715 (a) Both will hit the ground simultaneously 30 (b) Block A will have higher velocity than block B while hitting the ground P = 450 Watt (c) Block A will hit the ground earlier (d) Block B will hit the ground earlier 181. Torque acting on a body of moment of Inertia (I) and angular acceleration ( ) is: Gujarat PSC AE 2019 (a) 2 I\u03b1 (b) 22 I\u03b1 Ans : (a) : Initially both block have same potential (c) 23 I\u03b1 (d) I\u03b1 energy and at the lowest point both block have same kinetic energy in absence of friction. TRB Polytechnic Lecturer 2017 So both block have same velocity and both will hit the ground simultaneously. Ans. (d) : T = I\u03b1 417","186. Wheel has mass 100 kg and radius of gyration Ans. (b) : By using conservation of mechanical energy, of 0.2 m. The additional amount of energy 1 mv2 = \u2212 1 kx2...........(i) (\u2013Ve \u2192 compression of stored in flywheel, if its speed increases from 30 22 spring) rad\/s to 35 rad\/s, will be: (a) 65 J (b) 650 kJ 150\u00d71000 \u00d7 4 10 \u00d71000 \u00d7 x2 9.81 1.25 \u00d710\u22122 (c) 650 J (d) 65 kJ = (e) 65 MJ CGPSC AE 2014- I x = 27.65 cm Ans. (c) : Data given : 191. A body is pulled through a distance of 15 m m = 100kg k = 0.2 along a level track. The force applied is 400 N, \u03c91 = 30 rad\/s \u03c9 2 = 35 rad\/s acting at an angle of 60 to the direction of Then additional amount of energy & forced in flywheel, motion. Then the work done is \u2206E = E2 \u2212 E1 (a) 13.33 N-m TSPSC AEE 2015 (c) 5196.15 N-m (b) 3000 N-m \u2206E = 1 I\u03c9 2 \u2212 1 I\u03c912 (d) 26.66 N-m 2 2 2 Ans : (b) Given, = 1 \uf8f0\uf8eemK 2 \uf8f9 \uf8ee\uf8f0\u03c9 2 \u2212 \u03c912 \uf8f9 Distance = 15 m 2 \uf8fb 2 \uf8fb Applied force = 400 N. = 1 \u00d7100 \u00d7(0.2) 2 [35 \u2212 30][35 + 30] Acting angle = 60\u00b0 work done = Fd cos 60 2 W = 400 \u00d7 15 cos 60 \u2206E = 650 J W = 3000 N .m 187. A ball is thrown up. The sum of kinetic and potential energies will be maximum at 192. A body is moving with a velocity 1 m\/s and a force F is needed to stop it within a certain (a) the ground (b) the highest point distance. If the speed of the body becomes three (c) the centre (d) all the points times, the force needed to stop it within the same distance would be TNPSC AE 2017 Ans. (d) : A ball is thrown up. The sum of kinetic and (a) 1.5 F (b) 3.0 F potential energies will be maximum at all the points. (c) 6.0 F (d) 9.0 F 1 mv2 + mgh = constant. UKPSC AE 2012 Paper-I 2 Ans. (d) : 9.0 F Then we can say that summation of energy will remain 193. When a body is thrown up at an angle of 45\u00b0 constant at each point. with a velocity of 100 m\/sec, it describes a 188. The potential energy an elevator losses in parabola. Its velocity on point of return down coming down from the top of a building to stop will be (b) 50 m\/sec (a) zero at the ground floor is (c) 100 (a) lost to the driving motors 2m \/s (d) 100 2m \/ sec (b) converted into heat (c) lost in friction of the moving surfaces UKPSC AE 2012 Paper-I (d) used up in lifting the counter poise weight Ans. (c) : The velocity on point of return will be the TNPSC AE 2017 velocity at the maximum height but at the highest point Ans. (d) : The potential energy an elevator losses in coming down from the top of a building to stop at the only the constant horizontal velocity (4 cos\u03b8) ground floor is used up in lifting the counter poise = 100 cos 450 weight. = 100 189. For a conservative force, the work done is 2m \/ s independent of (b) time 194. The unit of energy in S.I unit is (a) path (d) All of the above (a) Dyne (b) Watt (c) distance (c) Newton (d) Joule APPSC-AE-2019 UKPSC AE 2012 Paper-I Ans. (d) : Conservative force are electrical force, Ans. (d) : Joule gravitational force and elastic force etc. 195. A thin circular ring of mass 100 kg and radius 190. A truck weighing 150 kN and travelling at 2 2 m resting on a smooth surface is subjected to m\/sec impacts with a buffer spring which a sudden application of a tangential force of compresses 1.25 cm per 10 kN. The maximum 300 N at a point on its periphery. The angular compression of the spring will be : acceleration of the ring will be (a) 26.6 cm (b) 27.6 cm (a) 1.0 rad\/sec2 (b) 1.5 rad\/sec2 (c) 28.6 cm (d) 30.6 cm (c) 2.0 rad\/sec2 (d) 2.5 rad\/sec2 OPSC Civil Services Pre. 2011 UKPSC AE 2012 Paper-I 418","Ans. (b) : 450 \u2013 30x + 450 \u2013 30x = 4.5 (15+x) (15\u2013x) 900 = 4.5 (152 \u2013 x2) 200 = 225 \u2013 x2 x2 = 25 x = 5 km\/hr 200. Identify the pair which has same dimensions : (a) Force and power (b) Energy and work Given F = 300 N, m = 100 kg, r=2m (c) Momentum and energy Angular Acceleration (a) = r\u03b1 (d) Impulse and momentum By the Newton's second law UKPSC AE 2012 Paper-I F = ma Ans. (b) : Energy and work F = mr\u03b1 300 = 100 \u00d7 2 \u00d7 \u03b1 201. 0.01 kilowatt is equal to (b) 1.0 J\/s (a) 10.01 J\/s \u03b1 = 1.5rad \/ sec2 (c) 0.10 J\/s (d) 0.01 J\/s 196. A train crosses a tunnel in 30 seconds time. The UKPSC AE 2007 Paper -I speed of the train at entry and at exit from the Ans. (a) : 10.01 J\/s tunnel are 36 and 54 km\/hour respectively. If 202. The wheels of a moving car possesses (a) kinetic energy of translation only acceleration remains constant, the length of the (b) kinetic energy of rotation only tunnel is (c) kinetic energy of translation and rotation both (a) 350 m (b) 360 m (c) 375 m (d) 400 m (d) strain energy UKPSC AE 2012 Paper-I UKPSC AE 2007 Paper -I Ans. (c) : 375 m Ans. (c) : Kinetic energy of translation and rotation 197. If T1 and T2 are the initial and final tensions of both an elastic string and x1 and x2 are the 203. The total energy possessed by moving bodies corresponding extensions, then the work done (a) remain constant at every instant is (b) varies from time to time (a) (T2 + T1) (x2 \u2013 x1) (b) (T2 \u2013 T1) (x2 + x1) (c) is maximum at the start (d) is minimum before stopping (c) (T2 \u2212 T1 )( x2 + x1 ) (d) (T2 + T1)( x2 \u2212 x1) UKPSC AE 2007 Paper -I 22 UKPSC AE 2012 Paper-I Ans. (a) : Remain constant at every instant Ans. (d) : (T2 + T1 )(x2 \u2212 x1 ) 204. The escape velocity on the surface of the earth is (a) 1.0 km\/s (b) 3.6 km\/s 2 (c) 8.8 km\/s (d) 11.2 km\/s 198. The escape velocity on the surface of the earth UKPSC AE 2007 Paper -I is Ans. (d) : 11.2 km\/s (a) 11.2 km\/s (b) 8.2 km\/s 205. Inertia force of a body is expressed as (c) 3.2 km\/s (d) 1.2 km\/s (a) product of mass of the body and the UKPSC AE 2012 Paper-I acceleration of its centre of gravity in the Ans. (a) : 11.2 km\/s direction of acceleration (b) product of mass of the body and the 199. A motor boat whose speed in still water is 15 acceleration of its centre of gravity acting in km\/hr goes 30 km downstream and comes back in a total time of four and half hours. The an opposite direction of acceleration (c) product of linear acceleration of the body and stream has a speed of (a) 3 km\/hr (b) 4 km\/hr its mass moment of inertia in the direction of acceleration of its centre of gravity (c) 5 km\/hr (d) 6 km\/hr (d) none of the above UKPSC AE 2012 Paper-I UKPSC AE 2007 Paper -I Ans. (c) : Data given Ans. (b) : Product of mass of the body and the Speed of boat = 15 km\/hr acceleration of its centre of gravity acting in an opposite Total time taken will 4.5 hr direction of acceleration Let x be the speed (T) of steam 206. Which one of the following is a scalar quantity? (T) = t1+ t2 (a) Force (b) Displacement T= 30 + 30 { }Where = SD SU SD = speed of down stream (c) Speed (d) Velocity SU Speed of upstream UKPSC AE 2012 Paper-I T = 30 + 30 Ans. (c) : Speed SB + SS SB \u2212 SS 207. A bullet of mass 0.03kg moving with a speed of 4.5 = 30 + 30 400 m\/s penetrates 12 cm into a fixed block of 15 + x 15 \u2212 x wood. The average force exerted by the wood on the bullet will be 419","(a) 30 kN (b) 20 kN 213. When a body of moment of inertial (I) is (c) 15 kN (d) 10 kN ESE 2017 rotated about that axis with an angular velocity, then the K.E. of rotation is Ans. (b) : m = 0.03 kg v = 400 m\/s (a) 0.5 I\u03c9 (b) I\u03c9 (c) 0.5 I\u03c92 (d) I\u03c92 K.E. of bullet = Work done TNPSC AE 2018 1 mv2 = Force \u00d7 distance 2 Ans. (c) : K.E. of rotation is given as 1 \u00d7 0.03 \u00d7 (400)2 = Force \u00d7 0.12 2 ( K.E.) Rotation = 1 I\u03c92 2 F = 20 kN 10. Principle of Virtual Work and 208. A cricket ball of mass 175 gm is moving with a Simple Machines velocity of 36 km\/hr. What average force will be required to stop the ball in 0.2 second? (a) \u20135.75 N (b) \u20136.75 N (c) \u20137.75 N 214. In actual machines mechanical advantage is (d) \u20138.75 N (a) unity UKPSC AE 2007 Paper -I (b) less than unity Ans. (d) : \u20138.75 N (c) less than velocity ratio 209. Which technique is utilized to find percent idle (d) greater than velocity ratio time for man or machine? UKPSC AE 2007 Paper -I (a) Work sampling (b) Time study Ans. (c) : Less than velocity ratio (c) Method study (d) ABC analysis 215. Which one of the following is not an example of UKPSC AE-2013, Paper-I plane motion ? (a) Motion of a duster on a black board. Ans. (a) : Work sampling technique is utilized to find (b) Motion of ball point of pen on the paper. (c) Motion of a cursor on the computer screen. percent idle time for man or machine. (d) Motion of a nut on a threaded bolt. 210. Dimensional formula ML2T-3 represents:- (a) Work (b) Force UKPSC AE 2012 Paper-I (c) Momentum (d) Power UKPSC AE-2013, Paper-I Ans. (d) : Dimensional formula ML2T-3 represents Ans. (d) : Motion of a nut on a threaded bolt. 216. The velocity ratio of a lifting machine is '8', power. which lifts a load 900 N by an effort of 150 N. 211. A bullet of 0.03 kg mass moving with a speed of 400 m\/s penetrates 12cm into a block of wood. Then, the efficiency of the machine is Force exerted by the wood block on the bullet is:- (a) 75% (b) 70% (c) 65% (d) 60% (a) 10 kN (b) 20 kN TSPSC AEE 2015 (c) 25 kN (d) 30 kN Ans : (a) Velocity ratio (V.R) = 8 UKPSC AE-2013, Paper-I lifted load (w) = 900N Ans. (b) : Change in kinetic energy of bullet = work Effort (p) = 150 N. done by bullet to penetrates into a block of wood 1 mv2 = Force\u00d7 distancea Efficiency (\u03b7) = mechanical advantage 2 velocity ratio. 1 \u00d7 0.03\u00d7(400)2 = Force \u00d7 12 M. A = w = 900 = 6 P 15 2 100 Force = 20 kN \u03b7= 6 8 212. A body moving with a velocity of 1 m\/s has \u03b7 = 0.75 OR 75% kinetic energy of 1.5 Joules. Mass of the body is:- 217. The velocity ratio of a lifting machine is 20 and (a) 0.75 kg (b) 1.5 kg an effort of 200 N is necessary to lift a load of (c) 3.0 kg (d) 30 kg 3000 N. The frictional load is UKPSC AE-2013, Paper-I (a) 7000N (b) 1000N Ans. (c) : We know that (c) 50 N (d) 350N K.E. = 1 mV2 TSPSC AEE 2015 Ans : (b) Velocity ratio (V.R) = 20 2 Effort (P) = 200 N. 1.5 = 1 \u00d7 m \u00d7 (1)2 Lift a load (w) = 3000 N. Frictional load = (P \u00d7 VR - lifted load) 2 m = 3.0 kg frictional load = (20\u00d7 200 - 3000) Frictional load = 1000 N. 420","218. In virtual work principle, the work done by the Mechanical Advantage (MA) = W = 10000 = 20 frictional force acting on wheel when it rolls P 500 without slip is : (a) Zero (b) Positive Velocity Ratio (VR) = D = 20 = 25 (c) Negative (d) None of these d 0.8 HPPSC W.S. Poly. 2016 Efficiency = MA = 20 = 0.8 = 80% VR 25 Ans : (a) In virtual work principle, the work done by the frictional force acting on wheel when it rolls without 224. In a lifting machine, an effort of 200 N is slip is zero. applied to raise a load of 800 N. what will be 219. In an ideal machine, the output as compared to input is the velocity ratio, if efficiency is 50% . (a) Less (a) 8 (b)6 (b) More (c) 7 (d)9 (c) Equal RPSC INSP. OF FACTORIES AND BOILER 2016 (d) May be less more depending of efficiency Ans : (a) Given Effort, Vizag Steel (MT) 2017 P = 200 N Ans. (c) :In an ideal machine, the output as compared to input is equal. W = 800 N Ideal machine efficiency is 100%. Mechanical Advantage = \u03b7 \u00d7 VR 220. If the algebraic sum of the virtual work for W = 800 = 0.5 \u00d7 VR every displacement is ......... the system is in P 200 equilibrium. (b) one V.R. = 8 (a) zero (d) none of these (c) infinity 225. Virtual work means 1. work done by real forces due to virtual (HPPSC LECT. 2016) displacement Ans : (a) Principle of virtual work:- If a particle is in 2. work done by virtual forces during real equilibrium, the total virtual work of forces acting on the particle is zero for any virtual displacement since displacement work done by internal forces [equal, opposite and (a) Both 1 and 2 are correct collinear) cancels each other. (b) Both 1 and 2 are wrong 221. Virtual work refers to : (c) 1 is correct and 2 is wrong (a) Virtual work done by Virtual forces (d) 2 is correct and 1 is wrong (b) Virtual work done by Actual forces APPSC-AE-2019 (c) Actual work done by Actual forces Ans. (a) : Virtual work = virtual force \u00d7 real (d) Actual work done by Virtual forces TRB Polytechnic Lecturer 2017 displacement (or) virtual displacement \u00d7 real force Ans. (b) : Virtual work refers to virtual work done by actual forces. 226. In virtual work equation some forces are neglected. Select the most appropriate answer 222. In the third order pulley system of the pulleys, from the following: the velocity ratio is given by (a) (n2 - 1) (b) (2n - 1) (a) Reaction of a rough surface on a body (c) nn (d) 2n which rolls on it without slipping. (e) 2n (b) Reaction of any smooth surface with which CGPSC 26th April 1st Shift the body is in contact. Ans. (b) : (c) Reaction at a point or on an axis, fixed in Velocity ratio of first order pulley system = 2n space, around which a body is constrained Velocity ratio of second order pulley system = n to turn Velocity ratio of third order pulley system = 2n - 1 (d) All of the above. 223. In a lifting machine, an effort of 500 N is to be UPPSC AE 12.04.2016 Paper-I moved by a distance of 20 m to raise a load of Ans : (d) All of the above 10,000 N by a distance of 0.8m. Determine the 227. In virtual work principle the work done by self efficiency of the machine. weight of body is taken into consideration when (a) 70% (b) 75% (a) centre of gravity moves vertically (b) centre of gravity moves horizontally (c) 80% (d) 85% (c) shear centre moves horizontally (e) 90% (d) shear centre moves vertically CGPSC 26th April 1st Shift TSPSC AEE 2015 Ans. (c) : Given, load (W) = 10,000 N Ans. (b) : In virtual work principle the work done by effort (P) = 500 N Distance moved by the effort (D) = 20 m self weight of body is taken into consideration when Distance moved by the load (d) = 0.8 m centre of gravity moves horizontally. 421","11. Impulse, Momentum and Ans. (d) : Given m = 20 kg, g = 10 m\/sec2 Collision Then velocity after 3 seconds 228. Which of the following statement is correct: (a) The kinetic energy of a body during impact remains constant (b) The kinetic energy of a body before impact is v = u + gt equal to the kinetic energy of a body after v = 0 + 10 \u00d7 3 impact v = 30 m \/ sec (c) The kinetic energy of a body before impact is less than the kinetic energy of a body after Then momentum after 3 seconds M = Final momentum \u2013 initial momentum impact = mv \u2013 mu (d) The kinetic energy of a body before impact is = 20 \u00d7 300 \u2013 20 \u00d7 0 more than the kinetic energy of a body after M = 600 N \u2212 sec. impact TNPSC AE 2013 Ans. (d) : The kinetic energy of a body before impact is 232. A stone of mass is tied to an inextensible more than the kinetic energy of a body after impact is massless string of the length l and rotated in correct statement. vertical circle. The minimum speed required at the top is 229. Coefficient of restitution of perfectly elastic (a) (0.5lg) (b) (lg ) body is (a) 0 (b) 1 (c) (2lg) (d) (3lg) (c) 0.5 (d) infinite (e) (4lg) Gujarat PSC AE 2019 CGPSC 26th April 1st Shift Ans. (b) : A particle of mass m is attached to a light and Ans : (b) : inextensible string. The other end of the string is fixed Coefficient of restitution of perfectly elastic body is 1 at O and the particle moves in a vertical circle of radius r equal to the length of the string as shown in the figure. (one). Coefficient of restitution of perfectly inelastic body is 0 (zero). 230. If (F) refers to force, (m) refers to mass, (v) refers to velocity and (t) refers to time, then which of the following equation is known as momentum principle? (a) F = d (m2v) (b) F = dv dt dt (c) F = d (mv) (d) F = d (mv) dt dt 2 TNPSC 2019 Consider the particle when it is at the point P and the Ans. (c) : Momentum principle- Momentum is the string makes an angle \u03b8 with vertical forces acting on the particle are quantity of motion of a moving body. T = tension in the string along its length From Newton's second law mg = weight of the particle vertically F=ma downwards = m \u00d7 dv m = constant Hence, net radial force on the particle dt FR = T - mg cos\u03b8 F= d (dmtv), mv2 = T - mgcos\u03b8 This above equation is known as momentum l principle equation. T = mv2 + mgcos\u03b8 l 231. A body of mass 20 kg falls freely under gravity. Since speed of the particle decrease with height, hence What will be its momentum after 3 seconds? (take g = 10 m\/s2) tension is maximum at bottom (i.e. \u03b8 = 0) and tension is (a) 300 N sec (b) 400 N sec minimum at top (i.e. \u03b8 = 180) (c) 500 N sec (d) 600 N sec Tmax = mv 2 + mg L (e) 700 N sec l (CGPCS Polytechnic Lecturer 2017) (cos 0\u00ba = 1) 422","Tmin = mvT2 - mg Ans. (b) : v 2 = u 2 + 2 \u00d7 g \u00d7 h l v 2 = 2 \u00d7 9.81\u00d7 6 (cos 180\u00ba = 1) For vT to be minimum, T \u2243 0 v \u2193= 10.8498 m \/ s mvT2 - mg = 0 l vT2 = lg vT = l g 233. A cube strikes a stationary ball exerting an average force of 50 N over at time of 10ms. The ball has mass of 0.20 kg. Its speed after the impact will be (b) 2.5 m\/s (vf )2 = u 2 \u2212 2gh (a) 3.5 m\/s (d) 0.5 m\/s i (c) 1.5 m\/s u 2 = 2\u00d7 9.81\u00d7 4 i ESE 2018 u i \u2191= 8.8588 m \/ s then, v final = u i \u2191 and v initial = V \u2193 Ans. (b) : Given data F = 50 N \u00b5=0 Then impulse m = 0.2 kg I = \u2206P t = 10 ms = 10 \u00d7 10\u22123 sec I = m [Vfinal \u2013 Vinitial] Favg = \u2206p = mv \u2212 mu = 0.1 \uf8ee\uf8f08.858 \u2212 ( \u221210.8498)\uf8f9\uf8fb \u2206t \u2206t 9.81 50 = mv = 0.2 \u00d7 v I = 0.2009 = 0.201 N \u2212 s \u2206t 10 \u00d710\u22123 237. Momentum equations are derived from v = 2.5 m\/s (a) First Law of Thermodynamics 234. A ball is dropped on a smooth horizontal (b) Newton's First Law surface from height 'h'. What will be the height (c) Newton's Second Law of Motion (d) Second Law of Thermodynamics of rebounce after second impact. If coefficient TNPSC AE 2013 of restitution between ball and surface is 'e'? (a) e2h Ans. (c) : Momentum equations are derived from (b) eh Newton's Second Law of Motion. (c) e3h (d) e4h UPRVUNL AE 2016 Fnet = ma = d ( mv) = m dv Ans. (d) : e = co-efficient of restitution dt dt e = V2 = 2gh2 = h2 = h3 V1 2gh1 h1 h2 = mv1 \u2212 mv2 = P1 \u2212 P2 = dP ee22hh21 v2 dt dt h2 = = e2 \u00d7 e2 h1 h3 = Equation (1) is known as momentum equation. h3 = e4h1 238. If v1 and v2 are the initial velocities of two 235. Principle of conservation of momentum is bodies making direct collision and if u1 and u2 are their respective velocities after collision (a) the initial momentum is greater than final momentum then the coefficient of restitution is given by: (b) the initial momentum is equal to final (a) (u1 \u2212 u2 ) (b) (u1 \u2212 u2 ) momentum ( v2 \u2212 v1 ) ( v1 \u2212 v2 ) (c) the initial momentum is smaller than final (c) (u1 + u2 ) (d) (u1 \u2212 v2 ) momentum (v1 \u2212 v2 ) (u1 \u2212 u2 ) (d) the initial momentum is equal to zero CIL MT 2017 2017 IInd shift TNPSC AE 2013 TNPSC AE 2014 Ans. (b) : The principle of conservation of momentum Ans. (a) : The energy dissipation during impact is states that when you have an isolated system with no external forces, the initial total momentum of objects called by the term, coefficient of restitution, a scalar before collision equals the final total momentum of the quantity objects after the collision. e = velocity of separation 236. A steel ball of weight 0.1 N falls a height 6 m velocity of approach and rebounds to a height of 4 m. The impulse is (a) 0.0201 N - s (b) 0.201 N - s e = u1 \u2212 u2 (c) 1.205 N - s v2 \u2212 v1 (d) 12.05 N - s TNPSC AE 2013 423","239. An elevator weighing 10 kN attains an upward 242. A particle is dropped from a height of 3m on a velocity of 4 m\/s in 2 sec with uniform acceleration. horizontal floor, which has a coefficient of The tension in the wire rope is nearly restitution with the ball of 1\/2. The height to (a) 6 kN (b) 8 kN which the ball will rebound after striking the (c) 10 kN (d) 12 kN floor is (a) 0.5 m (b) 0.75 m JWM 2017 (c) 1.0 m (d) 1.5 m Ans. (d) : Given, Elevator weight W = 10 kN = mg TSPSC AEE 2015 Ans. (b) : e = h2 = 1 h1 2 h2 = 1 34 upward velocity v = 4 m\/s h2 = 0.75 m time, t = 2 sec. 243. A bullet of mass 1 kg if fired with a velocity of For uniform acceleration- u m\/s from a gun of mass 10 kg. The ratio of v - u = at kinetic energies of bullet and gun is 4 = 2a (a) 10 (b) 11 Acceleration, a = 2 m\/s2 (c) 1.1 (d) 1.0 Fnet = ma = T \u2013 W = T - mg TSPSC AEE 2015 T = ma + mg Ans. (a) : Total initial momentum of gun and bullet = 2 + 10 = m1u1+ m2u2 = 0 \uf8ee\u2235 mg = 10kN\uf8f9 Total momentum of gun and bullet after firing- \uf8ef \uf8fa = m1 v1 + m2 v2 = 1 \u00d7 u + 10 \u00d7 v2 \uf8f0 m = 1kg \uf8fb Law of conservation of linear momentum Tension in wire rope, T = 12 kN Total momentum after firing = Total momentum before firing 240. If a constant force 'F' acts on a body of mass 'm' for time 't' and changes its velocity from u Then, v2 = \u2212u 10 to v under an acceleration of 'a' all in the same direction, then for equilibrium of the body It is recoil velocity of gun = \u2212 u m \/ s. 10 (a) F = mu (b) F = mv t t 1 1 2 2 (c) F = m \uf8eb v \u2212 u \uf8f6 (d) F = m \uf8eb v + u \uf8f6 Then, ( K.E.)B = m1v12 = \u00d7 1\u00d7 u2 \uf8ed\uf8ec t \uf8f8\uf8f7 \uf8ed\uf8ec t \uf8f8\uf8f7 1 1 \uf8eb u \uf8f62 ( K.E.)B TNPSC AE 2017 ( K.E.)G = 2 m 2v 2 = 2 \u00d710 \u00d7 \uf8ed\uf8ec 10 \uf8f7\uf8f8 = 2 10 Ans. (c) : We know that Newton's second law of motion- ( K.E.)B ( K.E.)G Fnet = ma = m dv = 10 dt F \u00d7 t = m \u00d7 [change in velocity] 244. Two balls of equal mass and of perfectly elastic F \u00d7 t = m [Vf \u2013 Vi] ...(1) material are lying on the floor. One of the ball with velocity 'v' is made to struck the second Then, we can say that product of force and time ball. Both the balls after impact will move with a velocity. equal to change in linear momentum in the direction of force. acting (v \u2212 u) (a) v (b) v\/2 F=m t (c) v\/4 (d) v\/8 TSPSC AEE 2015 241. The coefficient of restitution for inelastic bodies Ans. (b) : Given as, is (a) Zero m1 = m2 = m (b) between zero and one u1 = v, u2 = 0 (c) one e=1 (d) more than one According to momentum conservation principle- APPSC AEE 2016 m1 u1 + m2 u2 = m1 v1 + m2 v2 Ans. (a) : The coefficient of restitution, denoted by (e), mv + m.o = mv' + mv' is the ratio of the final to initial relative velocity v' \u00d7 2m = mv between two objects after they collide. A perfectly v'= v inelastic collision has a coefficient of zero, but a zero 2 value does not have to be perfectly inelastic. 424","245. The co-efficient of restitution of a perfectly (a) The velocity of ball A is 1 2gh 2 plastic impact is (b) The velocity of ball A is zero (a) 0 (b) 1 (c) The velocity of both the balls is 1 2gh (c) 2 (d) 3 2 TSPSC AEE 2015 (d) None of the above Ans. (a) : The coefficient of restitution of a perfectly plastic impact is zero (\u2208= 0) whereas for perfectly elastic impact is one(\u2208= 1). OPSC Civil Services Pre. 2011 246. A lead ball with a certain velocity is made to Ans. (b) : The velocity of ball A is zero strike a wall, it falls down, but rubber ball of Velocity of ball A before collision same mass and with same velocity strikes the same wall. It rebounds. Select the correct uA = 2gh reason from the following- Velocity of ball B before collision (a) both the balls undergo an equal change in uB = 0 For perfectly elastic impact momentum (b) the change in momentum suffered by rubber Velocity of Approach = velocity of separation ball is more than the lead ball uA\u2013uB = vB \u2013vA (c) the change in momentum suffered by rubber vB \u2212 vA = 2gh ................(i) is less than the lead ball Using momentum conservation (d) none of the above muA+ muB = mvA+ mvB TSPSC AEE 2015 Ans. (b) : A rubber ball which have same mass with m 2gh + 0 = m(vA + vB) same velocity strikes the same wall as lead ball, rubber vA + vB = 2gh ...............(ii) ball rebounds because of the change in momentum From (i) and (ii) suffered by rubber ball is more than the lead ball. 247. If u1 and u2 are the velocity of two moving vA = 0 vB = 2gh bodies in the same direction before impact and v1 and v2 are their velocities after impact, then 250. Rate of change of momentum takes place in the co-efficient of restitution is given by direction (a) v1 \u2212 v2 (b) v2 \u2212 v1 (a) of applied force u1 \u2212 u2 u1 \u2212 u2 (b) of motion (c) opposite to the direction of applied force (c) u1 \u2212 u2 (d) u2 \u2212 u1 (d) perpendicular to the direction of motion v1 \u2212 v2 v2 \u2212 v1 (KPSC AE 2015) TSPSC AEE 2015 Ans : (a) Rate of change of momentum takes place in TNPSC AE 2014 the direction of applied force. Ans. (b) : Co-efficient of restitution e = v2 \u2212 v1 = velocity of separation Rate changeof momentum = d ( mv) = m dv u1 \u2212 u2 velocity of approach dt dt 248. During inelastic collision of two particles, Ratechange of momentum = m dv which one of the following is conserved? dt (a) total kinetic energy only Fnet = ma = m dv (b) total linear momentum only dt (c) both linear momentum and kinetic energy 251. Two masses 2 kg, 8 kg are moving with equal (d) neither linear momentum nor kinetic kinetic energy. The ratio of magnitude of their momentum is. energy RPSC INSP. OF FACTORIES AND BOILER 2016 (a) 0.25 (b) 0.50 Ans : (b) During collision of two Particles (c) 0.625 (d) 1.00 1. The momentum is conserved in both elastic and UPPSC AE 12.04.2016 Paper-I in inelastic collision as there is no force applied externally. Ans : (b) Kinetic Energy = 1 mv2 2. The energy is conserved in elastic collisions only. 2 In case of energy is dissipated at each collision in ( KE )1 = p12 .............(i) form of heat and vibration, causing a heating and 2m1 deformation of bodies. ( KE )2 = p22 .............(ii) 249. A ball 'A' of mass 'm' falls under gravity from 2m2 a height 'h' and strikes another ball 'B' of mass 'm' which is supported at rest on a spring of Given, m1 = 2kg, m2 = 8kg stiffness 'k'. The impact between the balls is perfectly elastic. Immediately after the impact: 425","kinetic energy equal p12 = p22 258. When the coefficient of restitution is zero, the 2m1 2m2 bodies are : (a) Inelastic (b) Elastic p1 = m1 (c) Near elastic (d) None of the above p2 m2 OPSC Civil Services Pre. 2011 Ans. (a) : When the coefficient of restitution is zero the p1 = 0.50 bodies are inelastic. p2 When the coefficient of restitution is one (1) 252. When two bodies collide without the presence the bodies are perfect elastic. of any other force or force fields? 259. Impulse is:- (a) Minimum momentum (a) Their total kinetic energy must be conserved. (b) Maximum momentum (c) Average momentum (b) Their total momentum must be conserved. (d) Final momentum - Initial momentum (c) Their collision must be direct. (d) Both (a) and (b) UKPSC AE-2013, Paper-I UPPSC AE 12.04.2016 Paper-I Ans. (d) : Ans : (d) When two bodies collide without the presence Impulse = Final momentum \u2013 Initial momentum of any other force or force fields. 260. A ball of 2kg drops vertically onto the floor (i) Their total kinetic energy must be conserved. with a velocity of 20m\/s. It rebounds with an (ii) Their total momentum must be conserved. initial velocity of 10m\/s, impulse acting on the 253. If the momentum of a body is doubled, its ball during contact will be:- kinetic energy will be (a) 20 (b) 40 (a) doubled (b) quadrupled (c) 60 (d) 30 (c) same (d) halved UKPSC AE-2013, Paper-I UKPSC AE 2007 Paper -I Ans. (c) : We know that, impulse is equal to change in Ans. (b) : Quadrupled momentum 254. The bodies which rebound after impact are So, initial momentum called = 2 \u00d7 20 = 40 kg - m\/s (a) elastic (b) inelastic Final momentum (c) plastic (d) none of the above = \u2013 2 \u00d7 10 = \u2013 20 kg-m\/s UKPSC AE 2007 Paper -I Then impulse Ans. (a) : Elastic 255. The total momentum of a system of moving = [\u2013 20 \u2013 40] bodies in any one direction remains constant, = \u2013 60 kg-m\/s unless acted upon by an external force in that = 60 kg-m\/s direction. This statement is called (a) Principle of conservation of energy So, the impulse is 60 Ns acting upwards. (b) Newton's second law of motion (c) Newton's first law of motion 261. The loss of kinetic energy, during inelastic (d) Principle of conservation of momentum impact of two bodies having masses m1 and m2, UKPSC AE 2007 Paper -I which are moving with velocity v1 and v2 respectively, is given by Ans. (d) : Principle of conservation of momentum (a) 2 ( m1m 2 ) ( v1 \u2212 v2 )2 256. The velocity of a body on reaching the ground m1 + m2 from a height 'h', is given by (b) 2 ( m1 + m2 ) ( v1 \u2212 v2 )2 (b) v = 2gh2 m1m 2 (a) v = 2gh (c) v = 2gh (d) v = h2 m1m2 2g 2(m1 + m2 ) ( )(c) v12 \u2212 v22 UKPSC AE 2007 Paper -I 2(m1 + m2 ) m1m 2 Ans. (c) : v = 2gh ( )(d) 257. Which of the following equations is known as v12 \u2212 v22 momentum principle? UKPSC AE 2012 Paper-I (a) F = d (m2v) (b) F = dv Ans. (a) : 2 ( m1m 2 ) ( v1 \u2212 v2 )2 dt dt m1 + m2 (c) F = d (mv) d (mv) dt (d) F = dt 2 262. The force applied on a body of mass 100 kg to produce an acceleration of 5 m\/s2 is Ans. (c) : F = d (mv) UKPSC AE 2007 Paper -II dt TSPSC AEE 2015 (a) 500 N (b)100 N (c) 20 N (d)10 N 426","Ans : (a) Second law of Newton's :- 265. The period (T) for the pendulum with length (l) and placed at the gravitational acceleration (g) F = m. dv is given by: dt (a) T = 2\u03c0 \u2113 (b) T = 2\u03c0 \u2113 g F = ma g F = 500 N. 263.A rubber ball is dropped from a height of 2 m. if (c) T = 3\u03c0 \u2113 (d) T = 3\u03c0 \u2113 g g there is no loss of velocity after rebounding the ball will rise to a height of CIL MT 2017 2017 IInd shift (a) 1 m (b) 2 m Ans. (a) : Time period of pendulum T = 2\u03c0 \u2113 g (c) 3 m (d) 4 m Vizag Steel (MT) 2017 Ans. (b) : For rebounding height attain = e2x 266. In case of S.H.M. the period of oscillation is then h1 = e2x given by for x = 2 m No loss in kinetic energy so for elastic body e = 1 (a) T = 2\u03c9 (b) T = 2\u03c0 \u03c02 \u03c9 h1 = 1 \u00d7 2 for second rebounding (c) T = \u03c9 (d) T = \u03c0 2\u03c0 2\u03c9 h2 = e4x x =4m TNPSC 2019 h2 = 1\u00d74 h2 = 4 m Ans. (b) : In case of S.H.M. the period of oscillation is then h1 = 2 h2 4 given by T = 2\u03c0 \u03c9 h1 = 1 h2 2 267. In case of S.H.M. the period of oscillation is given by (a) T = 2\u03c9 (b) T = 2\u03c0 \u03c02 \u03c9 h2 = 2h1 (c) T = \u03c9 (d) T = \u03c0 2\u03c0 2\u03c9 12. Simple Harmonic Motion and Projectile Motion TNPSC 2019 264. In simple harmonic motion, acceleration is Ans. (b) : In case of S.H.M. the period of oscillation is given by T = 2\u03c0 \u03c9 proportional to 268. The radius of arc is measured by allowing a 20 (a) \u03c9 (frequency) mm diameter roller to oscillate to and fro on it (b) velocity and the time for 25 oscillations is noted at 56.25 s. The radius of arc will be (c) rate of change of velocity (a) 865 mm (b) 850 mm (d) displacement (c) 835 mm (d) 820 mm (e) \u03c9 CGPSC 26th April 1st Shift Ans. (b) : Given, ESE 2019 RPSC AE 2016 Radius of roller (HPPSC LECT. 2016) Ans. (d) : An object is undergoing simple harmonic motion (SHM) if the acceleration of the object is directly proportional to its displacement from its equilibrium position. In SHM, the displacement of particle at an instant is given by y = r sin \u03c9t r = 20 = 10 mm Velocity (v) = dy = r\u03c9cos\u03c9t 2 dt Time period of Acceleration (a) = dv = -\u03c92r sin\u03c9t Oscillation T = 56.25 = 2.25 sec dt 25 a = -\u03c92y \u03c9n = 2g a\u221dy 3(R \u2212 r) 427","T = 2\u03c0 3(R \u2212 r) Ans. (a) : At the highest point of the trajectory the shell 2g will have the only horizontal velocity that is \u2013v cos \u03b8. 2.25 = 2\u03c0 3(R \u2212 0.01) For its one part to retrace its parth. 2\u00d7 9.81 0.8395 = R \u2212 0.01 R = 0.849 m R = 850 mm 269. A car is travelling on a curved road of radius 300 m at speed of 15 m\/s. The normal and tangential components of acceleration respectively are given by: After explosion the piece that retraces its path is having (a) 0.75 m\/s2, zero (b) 0.75 m\/s2, 0.75 m\/s2 velocity, (c) zero, zero v1 = \u2013v cos \u03b8 (d) zero, 0.75 m\/s2 Since, there is no force acting on the shell in horizontal UPRVUNL AE 2016 direction, so its linear momentum remains constant, Ans. (a) : Radial\/normal components of acceleration m m 2 2 due to change in direction continually (ar) mv cos\u03b8 = v1 + v2 = v2 = (15)2 v2 = 2v cos \u03b8 \u2212 v1 r 300 v2 = 2v cos \u03b8 + (\u2212v cos \u03b8) ar = 0.75 m\/s2 v2 = 3v cos \u03b8 272. The particle is projected a point 'Q' with initial velocity 'u' inclined at ' ' to x-axis, x- component of initial velocity at point 'O' is (a) ux = u sin\u03b1 In this case no tangential acceleration because of no (b) ux = u cos\u03b1.sin \u03b1 (c) ux = u cos\u03b1 change in angular velocity due to unifrom circular (d) ux = u tan \u03b1 motion. TNPSC AE 2013 270. A particle is projected with an initial velocity of Ans. (c) : u x = u cos \u03b1 60 m\/sec at an angle of 75o with horizontal. The Time of flight (T) maximum height attained by the particle is (a) 171.19 m (b) 185.22 m T = 2 \u00d7 t peak = 2.u sin \u03b1 g (c) 221.11 m (d) 198.20 m TNPSC AE 2014 Ans. (a) : Given, u = 60 m\/s, \u03b8 = 750 The maximum height its, h max = u2 sin 2 \u03b8 2g h max = (60) 2 \u00d7 (sin 75\u00ba) 2 R = u 2 sin 2\u03b1 , u 2 sin 2 \u03b1 g 2g 2\u00d7 9.81 H max = h max = 171.194 m 271. A shell fired from cannon with a speed 'v' at an 273. The maximum acceleration of a particle moving with SHM is angle ' ' with the horizontal direction. At the (a) 2 highest point in its path it explodes into pieces (c) 2\/r (b) r of equal mass. One of the pieces retraces its (d) 2 r path to the cannon. The speed of other piece TSPSC AEE 2015 immediately after explosion is: Ans : (d) Velocity and acceleration of a particle moving (a) 3 v cos \u03b8 (b) 2 v cos \u03b8 with simple harmonic motion:- (c) 3\/2 v cos \u03b8 (d) 3\/ 2 v cos \u03b8 (i) Maximum velocity (vmax) = \u03c9r OPSC Civil Services Pre. 2011 (ii) Maximum acceleration ( \u03b1max ) = \u03c92r 428","274. If the velocity of projection is u m\/sec and the 276. A body is having a simple harmonic motion. angle of projection is a, the maximum height Product of its frequency and time period is of the projectile on a horizontal plane is : equal to:- (a) u2 cos2 \u03b1 (b) u2 sin 2 \u03b1 (a) Zero (b) One 2g 2g (c) Infinity (d) 0.5 UKPSC AE-2013, Paper-I (c) u2 tan 2 \u03b1 (d) u2 sin 2 \u03b1 Ans. (b) : We know that, 2g g motion. in the simple harmonic HPPSC W.S. Poly. 2016 Time period (T) = 1 Ans : (b) Frequency (f ) Then, T \u00d7 f = 1 277. A projectile on a level ground will have maximum range if the angle of projection is (a) 30\u00b0 (b) 45\u00b0 U = velocity of projection (c) 60\u00b0 (d) 75\u00b0 \u03b1 = angle of projection (i) Flight - time of Projectile:- UKPSC AE 2012 Paper-I T = 2u sin\u03b1 Ans. (b) : 45\u00b0 g 278. If the period of oscillation is to become double, (ii) Height of projectile:- then h = u2 sin2 \u03b1 2g (a) the length of simple pendulum should be doubled. (b) the length of simple pendulum should be quadrupled. (iii) Range of projectile:- (c) the mass of the pendulum should be doubled. R = u2 sin 2\u03b1 (d) the length and mass should be doubled. g UKPSC AE 2012 Paper-I 275. A particle is projected at such an angle with the Ans. (b) : the length of simple pendulum should be horizontal that the maximum height attained quadrupled. by the particle is one-fourth of the horizontal 279. For the maximum range of a projectile, the angle of projection should be range. The angle of projection should be:- (a) 30\u00ba (b) 45\u00ba (a) 30\u00ba (b) 45\u00ba (c) 60\u00ba (d) 90\u00ba (c) 60\u00ba (d) 75\u00ba UKPSC AE 2007 Paper -I UKPSC AE-2013, Paper-I Ans. (b) : In projection motion Ans. (b) : 45\u00ba h = u2 sin 2 \u03b8 280. If the period of oscillation is doubled 2g (a) the length of simple pendulum should be doubled and (b) the length of simple pendulum should be quadrupled Range (R ) = u2 sin 2\u03b8 (c) the mass of the pendulum should be doubled g According to question (d) the length and mass should be doubled R = h at \u03b8max . UKPSC AE 2007 Paper -I 4 Ans. (b) : The length of simple pendulum should be u2 sin 2\u03b8 = u2 sin2 \u03b8 quadrupled 4\u00d7 g 2g 281. The maximum displacement of a body moving sin \u03b8 cos \u03b8 = sin2 \u03b8 with simple harmonic motion from its mean position is called tan \u03b8 = 1 (a) oscillation (b) amplitude \u03b8 = tan\u22121 (1) (c) beat (d) none of the above Ans. (b) : Amplitude UKPSC AE 2007 Paper -I \u03b8 = 45o 429"]