Motion in a Plane – Physics | NEET Online Mock Test Leave a Comment / Blog / By admin Online Mock Test Description Mock TestMotion in a PlaneExamNEET Medical TestSubjectPhysicsMarks100No. Questions50 QuestionsTime50 Minutes 1 Sorry! Time up Motion in a Plane - NEET (Physics) Best of Luck! 1 / 50 1 2 3 4 2 / 50 A man who can swim at the rate of 2 km/h (in a still river) crosses a river to a point exactly opposite on the other bank by swimming in a direction of 120° to the flow of river water. The velocity of water current in km/h is 1 2 1/2 3/2 3 / 50 1 2 3 4 4 / 50 The trajectory of a projectile launched from the ground is given by the equation y = (- 0.025x^{2} + 0.5x) where x and y are the coordinates of the projectile on a rectangular system of axes. Find the angle at which the projectile is launched.θ = tan^{-1}(0.2)θ = tan^{-1}(0.4)θ = tan^{-1}(0.5)θ = tan^{-1}(0.6) 1 2 3 4 5 / 50 A projectile is thrown at an angle of 37° from the vertical. The angle of elevation of the highest point of the projectile from the point of projection istan^{-1}(3/2)tan^{-1}(2/3)tan^{-1}(3/8)tan^{-1}(8/3) 1 2 3 4 6 / 50 1 2 3 4 7 / 50 1 2 3 4 8 / 50 A man can swim at a speed of 5 km/h w.r.t. water. He wants to cross a 1.5 km wide river flowing at 3 km/h. He always keeps himself at an angle of 60° with the flow direction while swimming. The time taken by him to cross the river will be 0.25 h 0.35 h 0.45 h 0.55 h 9 / 50 1 2 3 4 10 / 50 The trajectory of a projectile in a vertical plane is y = ax-bx^{2}, where a and b are constants and x and y respectively are horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal areb^{2}/2a ,tan^{-1}(b)a^{2}/b ,tan^{-1}(2a)a^{2}/4b ,tan^{-1}(a)2a^{2}/b ,tan^{-1}(a) 1 2 3 4 11 / 50 The rain falls vertically on a man walking at 3km/h. When he increases his speed to 6km/h, it appears to meet him at an angle of 45° with vertical. The speed of rain is2√3 km/h√6 km/h3√2 km/h4√3 km/h 1 2 3 4 12 / 50 1 2 3 4 13 / 50 A boat that has a speed of 5km/h in still water crosses a river of width 1km along shortest possible path in 15 minutes. The velocity of river water in km/h is: 1 3 4 42 14 / 50 A particle is moving in X - Y plane such that V_{x} = 4 + 4t and v_{y} = 4t(t in second and v_{x}, v_{y} in m/s). If the initial position of a particle is (1, 2) m. Then the equation of trajectory isy^{2} = 4xy = 2xx^{2} = y/2None of these 1 2 3 4 15 / 50 A particle moves in a circle of radius 1.0 cm at a speed given by v 2.0t where v is in cm/s and t in seconds. The magnitude of total acceleration at = 1s is nearest to4.5 cm/s^{2}2.5 m/s^{2}6.2 cm/s^{2}8.5 cm/s^{2} 1 2 3 4 16 / 50 If the speed of an object revolving in a circular path is doubled and angular speed is reduced to half of the original value, then centripetal acceleration will become/remain Same Double Half Quadruple 17 / 50 1 2 3 4 18 / 50 A ball is moving in a circular path of radius 5 m. If tangential acceleration at an instant is 10 m s-2 and the net acceleration makes an angle of 30° with the centripetal acceleration, then the instantaneousspeed is50√3 ms^{-1}9.3 ms^{-1}6.6 ms^{-1}5.4 ms^{-1} 1 2 3 4 19 / 50 A swimmer heads directly across a river swimming at 1.6 m/s relative to water. She arrives at a point 40m downstream from the point directly across the river, which is 80 m wide. Find the time taken to cross the river ? 30 s 40 s 50 s 60 s 20 / 50 To a man walking due west with a speed of 4m/s, the wind appears to blow from the north-east, when he increases his speed to 7m/s the wind appears to blow from the north. Find the velocity of wind with respect to the ground√50√40√58√42 1 2 3 4 21 / 50 Rain is falling at a speed of 4m/s in a direction making an angle of 30° with vertical towards the south. What should be the magnitude and direction of the velocity of a cyclist to hold his umbrella exactly vertically, so that rain does not wet him? 2m/s towards north 4m/s towards south 2m/s towards south 4m/s towards north 22 / 50 A particle moves with a constant speed v along a parabolic path y = kx^{2}, where k is a positive constant. Find the acceleration of the particle at the point x=0.kv^{2}1/2kv^{2}2kv^{2}3kv^{2} 1 2 3 4 23 / 50 1 2 3 4 24 / 50 A man can swim at a speed of 10 km/h in still water. A river of width 800m is flowing at a speed of 6 km/h. Find the time taken to cross the river along the shortest path. 4.8 min 6 min 12 min 9 min 25 / 50 1 2 3 4 26 / 50 The speed of a particle moving in a circular path of radius 1m varies with time as v = 2t2 Here v and t are measured in SI units. Find the acceleration of the particle at t=1s.2√2m/s^{2}4√2m/s^{2}4 m/s^{2}2 m/s^{2} 1 2 3 4 27 / 50 Rain is falling vertically at a speed of 30 m/s. The wind starts blowing after some time with a speed of 15 m/s from east to west direction. In which direction should a boy waiting for the bus hold his umbrella?At an angle tan^{-1}(1/2) with vertical towards eastAt an angle tan^{-1}(1/2) with vertical towards westAt an angle tan^{-1}(2) with vertical towards eastAt an angle tan^{-1}(2) with vertical towards west 1 2 3 4 28 / 50 A particle projected at some angle with a velocity of 50 m/s crosses a 20 m high wall after 1s and 4 s from the time of projection. The angle of projection of the particle is30^{o}40^{o}60^{o}53^{o} 1 2 3 4 29 / 50 The coordinates of a projectile projected from the surface of a planet are given by y = (4t- 2t^{2}) and x = 4t, where t is in second, x and y in m, point of projection is taken as the origin, horizontal and vertical direction as x and y-axes. The initial direction of motion of the particle makes an angle of 0 with an x-axis where 0 is45^{o}37^{o}30^{o}60^{o} 1 2 3 4 30 / 50 A body is moving in a circular path of radius 16 m. It completes one revolution in 8 seconds. Considering the speed to be uniform, the value of centripetal acceleration in m/s^{2} is 484/49 22/7 44/7 89/7 31 / 50 The angle made by vector √3i + j with the x-axis is45^{o}30^{o}60^{o}90^{o} 1 2 3 4 32 / 50 The x and y coordinates of a particle at any time t are given by x = 7t + 4t^{2}, and y = 5t where x and y are in m and t in s. The acceleration of the particle at 5 s is4 m/s^{2}6 m/s^{2}8 m/s^{2}10 m/s^{2} 1 2 3 4 33 / 50 A motor car is travelling at 30 m/s on a circular road of radius 500 m. It is increasing its speed at the rate of 2 m/s^{2}, what is its acceleration at the instant?2 m/s^{2}1.8 m/s^{2}2.7 m/s^{2}3.8 m/s^{2} 1 2 3 4 34 / 50 1 2 3 4 35 / 50 The horizontal range and maximum height attained by a projectile are R and H respectively. If a constant horizontal acceleration a = g/2 is imparted to the projectile due to wind, then its new horizontal range will be (R + H/2) (R + 2H) (R + H) (R + H/3) 36 / 50 A projectile is fired from level ground at an angle 0 above the horizontal. The elevation angle ɸ of the highest point as seen from the launch point is related to the relationtanɸ = 1/2 tanθtanɸ = tanθtanɸ = 1/4 tanθtanɸ = 4tanθ 1 2 3 4 37 / 50 A boat is sent across a river with a velocity of 8 km/h (relative to water). If the velocity of the boat with respect to the ground is 10 km/h, the river is flowing with a velocity 4 km/h 6 km/h 8 km/h 9 km/h 38 / 50 1 2 3 4 39 / 50 The component of velocity of a body moving in a circle of radius R in the radial direction is (symbols have the usual meaningZeroRω^{2}v^{2}/Rvω 1 2 3 4 40 / 50 A particle is projected horizontally making an angle 60° with initial velocity u_{0}. If the time taken by the particle to make an angle 45° from horizontal is 1.5 s, then u_{0} is nearest to 40 m/s 50 m/s 60 m/s 20 m/s 41 / 50 1 2 3 4 42 / 50 Out of the following set of forces the resultant of which cannot be 4 Newton 2 N and 2 N 2 N and 4 N 2 N and 6 N 2 N and 8 N 43 / 50 A car with a vertical windshield moves horizontally in a rainstorm at a speed of 40 km/h. The raindrops fall vertically with a constant speed of 20 m/s with respect to the ground. The angle at which raindrops strike the windshield istan^{-1}(2/3)tan^{-1}(3/2)tan^{-1}(5/9)tan^{-1}(2) 1 2 3 4 44 / 50 A particle is thrown with the speed u at an angle α with the horizontal. When the particle makes an angle ẞ with the horizontal, its speed will beu tan^{2} αu sinα cosecẞu sin^{2} α /cosẞu cosα.secẞ 1 2 3 4 45 / 50 An artillery piece that consistently shoots its shells with the same muzzle speed has a maximum range of R. To hit a target that is at an R/2 distance from the gun, the angle of projection must be 15 Degree 30 Degree 45 Degree 60 Degree 46 / 50 A bomber plane moves horizontally with a speed of 600 m/s and a bomb released from it strikes the ground in 80 s. The angle at which it strikes the ground will be (g = 10 m/s^{2})tan^{-1}(1/2)tan^{-1}(4/3)tan^{-1}(3/5)tan^{-1}(1) 1 2 3 4 47 / 50 A particle is moving along a circular path of radius 1 m. If its angular speed is 2л rad/sec, the angle between the centripetal acceleration and angular velocity (with respect to the centre) - Is 90 degree Keeps on changing Increases then decreases Decreases then increases 48 / 50 The equation of projectile on a certain planet is y=√3x-5x^{2}. The ratio of maximum height attained to the horizontal range is √3 √3/4 4/√3 2/√3 49 / 50 1 2 3 4 50 / 50 A fan makes 2400 rpm. If after switching off, it comes to rest in 10 seconds, then find the number of times it will rotate before it comes to rest, after it is switched off. (Assume constant retardation) 100 200 400 800 Your score isThe average score is 38% 0% Restart quiz next chapter : Laws of motion