File No.06/06/21/07/2014 VI CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD ANDHRA PRADESH - TELANGANA 2014-2015 PROGRAM M E 2nd-CHAPTER-SOLUTION www.eabhyasacademy.com 2. KINEMATICS 1. The wheel of moving car has rotational, linear. 2. The motion of a striker across a carromboard is the example of curve linear. 3. The passengers sitting in a flying aeroplane are in rest with respect to clouds, but are in rest with respect to the interior of the aeroplane. 4. The displacement travelled by amrith 2 m 5. A person, seated in a train under motion, is at rest with reference to A person watching him from the front seat and a car moving in the same direction with same velocity of train. 7. An example of rectilinear motion is apple falling from a tree. 9. Oscillatory motion is executed by a pendulum of a wall clock. 10. Motion of a screw while going into the wood is an example of linear and spin motion. 11. Motion of earth has circular motion, periodic motion and rotational motion. 13. The motion caused is translatory 14. The motion of sea waves is Rectilinear, oscillatory. 15. The objects does not have more that one type of motion is child on a seesaw. 16. Any motion that repeats itself at regular intervals of time is periodic motion. 17. Motion of foot ball player on the ground is an example of Random motion 19. The actual length of the path covered by the moving object distance. 20. if a body moves in a circular path and reach back to its initial position then magnitude of displacement = 0. 1. When both the cars move at same speed, the relative motion between them is zero and they will not collide with each other. t = 1.5 min = 1.5 x 60 sec = 90 sec 2. 2014 - 2015 CHALLENGER 1 PHYSICS - CHAPTER SOLUTIONS - 2 File No.06/06/21/07/2014 3. VI CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD s = d/t = 30/90 = 1/3 = 0.33 m/s. Speed = distance/time time = 30 min = 1/2 hour 10km  S= 1 20 kmph. 2 4. 20 m ------ 5sec 80 m -----? 80  5  20 sec 20 5. The distance between the cities=260 km Time taken to travel the distance=4h  The speed of the car= 6. Speed of the first bus=90kmh-1  90  7. dis tance 260   65kmh1 time 4 5  25ms 1 18 Speed of second bus=2 ms-1  First bus is faster Total distance travelled=AB+BC+CD =4+6+4 =14 cm. Total displacement made by the body=straight line path between A and D C 4cm D 6cm 2014 - 2015 A 4cm B In AOC AD2  AO 2  OD2  AO2  OD2  (AB  BO)2  OD2  82  62 AD2  64  36 AD  10cm www.eabhyasacademy.com 8. Distance=5 km tme=10 min=10/60 h=1/6 h Speed= 9. 5 dis tance  30kmh1 . = 1/ 6 time Given time t = 2 hour d = 260 km Speed = 260 = 130 kmph 2 CHALLENGER 2 PHYSICS - CHAPTER SOLUTIONS - 2 File No.06/06/21/07/2014 10. VI CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD First bus travels with a speed of 36 kmph = 20 m/s second bus travels with a speed of 20 m/s Third bus travels with a speed 18 kmph = 18 x 11. d = 50 km = 50 x 103 m t = 10 min = 10 x 60 = 600 sec Speed = 12. 5 = 5 m/s. 18 50000 18 = 83.33 x =300 kmph. 600 5 t = 15 sec d = 100 m d = 5 km 100 m - 15 sec 5 km - ? t=? v= 6.66 = 100 = 6.66 m/s 15 5x103  t = 750 sec t = 12.5 min. 13. 14. 15. 17. d = 360 km t= 1 hour 2 v= 360 = 720 kmph 1/ 2 = 720 x 5/18=200 m/s. v = 30 m/s t = 1 min = 60 sec 2014 - 2015 d = v x t = 30 x 60 = 1800 m distance covered in half of circular path = r t = 3 min 20sec = 3 x 60 + 20 = 200 sec v = 6 m/s v= 18. speed = x m/s d = 1.2 km = 1.2 x 103m t = 80 sec x= 19. www.eabhyasacademy.com 20. d  d = v x t = 6 x 200 = 1200 m. t 1.2x103 1200  =15m/s. 80 80 v = 100 m/s, t = 25 min = 25 x 60 d = v x t = 100 x 25 x 60 = 150000 m = 150 km. d = 324 km t = 20 min 324 x10 3 18 v = 20 x 60 =270 x =972 kmph. 5 CHALLENGER 3 PHYSICS - CHAPTER SOLUTIONS - 2 File No.06/06/21/07/2014 VI CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD BRAIN TWISTERS 1. A stone is thrown horizontally by a body from a height, A javelin thrown by an athlete, firing of a bullet from a gun all are curvilinear motion. 2. Oscillatory motion is also known as to and fro motion, back of forth motion. 3. Motion of a ceiling fan is rotatory motion and circular motion. 4. cm/s (or) cm s-1 is the unit of velocity in C.G.S. system. 5. Metre (or) m is the unit of distance in M.K.S system. 6. a) Distance - path between two points b) displacement - shortest path between two points c) Speed - dis tance time d) Velocity - displacement time e) Motion - position changes w.r.t time 7. a) Motion of axle of moving fan is rotation b) Motion of the earth around its axis is rotation c) Motion of the blades of a moving fan -revolution 2014 2015 d) Motion of the earth going around the sun revolution. 8. a) The wheel of sewing machine - Rotatory motion b) Rolling drum on a straiaght road - Rotatory, Rectilinear c) Girl sitting on moving cradle - oscillatory, periodic d) Buzzing bee - Random motion. 9. a) A car moving on striaght road - translatory motion b) A javelin put thrown by an athlete - Curvilinear motion c) Motion of potter’s wheel - Rotatory motion d) Motion of heart beats - Periodic motion 10. a) Distanc - does not depend on path www.eabhyasacademy.com b) Displacement - depends on path c) Circumference of the circular path - 2  r d) Half of the circumference -  r 11. Both assertion and reason are correct but reason does not explain about assertion. CHALLENGER 4 PHYSICS - CHAPTER SOLUTIONS - 2 File No.06/06/21/07/2014 VI CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD 12. Both assertion and reason are correct and reason does not explain assertion. 13. velocity displacement 1  1. So, speed dis tance Both assertion and Reason are correct but reason does not explain about assertion. 14. Assertion is correct. Reason is incorrect. 15. Along path ABC distance = 10m. 16. Along path CDA displacement 17. Total distance ABCDA 20m. 18. Given room dimensions 4m x 3m displacement = 42  32 = 25 =5 m. 19. Total distance travelled = 6km + 6km = 12km. 20. Circumference of circle = 2  r.. 5. Total distance = 4+3+12 = 19m. Total displacement = 6. www.eabhyasacademy.com 52  52 = 50 =5 2 m. 2014 - 2015 42  32  122  169  13m . Distance covered in each pair of steps = 0.75 + 0.25 = 1m. At the end of 100 steps distance covered = 100 m Displacement of each pair of step = 0.75 - 0.25 = 0.5. Total displacement = 0.5 x 100 = 50 m along forward direction. 7. Total distance 2x 40   m/ s x x 3 Average speed = Total time .  10 20 8. Average speed = dis tance time 9. Average speed = dis tance time 10. Average speed = dis tance time 12. Change in velocity = (108-36) kmh–1 = 72 kmh–1 Change in velocity in ms–1 = 72  Acceleration in m/s2  CHALLENGER 5  20 18 Change in velocity 20  5 Time 4 5 PHYSICS - CHAPTER SOLUTIONS - 2 File No.06/06/21/07/2014 VI CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD 14. aresul tan t  a12  a22  42  32  5m / s2 . 15. v resul tan t  v12  v 22  162  92  337m / s . 16. Average speed = totaldis tance totaltime 3x 60 x 3  x x x 10 =   10 20 60 17. 18. = 18 km/h. Total time of motion on circular track = 2 x 60 + 20 = 140s. Time period of revolution (=3 x 40 = 120s) = zero as athelete will be reaching at the starting point. Thus displacement in 140 second = displacement in 20 seconds = 2R. Distance travelled in one rotation (lap) = 2  r Average speed = 2r 2 x 3.14 x100  =10m/s t 62.8 Net displacement in one lap = 0  Average velocity = = 19. 20. net displacement time 0 =zero. 62.8 When location of a particle has changed, it must have covered some distance and undergone some displacement. A body in uniform circular motion is moving in a plane and is two dimentional motion. www.eabhyasacademy.com 2014 - 2015 CHALLENGER 6 PHYSICS - CHAPTER SOLUTIONS - 2