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Physics NEET MCQ
Motion In Two And Three Dimensions Mcq
Quiz 9
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Q.1
Component of 3i+4j perpendicular to i+j and in the same plane as (3i + 4j) is
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a) (1/2) (j - i)
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b) (3/2) (j - i)
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c)(5/2) (j - i)
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d)(7/2) (j - i)
Explanation
Vector perpendicular to (i +j) is (i- j) Componenet of A along B is ∴ component of 3i+4j along i- j is Answer: (a)
Q.2
An aeroplane is flying at a constant speed v on a circular path. Change in velocity, after the aeroplane has described an angle of 60°
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a) v/3
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b) v/2
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c)v
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d)2v
Explanation
Initial velocity V1 have only vertical component Final velocity V2 have horizontal and vertical components Vsin60=(v√3)/ 2 and vcos60=v/2 respectively Thus chage in Vertical component of velocity=V/2 - v=-v/2 change in horizontal component of velocity=0 - (v√3)/ 2=-(v√3)/ 2 Thus magnitude of change in velcoity Answer:(c)
Q.3
A helicopter is flying at 36√2 km/hr towards North-East. The wind is blowing at 36 km/hr towards south. The displacement of helicopter in 2 hrs is
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a) 36 km
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b) 72 km
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c) 84 km
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d) 96 km
Explanation
from diagram resultant velocity of helicopter is 36km/hr . Thus distance travelled in 2 hr is=36 × 2=72 km Answer: (b)
Q.4
A is point on wheel rolling on a horizontal road. The radius of the wheel is R. Initially the point A is in contact with the ground. The wheel rolls through half the revolution. The displacement of point A is
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a)πR
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b) 2πR
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c)
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d)
Explanation
Initial and final position of the point A are shown in figureThe distance travelled in half revolution is=πR so displacement=Answer: (d)
Q.5
A ball is projected from the ground at a speed of 10 m/s making an angle of 30° with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the second ball if they collide at maximum height is
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a) 1.0 m
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b) 1.25 m
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c)2.0 m
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d)2.5 m
Explanation
Time taken by the first ball to reach maximum height=usinθ/g=10sin30/10=1/2 second Second ball is released for a point on the verical line , this second ball fall in (1/2)s. such that it collied first at maximum height . Thus Initial height of second ball=maximum height of first ball + height fallen by second ball= Answer: (d)
Q.6
A ball is thrown at different angles with same speed u and from the same point. It has same range in both the cases. If y1 and y2 are maximum height attained, then y1 + y2 is
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a) u2 / g
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b) 2u2 / g
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c)u2 / 2g
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d)u2 / 4g
Explanation
As range is same , the angle of projection must be θ and 90-θ. So Answer:(c)
Q.7
Two particles of same mass are projected from same place with same velocity u, such that their ranges are same. IF h1 and h2 are the maximum heights attained by them, then the ralation between h1, h2 and R is
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a) R=h1h2
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b) R2=16h1h2
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c) R2=h1/h2
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d) R2=(h1/h2)2
Explanation
Answer: (b)
Q.8
If vectors are functions of time, then the value of t at which they are orthogonal to each other is …. [ ReAIPMT 2016 ]
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a) t = 0
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b) π/4ω
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c) π/2ω
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d) π/ω
Explanation
When two vectors are orthogonal they are perpendicular A ∙ B = 0 From trigonometric identity Answer:(d)
Q.9
Two particles A and B, move with constant velocities v1 and vAt the initial moment their position vectors are r1 and r2 respectively. The condition for particle A and B for their collision is ..
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a)
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b)
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c)
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d)
Explanation
For two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle. Direction relative position of particle 1 with respect to 2 is given by Direction of relative velocity of particle 1 with respect to 2 is given by If particle has to collide direction of position and direction of relative velocity must be opposite Answer:(b)
Q.10
A particle of moving such that its position coordinates (x, y) are (2m, 3m) at time t = 0, (6m, 7m) at time t = 2s and (13m, 14m) at time t = 5s. Average velocity vector V from t = 0 to t = 5s is…..[AIPMT 2014 ]
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a)
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b)
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c)
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d)
Explanation
Total time 5 s Final position =(13m,14m) Initial position = (2m, 3m) Change in position = (13 -2, 14 -3) = (11m, 11 m) Velocity = change in position / time Answer:(b)
Q.11
The diagram shows the variation of 1/V ( here V is velocity of the particle) with respect to time. At time t=3s, using the details given in the graph, the instantaneous acceleration will be equal to
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a) -2 m/s2
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b) +3 m/s2
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c) +5m/s2
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d) -6 m/s2
Explanation
Motion is accelerated Derivative of 1/V with time which will be equal to slope of straight line, or tan45 = -1 ( since angle of 45 is with negative x axis) Now dV/dt = a , acceleration From equation eq(2) Answer:(b)
Q.12
The velocity of a projectile at the initial point A is (2i + 3j) m/s. Its velocity (in m/s) at point B is…… [AIPMT/NEET partI 2016]
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a) -2i-3j
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b) -2strong>i + 3j
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c) 2strong>i – 3j
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d) 2strong>i+3j
Explanation
At point B x-component remains same as of at A, but Y component is changed by 180° . Therefore velocity at B is 2i – 3j Answer:(c)
Q.13
A particle of moving such that its position coordinates (x, y) are (2m, 3m) at time t = 0 (6m, 7m) at time t = 2s and (13m, 14m) at time t = 5s. Average velocity vector vav from t = 0 to t = 5s is
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a)
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b)
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c)
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d)
Explanation
Displacement vector = 11i+11j Answer:(b)
Q.14
A projectile is fired from the surface of the earth with a velocity of 5ms–1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms–1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms–2) is (given g = 9.8ms–2) …. [ AIPMT 2014]
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a) 16.3
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b) 110.8
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c) 3.5
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d) 5.9
Explanation
Same angle , same trajectory means same range g’= 3.528 ≈ 3.5 ms-2 Answer:(c)
Q.15
A ship A is moving Westwards with a speed of 10 km h-1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h-The time after which the distance between them becomes shortest, is : …[ NEET 2015]
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a) 10√2 h
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b) 0 h
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c) 5h
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d) 5√2 h
Explanation
Let t be the time when distance between the ships be minimum, Thus ship travelled 10t distance from its initial point and ship is (100-10t) for minimum h, dh/dt =0 400t=2000 ⇒ t= 5hr Answer:(c)
Q.16
osition vector of a particle R ⃑ as a function of time is given by :- Where R is in meters, t is in seconds and i ̂ and j ̂ denote unit vectors along x and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?
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a) Path of the particle is a circle of radius 4 meter
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b) Acceleration vectors is along -R
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c) Magnitude of acceleration vector is v2/R where v is the velocity of particle
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d) Magnitude of the velocity of particle is 8 meter/second
Explanation
Is equation of circle having radius4 option a is correct Magnitude of v = 8π√2 option d incorrect Object is following circular mnotion thus option c is correct Answer:(d)
Q.17
A particle moves so that its position vector is given by . Where ω is a constant. Which of the following is true ?
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a) Velocity and acceleration both are perpendicular to r .
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b) Velocity and acceleration both are parallel to vector r
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c) Velocity is perpendicular to vectorr and acceleration is directed towards the origin
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d) Velocity is perpendicular to vector r and acceleration is directed away from the origin
Explanation
Question is about directions of velocity, acceleration perpendicular, towards thecentre By taking derivativeof "r" we will get velocity, and derivative of velocity will give acceleration Now if dot product of position vector and velcoity vector is zero then they are perpendicular If cross productis zero they are parallel since r . v =0 Thus vector r and vector v are perpendicular Negative sign of vector a shows acceleration is directed towards the origin Answer:(c)
Q.18
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is :-
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a) 0°
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b) 90°
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c) 45°
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d) 180°
Explanation
2ABcosθ =0 ⇒ θ = π/2 Answer:(b)
Q.19
If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is :- …[ AIPMT 2015]
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a)
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b)
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c)
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d)
Explanation
Answer:(c)
Q.20
The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is a) b) c) d)
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a) 0
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b) 5 m/s2
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c) –4 m/s2
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d) –8 m/s2
Explanation
Since acceleration is constant acceleration is -4m/2 Answer:(c)
Q.21
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time tOn other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2 . The time taken by her to walk up on the moving escalator will be …[ NEET 207]
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a)
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b)
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c)
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d)
Explanation
Velocity of Preeti with respect to elevator Vpe then Velocity of elevator with respect to observer Veo then All vectors have same directions thus Vpo = Vpe + Veo if t is the time taken by Preeti to walk on moving elevator then Answer:(c)
Q.22
A Projectile is given an initial velocity of ( i+2j) m/s, + where i is along the ground and j is along the vertical. If g = 10 m/s2, the equation of its trajectory is … [ IIT Mains 2013]
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a) y = x − 5x2 (2) (3) (4)
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b) y = 2x − 5x2
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c) 4y = 2x − 5x2
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d) 4y = 2x − 25x2
Explanation
Equation of trajectory y= 2x -5x2 Answer:(b)
Q.23
) Three vectors P ,Q and R are shown in the figure. Let S be any point on the vector R. The distance between the points P and S is bR. The general relation among vectors P,Q and S is …[ IIT Advance 2017]
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a)
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b)
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c)
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d)
Explanation
Answer:(c)
0 h : 0 m : 1 s
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