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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 3
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Q.1
The displacement of particle varies with time according to the relation:Then velocity of the particle is:
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a) k(e-bt)
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b) k(e-bt) / b2
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c) kb(e-bt)
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d)none of above
Explanation
We will take derivative of given equation to find velocity=k(e-bt) Answer:(a)
Q.2
A car runs between two station A and B. It goes from A to B at the speed of 30km/h and returns from B to A at 40km/hr. The average speed of the car between A and B is :
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a) 35 km/hr
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b) 34.3 km/hr
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c) 34.5 km/hr
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d) 10 km/hr
Explanation
Distance is same for both the paths thus we can use formula Here V1=30km/hr and V2=40km/hron substituting the values and solving we getAverage velocity=34.3 km/hr Answer: (b)
Q.3
The relation between time and distance x is t=αx2+βx where α and β are constant. The retardation is :
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a) -2αv3
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b) 2βv3
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c)2αβv3
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d)2α2βv3
Explanation
By taking darivative of given equation with x we getAnswer: (a)
Q.4
Two bodies of different masses ma and mb are dropped from different heights a and b then ratio of time taken to reach ground is ..
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a) a:b
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b) mb /ma : b/a
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c)√a : √b
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d)a2:b2
Explanation
Time taken is independent of mass when motion is free fallgravitation acceleration is same for both massesNow h=½ g t2For mass at height a , a=½ g (ta)2 For mass at height b , b=½ g (tb)2 Taking the ratios of above two equation and on solving we get ta /tb=√a : √bAnswer: (c)
Q.5
The distance x covered by in time t by a body having initial velocity vo and having acceleration a is given by x=vot + ½ a t2This result follows from:
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a) Newton's I law
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b) Newton's II law
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c)Newton's III law
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d)None of these
Explanation
Newton's laws explain force.. Answer:(d)
Q.6
The co-ordinates of a moving particle at any time are given by x=ct2 and y=bt2, the speed of the particle at time t is given by :
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a) 2t(c+b)
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b) [c2 + b2]1/2
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c) 2t[c2 + b2]1/2
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d) 2t[c2 - b2]1/2
Explanation
x=ct2 is time dependant taking derivative with respect to time we get velocity along x axis vx=2ct --(1)y=bt2 is time dependant taking derivative with respect to time we get velocity along y axis vy=2bt --(2)Magnitude of velocity=[ vx2 + vy2]1/2∴ Magnitude of velocity=[ (2ct)2 + (2bt)2]1/2Magnitude of velocity=2t[c2 + b2]1/2 Answer: (c)
Q.7
A body of mass 3 kg falls from the multi storied building 100 m high and buries it self 2 m deep in the sand. The time of penetration will br:
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a)9 sec
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b) 0.9sec
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c)0.09sec
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d)10 sec
Explanation
When body touch the ground its velocity v can be calculated using formula v2=u2 + 2gh --(1)Here u=0 thus v=[2×9.8×100]1/2=√1960 m/sBody penetrate 2 m deep and comes to rest thus v=0 and u=√1960 m/s substituting values of u and v in equation (1)0=1960 - a (2) Thus retardation a=490 m/s2Now using V=u+at equation we can find time tHere V=0 and u=√1960 0=√1960 -490 t t=√1960 / 490=0.09 sAnswer: (c)
Q.8
A body travelling with uniform acceleration crosses two points A and B with velocities 20 m/s and 30 m/s respectively. The speed of the body at mid-point of A and B is :
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a) 25 m/s
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b) 25.5 m/s
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c)24 m/s
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d)10√10 m/s
Explanation
from equation v2=u2 + 2as --(1)a=v2 - u2 / 2sHere s is the distance between A and B u=20m/s and v=30m/sa=302 - 202 / 2sa=500/2sDistance of mid point from A is s/2 let velocity be v' and distance is s/2, u=20m/s substituting new values in equation (1)v'2=202 + 2(500/2s) (s/2)V'2=400 + 250=650m/sV'=√650=25.495=25.5 m/sAnswer: (b)
Q.9
Drop of water falls from the roof of the building 9m height at regular intervals of time, the first drop reaching the ground at the instant fourth drop start to fall. What are the distances of the second and the third drop from the roof
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a) 6m and 2m
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b) 6m and 3m
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c)4m and 1m
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d)4m and 2m
Explanation
Let t be the time interval between two drop , the time take by first drop to reach ground be 3t then time period from second drop is 2t and third drop is tDistance travelled by third drop 9=½ g (t)2=½ g (3t)2 thus ½gt2=1 Distance travelled by second drop h=½ g(2t)2substituting value of ½ gt2 in above equation we get h=1(2)2 =4 m Distance travelled by third drop h=½ g(t2)=1×(1)=1 m Answer:(c)
Q.10
A stone dropped from a tower reaches the ground after 4 sec's. If it is stopped 2 sec's after its fall and then released again, it will reach the ground after how many more seconds?
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a) 2
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b) √8
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c) √12
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d) 3
Explanation
Height of tower h=½ gt2 h=½ g(4)4=½ g(16)=8g --(1) Distance traveled by the stone in two sec h'=½ g (t)2=½ g (4) From equation (1) h'=h/4 Thus balance distance is 3h/4 Let t be the time taken by the stone to travel remaining distance Now 3h/4=½ g(t)2 From equation 1 replacing value of h we get (3/4) ( 8g)=½ g(t)2 6=½ t2 t=√ 12 Since stone was stopped for two sec Extra time taken=√12 Answer: (c)
Q.11
A boy sitting on the topmost berth in the compartment of a train which is just going to stop on a railway station, drops an apple aiming at the open hand of the brother situated vertically below his hands at a distance of about 2 meters, The apple fall: [ CPMT 1986)
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a)precisely in the hand of his brother
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b) slightly away from the hand of his brother in the direction of motion of train
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c)slightly away from the hand of his brother in the direction opposite to the direction of motion of train
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d)none-of above
Explanation
Train is stopping thus acceleration is negative pseudo acceleration on apple will be in the direction of motion, thus option (b) is correctAnswer: (b)
Q.12
The area under acceleration-time graph represents .. [ MPPMT 1993]
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a) the displacement
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b) velocity
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c)change in velocity
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d)distance traveled
Explanation
Answer: (c)
Q.13
The position x of a particle varies with time t, as x=at2 - btThe acceleration of the particle will be zero, at time equal to .. [ CBSE 1997]
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a) 2a/ 3b
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b) a/b
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c)a/ (3b)
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d)zero
Explanation
By taking derivative of above equation two times we get equation for acceleration (A) as A=0=2a- 6bt Thus when t=a/ (3b) acceleration is zero Answer:(c)
Q.14
A particle starts from rest and travels a distance of 's' with uniform acceleration, then it travels a distance '2s' with uniform speed, finally it travels a distance '3s' with uniform retardation of the particle is a straight line then the ratio of its average velocity to maximum velocity is:
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a) 6/7
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b) 4/5
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c)3/5
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d)2/5
Explanation
Motion of particle can be divided in three1) particle moves Positive acceleration 'a' for a distance 's' Particle initial velocity u=0 Distance traveled=s Maximum velocity at end of 's' distance=Vm ∴ average velocity=Vm/2 Let time taken be=t Displacement s=Vm t /2 --(1) Also Vm=at --(2) 2) Particle moves with maximum velocity for distance '2s' Let particle takes time t' to cover distance of 2s displacement 2s=Vm t' From equation (1) replacing value of 's' in above equation 2(Vm t /2)=Vm t' t'=t thus 2s=Vm t Vm=2s/t --(3) 3) Particle covers 3s distance and comes to rest Since distance traveled is three times the distance while accelerating retardation a'=a/3Time taken will be=3tNow Average velocity=Total distance / total timeAverage velocity=( s+2s+3s) / (t+t+3t )=6s/5tFrom equation (3) maximum velocity=2s/t taking ratio=(6s/5t) / (2s/t) Ratio=3/5 Answer:(c)
Q.15
A particle at rest starts moving in a horizontal straight line with a uniform acceleration. The ratio of the distance covered during the fourth and the third second is .. [ CBSE 1993]
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a) 4/3
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b) 26/9
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c) 7/5
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d) 2
Explanation
Equation for displacement in nth second is- From above formula for u=0 and time n=4 we get S4=7a/2For u=0 and time n=3 we get S3=5a/2Taking ratio of S4 to S3 we get7/5 Answer: (c)
Q.16
A body moves from rest with a constant acceleration of 5m/sIts instantaneous speed in m/s at the end of 10 sec is ..[ UPSC 1994]
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a)50
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b) 5
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c)2
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d)0.5
Explanation
v=u + at v=0 + 5(10) v=50 m/sAnswer: (a)
Q.17
A particle moves in x-y plane according to equation x=4t2 +5t +16, and y=5tThe acceleration of the particle must be [ CPMT 1993
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a) 8 m/sec2
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b) 13 m/sec2
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c)14 m/sec2
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d)none of the above
Explanation
By taking derivative for two times with respect to time we get Acceleration along x-axis ax=8 m/sec2Acceleration along y-axis=0Now Magnitude of acceleration=[ ax2 +ay2]1/2 magnitude of acceleration=[82 + 0]1/2acceleration=8 m/sec2Answer: (a)
Q.18
The acceleration 'a' in m/sec2, of a particle is given by a=3t2 + 2t +2 where 't' is the time. If the particle starts out with velocity 2m/s, velocity at the end of 2 seconds is : [ UPSE 1994]
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a) 12 m/sec
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b) 14 m/s
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c)16 m/s
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d)18 m/sec
Explanation
By taking integrating of above equation we will get the equation for velocity v=∫ (3t2 + 2t +2) dt v=t3 + t2 + 2t + C When t=0 v=2 m/sec by substituting values in above equation we get value of integration constant C C=2 Thus equation for velocity is t3 + t2 + 2t + 2 By substituting t=2 in above equation we get v=23 + 22 + 2(2) + 2 v=8 +4+4+2=18 m/s Answer:(c)
Q.19
A body falling for 2 second covers a distance S equal to that covered in next second. Taking g=10 m/s2; S=? [ EAMCET 1995]
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a) 30 m
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b) 10 m
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c) 60 m
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d) 20 m
Explanation
Form equation for displacement in nth second is given by Taking value time t=n g=10m/sec2 and u=0 we get Displacement in nth sec Sn=5(2n-1) --(1)Now displacement in previous two seconds will betime n=(n-1) and g=10m/sec2 and u=0Sn-1=5(2n-3) --(2)time n=(n-2) and g=10m/sec2 and u=0Sn-2=5( 2n - 5) --(3)Given Sn-2 + Sn-1=SnThus from equation (1), (2), (3) we get 5(2n-5) +5(2n - 3 )=5( 2n - 1) On solving n=7/2 secSubstituting the value of n=7/2 in equation (1) we get S=30 m Answer: (a)
Q.20
A car has to to cover the distance of 60 km, if half of the total time it travel with velocity 80 km/h and in rest time its speed becomes 40 km/h the average speed if the car will be .. [ Raj. PET 1996]
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a)60km/h
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b) 80km/h
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c)120 km/h
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d)180 km/h
Explanation
segment first:let time taken be t thus distance travelled S1=80×tsegment secondtime taken=tDistance travelled S2=40×t Given total displacement=S=60 kmThus S=S1 + S260=80×t + 40 ×t∴ t=60/120=0.5 hTotal time taken to cover 60 km=t +t=1 hAverage velocity=total distance / total time Average velocity=60 /1=60 km/hrAnswer: (c)
Q.21
A car travels the first half of the distance between the places at a speed of 30 km/h and the second half of the distance at 50km/h. The average speed of the car for the whole journey is : [ UGET 1995]
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a) 42.5 k/h
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b) 40.0 km/h
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c)37.5 km/h
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d)35.0km/h
Explanation
Time taken to cover first half=S/ 30Time taken to cover second half=S/ 50Total time=S/ 30 + S/50=8S/150=4S/75Total displacement=2S Average velocity=Total displacement / total time=2S/ (4S/75) Average velocity=37.5 km/hAnswer: (c)
Q.22
P, Q, R are three balloons ascending with velocities U ,4U and 8U respectively. If stones of the same mass be dropped from each, when they are at the same height, then.. [ ISM Dhanabad 1994]
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a) they reach the ground at the same height
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b) stone from P reaches the ground first
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c)stone from R reaches the ground first
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d)stone from Q reaches the ground first
Explanation
Balloon P have the smallest upward velocity thus stone from balloon P will reach first Answer:(b)
Q.23
A body drop from height 'h' with an initial speed zero. Strikes the ground with velocity 3km/h. Another body of the same mass dropped from the same height 'h' with initial speed u'=4km/h. Find the final velocity of second mass, with which it strikes the ground .. [ CBSE 1996]
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a) 3 km/h
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b) 4km/h
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c) 5km/h
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d) 6 km/h
Explanation
Form first mass u=0 final velocity v2=2gh (3)2=2gh 9=2gh --(1) for second mass u=4 km/h v'2=(4)2 + 2gh from equation (1) 2gh=9 v'2=16 + 9 ∴ v=5 Answer: (c )
Q.24
The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity vo. The distance traveled by the particle in time t will be
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a)vot + (1/3) bt2
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b) vot + (1/3) bt3
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c)vot + (1/6) bt3
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d)vot + (1/2) bt2
Explanation
Acceleration a=btBy taking integration we will get equation for velocityBy integrating velocity equation we get equation for displacementAnswer: (c)
Q.25
The position of a particle moving in the xy-plane at any time t is given by x=(3t2 -6t) m , y=( t2 -2t) m. Select the correct statement about the moving particle from following .. [ PMT 1995]
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a) The acceleration of the particle is zero at t=0 second
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b) The velocity of the particle is zero at t=0 second
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c)The velocity of the particle is zero at t=1 second
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d)The velocity and acceleration of particle are never zero
Explanation
Derivative first order derivative of both equation we get vx=6t - 6 vy=2t -2 Now at t=1 both vx and vy are zeroAnswer: (c)
Q.26
A balloon is at height of 81 m and is ascending upwards with velocity of 12 m/s. A body of mass 2 kg weight is dropped from it. If g=10 m/s2, the body will reach the surface of the earth in.. [ MPPMT 1994]
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a) 1.5 s
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b) 4.025 s
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c)5.4 s
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d)6.75 s
Explanation
h=81 mu=-12 m/sg=10 m/s2h=ut + ½ g t281=-12t + ½ 10 × t2∴ 5t2 -12t -81=0 on solving t=-3s and t=27/5=5.4 s Answer:(c)
Q.27
The deceleration experienced by a moving motor boat after its engine is cut off, is given by dv/dt=-kv3, where k is constant. If vo is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time t after the cut-off is : [ CBSE 1994]
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a) vo
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b) vo / 2
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c) vo e-kt
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d)
Explanation
dv/dt=-kv3 dv / v3=-kdt Integrating we get Answer: (d)
Q.28
A body is projected up with speed 'u' and the time taken by it is 'T' to reach maximum height H. Pick out the correct statement.. [ EAMCT 1995]
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a)It reaches H/2 in T/2 time
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b) It acquires velocity u/2 in T/2 sec
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c)Its velocity is u/2 at height H/2
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d)same velocity at 2T
Explanation
At end of 2T speed will be same but direction will be opposite that of earlier hence option (d) is not correctv=u - gt at t=T v=0 thus u=gTAt time T/2 v=u - g(T/2)v=gT -(gT/2)=(gT/2)=u/2Answer: (b)
Q.29
The velocity of body depends up on time according to the equation v=20 + 0.1tthe body is undergoing : [ MNR 1995]
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a) uniform acceleration
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b) uniform retardation
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c)non-uniform acceleration
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d)zero acceleration
Explanation
by taking derivative of equation we will get equation for acceleration as a=0.2t which is time dependant thus option (c) is correctAnswer: (c)
Q.30
A particle is moving with uniform acceleration travels 24 m and 64 m in the first two consecutive intervals of 4 sec each. Its initial velocity is .. [ MPPMT 1995]
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a) 1 m/s
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b) 10 m/s
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c) 5 m/s
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d) 2m/s
Explanation
s=ut + ½ a(t)2 for first interval t=4 sec , S=24 24=u(4) + ½ (a) (4)2 24=4u + 8a or 2a+u=6 --(1) For next interval total time t=4+4=8 and distance S=24+ 64=88m 88=u(8) + ½ (a) 82 88=8u + 32a 11=u + 4a 4a +u=11 --(2) form equation (1) and (2) we get a=2.5 m/s2 putting value of a in equation (2) we get u=1 m/s Answer: (a)
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