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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 8
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Q.1
In the figure there is a graph of a → x for a moving particle. Hence da/dt=...v
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a)xo / ao
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b) -xo / ao
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c)-ao / xo
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d)ao / xo
Explanation
Equation for the graph a=(-ao/xo)x - ao thus da/dt=(-ao/xo)(dx/dt) da/dt=(-ao/xo)(v)Answer: (c)
Q.2
A particle is moving in a straight line with initial velocity of 10 m/s. A graph of acceleration → time of the particle is given in the figure. Find velocity at t=10 s
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a) 25
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b) 35
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c)45
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d)15
Explanation
Area covered under graph=change in velocity=25 m/s Final velocity=Initial + change in velocity=10 +25=35 m/sAnswer: (b)
Q.3
A graph of moving body with constant acceleration is given in the figure. What is the velocity after time t?
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a)
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b)
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c)
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d)
Explanation
Answer:(a)
Q.4
The graph given in the figure shows that the body is moving with .....
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a) increasing acceleration
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b) decreasing acceleration
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c) constant velocity
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d) increasing velocity
Explanation
Answer: (b)
Q.5
Here are the graphs of velocity → time of two cars A and B, Find the ratio of the acceleration after time t.
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a)1/√3
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b) 1/3
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c)√3
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d)3
Explanation
Answer: (b)
Q.6
Here is a velocity - time graph of a motorbike moving in one direction. Calculate the distance covered by it in last two seconds.
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a) 10 m
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b) 20 m
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c)50 m
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d)25 m
Explanation
Answer: (a)
Q.7
In the following figure acceleration (a) v/s time (t) graph is given. Hence V ∝ .....
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a)a
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b)√a
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c)a2
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d)a3
Explanation
Answer:(c)
Q.8
The graph of displacement (x) → time (t) for an object is given in the figure. In which part ofthe graph the acceleration of the particle is positive ?
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a) OA
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b) AB
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c) O - A - B
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d) acceleration is not positive at any part.
Explanation
Draw tangent at O ,A and B Slope of tangent gives velocity. Now Value of velocity at A is less than velocity O Velocity at B is negative Thus acceleration is never positive or always negative Answer: (d)
Q.9
An object moves in a straight line. It starts from the rest and its acceleration is 2ms–After reaching a certain point it comes back to the original point. In this movement its acceleration is -3ms-till it comes to rest.The total time taken for the movement is 5 second. Calculate the maximum velocity.
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a) 6 m/s
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b) 5 m/s
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c)10 m/s
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d)4 m/s
Explanation
Using formula v=u +at For fist case Vmax=2×t1 t1=Vmax /2 For second case v=o and u=Vmax Vmax=2×t2 t1=Vmax /3 given t1 + t2=5 sec. Thus Answer: (a)
Q.10
Particles A and B are released from the same height at an interval of 2 s. After some time the distance between A and B is 100m. Calculate time t.
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a)8 s
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b)6 s
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c)3 s
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d)12 s
Explanation
Distance travelled by A S=½ gt2 ( u=0 because free fall)Distance travelled by B S'=½ g(t-2)2 ( as B is dropped after 2 sec) Difference in position 100 m given S-S'=100 m Thus ½ gt2 - ½ g(t-2)2=100 take g=10 t2 - (t-2)2=20 4t - 4=20 t=6 secAnswer: (b)
Q.11
A particle is moving in a circle of radius R with constant speed. It covers an angle θ in sometime interval. Find displacement in this interval of time.
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a)2R sin(θ/2)
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b) 2Rcos(θ/2)
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c)2Rcosθ
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d)2Rsinθ
Explanation
Given Angle AOB=θ ans OA=OB=R thus angle AOM=θ/2From Trigonometric relation AM=Rsinθ/2Thus AB=2Rsinθ/2 Answer:(a)
Q.12
A freely falling stone crashes through a horizontal glass plate at time t and losses half of its velocity. After time t/2 it falls on the ground. The glass plate is 60 m high from the ground. Find the total distance travelled by the stone. [g=10ms–2]
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a) 120 m
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b) 80 m
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c) 100 m
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d) 140 m
Explanation
Velocity before crash=gt Distance travelled after crash=60 m given velocity becomes half=gt/2 and time t'=t/2 Thus 60=(gt/2)t' + ½ gt'2 60=gt2/4 + gt2/8 480=3gt2 t=4 sec Distance before crash h=½ gt2 h=½ 10 (4)2=80 m Total distance travelled=80 +60=140 m Answer: (d)
Q.13
A freely falling object travels distance H. Its velocity is V. Hence, in travelling further distance of 4H its velocity will become ....
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a)√3 V
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b) √5 V
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c)2V
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d)3V
Explanation
Using formula v2=u2 + 2gH case I : v2=2gH --eq(1) Case II: Furthere 4H v'2=v2 + 8gH From eq(1) v'2=v2 + 4v2 v'2=5v2 v'=√5 vAnswer: (b)
Q.14
Two particles P and Q get 5 m closer each second while travelling in opposite direction. They get 1 m closer each second while travelling in same direction. The speeds of P and Q are respectively ...
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a)5 m/s, 1 m/s
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b)3 m/s, 4 m/s
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c)3 m/s, 2 m/s
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d)10 m/s, 5 m/s
Explanation
Vp be the velocity of P and Vq be the velocity of Q 5 m closer each second give relative velocity Vp + Vq=5 --eq(1)and get 1 m closer each second while travelling in same direction Vp - Vq=1 --eq(2) Adding eq(1) and eq(2) 2Vp=6 m/s Vp=3 m/s Vq=2Answer: (c)
Q.15
Motion of a particle is described by following equation, where v, y and A are velocity distance and a constant respectively. Find the acceleration of the particle
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a)1 unit
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b) 2 unit
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c)1/2 unit
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d)3 unit
Explanation
Acceleration a=dv/dt Answer:(c)
Q.16
A goods train is moving with constant acceleration. when engine passes through a signal its speed is U. Midpoint of the train passes the signal with speed V. What will be the speed of the last wagon ?
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a)
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b)
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c)
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d)
Explanation
equation for mid point V2=U2 + 2ax x is the distance of mid point from engine 2ax=V2 - U2 ---eq(1) equation for last wagon V'2=V2 + 2ax From eq(1) V'2=V2 + V2 - U2 V'2=2 V2 - U2 Answer: (d)
Q.17
A ball is thrown vertically upward. What is the velocity and acceleration of the ball at the maximum height ?
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a)–gt m/s, 0
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b)0, –9 ms–2
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c)g ms, 0
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d)0, –g ms–2
Explanation
Answer: (d)
Q.18
The relation between velocity and position of a particle is V=Ax + B where A and B are constants.Acceleration of the particle is 10 ms–2 when its velocity is V, How much is the acceleration when its velocity is 2V.
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a)20 ms–2
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b)10 ms–2
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c)5 ms–2
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d)0
Explanation
Answer: (a)
Q.19
A particle moves on a plane along the path y=Ax3 + B in such a way that dx/dt=C. A, C, B are constant Calculate the acceleration of the particle.
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a) 3AxC jms–2
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b)6AxC2 jms–2
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c)3AxC2 jms–2
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d)(Ci + 3AxC2 j )ms–2
Explanation
Answer:(b)
Q.20
In the triangle AC=5, BC=8 and B=π/6 Find the value of angle A.
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a) sin-10.6
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b) sin-10.8
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c) sin-10.12
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d) sin-10.4
Explanation
Answer: (b)
Q.21
The SR-71 strategic reconnaissance aircraft, the Blackbird,set a world speed record by flying from London to Los Angeles (8,790 km) in 3 hr 47 min 36 s.Compute the average speed in meters per second (m/s).
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a)0.644 m/s
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b) 644 m/s
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c)648 m/s
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d)814 m/s
Explanation
Convert km in meters and time in seconds 3hr 47min 36 sec=13656Use formula average speed=distance / time Answer: (b)
Q.22
In 1976, the SR-71 strategic reconnaissance aircraft, the Blackbird, recaptured the 1000 km (1,000,000 m) closed-circuit-course record (previously held by the Russian MIG-25 Foxbat) at an average speed of 935 m/s. How much time it will take to for one round
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a) 1068.38 sec
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b) 17.8 min
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c)35.65 min
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d)1000 sec
Explanation
To and fro thus total distance=1000 km + 1000 km=2000 km=2000,000 m Average speed=935 m/s use formula time period=distance / average speed Answer: (c)
Q.23
Suppose you fire a bullet (speed 1600 m/s) in a shooting gallery and hear the gong on the target ring 0.731 s later. Taking the speed of sound to be 330 m/s and assuming the bullet travels straight downrange at a constant speed, how far does the bullet travel before hitting the target?
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a)1170 m
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b) 330 m
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c)241 m
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d)200 m
Explanation
Let 'x' is the distance between gun and target Let bullet traveled 'x' distance in t1 seconds given speed of bullet 1600 m/s thus time taken by bullet to heat the target t1=distance / average speed t1=x/ 1600 and sound traveled x distance in t2 seconds given speed of sound 330 m/s time taken by sound to reach the listener t2=distance / average speed t2=x/330 Here sound heard after 0.731 s after firing bullet. thus total time t=t1 +t2 seconds total time given 0.731 sec Answer:(d)
Q.24
Julie drives 100 km to Grandmother's house. On the way toGrandmother's, Julie drives half the distance at 35 km/hr and half the distance at 55 km/hr. What is Julie's average speed on the way to Grandmother's house?
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a) 42.73 km/hr
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b) 45 km/hr
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c) 60 km/hr
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d)55 km/hr
Explanation
Total distance=100 km Half distance 50 km traveled at 35 km/ hr thus time period t1=distance /speed t1=50/35 Remaining half distance (50km) traveled at speed 55km/hr thus t1=50/55 Total time taken to travel t=t1 + t2 Now average speed=total distance / total time Answer: (a)
Q.25
A motorcyclist rides onto a road at 22.2 m/s at 1200 noon and maintains that speed for the rest of her journey. At 1201 a car travelling at 27.8 m/s turns onto the road at the same point as did the motorcycle. How many seconds after the motorcycle started will it be passed by the car?
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a)48 sec
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b) 102 sec
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c) 298 sec
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d)78 sec
Explanation
Car started from same point as that of motorcyclist after 1 min as motorcyclist started at 1200 and car at 1201 If t is the time in seconds, when car overtake motorcycle then distance travelled by motorcycle d=speed × time period d=(22.2)tNow time taken by the car to travel distance same as motorcycle to overtake=(t -60) secdistance travelled by the car=27.8 ×(t-60)Since both car and motorcycle have travelled same distance we get(22.2) t=(27.8) (t-60)(22.2) t=(27.8 t) - 1668t=298 secAnswer: (c)
Q.26
A body is thrown with a velocity of 9.8 m/s in the vertical upward direction. Its velocity after one second is
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a) 9.8 m/s
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b) 4.5 m/s
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c)2.45 m/s
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d)zero
Explanation
v=u - gt v=9.8 - (9.8) × 1 v=0Answer: (d)
Q.27
A heavy and light body of same size are dropped from top of a tower in air, from the same height. Which will reach the ground first
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a) lighter body
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b) heavier body
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c)both will reach simultaneously
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d)not sure
Explanation
Since weight of both body is same volume of lighter body is more hence upward buoyant force is more for lighter body. Answer:(b)
Q.28
An object is projected up wards with velocity of 100 m/sec. 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
We know that time taken to go to maximum position is equal to time taken to return back at maximum height final velocity v=0 v=u -gt 0=100 - 10t t=10 sec is time taken to go to maximum height Thus total time of flight=2 ×10=20 sec Answer: (b)
Q.29
A : The speed of a body can be negative R : If the body is moving in the opposite direction of positive motion, then its speed is negative
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a) A and R both true, and R is correct explanation for A
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b) A and R both true and R is not a correct explanation for A
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c) A is true but R is false
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d) Both A and B are false
Explanation
Velocity is vector, can be negative but Speed is a restricted quantity in physics because it only tells how fast something is going Answer:(d)
Q.30
If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1s and 2s is ….[ AIPMT/ NEET part I -2016]
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a)
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b) 3A + 7B
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c)
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d)
Explanation
Displacement formula can be obtain by integration as v = ds/dt ds = (At+Bt2) dt Integrating form 0 to t we get formula for displacement Position at t=1 sec Position at t=2sec Displacement in 1s and 2s is S2-S1 Answer:(c)
0 h : 0 m : 1 s
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