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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 5
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Q.1
A particle is thrown vertically upwards. Its velocity at half of the height is 10 m/s, then the maximum height attended by it will be (g=10 m/s2 --[ AIIMS 1999]
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a)10 m
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b)20 m
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c)15 m
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d)25 m
Explanation
Let maximum height be hu2=2gh At half the height v=1010=u2-2g(h/2) 100=2gh - gh 100=gh 100=10 ×h h=10 m Answer:(a)
Q.2
a body is released from the top of the tower H m high. It takes 't' sec to reach the ground. Where is the body after 't/2' second after released? --[ AIIMS 2000]
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a) at 3H/4 metre from the ground
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b) at H/2 metre from ground
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c) at H/6 metre from the ground
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d) at H/4 metre from the ground
Explanation
from the equation for free fall H=½ gt2 --(1) After t/2 sec height H' is H'=½ g (t/2)2 H'=½ g (t)2× (1/4) --(2) from equation (1) and (2) H'=H/4 from top Thus distance of object from ground is 3H/4 Answer: (a)
Q.3
If car accelerated uniformly to a speed of 144 km/h in 20 sec it covers a distance .. [ AIIMS 1997]
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a)400 m
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b) 1440 m
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c)2880 m
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d)25 m
Explanation
Final speed=144 km/h=40 m/s u=0 and time t=20secv=u + at a=40 /20=2 m/s2From equation s=ut + ½ at2s=½ × 2 × (20) 2s=400 mAnswer: (a)
Q.4
A body starts from rest with an acceleration aAfter two seconds another body B starts from rest with an acceleration aIf they travel equal distance in fifth second after the starts of A, the ratio of a1 :a2 will be equal to [ AIIMS 2001]
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a) 9:5
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b) 5:7
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c)5:9
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d)7:9
Explanation
Body A gets 5 sec. thus displacement in fifth seconds S=(a1 /2 )( 2× 5 - 1)=a1 (9/2) --(1)Body gets 3 sec, thus displacement in third seconds S=(a2 /2) ( 2× 3 - 1)=a2 (5/2) --(2)from equation (1) and (2) we get (9a1/2)= (5a2/2)a1 : a2=5:9Answer: (c)
Q.5
A ball is dropped from a bridge 122.5m high. After the first ball has fallen, a second ball is thrown down after 2s it. What must be the initial velocity of the second ball be, so that both the balls hit the surface of water at the same time? [ AIIMS 1997]
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a)26.1 m/s
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b) 9.8 m/s
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c)55.5 m/s
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d)49 m/s
Explanation
time taken by first ball to reach water h=½ (g) t2 122.5=½ (9.8) t2 t=5 secSecond ball gets 3 sec to reach the water 122.5=u(3) + ½(9.8) (3)2122.5=3u + 44.1 u=26.1 m/s Answer:(a)
Q.6
A ball is thrown upwards. Its height varies with time as follows If the acceleration due to gravity is 7.5 m/s2, then the height h is : [ AIIMS 2011]
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a)10 m
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b) 15 m
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c)20 m
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d)25 m
Explanation
Velocity at highest point becomes zero∴ 0=u - at or u=at u=7.5 × 3.5 u=26.25 m/s displacement along y axis is given by y=ut - ½ × 7.5 × t2 dispalcement along y axis at t=1 sec y1=26.25× 1 - ½ 7.5 × (1)=22.50 m dispalcement along y axis at t=2 sec y2=26.25 × 2 -½ × 7.5 × ×22=37.5 m h=y2 - y1h=37.5-22.5=15 m Answer: (b)
Q.7
the position (x) of a particle at any time (t) is given by x(t)=4 t3 - 3t2 + 2 The acceleration and velocity of the particle at any time t=2 sec are respectively
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a) 16 m/s2 and 22 m/s
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b) 42 m/s2 and 36 m/s
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c)48 m/s2 and 36 m/s
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d)12 m/s2 and 25 m/s
Explanation
We have x(t)=4t3 -3t2 + 2 v=dx/dt=12t2 - 6t a=dv/dt=24t-6 v at t=2s is 12(2) 2 - 6(2)=36 m/sand a=24×2-6=42 m/s2Answer: (b)
Q.8
a ball is thrown vertically upwards. Which of the following plot represents the speed-time graph of the ball during it flight if the air resistance is not ignored? [ AIIMS 2003]
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a)
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b)
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c)
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d)
Explanation
For a body going in upward direction v=u - gtslope of the graph=-g constantBut when we take into account the effect of air resistance it will have sharper slope. Answer:(d)
Q.9
the velocity-displacement graph of a particle moving along a straight line is shownwhich graph correctly represent acceleration Vs time graph [IIT 2005]
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a)
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b)
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c)
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d)
Explanation
The equation for the v-x graph is Differentiating the above equation w.r.t t, we get comparing equation with equation of straight line y=mx + cHere slope of graph is positive and y-intercept c is negativethe above conditions are satisfied by option(a) Answer: (a)
Q.10
An object is thrown in vertically upward direction. Which of the following velocity time graph is appropriate for it?
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a)
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b)
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c)
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d)
Explanation
for upward motion equation is v=u - gt graph is straight line and y intercept is positive slope of line is negative at maximum height velocity becomes zero and object starts downward motion for free fall equation is v=gtis again a strain line with positive slope and no intercept on y axis If upward direction velocity is considered positive then down ward direction velocity will be negative only option (d) satisfy all conditionsAnswer: (c)
Q.11
A car moving over a straight path, covers a distance x with constant speed of v1 and the same distance with constant speed of vAverage speed of a car would be...
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a)
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b)
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c)
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d)
Explanation
from given time t1 for first section of path=x/v1time t1 for second section of path=x/v2Total time taken t Total displacement=2xAverage velocity=Answer: (c)
Q.12
A ball is thrown in vertically upward direction. Neglecting the air resistance acceleration of a ball in will be..
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a) zero
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b) continuously increasing
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c)remains constant
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d)increases when ball is going up and will decrease when it is coming down
Explanation
Answer:(c)
Q.13
Figure shows the x-t graphs of car A and B. The velocity of car A with respect to car be will be...
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a) +5 m/s
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b) -2.5 m/s
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c) -5 m/s
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d) +2.5 m/s
Explanation
Slop of graph give velocity velocity of car A VA=(15-0) / (6-0)=2.5 m/s velocity of car B VB=(15-0) /( 6 -3)=5 m/s Relative velocity of Car A with respect to B VAB=VA - VB VAB=2.5 - 5=-2.5 m/s Answer: (b)
Q.14
A car moving with velocity of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car moving at a speed of 100km/hr, the minimum stopping distance is [ AIEEE 2003]
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a)12 m
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b) 18 m
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c)24 m
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d)6 m
Explanation
Retardation is constant in both cases and final velocity is zerofrom formula v2=u2 -2asu12=2as1 --(1)u22=2as2 --(2)Taking ratio of equation (1) and (2)u1=50 km/hr, u2=100 km/hr, s1=6 ms2=4 × 6=24 mAnswer: (c)
Q.15
An automobile traveling with a speed of 60 km/hr, can brake to stop within a distance of 20m. If car is going twice as fast i.e. 120 km/hr, the stopping distance will be ..[ AIEEE 2004]
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a) 60 m
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b) 40 m
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c)20 m
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d)80 m
Explanation
Solution same as question Q133Answer: (d)
Q.16
The relation between time t and distance x is t=ax2 + bxwhere a and b are constants. the acceleration is ... [ AIEEE 2005]
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a) 2bv3
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b) -2abv2
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c)2av2
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d)-2av3
Explanation
take first order derivative with respect to x reciprocal of equation is v taking derivative of above equation with time we get accelerationnow v=(2ax+b)-1 and dx/dt=v Answer:(d)
Q.17
A ball is released from the top of a tower of height h metre. It takes T sec to reach the ground. What is the position of the ball at T/3 .. [ AIEEE 2004]
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a) (8/9) h
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b) (7/9) h
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ac) h/ 9
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d) (17/18) h
Explanation
From the equation h=½ g T2 h'=½ g (T/3)2 ∴ h'=h/9 from top of tower ∴ height form ground=h - (h/9)=(8/9) hAnswer: (a)
Q.18
Two bodies of different masses m1 and m2 are dropped from different heights 'a' and 'b'. the ratio of times taken by the two to drop through these distance is
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a)a:b
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b) m1 :m2 : b/a
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c)√a : √b
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d)a2 : b2
Explanation
Use formula s=ut + ½ gt2Answer: (c)
Q.19
A body is released from the great height and falls freely towards the earth. Another body is released from the same height, one second later. Then the separation between the two bodies, two seconds after release of the second body is
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a) 4.9 m
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b) 9. m
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c)19.6 m
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d)24.5 m
Explanation
Answer: (d)
Q.20
The initial velocity of the particle is 10m/s and its retardation is 2m/secThe distance travelled by the particle in 5th second of its motion is
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a) 1 m
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b) 19 m
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c)50 m
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d)75 m
Explanation
Answer:(a)
Q.21
A ball is dropped from a bridge 122.5 m high. After the first ball has fallen for 2 sec, a second ball is thrown straight down after it. What must be the initial; velocity of the second ball, so that both the balls hit the surface of water at same time :
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a) 49 m/s
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b) 55.5 m/sec
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c) 26.1 m/s
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d) 9.8 m/s
Explanation
First ball s=½ gt2 122.5=½( 9.8) (t2) t=5sec second ball time=5-2= 3 sec 122.5=3u + ½ (9.8) (3)2 u=26.1 m/sec Answer: (c)
Q.22
The distance x covered in time t by a body having initial velocity u and having a constant acceleration a is given by x=ut + ½ at2 This result follow 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
Answer: (d)
Q.23
The displacement y ( in metres) of a body varies with time ( in seconds) according to the equation y=-(2/3)t2 + 16t +2after how long does the body comes to rest?
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a) 8 sec
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b) 10 sec
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c)12 sec
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d)14 sec
Explanation
Answer: (c)
Q.24
A stone thrown vertically upwards with a speed of 5m/s attains a hight HAnother stone thrown upwards from the same point with speed of 10m/s attains height HThe correct relation between H1 and H2 is
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a)H2=4 H1
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b) H2=3 H1
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c) H1=2 H2
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d) H1=H2
Explanation
Answer:(a)
Q.25
A body is dropped from a height of 490 m from the ground. It will hit the ground after
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a) 30 sec
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b) 20 sec
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c) 10 sec
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d) 5 sec
Explanation
Answer: (c)
Q.26
A body of mass 3kg falls from the multi storey building 100 m high and buries itself 2 m deep in the sand. The time of penitaration will be
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a)9 sec
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b) 0.9 sec
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c)0.09 sec
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d)10 sec
Explanation
Use v2=u2 + 2ghon solving we get v=√1960 m/sThis is the velocity with which the stone hits the ground To find retardation use v2=u2 + 2ah but v=0 and u=√1960 m/s on solving we get a=-490 m/s2Now using v=u + at here v=0 and u=√1960 m/s on solving we get t=0.09 secAnswer: (c)
Q.27
A stone is dropped from the top of a toll tower and after 1 second another stone is dropped from a balcony 20 m below the top. If both the stones reaches the ground at the same time then the height of the tower is
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a) 11.2 m
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b) 312.5
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c)31.25 m
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d)3.125 m
Explanation
Lt H be the height of tower then ball dropped from top H=½ gt2 --(1) ball dropped for 20 m below the top after 1second H-20=½ g ( t-1)2 --(2) Putting value of H from eq(1) in eq(2) we get ½ gt2 - 20=½ (t2 -2t -1) t=249.9/9.8 sec putting the value of t in eq(1) we get h=31.25 m Answer: (c)
Q.28
Which of the following equation represents the motion of a body with finite constant acceleration. In this equation y represents displacement and a,b,c are the constant of the maotion
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a) y=a/t + bt
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b) y=at
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c)y=at + bt2
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d)y=at + bt2 + ct3
Explanation
take derivative of y with respect to time for two times to determine acceleration Answer:(c)
Q.29
An object is projected with a velocity of 100m/s. It will strike the ground in approximately ( g=10 m/sec2)
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a) 10 sec
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b) 20 sec
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c) 15 sec
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d) 5 sec
Explanation
Answer: (b)
Q.30
A train of length 100 is crossing a bridge 200 m in length at the speed of 72 km/hr. the time taken by the train to cross the bridge is
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a)24 sec
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b) 15 sec
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c)10 min
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d)10 sec
Explanation
When last bogey of train crosses the bridge we say train has crossed the bridge thus total distance=200 +100=300 speed of train=72 km/hr=20 m/secAnswer: (b)
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