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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 4
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Q.1
A ball is thrown vertically upwards from the top of tower at 4.9 m/s. It strikes the ground near the base of the tower in 3 sec height of tower is
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a)73.5 m
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b) 44.1 m
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c)29.4 m
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d)14.7 m
Explanation
h=ut + ½ g t2 here u=-4.9 m/s as ball is thrown opposite to gravitation , t=3 sech=-4.9(3) + ½ (9.8) 9 h=29.4 m Answer: (c)
Q.2
If a car at rest accelerates uniformly to speed of car 144 km/h in 20 sec , it covers a distance of .. [ CBSE 1997]
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a) 20 m
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b) 800 m
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c)400 m
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d)1200 m
Explanation
Final speed is 144 km/h=40 m/s Initial speed is=0 Average speed=(Final speed + Initial speed ) / 2=20 m/s Distance traveled=Average speed × time Distance=20 × 20=400 mAnswer: (c)
Q.3
Velocity -time curve for a body projected vertically upward is .. [ EAMCET 1995]
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a) parabola
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b) ellipse
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c)hyperbola
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d)straight line
Explanation
Answer:(d)
Q.4
A person throws balls into air after every second. The next ball is thrown when the velocity of the first ball is zero. How high do the ball rise above his hand
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a) 2 m
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b) 5m
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c) 8m
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d) 10 m
Explanation
Ball is thrown upward v=u - gt When final velocity v=0, time is t=1 sec u=0 +10×1=10 from equation h=ut - ½ g t2 h=10 - ½ 10 (1) h=5 mAnswer: (b)
Q.5
A ball is dropped downwards, after 1 sec another ball is dropped downwards from same point. What is the distance between them after 3 sec.. [ BHU 1998]
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a)25 m
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b) 20 m
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c)50 m
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d)9.8 m
Explanation
Time for first ball is t=3 secfrom equation h=ut + ½ gt2Time for first ball is t=3 sec and u=0 , g=10 m/s2h=½ (10) 9=45 mFor second ball time t=2s , u=0 and g=10 h'=½ (10) 4=20 >Thus distance between both ball=h-h'=45-20=25 mAnswer: (a)
Q.6
Two trains are each 50 m long moving parallel towards each other at speeds 10 m/s and 15 m/s respectively, at what time they will pass each other.. [ CPMT 1999]
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a) 5 √2
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b) 4 sec
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c)2 sec
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d)6 sec
Explanation
Relative speed of each train with respect to each other be v=10+15=25 m/sHere distance covered by each train=sum of their length =50+50=100 m∴ Required time=100 /25=4 secAnswer: (b)
Q.7
The position vector of a particle is given by r=ro(1-at)t, where t is the time and a and ro are constants. After what time the particle returns to the starting point and distance traveled by the
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a) a, ro / a
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b) 1/a , ro / 2a
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c)a2 , 2ro / a
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d)1/a2 , ro / 4a
Explanation
Initial position of particle time t=0r=ro(1-a×0)× 0 r=0 Thus 0=ro(1-a×t)× t 0=(1-a×t) t=1/a Total time to return=1/a ∴ Time of forward journey=time of backward journey=1/ 2a Total distance covered=Forward distance + return distance S=ro( 1- a × 1/ 2a) + ro( 1- a × 1/ 2a) S=ro / 4a + ro / 4a=ro / 2a Answer:(b)
Q.8
Three different balls of mass m1, m2 andm3 are allowed to slide from rest on three frictionless paths OA, OB, OC respectively from O. The height O above ground is h. The respective speed S1 S2,S3 at the bottom A, B and C are
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a) S1=S2=S3
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b) S1 < S2 < S3
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c) S2
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d)
Explanation
All the paths are frictionless hence no loss in energy According to law of conservation of energy Potential energy=kinetic energy mgh=½ mv2 v2=2gh from above velocity is in dependant of mass and depends on height for all balls height is same thus velocity will be same Answer: (a)
Q.9
Two trains travelling on the same track are approaching each other with equal speed of 40 m/s. The driver of the trains begin to decelerate. Simultaneously when they are just 2.0 km apart. Assuming the decelerations to be uniform and equal, the value of the deceleration to barely avoid collision should be .. [ AMU 1995]
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a)11.8 m/s2
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b) 11.0 m/s2
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c)2.1 m/s2
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d)0.8 m/s2
Explanation
Trains are moving towards each other thus relative velocity=40-(-40)=80 m/sTo avoid collision final velocity should be zero∴ a=v2 / 2Sa=(80×80) / (2000 × 2)=1.6 m/s2 This retardation is the relative retardation. therefore the retardation of each train should be 0.8m/s2Answer: (d)
Q.10
A car moving with a speed 40 km/h can be stopped by applying brakes after 2m. If the same car is moving with speed of 80 km/h, what is the minimum stopping distance
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a) 8 m
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b) 2m
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c)4m
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d)6 m
Explanation
from formula v2=u2- 2aS If u=40 k/h S=2m = 2×10-3 0=402 -2a× 2×10-3 ∴ 2a=800×103 km/h Now u=80 km/h0=802- (800×103) SS=80 × 80 / 800=8 mAnswer: (a)
Q.11
Which of the following four statements is false:
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a) A body can have zero velocity and still be accelerated
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b) A body can have constant velocity and still have a varying speed
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c)A body can have a constant speed and still have a varying velocity
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d)The direction of the velocity of a body cn change when its acceleration is constant
Explanation
Answer:(b)
Q.12
A boat at anchor is rocked by waves whose crests are 100 m apart and whose velocity is 25 m/sec. These waves reach the boat after every
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a) 2500 sec
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b) 75 sec
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c) 4 sec
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d) 0.25 sec
Explanation
times=distance / velocity times=100 /25=4 sec Answer: (c)
Q.13
The displacement of a particle as a function of time is shown in figure. the figure indicates that ..[ CPMT 86]
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a)the particle starts with a certain velocity but the motion is retarded and finally the particle stops
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b) the velocity of the particle is constant throughout
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c)the acceleration of the particle is constant throughout
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d)the particle starts with constant velocity, the motion is accelerated and finally the particle moves with another constant velocity
Explanation
Answer: (a)
Q.14
Figure shows the v-t graph for two particles P and Q. Which of the following statements regarding their relative motion is true : There relative velocity is ..
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a) is zero
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b) is non zero but constant
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c)continuously decreases
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d)continuously increases
Explanation
Answer: (d)
Q.15
A man in a balloon rising vertically with an acceleration of 4.9 m/s2 releases a ball 2 seconds after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (g=9.8 m/s2 [ MLNR 1986]
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a) 14.7 m
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b) 19.6 m
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c)9.8 m
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d)24.5 m
Explanation
After two second velocity of balloon and ball will be v=u + at v=0 +4.9 × 2v=9.8 m/sHeight of balloon when ball is released h=½ at2h=½ 4.9 × 4=9.8 m --(1) Upward displacement of ball after release h' Final velocity of ball will be zero after reaching maximum heightUsing v2=u2-2gh'0=(9.8)2 - 2 × 9.8 × h' h'=4.9 m --(2)Thus, the maximum height of the ball above the ground level=(9.8 +4.9 )=14.7 m Answer:(a)
Q.16
A stone is dropped into a lake from 500 meter height. the sound of splash will be heard after .. [ Manipal 2002]
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a) 10s
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b) 11.5 s
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c) 14 s
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d) 21s
Explanation
Time taken by the stone to reach lake h=ut + ½ g t2 500=½ (10) t2 t=10 sec --(1) Velocity of sound in air 350 m/s time taken by sound to traveled 500 m t'=distance / speed t'=500 /350=1.5s total time=t + t'=10+ 1.5=11.5 s Answer: (b)
Q.17
A body freely falling from rest has velocity v after it falls through a height h. The distance it has fall down further for its velocity to become double is .. [ B.H.U 1999]
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a)4h
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b) 6h
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c)8h
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d)10h
Explanation
For free fall v=√(2gh) and v'=√(2gh') so v'/v=√ (h'/h)given v'=2v2=√ (h'/h)h'=4hAnswer: (a)
Q.18
By what velocity a ball be projected vertically upward so that the distance covered in 5th second is twice that covered by 6th second (g=1o ms-2 ) -- [CPMT 1997]
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a) 19.6 m/s
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b) 58.8 m/s
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c)49.0 m/s
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d)65 m/s
Explanation
Here ball is going up thus g is negative Now by formula S5=u - 5(2×5 - 1) S5=u - 45 S6=5(2 ×6 - 1) S6=u - 55 Given S5=2×S6 u - 45=2× (u - 55) u=65 m/sAnswer: (d)
Q.19
A ball A is thrown up vertically with speed u, At the same time another ball B is released from rest at height, At time, the speed of A relative to B is
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a) u
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b) u - 2gt
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c)u - gt
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d)(u2 - 2gh)1/2
Explanation
Speed of ball (A) thrown up with initial velocity at time t is v=u -gtSpeed of ball (B) released from height h at time t us v'=gt V' will be negative since up going ball velocity is considered positive thus v' - gtSpeed of boll A relative to ball B=v-v'=(u-gt) -(-gt )=u Answer:(a)
Q.20
A balloon starts rising from the ground with an acceleration of 1.25 m/sAfter 8 seconds, a stone is released from the balloon. the stone will .. (g=10 m/s2)
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a) covers a distance of 40 m
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b) have displacement of 50 m
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c) reaches the ground in 4 sec
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d) begin to move downwards after being released
Explanation
Velocity of balloon after 8 sec v=u + at u=0 a=1.25 m/s2 t=8s v=1.25× 8=10 m/s Height of balloon after 8 sec h=ut + ½ at2 h=½ (1.25) 82 h=40 m Taking downward motion of stone released from balloon at height h=40 h=ut - ½ gt2 here u=-10 m/s g=10 m/s2 and h=40 40=-10× t + ½ (10) t25t2 - 10t + 40=0 t2 - 2t - 8=0on solving t=4Answer: (c)
Q.21
Two particle, one with constant velocity 50 m/s and the other with uniform acceleration 10 ms-2, start moving simultaneously from the same place in the same direction. they will be at a distance of 125m from each other after .. [ Roorkee 1998]
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a)5 sec
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b) 5( 1 +√2) sec
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c)10 sec
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d)10( √2 + 1) sec
Explanation
For particle moving with constant velocity S1=50 t For particle 2 moving with constant accelerationS2=½ a t2=½ × 10 t2=5 t2 Given in question S1 - S2=12550t-5t2=125 5t2 -50t +125 =0 on solving we gett=5 secAnswer: (a)
Q.22
In 1.0 sec a particle goes from A to point B moving in a semicircle of radius 1.0 m. the magnitude of the average velocity is ...[ IIT 1999]
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a) 3.14 m/s
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b) 2.0 m/s
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c)1.0 m/s
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d)zero
Explanation
Average velocity=Total displacement / timeAverage velocity=2 /1=2 m/sAnswer: (b)
Q.23
A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t=At this instant of time, the horizontal component of its velocity is v. A bead of the same mass as P ejected from A at t=0 along the horizontal direction, with the speed v. Friction between the bead and string may be neglected. Let tp and tQ be the respective times taken by P and Q to reach the point B. Then ..[ IIT 1993]
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a)tP < tQ
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b)tP=tQ
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c)tP > tQ
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d)
Explanation
Horizontal velocity of particle P increases from v up to lowest point, after that it start to decreasing and again becomes v at point B. It means the average velocity of particle P is more than v where as the particle Q moves with constant velocity v only. Therefore to cover the same horizontal distance, the particle P will take smaller time than that of particle Q option (a) correct Answer:(a)
Q.24
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to height d/Neglecting subsequent motion and air resistance, its velocity v varies with the height oh above ground as.. [ IIT 2000]
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a)
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b)
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c)
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d)
Explanation
If we consider the velocity of ball negative when coming down can be calculated as v=√2gd And velocity of ball will be positive when rising up final velocity is zero u=√gd With the above considerations, we note that the variation of v and h will be given by option(a)Answer: (a)
Q.25
A particle moving in a straight line with initial velocity u and uniform acceleration f. If sum of the distance covered in tth and (t+1)th second is 100cm, then its velocity after t seconds is cm/s is .. [ Roorkee 1998]
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a)20
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b) 30
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c)50
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d)80
Explanation
From the formula for displacement in nth secVelocity after t sec v=u + at v=50 cm/sAnswer: (c)
Q.26
A car accelerates from rest at a constant rate of 3ms-1 for some time. Then it retards at constant rate of 6ms-2 for some time. and comes to rest. If total time for which it remains in motion is 3 sec, what is the total distance traveled?
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a) 3 m
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b) 4.5 m
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c)6 m
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d)9 m
Explanation
First body has positive acceleration reaches max speed in time t1from rest u=0V=u +at V=3t1 --(1)After time t1 body have negative acceleration and final velocity is zero after time t2 thus 0=V - 6t2V=6t2 --(2)from equation (1) and (2)3t1=6t2 t1=2 t2given t1 + t2=3 secthus t1=2 secand t2=1 secfrom equation (2) V=6 m/sform formula S=ut + ½ at2Displacement in t1S1=0 +½ (3)(4)=6 mDisplacement in t2 u=6m/sS2=6 -½ 6 (1)=3 mTotal displacement S=S1 + S2S=6 + 3=9 mAnswer: (d)
Q.27
A stone is dropped from a height of 45m. What will be the distance traveled by it during the last one second of its motion? [ g=10 m/s2]
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a)35 m
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b)25m
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c)12.5 m
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d)10 m
Explanation
total time taken by the stone to reach groundfrom equation for free fall h=½ gt245=½ (10) t2t=3 secfrom the formula for displacement in nth secS=0 + (10/2) (2×3 -1)=25 m Answer:(b)
Q.28
A student is standing at a distance of 50m from bus. As soon as the bus begins its motion with an acceleration of 1 m/s2, the student starts running towards the bus with uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is .. [AIIMS 2010]
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a) 8 m/s
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b) 5 m/s
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c) 12 m/s
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d) 10 m/s
Explanation
Let the student catch the bus in time 't' at 'x' m from the starting point of bus 50+x=ut --(1) Distance traveled by bus in time t x=½ (1) t2 --(2) from equation (1) and (2) 50 + ½t2=ut t2 - 2ut + 100 Root of equation will be For root to be real u2 ≥ 100 u=10 m/s Answer: (d)
Q.29
A body is thrown vertically upwards with a velocity of 19.6 m/s. The position of the body after 4s will be [ AIIMS 2009]
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a)at the highest point
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b) at the mid-pint of the line joining the starting point and the highest point
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c)at the starting point
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d)non of the above
Explanation
from equation v=u - gt v=19.6 - 9.8 ×(4) v=-19.6velocity at end of 4sec is negative of initial hence object is at starting pointAnswer: (c)
Q.30
A particle is thrown vertically upwards with a velocity of 4m/s. The ratio of its accelerations after 1s and 2s of its motion is .. [AIIMS 2009]
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a) 2
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b) 9.8
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c)1
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d)4.9
Explanation
Gravitational acceleration is independent of time and is constant hence ratio will be 1Answer: (c)
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