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Physics NEET MCQ
Motion In Two And Three Dimensions Mcq
Quiz 5
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Q.1
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.2
The resultant of two vectors (A+B) and (A - B) is (A2 + B2 )1/2 , then the angle between the vectors is
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a)
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b)
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c)
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d)
Explanation
Let P=A + B P2=A2 + B2 + 2ABcosΦ Q=A - B Q2=A2 + B2 - 2ABcosΦ PQ=A2 - B2 Given resultant of P and Q is (A2 + B2 )1/2 P2 + Q2 +2PQcosθ=A2 + B2 Substituting values we get A2 + B2 + 2ABcosΦ+A2 + B2 - 2ABcosΦ +2PQcosθ=A2 + B2 2(A2 + B2) + 2PQcosθ=A2 + B2 2PQcosθ=-(A2 + B2)cosθ=-(A2 + B2) / 2PQSubstituting value pf PQ we get cosθ=-(A2 + B2) / [2(A2-B2)]Answer: (d)
Q.3
A ball rolls off the top of a staircase with a horizontal velocity u m/s. if the steps are h m high and W m wide the ball will hit the edge of the nth step is
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a)
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b)
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c)
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d)
Explanation
let the boll hit on n step Horizontal displacement=nWHorizontal displacement=utut=nw t=nw/u Vertical displacement=nh vertical displacement=½ (g) t2 nh=½ (g) t2substituting value of time t nh=½ (g) (nw/u)2on solving for n we get option(b)Answer: (b)
Q.4
A force F=- K(yi + xj) where K is positive constant ) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a,0) and then parallel to y-axis to the point (a,a). The total work done by the force F on the particle is [ IIT 1998]
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a)-2Ka2
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b) 2Ka2
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c)-Ka2
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d)Ka2
Explanation
Force is position dependent thus particle starting point is (0,0) and end pint is (a,a) thus equation for line is x=y and dx=dy substituting above values in equation we getAnswer: (c)
Q.5
Two like parallel forces P and 3P are 40cm apart. If the directions of P is reversed, then their resultant shifts through a distance .. [ Roorkee 1998]
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a) 30 cm
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b) 40 cm
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c)50 cm
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d)60 cm
Explanation
Case I when forces are parallel let resultant force be at distance x from force Pp(x)=3P(40-x)x=30 ( 30cm from P)Now force P is reversed let the resultant force be at distance y from force P -p(y)=3P(40-y) y=60 cm resultant shift through distance of (60-30)=30cm Answer: (a)
Q.6
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.7
Two balls are projected making an angle of 30° and 45° respectively with the horizontal. if both have same velocity at the highest points of their path, then the ratio of their velocities of projection is [ C.S.E.P 2001]
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a) √3 : √2
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b) √2 : 1
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c) √2 : √ 3
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d) √3 : 2
Explanation
let u be the velocity of first ball horizontal component=ucos30 let v be the velocity of second ball horizontal component=vcos45 given ucos30=vcos45 thus u/v=cos45/cos30 Answer: (c)
Q.8
If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the stone starts, then what is the value of initial speed u if stone is launched at an angle θ [ Hariyana C.E.T 2002]
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a)
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b)
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c)
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d)
Explanation
equation for path of projectile substituting y=h in above equation and solving for u we getAnswer: (b)
Q.9
Let F be the force acting on a particle having position vector r and T be the torque of this force about the origin . Then.. [ AIEEE2003]
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a) r⋅T=0 and F⋅T ≠ 0
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b) r⋅T ≠ 0 and F ⋅ T=0
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c)r⋅T ≠ 0 and F⋅T ≠ 0
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d)r⋅T=0 and F⋅T=0
Explanation
Torque is perpendicular to plane containing r and F thus dot product of r and T is zero and dot product of F and T is zero. option (d) is correctAnswer: (d)
Q.10
Three forces start acting simultaneously on a particle moving with velocity. These forces are represented in magnitude and direction by three sides of triangle as shown in figure. The particle will move with velocity
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a) less than v
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b) greater than v
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c)|v| in the direction of the largest force
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d)v remains unchanged
Explanation
Since three forces acting on the particle are represented by the three sides of a triangle taken in one order, their resultant is zero. So the particle velocity remains unchanged Answer:(d)
Q.11
A block is dragged on a smooth plane with the help of a rope which moves with velocity v as shown in figure. The horizontal velocity of the block is
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a) v
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b) vsinθ
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c) v/sinθ
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d) v/cosθ
Explanation
As shown in adjacent figure. h2 = x2+y2 Note y is constant. By taking derivative of above equation we get Answer: (c)
Q.12
A body is projected at time t=0 from a certain point on a planet surface with a certain velocity at a certain angle with the planet,s surface ( assumed horizontal ). The horizontal and vertical displacement x and y (in meters) respectively vary with time t (in second) as x=10t√3 and y=10t-tWhat is the magnitude and direction of the velocity with which the body is projected?
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a)20m/s at an angle of 30° with the horizontal
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b)20m/s at an angle of 60° with the horizontal
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c)10m/s at an angle of 30° with the horizontal
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d)10m/s at an angle of 60° with the horizontal
Explanation
we know that the position co-ordinates x and y are given by x=(vocosθ)t comparing above equation with given equation x=10√3 (t) we get vocosθ=10√3 --(1) y=vosinθt - ½ gt2 comparing given equation y=10t - t2 with standard equation we get vosinθ=10 --(2) by taking ratio of equation 2 to equation 1 we get tanθ=1/√3 θ=30° Substituting value of θ in equation (1) we get vocos30=10√3 vo (&radic3 /2)=10√3 vo=20 m/s Answer: (a)
Q.13
The area of the parallelogram formed from the vectorsA=i -2j + 3k and B=3i - 2j + k as adjacent sides is
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a) 4√3
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b) 4√6
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c)8√3
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d)8√6
Explanation
We know that |A×B|=Area of parallelogram Answer: (b)
Q.14
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.15
The angle between A and B is θ. R=A + B makes an angle θ/2 with A. Which of the following is true?
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a) A=2B
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b) 2A=B
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c) AB=1
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d) none of the above
Explanation
Angle made by the resultant vector with A given by None of the above options , β comes out to θ/2 Answer: (d)
Q.16
A particle is projected from a point O with velocity u in a direction making an angle α upward with the horizontal. At P, it is moving at right angle to its initial direction of projection, its velocity at P is ..
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a)u tanα
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b) ucotα
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c)ucosecα
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d)usecα
Explanation
As shown in figure component of u along x-axis is ucosα At point P it changes its direction by 90° let velocity be v Then component of v along x-axis is vcos(90-α) = vsinα Thus ucosα = vsinα V = ucotα Answer: (b)
Q.17
The velocity of a particle P moving freely under gravity is 4.9 m/s, the direction being 30° with the downward normal
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a) its acceleration normal to the direction of motion at P=9.8 m/s2
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b) the radius of curvature of P of the parabolic trajectory of particle is 4.9m
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c)the particle has no acceleration normal to the direction of motion
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d)the radius of curvature at P of the path depends upon the initial velocity of projection
Explanation
As shown in figure centripetal acceleration ac=gcos60 ac=9.8( 1/2)=4.9 m/sec2Centripetal acceleration is given by V2 /r V2 /r=4.9 (4.9)2 / r=4.9r=4.9 mAnswer: (b)
Q.18
A particle is projected up an inclined as shown in figure. For maximum range over the inclined plane the value of θ should be
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a) 45°
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b) 15°
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c)30°
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d)60°
Explanation
Angle made by u with horizontal α for maximum range α=(45 + β/2) Here β is angle made by the inclined plane with horizontal In given problem α=θ +30 and β=30 (45 + 30/2)=θ +30 θ=30° Answer:(c)
Q.19
A particle located at x=0 at time t=0, starts moving along with the positive x-direction with velocity 'v' that varies as v=α√x. The displacement of the particle varies with time as [ AIEEE 2006]
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a)t2
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b) t
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c)√t
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d)t3
Explanation
Answer: (a)
Q.20
A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is [ AIEEE 2007]
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a) K/2
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b) K
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c)zero
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d)K/4
Explanation
Horizontal velocity v'=vcos60=v/2Thus kinetic energy at top=K/4Answer: (d)
Q.21
The velocity of a particle is v=vo +gt + ftIf its position is x=0 at t=0, then its displacement after unit time (t=1) is [ AIEEE 2007]
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a) vo + g/2 + f
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b) vo + 2g + 3f
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c)vo + g/2 + f/3
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d)vo + g + f
Explanation
Answer:(c)
Q.22
A particle has an initial velocity of 3i + 4j and an acceleration of 0.4i+0.3j . Its speed after 10s is [ AIEEE 2009]
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a) 7√2 units
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b) 7 units
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c) 8.5 units
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d) 10 units
Explanation
Given u=3i + 4j, a=0.4i+0.3j, t=10 s v=u + at v=3i + 4j + (0.4i+0.3j)×10=7i + 7j |v|=7√2 units Answer: (a)
Q.23
If the magnitudes of A, B and C are 12, 5 and 13 units respectively and A + B=C, then the angle between A and B is
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a)0
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b) π
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c)π/2
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d)π/4
Explanation
Answer: (c)
Q.24
If A=B + C and the magnitude of A, B and C are 5,4,3 units, the angle between A and C is .. [ CBSE 1990]
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a) cos-1 (3/5)
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b) cos-1(4/5)
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c)π/2
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d)sin-1(3/4)
Explanation
Answer: (a)
Q.25
Out of the following sets of forces, the resultant of which cannot be zero [ CPMT 1985]
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a) 10,10,10
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b) 10,20,40
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c)10,10,20
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d)10,20,20
Explanation
Answer:(b)
Q.26
Out of following pairs, the resultant of which cannot be 4 Newton [ CPMT 1985]
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a) 2N and 2N
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b) 2N and 6 N
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c) 2N and 4N
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d) 2N and 8N
Explanation
Answer: (d)
Q.27
A person moves 30 m North, then 20m East then 30√2 m South-West. His displacement from the original position is [ CPMT 1989]
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a)15 m East
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b) 28 m South
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c)10 m West
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d)14 m South-West
Explanation
Answer: (c)
Q.28
The length of seconds hand of watch is 1cm. the change in velocity of its tip in 15 seconds is [ MPPMT 1987]
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a) zero
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b) π/ (30√2) cm/sec
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c)(π/30) cm/sec
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d)2π/(30√2) cm/sec
Explanation
Answer: (d)
Q.29
Consider a F=4 i - 3 j. Another vector perpendicular to F is ..[ MPPMT 1987]
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a) 4i+3j
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b) 6i
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c)7k
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d)3i - 4j
Explanation
Answer:(c)
Q.30
Consider a F1=4 i + 5 k and F2=3 j - 4 k. The magnitude of the scalar product of these vector is [ MPPMT 1987]
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a) 20
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b) 23
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c) 5√(33)
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d) 26
Explanation
Answer: (a)
0 h : 0 m : 1 s
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