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Physics NEET MCQ
Motion In Two And Three Dimensions Mcq
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Q.1
A particle moves in a plane with a constant acceleration in a direction different from the initial velocity. The path of the particle is a
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a) straight line
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b) arc of a circle
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c)parabola
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d)ellipse
Explanation
Answer: (c)
Q.2
At the top of the trajectory of a projectile the directions of its velocity and acceleration are
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a) parallel to each other
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b) inclined at an angle of 45° to the horizontal
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c)perpendicular to each other
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d)none of the above statement is correct
Explanation
Answer:(c)
Q.3
The angle between two vectors of magnitude 12 and 18 units when their resultant is 24 units is ... [ CBSE-PMT 1999]
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a) 63°51'
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b) 75°52'
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c)82°31'
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d)89°16'
Explanation
We know that resultant vector R2=A2 + B2 + 2ABcosθ242=122 + 182 + 2(18)(12)cosθ cosθ=108/432 θ=75°52'Answer: (b)
Q.4
If A=2i + 3j and B=i + 4j + k , then unit vector along (A+B) is
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a)(3i + 7j + k) / √59
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b) (2i + 3j ) / √59
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c)(i + 4j + k )/ √18
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d)(2i + 3j)/ √13
Explanation
Unit vector along (A+B)=(A+B) / |A+B|Answer: (a)
Q.5
if a unit vector is represented by 0.5i + 0.8j + ck, the value of c is .. [ CBSE-PMT 1999]
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a) 1
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b)√(0.11)
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c)√(0.01)
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d)0.39
Explanation
Magnitude of unit vector=1 1=[ (0.5)2 + (0.8)2 + c2]1/2 c2=0.11 c=√(0.11) Answer:(b)
Q.6
A 5000 kg rocket is set for firing. The exhaust speed is 800m/s. To give an initial upward acceleration of 20 m/s2, the amount of gas ejected per second to supply the needed thrust will be [AFMC1997]
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a) 137.5 kg/sec
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b) 185.5 kg/sec
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c) 125 kg/sec
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d) 187.5kg/sec
Explanation
In given case mass of rocket is changing with time, but velocity of gases remained constant. Therefore Thrust=v (dm/dt) -eq(1) Where dm/dt is rate at which mass of gas is ejected. Resultant acceleration = Acceleration due to thrust (up) - acceleration due to gravity(down) 20 = a - 10 Or a = 30 is the acceleration produced by thrust According newton's second law of thrust=(mass)×(acceleration)...(2) From equation (1) and (2) we get ma=v ( dm/dt) 5000 × 30=800 ( dm/dt ) On solving equation for (dm/dt)we get (dm/dt)=187.5 kg/sec Answer:(d)
Q.7
A particle revolves around a circular path, the centripetal acceleration of the particle is inversely proportion to [AFMC1997]
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a) mass of particle
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b) radius of particle
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c) velocity of particle
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d) both (a) and (c)
Explanation
Centripetal acceleration acac=v2 / r So acceleration is inversely proportional to radius of the circular path. Answer:(b)
Q.8
What is the dot product of two vectors of magnitude 3 and 5, if the angle between them is 60o? [AFMC 1997]
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a) 5.2
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b) 7.5
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c) 8.4
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d) 8.6
Explanation
Dot product of vector A and B is Dot product=3×5×cos60=7.5 Answer:(b)
Q.9
If an iron ball and wooden ball of the same radius are released from a height h in vacuum the time taken by both of these to reach to ground is [ AFMC1998]
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a) roughly equal
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b) exactly equal
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c) unequal
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d) zero
Explanation
Both the balls are released from same height and in vacuum. Hence there is no buoyant force Hence acceleration for both the balls will be same, both have same initial velocity zero. Therefore both the balls will reach ground at the same same Answer :(b)
Q.10
If two numerically equal forces P and Q acting at a point produce a resultant force of magnitude P then the angle between the two original forces is [AFMC1998]
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a)120o
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b) 90o
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c) 0o
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d) 45o
Explanation
Since magnitude of all vectors is same let it be R then R2=R2+ R2+2R×Rcosθ R2=2R2+2R2cosθSolving above equation we get cosθ=(-1/2)θ=120o Answer:(a)
Q.11
A stone is dropped from a certain height which can reach the ground in 5 seconds. If stone is dropped and stopped after 3 sec of fall, and then allowed to fall again, then the time taken by the stone to reach the ground for the remaining distance is.. [AFMC1998]
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a) 3 sec
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b) 4 sec
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c) 2 sec
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d) none of these
Explanation
Initial velocity of stone 'u'=0 Time taken to reach ground 't'=5 sec gravitational acceleration 'g'=10 m/s2 Using formula for free fall we can calculate total height 'h' h=½(gt2) -eq(1) By substituting above quantities in equation for 'h' and solving for 'h' we get h=125 m Similarly distance traveled by stone in 3 sec after the fall=½(10×9)=45 m Now stone is drop again with initial velocity zero, from the height=125-45=80m Let time taken by the stone to reach the ground be 't' Substituting value of h=80m and g=10 we get 80=½(10×t2) t=4 sec Answer:(b)
Q.12
An aeroplane is moving with a horizontal velocity u at a height h . The velocity of a packet dropped from it on the earth`s surface will be [AFMC 1999]
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a)(u2 - 2gh )½
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b) 2gh
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c)( gh )½
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d) (u2 + 2gh )½
Explanation
When packet is dropped from the plane, packet will have only horizontal velocity and acceleration along horizontal direction zero packet will have vertical velocity zero, but acceleration will be vertically down 'g' When packet strikes the surface velocity along horizontal will not change, as there is no air resistance it it be 'u' Vertically downward velocity can be calculated by using formula for velocity v2=u2+2gh Vertical initial velocity u=0, height is h substituting values in equation for velocity we get V=(2gh)½ Hence when packet strikes the surface it will have two component of velocity u along horizontal direction And vertical component v Resultant velocity=(u2 + v2 )½ On substituting the value of v in above equation Resultant velocity=(u2 + 2gh )½ Answer :(d)
Q.13
A bomb is dropped from an aeroplane moving horizontally at constant speed. If air resistance is taken in to consideration , then the bomb [AFMC1999]
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a) falls on earth exactly below the aeroplane
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b) falls on the earth exactly behind the aeroplane
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c) falls on the earth ahead of aeroplane
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d) flies with the aeroplane
Explanation
If there is no air resistance bomb will drop at a place exactly below the flying aeroplane, it is flying with constant velocity When we take air resistance in account horizontal velocity of bomb will reduce. So it will fall on surface behind the aeroplane Answer:(b)
Q.14
A stone is thrown with an initial speed of 4.9 m/s from a bridge in vertically upward direction . It falls down in water after 2 sec. The height of the bridge is [AFMC1999]
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a) 24.7 m
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b) 19.8 m
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c) 9.8 m
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d) 4.9 m
Explanation
Stone is thrown up against the gravitational acceleration we will take initial speed of stone 'u' negative gravitational acceleration positive Let h be the height of bridge h=ut + ½(gt2) h=-4.9(2) +½(9.8×22) h=9.8 m Answer :(c)
Q.15
Two vectors A and B are such that |C|=|A| + |B| If θ is the angle between positive direction of vector A and vector B then the correct statement is [AFMC 1999]
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a) θ=π
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b)2π/3
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c) θ=0
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d) θ=π/2
Explanation
Given that vector C is resultant of vector A and B According to formula for vector addition C2=A2 + B2 + 2ABcosθ Comparing C2=A2 + B2 with above equation cosθ=0 since magnitude of A and B is not zero ⇒θ=π/2 Answer: (d)
Q.16
The speed of a boat is 5 km/hr in still water. If it crosses river of width 1km along the shortest possible path in 15 minutes, the velocity of the river's water is
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a) 1km/hr
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b) 2km/hr
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c) 3km/hr
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d) 4km/hr
Explanation
Shortest possible path is crossing across width As shown in figure VRO is velocity of river with respect to observer VBO is velocity of boat with respect to observerVBR is velocity of boat with respect to river=5km/hrNow according to law of vector addition Here angle between VBO and VRO is 90o Now according to observer boat crossed river of width 1 km in 15 min or 0.25hr VBO=1/0.25=4km/hr By substituting the values in above equation we get 5 2=42 + VRO2 VRO=3 km/hr Answer: (c)
Q.17
A missile moves is fired for maximum range with initial velocity 20 m/s. If g=10 m/s2, the range of the missile is ... [ CBSE-PMT 2011]
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a)40 m
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b) 50m
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c)60 m
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d)20 m
Explanation
For maximum range, the angle of projection θ=45°Answer: (a)
Q.18
If a body A of mass M is thrown with velocity 'a' at an angle of 30° to the horizontal and another body B of same mass is thrown with same speed at angle of 60° to the horizontal the ratio of horizontal range of A and B will be.. [ CBSE-PMT 1992]
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a) 1 : 3
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b) 1:1
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c)1:√3
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d)√3 : 1
Explanation
Horizontal range is same when the angle of projection with horizontal is θ and (90° - θ)Answer: (b)
Q.19
From a 10 m high building a stone 'A' is dropped and simultaneously another identical stone 'B' is thrown horizontally with an initial speed of 5 m/s. which one of the following statement is true? [ CBSE-PMT 2002]
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a)It is not possible to calculate which one of the two stones ill reach the ground first
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b) Both the stones ('A' and 'B') will reach the ground simultaneously.
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c)'A' stone reach the ground earlier than 'B'
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d)'B' stone reach the ground earlier than 'A'
Explanation
In both the cases downward component of initial velocity is zero, thus both are free fall.Both A and B will reach ground at same time Answer:(b)
Q.20
A particle of mass m is projected with velocity 'v' making an angle of 45° with the horizontal . When the particle lands on the level ground the magnitude of the change in its momentum will be.. [ CBSE- pmt 2008]
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a) 2mv
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b) (mv) / √2
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c) mv√2
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d) zero
Explanation
When projectile reaches the ground direction of horizontal component of velocity don't change but direction of vertical component of velocity changes by 180° ∴ no change in horizontal component of momentum Now initial vertical component of momentum=mvsinθ=(mv)/ √2 Final vertical component ( now down word) momentum=-mvsin θ=(-mv)/ √2 Change in momentum=Final momentum - Initial momentum Change momentum=(mv)/ √2 - [ (-mv)/ √2] Change momentum=mv√2 Answer: (c)
Q.21
The circular motion of a particle with constant speed is .. [ CBSE-PMT 2005]
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a)periodic but not simple harmonic
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b) simple harmonic but not periodic
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c)periodic and simple harmonic
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d)neither periodic nor simple harmonic
Explanation
In circular motion of a particle with constant speed, particle repeats its motion after regular interval of time but does not oscillate about fix point. So, motion of particle is periodic but not simple harmonic.Answer: (a)
Q.22
A particle starting from the origin (0,0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are (√3, 3). The path of the particle makes with the x-axis an angle of .. [ CBSE-PMT 2007]
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a) 45°
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b) 60°
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c)0°
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d)30°
Explanation
Let θ be the angle made by the particle with x-axis as shown in figuretan θ=y co-ordinates / x - coordinate tanθ=3 / √3=√3thus θ=60°Answer: (b)
Q.23
if vector A and B are perpendicular to each other then value of α is .., [ CBSE-PMT 205]
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a)1/2
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b) -1/2
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c)1
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d)-1
Explanation
Given two vectors are perpendicular to each other therefore dot product of vector A and Vector B will be zero-8+12+8α=0α=-1/2 Answer:(b)
Q.24
Two boys are standing at the ends A and B of ground where AB=a. The boy at B starts running in a direction perpendicular to AB with velocity vThe boy at A starts running simultaneously with velocity 'v' and catches the other boy in time t, where t is.. [ CBSE-PMT 2005]
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a)
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b)
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c)
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d)
Explanation
Velocity of A relative to B is given by Taking x-component of above equation0=vsinθ - v1 sin θ=v1 / v Taking y components vy=v cosθ Time taken by boy at A to catch the boy at B is given by equationt=Relative displacement along Y axis / Relative velocity along Y axis t=a / v cosθ --eq(1) now cosθ=√(1-sin2θ) but sinθ=v1 / v Thus sin2θ=v12 / v2 substituting in eq(1) we get Answer: (d)
Q.25
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces... [ CBSE-PMT 2003]
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a)can not predicted
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b) are equal to each other
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c)are equal to each other in magnitude
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d)are not equal to each other in magnitude
Explanation
let vector P be the vector sum of vector A and Vector BLet Vector Q be the vector difference of vector A and Vector B Since vector P and Q are perpendicular there dot product will be zeroAnswer: (c)
Q.26
the position vector of a particle is r=(acosωt)i + (asinωt)j. The velocity of the particle is .. [ CBSE-PMT 1995]
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a) directed towards the origin
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b) directed away from origin
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c)parallel to the position vector
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d)perpendicular to the position vector
Explanation
v=dr/dt=(-aωsinωt)i + ( aωsinωt)j Now slope of position vector=aωsinωt / aωcosωt=tanωtslope of velocity vector=-aωcosωt / aωsinωt=-1/ tanωt ∴ velocity vector is perpendicular to displacement vector Answer: (d)
Q.27
Two projectiles are fired from the same point with same speed at an angle of projection 60° and 30° respectively. Which one of the following is true? [ CBSE-PMT 200]
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a) Their maximum height will be same
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b) Their range will be same
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c)Their landing velocity will be same
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d)There time of flight will be same
Explanation
We know that range of two projectile having angle of projection θ and (90-θ) is same. Answer:(b)
Q.28
A ball whose kinetic energy is E, is thrown at an angle of 45° with the horizontal. It's kinetic energy at the highest point of its flight will be..[ CBSE-PMT 1997]
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a) E
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b) E/2
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c) E/ √2
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d) 0
Explanation
Let v be the initial velocity Thus Initial kinetic energy E=½ m v2 At highest point Vertical velocity of ball will be zero. horizontal velocity will be vcosθ=v/(√2) Now kinetic energy at the highest point E'=½ m (v /√2) 2 E'=½ m v × (1/2) E'=E/2 Answer: (b)
Q.29
A person swims in river aiming to reach exactly opposite point on the bank of river. His speed of swimming is 0.5 m/s at an angle 120° with the direction of flow of water. The speed of water in stream is ... [ CBSE-PMT 1999]
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a)1.0 m/s
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b) 0.5 m/s
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c)0.25 m/s
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d)0.43 m/s
Explanation
Let VSR be the velocity of swimmer with respect to river=0.5 m/s VSO be the velocity of swimmer with respect to observer on bankVRO be the velocity of river with respect to observer on bank Now from the geometry of figure sin30=VRO / VSR1/2=VRO / 0.5 VRO=0.5/2=0.25 m/sAnswer: (c)
Q.30
A particle moves with a velocity given by vector v=(6i -4j +3k) m/s, under the influence of a constant force F=20i + 15j - 5k N. The instantaneous power applied to the particle is .. [ CBSE-PMT 2000]
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a) 45 J/s
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b) 35 J/s
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c) 25 J/s
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d) 195 J/s
Explanation
P=F. v=(6i-4j+3k) . (20i + 15j - 5k) P=45 J/s Answer: (a)
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