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Motion In Two And Three Dimensions Mcq
Quiz 3
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Q.1
A projectile can have the same range 'R' for two angles of projection. If T1 and T2 to be time of flight in two cases, then the product of the two time of flights is directly proportional to .. [ AIEEE 2004]
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a) R
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b) 1/R
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c)1 /R2
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d)R2
Explanation
Equation for time of flight is=The angle of projection is different but range is same , then angle of projections are complimentary Let one angle be θ then other is (90 - θ) Taking product of T1 and T2 we get2sinθcosθ=sin2θformula of range is from above T1T2=2R/g Answer: (a)
Q.2
A ball is thrown from a point with speed vo at an elevation angle of θ, From the same point and at the same instant, a person starts running with constant speed vo / 2 to catch the ball. Will the person be able to catch the ball? if yes what should be the angle of projection [ AIEEE 2004]
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a) No
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b) yes, 30°
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c)yes, 60°
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d)yes, 45°
Explanation
To catch the ball range of the ball=displacement of person T is time of flight from above equations Answer:(c)
Q.3
Which of the following statements is FALSE for a particle moving in a circle with constant angular speed? .. [ AIEEE 2004]
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a) The acceleration vector points to the center of the circle
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b) The acceleration vector is tangent to the circle
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c) The velocity vector is tangent to the circle
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d) The velocity and acceleration vectors are perpendicular to each other
Explanation
Answer: (b)
Q.4
A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at 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
During the flight horizontal component remain unchanged, and at maximum height vertical component becomes zeroif u is velocity at t=0 then at maximum height velocity=ucons60=u/2 K ∝ v2 K' / K=(u/2)2 / u2 K'=K/4Answer: (d)
Q.5
A particle has an initial velocity 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 unit
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c)8.5 units
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d)10 unit
Explanation
v=u + a t v=3i + 4j + ( 0.4i + 0.3j)10 v=(3+4)i + (4+3)jv=(7)i + (7)jAnswer: (a)
Q.6
A river is flowing from west to east at a speed of 5 meters per minute. A man on south bank of the river, capable of swimming at 10 meters per minute in still water, wants to swim across the river in shortest time. He should swim in a direction
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a) due north
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b) 30° east of north
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c)30° west of north
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d)60° east of north
Explanation
Time taken to cross river t=d/ vcosθ form minimum time cosθ=1 therefore θ=0 due north Answer:(a)
Q.7
The vector P=ai+aj+3k and Q=ai -2j-k are perpendicular to each other. The positive value of 'a' is .. [ AFMC 2000]
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a) 23
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b) 9
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c) 8
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d) 3
Explanation
Given P and Q are perpendicular for perpendicular vector P⋅Q=0 (ai+aj+3k) ⋅ ai -2j-k)=0 a2 -2a -3 =0 a = 3 Answer: (d)
Q.8
If an iron ball and wooden ball of same radius are released from a height h in vacuum the time taken by both of these to reach to ground is .. [ AFMC 1998]
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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
In absence of air magnitude of acceleration is same for both the ballsAnswer: (b)
Q.9
Two vectors A and B are such that A + B=C and A2 + B2=C2 If θ is the angle between positive direction of A and B then the correct statement is .. [ AFMC 1999]
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a) θ=π
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b) θ=2π/3
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c)θ=0
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d)θ=π/2
Explanation
Since A2 + B2=C2 Therefore A and B are the sides of right angle triangle. Thus angle between A and B is π/2Answer: (d)
Q.10
A body of mass 0.5kg is projected under gravity with a speed of 98 m/s at angle of 60° with the horizontal. The change in momentum ( in magnitude) of the body is ... [ AFC 2010]
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a) 50.0 N-s
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b) 98.0N-s
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c)24.5N-s
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d)49.0N-s
Explanation
Initial momentum pi=mu=0 (u=0)When projected with velocity v=98 m/s at the time of projection momentum p=mv=0.5×98=49.0N-s Answer:(d)
Q.11
A particle starting from the origin (0, 0) moves in a straight line 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 .. [ AFMC 2007]
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a) 30°
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b) 45°
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c) 60°
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d) 0°
Explanation
The final coordinate of the particle is (√3, 3). If θ be the angle made by the position vector of the particle with x-axis then tanθ=y/x=3/√3=√3 tan60=√3 Thus θ=60°Answer: (c)
Q.12
A car is moving along a circular road at speed of 20m/s. The radius of the circular road is 10m. If the speed in increased at the rate of 30m/s2, what is the resultant acceleration [ AFMC 2009]
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a)10 m/s2
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b) 50 m/s2
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c)250 m/s2
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d)80 m/s2
Explanation
In no-uniform circular motion linear acceleration have two components one is towards the center and one is tangential relation is given by Where ac=centripetal acceleration at=tangential acceleration Here ac=v2 / r ac=(20)2 / 10=40 m/s2at=30 m/s2 (given)∴ Answer: (b)
Q.13
Two stones are projected with same velocity but makes different angles with the horizontal. Their ranges are equal. If the angle of projection of one is π/3 and its maximum height is h, then the maximum height of the other will be [ AFMC 2010]
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a) h/6
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b) h/3
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c)h/2
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d)h/√2
Explanation
If range is same but angle of projection are different then θ1 + θ2=π/2 If θ1=π/3 then θ2=π/6 formula for maximum height isform formula for HH/H'=sin2 (π/3) / sin2(π/6) H/H'=(3/4) / (1/4)=3 H'=H/3Answer: (b)
Q.14
A bullet is dropped from the same height when another bullet is fired horizontally. They will hit the ground .. [ AFMC 2002]
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a) depends upon mass of bullet
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b) depends upon the observer
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c)one after another
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d)simultaneously
Explanation
When bullet is fired horizontally there is no component of velocity along vertical direction therefore both the bullets will hit ground simultaneously Answer:(d)
Q.15
A bomb is dropped from an areoplane moving horizontally at constant speed. If air resistance is taken into consideration, then the bomb .. [ AFMC 1999]
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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 the aeroplane
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d) flies with aeroplane
Explanation
If there is no resistance, bomb will drop at a place exactly below the flying aeroplane. But when take into account air resistance, bomb will face deceleration in its velocity. So, it will fall on the earth exactly behind the aeroplane. Answer: (b)
Q.16
An aeroplane moving horizontally with a speed of 720 km/hr drops a food packet, while flying at a height of 396.9m. The time taken by a food packet to reach the ground and its horizontal range is ( 9.8 m/s2) [ AFMC 2001]
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a)9 sec and 1800 m
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b) 8 sec and 1500 m
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c)3 sec and 2000 m
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d)5 sec and 500 m
Explanation
at time t=0 , velocity of packet along horizontal vx=720 km/hr=200 m/s and velocity along vertical down is 0m/s Thus it will fall down with initial speed of zero m/s and acceleration is 'g'From formula h=ut + ½ g(t)2 h=396.9 m, u=0 , g=9.8 m/s2 396.9=½ (9.8) 2t2=81 t=9 sec ( time of flight)Range=velocity along horizontal × time of flightRange=200 × 9=1800 mAnswer: (a)
Q.17
The maximum horizontal range of a projectile is 400m. The maximum height attended by it will be .. [ AFMC 2005]
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a) 100 m
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b) 200 m
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c)400 m
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d)800 m
Explanation
relation between range and maximum height is given by equation for maximum range θ=45 Thus R=H/4 R=400 /4=100 mAnswer: (a)
Q.18
What determines the nature of the path followed by the particle? [ AIEE2005]
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a) Speed
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b) Velocity
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c)Acceleration
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d)None of above
Explanation
When velocity changes, direction of path or nature of path changes,so, acceleration determines the nature of path Answer:(c)
Q.19
Force responsible for circular motion of body is .. [ AIEEE 2003]
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a) centripetal force
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b) centrifugal force
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c) gravitational force
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d) non of these
Explanation
Answer: (a)
Q.20
A projectile is thrown in upward direction making an angle of 60° with horizontal direction with a velocity of 147 m/s. Then the time after which its inclination with the horizontal is 45°, is ..[ AFMC 2006]
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a)15 s
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b) 10.98 s
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c)4.49 s
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d)2.745 s
Explanation
Horizontal component of velocity at time t=0 u=147 m/s and angle of projection=60° Horizontal component uxo=147cos60=147 ×( 1/2) --eq(1) at time t=t Horizontal component of velocity don't change with time let v be the velocity at time t when angle made by projectile with horizontal is 45° Thus vx=vcos45=v/√2 --eq(2) from equation (1) and (2) we get 147× (1/2)=v/√2 v=147 / √2 --eq(3) Vertical component of velocity Vertical component of velocity at time t=0 vy0=using 60=147 × (√3/2) V is the velocity at time t when angle of projectile with horizontal is 45° Vertical component v=sin45 As velocity component along vertical changes with time usin60 -gt=vsin45 Answer: (c)
Q.21
A river 4.0 mile wide is flowing at the rate of 2 miles/hr. The minimum time taken by a boat cross the river with speed v=4 mile/he (in still water) is approximately [ AFMC 2011]
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a) 1hr
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b) 2 hr and 7 minutes
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c)1 hr and 13 minutes
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d)2 hr and 25 minutes
Explanation
time taken to cross river t=d/ ucosθ; θ is the angle between flow and direction of boat According to question θ is not given. Therefore minimum time taken=4/4=1 hr, when boat move right across the flowAnswer: (a)
Q.22
An aeroplane is moving with a horizontal velocity u at a height h. The velocity of packet dropped from it on the earth's surface will be .. [ AFMC 1999]
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a)
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b) 2gh
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c)√(gh)
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d)
Explanation
At time t=0 horizontal velocity of packet=u Vertical velocity of packet=0 At t=t when packet hits the ground No gravitational acceleration along horizontal hence velocity horizontal=u Velocity component along vertical v2=2gh Thus velocity when packet hits the ground Answer:(d)
Q.23
If two numerically equal force P and Q acting at a point produces a resultant force of magnitude P then the angle between the two original forces is [AFMC 1998]
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a) 120°
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b) 90°
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c) 0°
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d) 45°
Explanation
Let R=P +Q Given that |R|=|P| and |P|=|Q| Answer: (a)
Q.24
What is the dot product of two vectors of magnitude 3 and 5, if angle between them is 60°? [ 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
A.B=ABcosθ Dot product=(3)(5)cos60Dot product=(3)(5)(1/2)=7.5Answer: (b)
Q.25
If A=B + C and the magnitude of A, B, 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(3/5)
Explanation
|A|2=|B|2+|C|2 + 2|B||C| coθ25=16 +9 + 2(4)(3) cosθ25=25 + 24 cosθ ⇒ cosθ=0 ∴ θ=π/2 From diagram Angle between A and C is β= cos-13/5 Answer:(a)
Q.26
Out of the following pairs, the resultant of which cannot be zero [ CPMT 1985]
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a) 10, 10, 10
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b) 10, 10, 20
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c) 10,20, 20
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d) 10, 20 , 40
Explanation
Alternative (d) is correct because the triangle is not formed if third side is greater than the sum of first two Answer: (d)
Q.27
If |V1 + V2|=|V1 - V2| and |V2| is finite then [ CPMT 1999]
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a)V1 is parallel to V2
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b) V1 is perpendicular to V2
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c)V1=V2
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d)V1=V2
Explanation
From given 2V1V2cosθ=- 2V1V2cosθ ⇒ cosθ=0 ∴ θ=π/2Both the vectors are perpendicularAnswer: (b)
Q.28
A boat is set across a river with velocity of 8 km/h. If the resultant velocity of the boat is 10 km/h., the river is flowing with a velocity of ..[ CBSE1994]s
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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)10km/h
Explanation
As shown in figureVriver2=Vresultant2 - Vboat2River=[ 102 - 82]1/2=6 km/hAnswer: (b)
Q.29
A particle is acted up on by two forces of 3N and 4N simultaneously. Which of the following is most correct:
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a) The resultant of these forces is 7N
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b) The resultant of these forces is 1N
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c)The resultant of these forces is 4N
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d)The resultant of these forces lie between 1 N and 7 N
Explanation
cos of angle between them may be between -1 to +1 thus correct option is (d) Answer:(d)
Q.30
A bird is flying towards south with velocity 40km/h and train is moving with velocity 40 km/h towards east . What is the velocity of the bird with respect to an observer in train
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a) 40√2 km/h north-east
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b) 40√2 km/h south-east
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c) 40√2 km/h south-west
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d) 40√2 km/h north-west
Explanation
To find the relative velocity of the bird with respect to train, reverse the direction of train velocity and add it in to the velocity of bird . As shown in figure. Therefore the relative velocity of bird with respect to train=40√South-west Answer: (c)
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