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Physics NEET MCQ
Quiz 9
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
) 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)
Q.17
A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is .. [ CBSE-PMT 2011]
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a)60°
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d)45°
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b) tan⁻¹ (1/2)
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c)tan⁻¹(√3 / 2)
Explanation
As shown in figure α be the angle of elevation.According maximum height formula using formula for Range ∴ R/2=u2 / 2g Now tanα=H / ( R/2) Substituting values of H and R/2 we get tanα=1/2 α=tan⁻¹(1/2)Answer: (b)
Q.18
The height y and the distance x along the horizontal plane of projectile on a certain planet ( with no surrounding atmosphere) are given byy=8t - 5t2 meter and x=6t meter, where t is in seconds. the angle with which the projectile is projected is
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a) tan⁻¹(3/4)
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d)not obtainable form given data
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b) tan⁻¹(4/3)
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c)sin⁻¹(3/4)
Explanation
vx is component of velocity along x axis at time to=0 dx/dt=vx=6 vy is component of velocity along y axis at time to=0 dy/dt=vy=8-10t for t=0 vy=8 tanθ=vy / vxtanθ=8/6=4/3θ=tan⁻¹ (4/3)Answer: (b)
Q.19
A body has an initial velocity of 3m/s and acceleration of 1m/s2, normal to the direction of initial velocity. Then the velocity of the body 4 seconds after start is
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a) 7 m/s along initial velocity
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b) 7 m/s in the normal direction to initial velocity
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c)7 m/s midway between initial and normal direction
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d)5 m/sec at an angle of tan⁻¹(4/3) with the direction of initial velocity
Explanation
Initial velocity is in horizontal direction Vx=3 m/sInitial velocity along vertical=0 Acceleration is vertical, Final Velocity vertical=Vy=at=1×4=4 m/sThus magnitude of velocity at end of 4 sec=[ Vx2 +Vy2]1/2velocity=5 m/s direction is tanθ=Vy / Vxtanθ=4/3direction=tan⁻¹(4/3) Answer:(d)
Q.20
If vectors P, Qand R have magnitude 5,12 and 13 units and P + Q=R, the angle between Q and R is ..[ Hariyana CEET 1998]
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a)cos-15/12
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b)cos-15/13
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c)cos-112/13
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d)cos-17/13
Explanation
P + Q=RP=R - QP2=R2 + Q2 - 2PQcosΘ52=132 + 122 -2(13)(12)cosθ 25=169 + 144 - 312cosΘ-288=-312 cosΘcosΘ=288/ 312cosΘ=12/13Θ=cos⁻¹ (12/13) Answer:(c)
Q.21
If at a height of 40 m, the direction of motion of a projectile makes an angle of π/4 with horizontal, then its initial velocity and angle of projection are respectively [ Roorkee 1998]
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a) 30, ½ cos⁻¹(-4/5)
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b) 30, ½ cos⁻¹(-1/2)
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c) 50, ½ cos⁻¹(-8/45)
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d) 60, ½ cos⁻¹(-1/4)
Explanation
Let initial speed be u, and angle of projection be θ component of u along horizontal ux=uscosθcomponent of u along vertical uy=usinθAt height h=40 let velocity be Vcomponent of v along horizontal vx=vcosθcomponent of v along vertical vy=vsinθgiven angle π/4 Vy2=uy2 -2g(40) Now tanπ/4=vy / vx∴ vx=vyHorizontal component do not change thus ux=vx=vy∴ ux2=uy2 -2g(40)ux2 - uy2=-2g(40)u2cos2θ - u2sin2θ=-2ghu2 ( cos2θ - sin2θ)=-2gh( from identity cos2θ - sin2θ=cos2θ)u2 cos2θ=-2gh cos2θ=-2gh / u2 Now |cos2θ|≤ 1 -2gh ≤ u2 from given options u=50 m/s satisfy condition cos2θ=-2(10)(40) / (2500) cos2θ=-8/25 θ=½ cos⁻¹(-8/25) Answer:(c)
Q.22
If A=4i - 2j 6k and B=i -2j-3k, the angle which the A + B makes with the x-axis is
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a) cos⁻¹ ( 3/ √50)
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b) cos⁻¹ ( 4/ √50)
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c) cos⁻¹ ( 5/ √50)
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d) cos⁻¹ ( 12/ √50)
Explanation
R=A+B=5i -4j +3k Let θ is angle made by R with x-axis then cosθ=x/|R||R|=[25+16+9]1/2|R|=√50cosθ=5 / √50 θ=cos⁻¹ ( 5 / √50) Answer:(c)
Q.23
A boy is running on the plane road with velocity v with a long hollow tube in his hand. The water is falling vertically downwards with velocity u. At what angle to the vertical he must incline the tube so that the water drops enter it without touching its sides?
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a) tan- (v/u)
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b) sin⁻¹(v/u)
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c)tan⁻¹(u/v)
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d)cos⁻¹(v/u)
Explanation
Velocity of rain=V , vertically down velocity of boy=u , horizontal ( say towards right) From vector diagram the boy should inclined the tube in a direction θ so that rain drops enter the tube. Then angle θ=tan⁻¹ (u/V)Answer: (c)
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