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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 2
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Q.1
If a ball is thrown vertically upwards with speed of 'u', the distance covered during the last 't' second of its ascent is .. [ CBSE-PMT 2003]
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a)( u+gt) t
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b) ut
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c)½gt2
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d)ut - ½ gt2
Explanation
Let body takes T sec. to reach maximum height. final velocity v=0 Thus 0=u -gT T=u/g u is initial velocity, g is negative as object is moving against the direction of gradational acceleration Velocity attended by the body in( T-t) sec is V=u -g(T-t) V=u -gT -gt V=u -g(u/g ) + gt V=gt Distance traveled in last t' seconds will be S=ut -½gt2 S=(gt)t - ½gt2 S=½gt2 Answer: (c)
Q.2
The water drops fall at regular intervals from a tap 5m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? take g=10 m/s2 [ CBSE-PMT 1995]
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a)1.25 m
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b) 2.5 m
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c)3.75 m
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d)5.0 m
Explanation
Height of tap is 5m and g=10 m/s2 Time taken for first drop to reach the groundS=ut +½gt25=½(10) t2t=1 secIt means third drop leaves 1 sec after first, since interval is regular thus Second drop leaves after 0.5seconds after firstDistance traveled by second drop in 0.5 secS=ut +½gt2S=½(10)(0.5)2S=1.25 mThere fore distance of Second drop from ground=5-1.25=3.75m Answer:(c)
Q.3
If a ball is thrown vertically upwards with velocity of 40m/s, then velocity of the ball after two seconds will be ( g=10 m/s2) [ CBSE-PMT 1996]
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a) 15 m/s
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b) 20 m/s
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c) 25 m/s
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d) 28 m/s
Explanation
According to equation of motion v=u -gt here initial velocity u=40 m/s g=10 m/s2 time t=2 sec v=40-10(2)=20 m/s Answer: (b)
Q.4
A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height. How high does the ball rise? take g=10 ms2[ CBSE-PMT 2005]
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a)10 m
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b) 5 m
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c)15 m
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d)20m
Explanation
Ball at the half of maximum height H:velocity at H/2 is 10 m/sAccording to equation of motion v2=u2 -2gshere initial velocity be u , final velocity 10 m/s and s=H/2(10)2=u2 -2g(H/2) ∴ 100 + gH=u2 --equ(1) Ball at the maximum height H At maximum height final velocity v=0 0=u2 -2g(H) Thus u2=2g(H) --eq(2) From equation 1 and 2 100 + gH=2gH g=10 m/s2 ∴ H=10 mAnswer: (a)
Q.5
A man throws balls with the same speed vertically upward one after the other at an interval of 2 seconds. what should be the speed of the throw so that more than two balls are in the sky at any time? g=m/s2
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a) only with speed 1.6 m/s
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b) more than 19.6 m/s
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c)At least 9.8 m/s
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d)Any speed less than 19.6 m/s
Explanation
Let the required speed of through be u m/s. The time taken to reach maximum height t=u/gFor two balls to remain in air at any time, t must be greater than 2.∴ u/g >2 ⇒ u > 2(g) u > 19.6 m/sAnswer: (b)
Q.6
A bus travells between two points A ans B. V1 and V2 are it average speed and average velocity then
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a) v1 > v2
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b) v1 < v2
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c) v1=v2
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d) depends on situation
Explanation
Answer: (d)
Q.7
Two bodies A of mass 1 kg and B of mass 3 kg, are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is .. [ CBSE-PMT 2006]
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a) 12/5
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b) 5/12
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c)4/5
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d)5/4
Explanation
From equations for free fallh=½gt2let t1 and t2 be the time taken by A and B respectively to reach ground For first body, 16=½gt12 For second body 25=½gt22 Answer:(c)
Q.8
A body is thrown vertically upward from the ground. It reaches a maximum height of 20 m in 5 sec. After what time, it will reach the the ground from its maximum height position? [ CBSE - PMT 1995]
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a) 2.5 sec
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b) 5 sec
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c) 10 sec
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d) 25 sec
Explanation
Time taken by the body to reach the ground from some height is the same as time taken to reach that height. Therefore, time taken to reach the ground from its maximum height is 5 sec. Answer: (b)
Q.9
A stone is released with zero velocity from the top of a tower, reaches the ground in 4 sec. The height of the tower is ( g=10 m / s2)
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a)20 m
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b) 40 m
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c)80 m
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d)160 m
Explanation
h=ut + ½gt2u=0, time t=4 sec, gravitational acceleration g=10 m/s2h=½×10×(4)2h=80 mAnswer: (c)
Q.10
The displacement of a particle is represented by the following equations=3t3 + 7t2 + 5t + 8s is in meters and t in second. The acceleration of particle at t=1 s is [CBSE-PMT 2000]
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a) 14 m/s2
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b) 18 m/s2
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c)32 m/s2
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d)zero
Explanation
v=dx/dt=9t2 + 14t + 5 a=dv/dt=18t + 14 at t=1 s acceleration a=18 + 14=32 m/s2 Answer: (c)
Q.11
The displacement of a particle varies with time(t) as s=at2 - btThe acceleration of the particle is zero at time ...[ CBSE-PMT 1997]
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a) a/b
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b) a/(3b)
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c)(3b)/ a
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d)(2a) / (3b)
Explanation
velocity v=dx/dt=2at -3bt2acceleration a=dv/dt=2a -6bt0=2a -6bt ∴ t=a/ (3b) Answer:(b)
Q.12
If a car rest accelerates uniformly to speed of 144km/h in 20s, it covers a distance of ... [ CBSE-pmt 1997]
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a) 2880 m
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b) 1440 m
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c) 400 m
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d) 20 m
Explanation
Initial velocity of car (u)=0 Final velocity of car (v)=144 km/h=40 m/s time taken=20s We know that , v=u + at 40=a × 20 a=2 m/s2 According to equations of motion v2=u2 + 2as Answer: (c)
Q.13
The displacement x of a particle varies with time t as x=ae-αt + beβt, where a, b, α and β are positive constants. The velocity of the particle will .. [ CBSE-PMT 2005]
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a)be independent of α and β
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b) drop to zero when α=β
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c)go on decreasing with time
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d)go on increasing with time
Explanation
Velocity v=dx/dt=-aαe-αt + bβeβt Since second derivative of x is positive greater than zero or acceleration is positive we can conclude that velocity goes on increasing with timeAnswer: (d)
Q.14
A particle moves along a straight line OX. At time t(in seconds) the distance x(in meters) of the particle from O is given by x=40 + 12t - tHow long would the particle travel before coming to rest? [ CBSE-PMT 2006]
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a) 40 m
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b) 56 m
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c)16 m
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d)24 m
Explanation
particle comes to rest then v=0v=dv/dx=12 - 3t20=12 -3t2t=2 sec, so particle will come to rest after 2 sec. Distance traveled in two seconds can be calculated by substituting t=2 in the given equation of displacementx=40 +12(2) - 23x=56 m At t=0 , particle is at , let's say x distance ,from O ; then putting t=0 in the given displacement-time equation we get; x = 40 m thus distance travelled = 56-40=16m Answer: (c)
Q.15
The position of a particle with respect to time t along x-axis is given by x=9t2 - t3, where x is in meters and t in second. What will be the position of this particle when it achieve maximum speed along the + ve X direction? [ CBSE-PMT 2007]
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a)54 m
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b)81 m
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c)24 m
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d)32 m
Explanation
Speed v=dx/dt=18t -3t2 For maximum speed dv/dt=0 dv/dt=18 -6 t 0=18 -6t t=3 second∴ Xmax=81 -27=54 m Answer:(a)
Q.16
A particle moving along x-axis has acceleration f, at time given by fo and T are constant. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0 , the particle’s velocity (vx ) is [ CBSE - PMT 2007]
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a) ½ foT2
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b) foT2
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c) ½ foT
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d) foT
Explanation
here f=dv/dt ( acceleration) Answer: (c)
Q.17
the distance traveled by a particle starting from rest and moving with an acceleration (4/3) m/s2 in the third second is .. [ CBSE-PMT 2008]
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a)6 m
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b) 4 m
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c)(10/3)m
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d)(19/3) m
Explanation
Distance traveled in nth second is given by Substituting u=0, a=4/3, and n=3 on solving equation we get s=10/3 mAnswer: (c)
Q.18
The displacement time graph of moving particle is shown belowThe instantaneous velocity of the particle is negative at the point... [ CBSE-PMT 1994]
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a) D
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b) F
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c)C
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d)E
Explanation
At point E slope of the curve negativeAnswer: (d)
Q.19
A car moves a distance of 200m. it covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. find the value of v. [ CBSE-PMT 1991]
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a)56 km/h
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b) 60 km/h
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c)50 km/h
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d)48 km/h
Explanation
Since distance is same in both the half we can use following formula Answer:(b)
Q.20
A body dropped from top of a tower falls through 40m during the last two seconds of its fall. The height of tower is ( g=10 m/s2) [ CBSE-PMT 1991]
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a) 60 m
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b) 45 m
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c) 80 m
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d) 50 m
Explanation
Let the body falls through the height of tower in n seconds. Formula for displacement in nth secondTotal distance in last 2 seconds S=Sn + S(n-1) Distance traveled in 'n' second is the height of tower H=ut + ½ at2H=½ × 10×32H=45 mAnswer: (b)
Q.21
Which one of the following equation represents the motion of a body moving with constant finite acceleration? in these equation, y denotes the displacement in time t and p, q. and r are constant... [ CBSE-PMT 2011]
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a)y=(p+qt)(t+pt)
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b) y=p + t/r
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c)y=(p+t)(q+t)(r+t)
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d)y=(p+qt) / rt
Explanation
If equation of displacement is quadric then second derivative will be constant. Hence motion with constant acceleration is represented by a quadric equation of tY=(p+qt)(r+pt)=pr +qrt+p2t + pqt2 Second derivative of above equation gies acceleration a=pq=constant Answer: (a)
Q.22
the slope of the velocity time graph for retarded motion is .. [ CBSE-PMt 2007]
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a) positive
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b) negative
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c)zero
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d)can be +ve , -ve or zero
Explanation
The slope of velocity-time graph denotes acceleration. For a retarded motion acceleration is negativeAnswer: (b)
Q.23
A train A which is 120 m long is running with velocity 20 m/s while train B which is 130 m long is running in opposite direction with velocity 30 m/s. What is the time taken by train B to cross the train A? [ CBSE-PMT 2009]
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a) 5 sec
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b) 25 sec
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c)10 sec
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d)100 sec
Explanation
Guard of train A has to cross the guard of second train B. thus total distance=sum of the length of trains=130+120=250 mSince both the trains are approaching velocities will get added Velocity of train A with respect to train B=20+30=50 m/s time taken=total distance / velocitytime taken=250 / 50=5 sec Answer:(a)
Q.24
The driver of a car traveling at velocity 'v' suddenly sees a broad wall in front of him at a distance d. He should.. [ CBSE-PMT 2007]
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a) brake sharply
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b) turn sharply
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c) (a) and (b) both
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d) none of the above
Explanation
Answer: (a)
Q.25
Choose the wrong option
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a)Speed can never be negative
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b) When particle returns to the starting point, its average velocity is zero but average speed is not zero
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c)Displacement does not tell the nature of the actual motion of the particle
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d)If the velocity of a particle is zero at an instant, its acceleration should also be zero at that instant
Explanation
When object attends maximum height its velocity momentarily zero but acceleration is not zeroAnswer: (d)
Q.26
A car is moving on a road and rain is falling vertically, then, select the correct option
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a) The rain will strike the hind screen only
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b) The rain will strike the front screen only
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c)The rain will strike the both screen only
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d)The rain will not strike both the screen
Explanation
green line indicates the relative velocity and direction of rain with respect to car, rain will hit the front screen onlyAnswer: (b)
Q.27
A bus traveled the first one third distance at the speed of 10km/hr, the next one third distance at 20km/hr and the last one third at 60km/hr. The average speed of the bus is
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a) 18 km/hr
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b) 16 km/hr
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c)9 km/hr
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d)48 km/hr
Explanation
Let the total distance be 3dAccording to formula t=S/v time taken to travel first one third distance t1=d/10 hrtime taken to travel second one third distance t2=d/20 hrtime taken to travel third one third distance t3=d/60 hrTotal time taken=t1 + t2+t3 t=(d/10)+ (d/20) + (d/60)=d/6Average speed=Total distance / total timeAverage speed=3d /( d/6)=18km/hr Answer:(a)
Q.28
A body is released from the top of the tower H meter high. It takes t seconds to reach the ground. Where is the body t/2 seconds after release
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a) At H/2 meters from ground
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b) At H/4 meters from ground
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c) At 3H/4 meters from ground
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d) At H/6 meters from the ground
Explanation
h=ut + ½ a t2 given u=0 H=½ g t2 Distance in t/2 time from top H'=½ g(t/2)2Taking ratio of H'/H=1/4Thus H'=H/4 from top∴ distance from ground=H-H'=3H/4 Answer: (c)
Q.29
A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to .. [ BHU 1995]
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a)t1/2
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b) t3/4
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c)t3/2
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d)t2
Explanation
Power P=F×V=maVNow S=½ a t2 a=2S/t2 V=S/t Thus P=m (2S/t2) S/t) P=m 2S2 / t3 P is constant S2∝ t3 S ∝ t 3/2Answer: (c)
Q.30
If 'a', 'b' and 'c' be the distances traveled by the body during xth ,yth , zth second from start, then which of the following relations is true
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a) a(y-z) + b(z-x)+ c(x-y)=0
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b) a(x-y)+b(y-z)+c(z-x)=0
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c)a(z-x)+b(x-y)+c(y-z)=0
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d)ax+by+cz=0
Explanation
From the equation for displacement in nth seconds is a=u + (2x-1)(f/2)b=u + (2y-1)(f/2)c=u + (2z-1)(f/2)a-b=(x-y)f -- (1)b-c=(y-z)f -- (2)c-a=(z-x)f --(3)Multiplying equation 1 by c, equation 2 by a, equation by 3 by b we getc(a-b)=c(x-y)f --(4)a(b-c )=a(y-z)f --(5)b(c-a)=c(z-x)f --(6) Adding above equation 4,5,6 we get ca-cb+ab-ca+bc-ba=c(x-y)f +a(y-z)f + c(z-x)f c(x-y)f +a(y-z)f + c(z-x)f=0Answer: (a)
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