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Page 1 of 20, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Q1 In which of the following examples of motion, can the body be considered, approximately a point object:, (a) a railway carriage moving without jerks between two stations., (b) a monkey sitting on top of a man cycling smoothly on a circular track., (c) a spinning cricket ball that turns sharply on hitting the ground., (d) a tumbling beaker that has slipped off the edge of a table., Solution:, (a), (b) The size of the train carriage and the cap is very small as compared to the distance they've, travelled, i.e. the distance between the two stations and the length of the race track, respectively., Therefore, the cap and the carriage can be considered as point objects., The size of the basketball is comparable to the distance through which it bounces off after hitting the, floor. Thus, basketball cannot be treated as a point object. Likewise, the size of the bottle is comparable, to the height of the chair from which it drops. Thus, the bottle cannot be treated as a point object., Q2. The position-time (x-t) graphs for two children A and B returning from their school O to their, homes P and Q respectively are shown in Fig. Choose the correct, entries in the brackets below:, (a) (A/B) lives closer to the school than (B/A), (b) (A/B) starts from the school earlier than (B/A), (c) (A/B) walks faster than (B/A), (d) A and B reach home at the (same/different) time, (e) (A/B) overtakes (B/A) on the road (ond, Qt, A, B, FUNDAMENT ORPYSICS, Solution:, (a) A lives closer to the school than B, because A has to cover shorter distances [OP < OQI,, (b) A starts from school earlier than B, because for x= 0, t = 0 for A but for B, t has some finite time., (c) The slope of B is greater than that of A, therefore B walks faster than A., (d) Both A and B will reach their home at the same time., (e) At the point of intersection, B overtakes A on the roads once., for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 2 of 20, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Q3. A woman starts from her home at 9.00 am, walks at a speed of 5 km/h on a straight road up, to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by auto with a, speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion., Solution:, 3.0-, A, 2.5+, В, 2.0+, 目 1.5十, 1.0+, 0.5, 9.30am, 5.06 pm, 9am 10, 11, 12, 1, Distance till her office = 2.5 km., Walking speed the woman= 5 km/h, Time taken to reach office while walking = (2.5/5 ) h = (1/2) h = 30 minutes, Speed of auto = 25 km/h, Time taken to reach home in auto, 2.5/25 = (1/10) h = 0.1 h = 6 minutes, In the graph, O is taken as the origin of the distance and the time, then at t = 9.00 am, x = 0, and at t = 9.30 am, x = 2.5, OA is the portion on the, time of stay in the offic, graph that represents her walk from home to the office. AB represents her, from 9.30 to 5. Her return journey is represented by BC., Q4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards, followed, again by 5 steps forward and 3 steps backwards, and so on. Each step is 1m long and requires, 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard, takes to fall in a pit 13 m away from the start., Solution:, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734), (uy ui)x, DARENTAL PHYSICS

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Page 3 of 20, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, x(m)+, 14+, 12+, 1아, 8+, 6, 4, 2+, 5, 10 13 1516, 20 21 24 25, 35 37 40 r(s), The time taken to go one step is 1 second. In 5s he moves forward through a distance of 5m and then, in next 3s he comes back by 3m. Therefore, in 8s he covers 2m. So, to cover a distance of 8m he takes, 32s. He must take another 5steps forward to fall into the pit. So, the total time taken is 32s + 5s = 37s, to fall into a pit 13 m away., OPNYSICS, Q.5. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the, speed of 1500 km/h relative to the jet plan, observer on the ground?, What is the speed of the latter with respect to an, Solution:, Speed of the jet airplane, VA= 500 km/h, Speed at which the combustion products are ejected relative to the jet plane, VB - VA, =- 1500 km/h, (The negative sign indicates that the combustion products move in a direction opposite to that of jet), Speed of combustion products w.r.t. observer on the ground, VB - 500 = - 1500, VB = - 1500 + 500 =, 1000 km/h, Q6 A car moving along a straight highway with a speed of 126 km h-1 is brought to a stop within, a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it, NDANENT, Solution:, The initial velocity of the car u, Final velocity of the car = v, Distance covered by the car before coming to rest = 200 m, Using the equation,, V = u + at, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 4 of 20, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, t = (v – u)/a = 11.44 sec., Therefore, it takes 11.44 sec for the car to stop., Q.7. Two trains A and B of length 400 m each are moving on two parallel tracks with a, uniform speed of 72 km h-1 in the same direction, with A ahead of B. The driver of, B decides to overtake A and accelerates by 1 m s-. If after 50 s, the guard of B just, brushes past the driver of A, what was the original distance between them?, Solution:, Length of the train A and B = 400 m, Speed of both the trains = 72 km/h = 72 x (5/18) = 20m/s, Using the relation, s = ut + (1/2) at?, Distance covered by the train B, SB = Ust + (1/2) at?, Acceleration, a = 1 m/s, Time = 50 s, SB = (20 x 50) + (1/2) x 1 x (50)2, = 2250 m, Distance covered by the train A, SA = Uat + (1/2) at?, Acceleration, a = 0, SA = Uat = 20 x 50 = 1000 m, Therefore, the original dis, ce between the two trains = SB - SA = 2250 – 1000 = 1250 m, Q. 8. On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars B and C approach, car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the, distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What, minimum acceleration of car B is required to avoid an accident?, AMENTAL OF PHYSICS, Solution:, The speed of car A = 36 km/h = 36 x (5/8) = 10 m/s, Speed of car B = 54 km/h = 54 x (5/18) = 15 m/s, Speed of car C = - 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are in opposite, direction), Relative speed of A w.r.t C, VAC= VA – VB = 10 - (-15) = 25 m/s, Relative speed of B w.r.t A, VBA = VB – VA = 15 - 10 = 5 m/s, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)

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Page 5 of 20, FUNDAMENTAL OF PHYSICS, for IIT-JEE, NEET, 11th & 12th, Distance between AB = Distance between AC = 1 km = 1000 m, Time taken by the car C to cover the distance AC, t = 1000/VAC = 1000/ 25 = 40 s, If a is the acceleration, then, S = ut + (1/2) at?, 1000 = (5 x 40) + (1/2) a (40) 2, %3D, a = (1000 – 200)/ 800 = 1 m/s?, Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s?, Q. 9. Two towns A and B are connected by regular bus service with a bus leaving in either, direction every T minutes. A man cycling with a speed of 20 km h- in the direction A to B, notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in, the opposite direction. What is the period T of the bus service and with what speed (assumed, constant) do the buses ply on the road?, Solution:, Speed of each bus = VD, Speed of the cyclist Vc= 20 km/h, The relative velocity of the buses plying in the direction of motion of cyclist is Vp - Vc, The buses playing in the direction of motion of the cyclist go past him after every 18 minutes i.e.(18/60), S., Distance covered = (Vb - Vc ) x 18/60, Since the buses are leaving every, minutes. Therefore, the distance is equal to Vb x (T/60), (Vb – Vc ) x 18/60 = Vb x (T/60, (1), The relative velocity of the buses plying in the direction opposite to the motion of cyclist is Vp + Vc, The buses go past the cyclist every 6 minutes i.e.(6/60) s., Distance covered = (Vb+ Vc ) x 6/60, Therefore,, +Vc)x 6/60 = Vb x (T/60), (2), 18/60]/ [(Vb + V) x 6/60 ]= [Vb X (T/60)] /[Vb x (T/60)], (Vb- Vc) 18/(Vb +Vc) 6 = 1, (Vb - Vc )3 = (Vb +Vc ), Substituting the value of Ve, (Vb – 20 )3= (Vb + 20 ), 3V, – 60 = Vp + 20, 2Vb = 80, for IIT-JEE, NEET, 11th & 12th, R.K.Sharma (8507152734)