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Motion In Two And Three Dimensions Mcq
Quiz 2
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Q.1
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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b) tan-1 (1/2)
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c)tan-1(√3 / 2)
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d)45°
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.2
The vector A and B are such that|A + B |=| A - B | The angle between the two vectors is .. [ CBSE-PMT 2006]
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a) 60°
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b) 75°
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c)45°
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d)90°
Explanation
|A + B |2=| A - B |2 A2 + B2 + 2ABcosθ=A2 + B2 - 2ABcosθ 4ABcosθ=0 ∴ θ=90° So angle between vector A and B is 90°Answer: (d)
Q.3
The magnitude of vectors A, B, C are 3, 4 and 5 units respectively. If A + B=C then the angle between A and B is .. [ CBSE-PMT 1998]
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a) π /2
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b) cos-10.6
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c)tan-1 7/5
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d)π/4
Explanation
from given (A + B)2=C2A2 + B2 +2ABcosθ=C232 + 42 +2 (3)(4) cosθ=52∴ 2ABcosθ=0 cosθ=0thus angle between A and B is 90° Answer:(a)
Q.4
A × B|=AB √3, then the value of |A+B| is .. [ CBSE-PMT 2004]
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a) (A2 + B2 + AB√3) 1/2
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b) (A2 + B2 + AB) 1/2
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c) (A2 + B2 + AB/√3) 1/2
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d) A + B
Explanation
| A ×B|=ABsinθ A . B ABcosθ given |A × B|=AB √3ABsinθ=√3 ( ABcosθ)tanθ=√3 ∴ θ=60° ∴ | A + B|=(A2 + B2 + 2ABcos60) 1/2| A + B|=(A2 + B2 + AB) 1/2 Answer: (b)
Q.5
If the angle between the vectors A and B is θ, the value of the product (B ×A) . A is equal to ... [ CBSE-PMT 2005]
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a)BA2sinθ
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b) BA2cosθ
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c)BA2sinθcosθ
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d)zero
Explanation
B ×A=C Direction of vector C is perpendicular to both A And B Now C . A=CBCos90Thus (B ×A) . A=0Answer: (d)
Q.6
A bullet is fired from a gun with speed of 1000 m/s in order to fit a target 100 m away. At what height above the target should the gun be aimed? ( resistance of air is negligible and g=10 m/s2 [CBSE-PMT 1995]
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a) 5cm
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b) 10cm
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c)15cm
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d)20cm
Explanation
initial velocity of Bullet along vertical direction is zeroTime taken by bullet to cover horizontal distance t=Distance / speed of Bullett=100 / 1000=0.1secDuring this time, bullet will fall down due to gravitational accelerationheight 'h'=ut + ½ g t2h=½ 10(0.1) 2 h=0.05 m=5cmAnswer: (a)
Q.7
for angle of projection of a particle (45° - θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio [ CBSE-PMT 2006]
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a) 1 : 3
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b) 1 : 2
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c)2 : 1
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d)1 : 1
Explanation
(45° - θ) and (45° + θ) are complementaryWe know that angle of projection of two projectile makes complementary angles, their ranges are equal. In this case, the range will be same the ratio is 1:1 Answer:(d)
Q.8
A bus is moving on a straight road towards north with uniform speed of 50 km/h turns through 90° . If the speed remains un changes after turning the increase in the velocity of bus in the turning process is .. [ CBSE-PMT 1989]
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a) 70.7 km/h along south-west direction
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b) zero
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c) 50 km/h along west
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d) 70.7 km/h along north-west direction
Explanation
Let v1 due North Let V2 due west Angle between velocity is 90° Change in velocity| V2 - V1| change in velocity=( v22 + v12) 1/2 change in velocity=(502 + 502) 1/2Change in velocity=70.7Direction of this change in velocity is in South-West Answer: (a)
Q.9
A boat is sent across a river with velocity of 8 km/h. If the resultant velocity of boat is 10km/h, then the velocity of the river is .. [CBSE-PMT 1993]
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a)12.8 km/h
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b) 6 km/h
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c)8 km/h
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d)10 km/h
Explanation
Let Velocity of boat with respect to observer VBO=10 km/hVelocity of boat with respect to river VBR=8 km/hVelocity of River with respect to Observer VRO Now VBO=VBR + VROAngle between VBO and VRO is 90°thus V2BR=V2BO - V2RO82=102 - V2RO64=100 - V2ROV2RO=36VRO=6 km/hAnswer: (b)
Q.10
The maximum range of a gun of horizontal terrain is 16 km. If g=10 m/s2, then muzzle velocity of a shell must be.. [ CBSE-PMT 1990]
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a) 160 m/s
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b) 200√2 m/s
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c)400 m/s
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d)800 m/s
Explanation
Formula for maximum range=V2 / g 16000 (10)=V2∴ v=400 m/sAnswer: (c)
Q.11
A body constrained to move in y-direction , is subjected to force given byF=( -2i + 15j + 6k ) N. What is the work done by this force in moving the body through a distance of 10m along y-axis [ CBSE-PMT 1994]
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a)190 J
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b) 160 J
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c)150 J
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d)20 J
Explanation
Since displacement is along y-direction hence displacement vector S=10j work done=F . S Work done=( -2i + 15j + 6k )( 10j) Work Done=150J Answer:(c)
Q.12
Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A to the left is 10 m/s. What is the velocity of B when angle α=60°? [ CBSE -PMT 1998]
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a) 5.8 m/s
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b) 9.8 m/s
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c) 10 m/s
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d) 17.3 m/s
Explanation
In one second angle become 60°. When the end A moves by 10m left, the end B moves up in one second=10×tan60=17.3 m/s Answer: (d)
Q.13
The result of (A × 0) will be equal to .. [ CBSE-PMT 1992]
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a)zero
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b) A
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c)zero vector
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d)unit vector
Explanation
When a vector is multiplied by a scalar result is vectorAnswer: (c)
Q.14
A particle moving with a velocity v=6i -4j+3k m/s under the influence of a constant force F=20i + 15j - 5kThe 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
power P=F . SP=(6i -4j+3k ) . ( 20i + 15j - 5k )P=45 J/sAnswer: (a)
Q.15
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. the velocity of the river water in km/h is .. [ CBSE-PMT 2000]
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a) 3
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b) 4
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c)√21
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d)1
Explanation
Speed along the shortest path=distance / time Speed along shortest path=1/ (15/60)=4 km/hFrom geometry of figurevelocity of revere v2=52 - 42v=3 km/h Answer:(a)
Q.16
The angle between the two vectors A=3i+4j+5k and B=3i + 4j -5k will be.. [ CBSE-PMT 2001]
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a) zero
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b) 45°
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c) 90°
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d) 180°
Explanation
cos θ=A . B / (|A| |B|) Now A . B=(3i+4j+5k) . (3i+4j-5k) A . B=9+16-25=0 |A| and |B| are nonzero ∴ cosθ=0 ∴θ=90° Answer: (c)
Q.17
A body of 3kg moves in the XY plane under the action of force given by 6ti + 4tj . Assuming that the body is at rest at time t=0, the velocity of the body at t=3s is... [ CBSE - PMT 2002]
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a)6i + 6j
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b) 18i + 6j
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c)18i + 12j
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d)12i + 18j
Explanation
a=F / ma=(6ti + 4tj) /3a=2ti + (4/3)tjVelocity equation v=u +atv=0 + (2ti + (4/3)tj)tv=(2t2i + (4/3)t2j)v=2(3)2i + (4/3) (3)2j v=18i + 12jAnswer: (c)
Q.18
three forces acting on a body are shown in the figure. To have the resultant force only along the y direction, the magnitude of the minimum additional force needed is ... [ CBSE-PMT 2008]
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a) 0.5 N
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b) 1.5 N
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c)[(√3) / 4] N
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d)√3 N
Explanation
As shown in figure the component of 1N and 2 N forces along positive x-axis=1cos60+ 2sin30=1.5Ncomponent of 4 N force along negative x-axis is=4sin30=2NTherefore, if force of 0.5N is applied along x-axis, the resultant force along x-axis will become zero and the resultant force will be obtained only along y-axisAnswer: (a)
Q.19
Consider a F=4i - 3j. 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
If vectors are perpendicular dot product is zero, Take dot product from the options, it is zero then that option is correctAnswer: (c)
Q.20
An aeroplane moves 400 m towards the north, 300 m towards west and then 1200m vertically upwards, then its displacement from the initial position is [ CBSE-PMT 1998]
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a) 1600 m
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b) 1800 m
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c)1500 m
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d)1300 m
Explanation
Here CD is perpendicular to the plane of page. Required distance DB From figure BD2=CB2 + CD2BD2=5002 + 12002BD=1300 m Answer:(d)
Q.21
for ordinary terrestrial experiments, the observer in an inertial frame in the following cases is .. [ AIIMS 2010]
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a) a child revolving in a giant wheel
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b) a driver in a sports car moving with constant speed of 200 km/h on a straight road
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c) the pilot of an aeroplane which is taking off
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d) a cyclist negotiating a sharp curve
Explanation
The car moving with a constant velocity has no acceleration. Hence, it is an inertial frame Answer: (b)
Q.22
Rain is falling vertically downwards with a velocity of 3km/h. A man walks in the rain with a velocity of 4km/h. The rain drop will fall on the man with a velocity... [ AIIMS 1997]
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a)5 km/h
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b) 4 km/h
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c)1 km/h
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d)3 km/h
Explanation
Let velocity of rain with respect to stationary observer be VRO Velocity of Man with respect to observer be VMOVelocity of rain with respect to man be VRMnow VRM=VRO + VOMbut VOM=- VMO∴ VRM=VRO - VMO Angle between VRO and VMO is 90°thus V2RM=V2RO + V2MOV2RM=9+16=25VRM=5Answer: (a)
Q.23
At the uppermost point of projectile, its velocity and acceleration are at an angle of ...[ AIIMS 2002]
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a) 180°
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b) 90°
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c)60°
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d)150°
Explanation
At uppermost point of projectile, velocity along vertical direction in zero, but have horizontal Velocity.Acceleration is gravitational and vertically down Thus angle is 90°Answer: (b)
Q.24
If R and H represent the horizontal range and the maximum height achieved by a projectile then which of the relation exists?
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a) H/R=4cotθ
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b) R/H=4cotθ
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c)H/R=4tanθ
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d)R/H=4tanθ
Explanation
Equation for range is : Equation for height By taking ratio of R/H we get Answer:(b)
Q.25
A projectile can have the same range R for two angle of projection. If t1 and t2 be the time of flights in the two cases, then the product of two time of flights is proportional to .. [ AIIMS 2008]
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a) 1/ R2
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b) R2
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c) R
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d) 1/R
Explanation
according to formula for same range different angles of projection Answer: (c)
Q.26
If A ×B=B×A , then angle between A and B is .. [ AIEEE 2004]
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a)π /2
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b) π/3
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c)π
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d)π/4
Explanation
given A ×B=B×AA ×B - B×A=0 Now - B×A=A×B ∴ A ×B +A ×B=0 ⇒ A ×B=0 ⇒ θ=0, π and 2π from given option θ=π Answer: (c)
Q.27
A ball whose kinetic energy is E, projected at an angle of 60° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be .. [ AIEEE 2002]
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a) E
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b) E/ √2
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c)E/4
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d)zero
Explanation
let initial velocity be u at time t=0vertical component of velocity usinθ=0 at max heighthorizontal component will be ucosθ=u/2 which remains constant Now E ∝ u2Thus at max height E'=E/4Answer: (c)
Q.28
A boy playing on the roof of a 10 m height building throws a ball with speed of 10 m/s at an angle of 30° with the horizontal. How far from the throwing point will the ball be at height of 10m from the ground? [ AIEEE 2003][g=10 m/s2, sin30=1/2, cos30=√3 /2]
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a) 5.2 m
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b) 4.33 m
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c)2.6 m
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d)8.66 m
Explanation
As shown in figure we have to find the range of projectileFrom the equation for projectile u=10 m/s , θ=30° g=10 m/s2 Answer:(d)
Q.29
A particle is moving eastwards with a velocity of 5m/s. In 10 sec the velocity changes to 5m/s northwards. The average velocity in this time is ..[ AIEEE 2005]
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a) (1/2) ms-2 towards north
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b) (1/√2) ms-2 towards north-east
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c) (1/√2) ms-2 towards north-west
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d) zero
Explanation
From figure Final velocity=5j Initial velocity=5i Change in velocity Δv=5j - 5i acceleration=Δv/t=(5j - 5i) / 10 a=(1/2)j - (1/2)i |a|=[ (1/2)2 - (1/2)2]1/2 |a|=1 /√2 tanθ=-1, x co-ordinate is negative hence direction is north west Answer: (c)
Q.30
The coordinates of moving particle at any time 't' are given by x=αt3 and y=βtthe speed of the particle at time 't' is given by ... [ AIEE 2003]
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a)
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b)
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c)
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d)
Explanation
By taking first order derivative of both equations we get velocity along x-axis and y-axis vx=3αt2vy=3βt2Thus resultant velocity=Answer: (b)
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