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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 7
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Q.1
If two balls of same density but different masses are dropped from a highest of 100 then:
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a) both will come together on earth
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b) both will come late on the earth
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c) first will come first and second will after that
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d) second will come first and first after
Explanation
Answer: (a)
Q.2
A body starts from rest and move with constant acceleration. ThE ratio of distance covered by the body in nth second to that covered in n second is
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a)1 : n
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b) (2n-1)/ n2
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c)n2 / (2n-1)
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d)(2n-1) / 2n2
Explanation
Answer: (b)
Q.3
Drops of water fall from the roof of the building 9m heigh at regular intervals of time, the first drop reaching the groung at the instant fourth drop starts to fall. What are the distances of the second and third drop from the roof
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a) 6 m and 2 m
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b)4 m and 1 m
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c)6 m and 3 m
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d)4 m and 2 m
Explanation
Answer: (b)
Q.4
A car runs at constant speed on a circular track of radius 100m taking 62.8 sec on each lap. What is the average speed and average velocity on each complete lap
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a) velocity 10 m/sec, speed 10m/sec
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b) velocity zero, speed 10 m/sec
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c)velocity zero and speed zero
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d)velocity 10m/s, speed zero
Explanation
Answer:(b)
Q.5
If x denotes displacement in time t and x=a cos t, then acceleration is
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a) acost
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b) -acost
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c) -asint
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d) asint
Explanation
Answer: (b)
Q.6
Drops of water fall from the roof of a building 9 m heigh at regular intervals of time, the first drop reaching the ground at the same intervals of time, the first drop reaching the ground at the same instant fourth drop starts its fall. What are the distances of the second and third drops from the roof?
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a)4 m and 2m
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b) 6m and 3 m
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c)4 m and 1 m
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d)6 m and 2 m
Explanation
Answer: (c)
Q.7
A car covers the first half distance at 40 km/hr and second half distance at 60km/hr along a straight road. The average speed is
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a) 60km/hr
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b) 52 km/hr
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c)50 km/hr
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d)48 km/hr
Explanation
Answer: (d)
Q.8
A body moving with uniform acceleration has velocities of 30 m/sec and 40 m/sec when passing points P and Q in its path. What is the velocity midway between P and Q
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a) 34.36 m/sec
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b) 35.35m/sec
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c)38.72 m/sec
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d)35.0 m/sec
Explanation
Answer:(b)
Q.9
With what speed must a ball be thrown down for it to bounce 10m higher than its original level? Neglect any loss of energy in striking the ground
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a) 14 m/sec
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b) 20 m/sec
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c) 5 m/sec
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d) none
Explanation
Answer: (a)
Q.10
On a clam day a boat can go across a lake and returned in time To at a speed V. On a rough day there is uniform current speed v to help the onward journey and impede the return journey. If the time taken to go across and return on the rough day be T, then T/To is
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a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (b)
Q.11
A particle moves in the x-y plane according to the law x=kt, y=kt( 1 -αt) where k and α are positive constants and t is time. The trajectory of the particle is
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a) y=kx
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b) y=x - (αx2)/k
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c)y=-(αx2)/k
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d)y=αx
Explanation
Answer: (b)
Q.12
A particle moving with uniform acceleration is found to travel 35 m in 8th second and 51 m in the 12th second. Its velocity is m/s at the begining of 11th second is
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a) 49
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b) 45
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c)47
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d)51
Explanation
Distance travelled in nth second on sloving equation(1) and (2) we gat a=4 m/s2 and u=5 m/s Now V=u + at=5 +4×10=45 m/s Answer:(b)
Q.13
A rubber ball is dropped from a height of 5 metre on a plane, where the acceleration due to gravity is not known. On bouncing it rises to a height of 1.8 metre. The ball loses its velocity on bouncing by a factor of
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a) 16/25
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b) 2/5
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c) 3/5
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d) 9/25
Explanation
Free fall USe V1 = 0 +2g(5) For bounce up use 0 = V2 -2g(1.8) Loss in velocity during bounce = V1 - v2 Divide loss in velocity by V1 So fraction of velocity lost = 1 - (v2 /v1) Answer: (b)
Q.14
As shown in the figure a particle starts its motion from 0 to A. And then it moves from A to B. AB is an arc find the Path length
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a)2r
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b) r + π/3
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c)r( 1+π/3)
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d)π(r+1)/3
Explanation
Distance from O to A=r and length of arc=πr/3 Thus option "c" is correctAnswer: (c)
Q.15
In above problem Q193 What is displacement from A to B
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a) r
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b) r/2
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c)πr/3
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d)2r
Explanation
In given problem angle is π/3 thus if we join AB by a straight line it will be a equilateral triangle of side r . Thus displacement B=rAnswer: (a)
Q.16
Here is a cube made from twelve wire each of length l. An ant goes from A to G through path A-B-C-G. Calculate the displacement.
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a)3l
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b) 2l
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c)(√3)l
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d)l/√3
Explanation
From figure BG=√2 lThus AG=(√3)l Answer:(c)
Q.17
As shown in the figure particle P moves from A to B and particle Q moves from C to D. Displacements for P and Q are x and y respectively then
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a) x > y
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b) x < y
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c) x=y
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d) x ≥ y
Explanation
Answer: (a)
Q.18
The graph of position Vs time shown in the figure for a particle is not possible because ...
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a) velocity can not have two values on one time
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b)Displacement can not have two values at one time
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c)Acceleration can not have two values at one time
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d)A, B and c are true
Explanation
Answer: (d)
Q.19
An ant goes from P to R on a circular path in 20 second Radius OP=10m. What is the average speed and average velocity of it ?
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a) (π/6) m/s, 3 m/s
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b) (π/3) m/s, √3/2
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c)(π/3) m/s,√3
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d)π √6
Explanation
Length of arc=(2π/3) × (op) Speed=Length of arc/ time Displacement PR From figure angle POM=60° thus PM=OPsin60 PM=10 (√3 /2) Thus PR=2PM=10( √3) Velocity=PR/time=√3/2Answer: (b)
Q.20
A car moving over a straight path covers a distance x with constant speed 10 m/s and then the same distance with constant speed of VIf average speed of the car is 16m/s, then V2=....
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a) 30 m/s
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b)20 m/s
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c)40 m/s
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d)25 m/s
Explanation
Since in both the cases distance is same we can use formula On substituting V1=10 and
=16 we get V2= Answer:(c)
Q.21
A car covers one third part of its straight path with speed V1 and the rest with speed VWhat is its average speed ?
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a)
0%
b)
0%
c)
0%
d)
Explanation
Let total distance be 3x One third distance = x with velocity v1 remaining distance 2x with velocity V2 Thus total time taken average velocity=Total distance / time Answer: (a)
Q.22
A particle moves 4m in the south direction. Then it moves 3m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity
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a) 5/7
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b) 7/5
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c)14/5
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d)5/14
Explanation
Answer: (b)
Q.23
A particle is projected vertically upwards with velocity 30ms–Find the ratio of average speeda nd instantaneous velocity after 6s. [g=10m/s2]
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a)2
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b)1/2
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c)3
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d)4
Explanation
From formula V=-u + gt ( particle is going up against gravity u negative g is positive)v=-30 + 10(6) v=-30 m/s it indicates particle has returned to point of projection.thus instantaneous velocity=30 m/s Average velocity Particle have attended maximum height in 3 sec and net three second it returned back maximum height be h then V2=u2 + 2g h 0=(30)2 - 2(10) h h=45 m Particle travelled=45 +45 m=90 m in 6 sec average speed=total distance / time=90 /6=15average speed/ average velocity=15/ 30=1/2 Answer:(b)
Q.24
Given figure shows a graph at acceleration → time for a rectilinear motion. Find average acceleration in first 10 seconds
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a) 10 m/s2
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b) 15 m/s2
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c) 7.5 m/s2
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d) 30 m/s2
Explanation
average acceleration=Change in velocity / time Change in velocity=area under graph=75 /10 option 'c' is correct Answer: (c)
Q.25
A body is moving in x direction with constant acceleration α . Find the difference of the displacement covered by it in nth second and (n–1)th second
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a)α
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b)α/2
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c)3α
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d)3α/2
Explanation
Use formula displacement in nth second and displacement in (n-1) seconds replace n by n-1 find difference Answer: (a)
Q.26
A freely falling particle covers a building of 45m height in one second. Find the height of the point from where the particle was released. [g=10ms–2]
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a) 120m
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b) 125m
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c)25m
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d)80m
Explanation
let u be the velocity when particle is at height 45m Thus from formula d=ut + ½ gt2 45=u(1) + ½(10) (1)2 u=40 m/s Object is free falling from heigh h above building initial velocity u=0 and v=40m/s from formula v2=u2 +2gh (40)2=0 +2(10) h h=80 m height of the point where the particle was released=45 +80=125 mAnswer: (b)
Q.27
The distance travelled by a particle is given by s=3 + 2t + 5t2 The initial velocity of the particle is ...
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a) 2 unit
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b) 3 unit
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c)10 unit
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d) 5 unit
Explanation
Instantaneous velocity v=ds/dt=2+10t at t=0 v=2 units Answer:(a)
Q.28
A particle is thrown in upward direction with Velocity Vo. It passes through a point p of height h at time t1 and t2 so t1 + t2=....
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a) vo / g
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b) 2vo / g
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c) 2h / g
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d) h / 2g
Explanation
h=Vot - ½ gt2 2h=2Vot - gt2 t2 - (2Vo/g)t +2h=0 is a quadratic equation thus have two real values thus t1 + t2=-b/a Here b=-Vo and a=1 t1 + t2=2Vo/gAnswer: (b)
Q.29
Ball A is thrown in upward from the top of a tower of height h. At the same time ball B starts to fall from that point. When A comes to the top of the tower, B reaches the ground. Find the the time to reach maximum height for A.
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a)
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b)
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c)
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d)
Explanation
Time taken by ball A to reach the maximum heigh = ½ ( time taken by ball B to reach ground) time taken by ball B to reach ground h = ½ gt2 t = √ ( 2h/g) Thus Time taken by ball A to reach the maximum heigh = ½ t = √ ( h/2g) Answer: (c)
Q.30
In the figure Velocity (V) → position graph is given. Find the true equation.
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a)
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b)
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c)
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d)
Explanation
Answer: (d)
0 h : 0 m : 1 s
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