MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 14

Two cyclists are travelling the same distance at the speed of

$10$ km\hr and

$15$ km\hr respectively. If one cyclist takes

$40$ minutes or more to cover the distance, then what distance (in km) are they travelling?

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$18$ km
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$20$ km
0%
$15$ km
0%
$12$ km

A system is shown in the figure and man is pulling the rope from both sides with constant speed

$'u'$ . Then the speed of the block will be (

$M$ moves vertically):

0%
$\dfrac{3u}{4}$
0%
$\dfrac{3u}{2}$
0%
$\dfrac{u}{4}$
0%
None of these

Explanation

Imagine block A and B is being pulled at the speed u instead of man pullling the string at speed $u$ .

By Virtual Method,

$\sum { T.V=0 }$

Let the velocity of mass $M$ and $$V$ in upward direction

applying the law

$-Tu-\frac { Tu }{ 2 } +2Tv=0$

$v=\frac { 3u }{ 4 }$

A ball is dropped from a height of

$7.2$ m. It bounces back to

$3.2$ m after striking the floor. The ball remains in contact with the floor for

$20$ ms. Given that

$g=10 ms^{-2}$ , the average acceleration of the ball during the contact is:

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$100 ms^{-2}$
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$200 ms^{-2}$
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$600 ms^{-2}$
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$1000 ms^{-2}$

An object is tossed vertically into the air with an initial velocity of 8 m/s. Using the sign convention upwards as positive, how does the vertical component of the acceleration

$a_y$ of the object (after leaving the hand) vary during the flight of the object?
o

0%
On the way up $a_y > 0$ , on the way down $a_y > 0$
0%
On the way up $a_y < 0$ , on the way down $a_y > 0$
0%
On the way up $a_y > 0,$ on the way down $a_y < 0$
0%
On the way up $a_y <0$ , on the way down $a_y < 0$

A scooterist is moving on a straight road with speed

$10\;m{s^{ - 1}}$ . A bus ahead of scooterist by 32 m, starts from rest with an acceleration

$1\;m{s^{ - 2}}$ , if both are moving in the same direction then the time after which scooter overtakes the bus is:

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$16$ s
0%
$4$ s
0%
$8$ s
0%
It cannot overtake the bus

A particle initially at rest starts moving in a straight line with constant accelration. it attains a speed of

$10m/s$ in 2s after which it deaccelerates with constant rate and comes to rest in the next 3 seconds.An average speed of the particle during the entire motion will be

0%
$5m/s$
0%
$10m/s$
0%
$15m/s$
0%
$20m/s$

A balloon is rising with unifrom speed

$14\;m{s^{ - 1}}$ .At a height

$98 m$ from the ground a stone is dropped from the balloon.The time taken by the stone to strike the ground is

$\left( {g = 10\;m/{s^2}} \right)$ :

0%
$40 s$
0%
$57.8 s$
0%
$60 s$
0%
$36 s$

A rocket of initial mass

$5000 kg$ ejects gets at a constant rate of

$60kg/s$ with a relative speed of

$2050 m/s$ . Acceleration of the rocket 15 second after it is blasted off from the surface of earth will be (

$g = 10m/{s}^{2}$ ?)

0%
$10m/{s}^{2}$
0%
$20m/{s}^{2}$
0%
$30m/{s}^{2}$
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$40m/{s}^{2}$

A stone

$A$ is thrown vertically upward with speed

$u$ and another stone

$B$ is thrown vertically downward with same speed

$u$ from a height

$h$ . The time taken by stone

$A$ and

$B$ to reach the ground are

$16\ s$ and

$4\ s$ respectively. The time taken by another stone to reach the ground after it has been dropped from height

$H$ is [Take

$g=10\ m/s^{2}$ ]

0%
$2\sqrt{10}\ s$
0%
$10\ s$
0%
$8\ s$
0%
$40\ s$

A particle slides down a smooth inclined plane of elevation

$\theta$ fixed in the elevator going up with an acceleration

$a_0$ as shown. The base of the incline has length

$L$ . Then the time taken by the particle to reach the bottom of the plane is given by:

0%
$\sqrt{\dfrac{2L}{(g+a_0)\sin\theta}}$
0%
$\sqrt{\dfrac{2L}{(g+a_0)\sin\theta \cos\theta}}$
0%
$\sqrt{\dfrac{2L}{g\sin\theta \cos\theta}}$
0%
None of the above

Explanation

Let a be the acceleration of the particle with respect to the incline.

Taking components of the forces parallel to the incline and applying Newton’s law, we get,

$mg sinθ + ma_0 sinθ = ma$

or $a$ = $(g + a_0) sinθ$

This is the acceleration with respect to the elevator.

Now, $cosθ = L/x$ , where $x$ is the distance moved by the particle.

$x = L/cosθ$

From the equation of motion

$S=ut+\frac { 1 }{ 2 } { at }^{ 2 }$

u is 0

$S=\frac { 1 }{ 2 } { at }^{ 2 }$

$\dfrac { L }{ cos\theta } =\dfrac { 1 }{ 2 } { \left( g+{ a }_{ 0 } \right) sin\theta \quad t }^{ 2 }$

$t=\left( \dfrac { 2L }{ { \left( g+{ a }_{ 0 } \right) sin\theta \quad }cos\theta ^{ } } \right) { }^{ 1/2 }$

Balls are dropped from the roof of a tower at a fixed interval of time. At the moment when

$9 ^ { \text { th } }$ ball reaches the ground

$\mathrm { n } ^ { \text { th } }$ ball is at

$3 / 4 ^ { \text { th } }$ height of tower. The value of

$n$ is:

0%
$13$
0%
$7$
0%
$6$
0%
$5$

A stone thrown upwards with a velocity u reaches upto a height h. If the initial velocity is 2u the height attained would be

0%
$2h$
0%
$4h$
0%
$8h$
0%
$16h$

A clean body of mass

$100\ g$ starts moving with a velocity of

$2\ m/s$ on a smooth horizontal plane and it accumulates dust at the rate of

$5\ g/s$ . The velocity of body after

$20\ s$ will be

0%
$0.5\ m/s$
0%
$1\ m/s$
0%
$2\ m/s$
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$4\ m/s$

A body moves with a velocity of 3 m/s due east and then turns due north to travel with the same velocity. If the total time of travel is 6s, the acceleration of the body

0%
$\sqrt 3 m/s^2$ towards north west
0%
$\frac{1}{\sqrt 2} m/s^2$ towards north west
0%
$\sqrt 2 m/s^2$ towards north east
0%
all the above

Explanation

$\textbf{Hint}$ : Rate of change of velocity is called as Acceleration.

$\textbf{Step 1}:\textbf{Change in velocity}$ :

Given The body is travelling with a velocity of 3m/s due East and after turning due north it is travelling with the same velocity.

Total time taken by the body is 6 seconds.

$Initial \quad velocity=3 \hat{i}$ (East or towards x-direction)

$Final \quad velocity=3\hat{j}$ (North or towards y-direction)

Change in velocity= Final Velocity-Initial Velocity

$=3\hat{j}-3\hat{i}=3(\hat{j}-\hat{i})$

Total time=6 seconds

$\textbf{Step 2}:\textbf{Acceleration}$ :

$Acceleration=\dfrac{|3(\hat{j}-\hat{i})}{6}|$

$=\dfrac{|(\hat{j}-\hat{i})|}{2}=\dfrac{\sqrt{(1)^2+(-1)^2}}{2}=\dfrac{\sqrt(2)}{2}$

$=\dfrac{1}{\sqrt{2}}m/s^{2}$ in the North-west direction.

${\textbf{Correct option: B}}$

Two persons start running towards each other from two points that are 120 m apart. First person runs with a speed of

$5\;m{s^{ - 1}}$ and the other with a speed of

$7\;m{s^{ - 1}}$ .Both the persons meet after

0%
10 s
0%
24 s
0%
1 min
0%
48 s

Burglars, after looting a bank, jump into a waiting van at rest and speed up along a straight road at a constant rate of $2.0{\text{ m}}{{\text{s}}^{ - 2}}$ . At that instant, a police jeep moving at a constant speed of 30 m/s is 300m behind the burglars moving in the same direction. Choose the correct statement from the following;

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the police with not be able to catch the burglars
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the shortest distance between the police and the burglars is 75m
0%
The shortest distance between the police and the burglars is 125m
0%
The police can catch the burglars if they accelerate at a constant rate of $0.5{\text{ m}}{{\text{s}}^{ - 2}}$

A ball is dropped freely while another is thrown vertically downward with an initial velocity of

$\dfrac{v}{2}$ from the same point simultaneously. After 't' second they are separated by a distance of?

0%
$\dfrac{vt}{2}$
0%
$\dfrac{1}{2}gt^2$
0%
$vt$
0%
$\dfrac{vt+gt^2}{2}$

Explanation

ball 1:

$u=0$

$v=u+at$

$v=gt$

$S1=\dfrac { ut+{ at }^{ 2 } }{ 2 }$

$S1=\dfrac { { gt }^{ 2 } }{ 2 }$

ball 2:

$u=v$

$vf=u+at$

$vf=v+gt$

$S2=\dfrac { vt+{ gt }^{ 2 } }{ 2 }$

distance of separation

$=S2-S1=vt.$

A rocket is projected at the surface of earth like that in which acceleration 19.6

$m/s^{2}$ produces. After 5 seconds if stops its engine then calculate the following before to come at the surface of the earth.

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Maximum height obtained by rocket
0%
The velocity of the rocket at the time of reach it in the surface.
0%
Total time of journey
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The graph between velocity-time and acceleration - time.

A stone dropped from a building of height

$h$ and it reaches after

$h$ seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity

$u$ and they reach the earth surface after

$t_1$ and

$t_2$ seconds respectively, then

0%
$t = t _ { 1 } - t _ { 2 }$
0%
$t = \dfrac { t _ { 1 } + t _ { 2 } } { 2 }$
0%
$t = \sqrt { t _ { 1 } t _ { 2 } }$
0%
$t = \sqrt { t _ { 1 } ^ { 2 } + t _ { 2 } ^ { 2 } }$

A particle is moving in a straight line with initial velocity u and uniform acceleration a .If the sum of the distances travelled in

$t^{th}$ and

$(t+1)^{th}$ seconds is 100 cm, then its velocity after t seconds in

$cm s^{-1}$ is

0%
20
0%
30
0%
50
0%
80

A balloon of total mass 1000kg float motionless over the earth's surface. If 100kg or sand ballast are thrown over board, the balloon starts to rise with an acceleration of

0%
$10 m/s^{2}$
0%
$9.8 m/s^{2}$
0%
$1.09 m/s^{2}$
0%
$4.9 m/s^{2}$

When a stone falling from the top of the vertical tower has fallen a distance x m, another is let fall from a point y m below the top. If the fall from rest and reach the ground together, then the height of the tower is

0%
$\dfrac{{{{\left( {x + y} \right)}^2}}}{{4x}}{\text{m}}$
0%
$\dfrac{{4{{\left( {x + y} \right)}^2}}}{{x}}{\text{m}}$
0%
$\dfrac{{4x}}{{{{\left( {x + y} \right)}^2}}}{\text{m}}$
0%
$4x{\left( {x + y} \right)^2}{\text{m}}$

A particle is falling under gravity. In first

$t$ second it covers

$s_{1}$ and in the next

$t$ seconds it covers

$s_{2}$ then

$t$ is given by

0%
$\sqrt{\dfrac{s_{2}-s_{1}}{2g}}$
0%
$\sqrt{\dfrac{s_{2}-s_{1}}{g}}$
0%
$\sqrt{\dfrac{s_{2}--s_{1}}{2g}}$
0%
$\sqrt{\dfrac{s_{2}^{2}-s_{1}2}{2g}}$

Two bodies of masses

$1 kg$ and

$3 kg$ obliquely projected in opposite direction towards each other. Their velocities are

$15 ms^{-1}$ and

$5 \sqrt 3 ms^{-1}$ and angle made with horizontal are

$30^o$ and

$60^o$ respectively. They collide and stick to each other at a point when both of them reach maximum height. What is the final velocity with which the combined mass hits the ground ?

$(g=10 ms^{-2})$

0%
$2.5 ms^{-1}$
0%
$7.5 ms^{-1}$
0%
$5 ms^{-2}$
0%
$15 ms^{-1}$

A tiger chases a deer

$30m$ ahead of it and gains

$3\ m$ in

$5\ s$ after the chase began. The distance gained by the tiger in

$10\ s$ is

0%
$6\ m$
0%
$12\ m$
0%
$18\ m$
0%
$20\ m$

A body travels

$200\mathrm { cm }$ in the first two seconds and

$220\mathrm { cm }$ in the next four second. The velocity at the end of the seventh second from the start will be (acceleration is constant)

0%
$10 \mathrm { cm } / \mathrm { s }$
0%
$5 \mathrm { cm } / \mathrm { s }$
0%
$15 \mathrm { cm } / \mathrm { s }$
0%
$20 \mathrm { cm } / \mathrm { s }$

To an observer moving along East, the wind appear to blow from North. If the doubles his speed, the air would appear to come from-

0%
North
0%
East
0%
North-East
0%
North-West

A body starting from rest is moving with a uniform acceleration of 8

$m/{s^2}.$ Then the distance travelled by it in 5th second will be.

0%
40 m
0%
36 m
0%
100 m
0%
zero.

Two particles

$A$ and

$B$ are initially located on

$y-axis$ with distance

$l$ apart. They start moving at time

$t = 0$ such that the velocity

$\vec u$ of

$B$ is always along the

$x-axis$ and velocity

$\vec v$ of

$A$ is continuously aimed at

$B.$ At

$t = 0, \vec { u }$ is perpendicular to

$\vec { v }.$ The particles will meet after time

0%
$\dfrac { u l } { v ^ { 2 } - u ^ { 2 } }$
0%
$\dfrac { v l } { v ^ { 2 } - u ^ { 2 } }$
0%
$\dfrac { u l } { v ^ { 2 } + u ^ { 2 } }$
0%
$\dfrac { v l } { v ^ { 2 } + u ^ { 2 } }$

An object of mass

$m$ is released from rest form the top of a smooth inclined plane of height

$h$ . Its speed at the bottom of the plane is proportional to

0%
$m^{0}$
0%
$m$
0%
$m^{2}$
0%
$m^{-1}$

An automobile and a truck start from rest at the same instant, the automobile initially at some distance behind the truck. The truck has a constant acceleration of

$2.2m/s^{2}$ and the automobile has an acceleration of

$3.5m/s^{2}$ . The automobile overtakes the truck when it (truck) has moved

$60m$ . How much time does it take the automobile to overtake the truck and how far behind the automobile the truck initially was ?

0%
$7.39s.35.5m$
0%
$1.35s.15.5m$
0%
$5.9s.15.5m$
0%
$9.39s.35.5m$

Two particles start moving from the same point along the same straight line .the first constant veiocity v and the second with constant accleration a. During the time that the second catch the first the greatres distance between the particles is

0%
$\frac { { v }^{ 2 } }{ a }$
0%
$\frac { { v }^{ 2 } }{ 2a }$
0%
$\frac { { 2v }^{ 2 } }{ a }$
0%
$\frac { { v }^{ 2 } }{ 4a }$

Two bodies of different masses are dropped simultaneously from same height,If air friction acting them is directly proportional to the square of their mass then

0%
Lighter body reaches the ground earlier
0%
Heavier body reaches the ground earlier
0%
both reach the ground n the same time
0%
Lighter body do not reach the ground

A body is moving with a uniform acceleration covers 200 m in the first 2s and 220 m in the next 4s. find the velocity in

${ms}^{-1}$ after 7s.

0%
10
0%
15
0%
20
0%
30

A body falling freely from rest covers

$\dfrac { 7 }{ 16 }$ of the total height in the last second of its fall. The height from which it falls is

0%
$38.4 m$
0%
$78.4 m$
0%
$9.8 m$
0%
$19.6 m$

Particle

$A$ and

$B$ are moving in coplanar circular paths centred at

$O$ . They are rotating in the same sense. Time periods of rotation of

$A$ and

$B$ around

$O$ are

$T_A$ and

$T_B$ , respectively, with

$T_B > T_ A$ . Time required for

$B$ to make one rotation around

$O$ relative to

$A$ is :

0%
$T _ { B } - T _ { A }$
0%
$T _ { B } + T _ { A }$
0%
$\frac { T _ { B } T _ { A } } { T _ { B } + T _ { A } }$
0%
$\frac { T _ { B } T _ { A } } { T _ { A } - T _ { B} }$

A trolley of mass 300 kg carrying sand 200 kg (in it) is moving uniformly with speed of 40km/h on a frictionless track. After a while sand starts leaking out of a hole on the track. What is the speed of the trolley after entire sand is leaked out?

0%
24km/h
0%
30km/h
0%
66km/h
0%
40km/h

A boat takes 2 hours to travel 8 km and back in still water. If the velocity of the water is 4 km/h then the time taken for going upstream of 8 km and then coming back would be

0%
2 hours
0%
2 hours and 40 min
0%
1 hours 20 min
0%
Cannot be found with the information given

A 75.0 kg man stands on a platform scale in an elevator. starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 1.00 s. It travels with this constant speed for the next 10.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.70 s and comes to rest. What does the scale register before the elevator starts to move?

0%
84.18 kg
0%
81.48 kg
0%
75 kg
0%
65 kg

A particle starts from the origin with a velocity of

$10 ms^{-1}$ and moves with a constant acceleration till the velocity increases to

$50 ms^{-1}$ . at that instant, the acceleration is suddenly reversed. what will be the velocity of the particle, when it returns to the starting point?

0%
Zero
0%
$10 ms^{-1}$
0%
$50 ms^{-1}$
0%
$70 ms^{-1}$

An object is dropped from the top of a tower. the distance it covers in the last second is seven time the distance it covered in the first second . find the time of flight

0%
2 sec
0%
3 sec
0%
4 sec
0%
5 sec

A particle is moving in x-y plane . At certain instant of time. The components of its velocity and acceleration are as follows

$V_x = 3\ m/s$ ,

$V_y = 4\ m/s$ ,

${ a }_{ x }=2\ m/s^{ 2 }$ and

${ a }_{ y }=1\ m/s^{ 2 }$ . The rate of change of speed at this moment is

0%
$\sqrt { 10 } m/{ s }^{ 2 }$
0%
$\sqrt { 4 } m/{ s }^{ 2 }$
0%
$\sqrt { 5 } m/{ s }^{ 2 }$
0%
$\sqrt { 2 } m/{ s }^{ 2 }$

A lift coming down is just about to reach the ground floor. Taking the ground floor as origin and positive direction upwards for all quantities, which of the following is correct (

$x$ =displacement,

$v$ = velocity,

$a$ =acceleration):

0%
$x < o, v<0, a>0$
0%
$x > 0, v < 0, a < 0$
0%
$x > 0, v < 0, a> 0$
0%
$x > 0, v >0,a >0$

Is it possible for an object's velocity to increase while its acceleration decreases?

0%
No, this is impossible because of the way in which acceleration is defined.
0%
No, because if acceleration is decreasing the object will be slowing down.
0%
No, because velocity and acceleration must always be in the same direction.
0%
Yes, an example would be a falling object in a viscous medium, where the acceleration continuously decreases but velocity increases until a certain point.

The position-time

$(x - t)$ graph for a body thrown vertically upwards from ground is best shown by

A body falls from rest under gravity and covers a distance

$44.1\ m$ in last one second of journey. If

$g = 9.8\ m/s^2$ then total time of fall is.

0%
$2s$
0%
$3.18s$
0%
$4.91s$
0%
$5s$

The motion of a particle is described by the law

$a = t ^ { 3 } - 3 t ^ { 2 } + 5$ where

$a$ is the acceleration is

$\mathrm { m } / \mathrm { s } ^ { 2 }$ and

$t$ is time in second. The position and velocity of the particle at

$t = 1s$ are

$8.30m$ and

$6.25 m/s$ . Calculate the position at

$t = 2s$ .

0%
$10 m$
0%
$12 m$
0%
$15.6 m$
0%
$14 m$

A body moves

$4 m \text { in } 3 ^ { r d }$ and

$12 m \text { in } 5 ^ { th }$ second. If the motion is uniformly accelerated, how far will it travel in next

$3$ seconds to be constant during entire motion.

0%
$60 \mathrm { m }$
0%
$40 \mathrm { m }$
0%
$20 \mathrm { m }$
0%
$10 \mathrm { m }$

A particle is moving in x-y plane . At certain instant of time. The components of its velocity and acceleration are as follows

$V_x=3\ m/s$ ,

$V_y = 4\ m/s$ ,

${ a }_{ x }=2\ m/s^{ 2 }$ and

${ a }_{ y }=1\ m/s^{ 2 }$ The rate of change of speed at this moment is

0%
$\sqrt { 10 }\ m/{ s }^{ 2 }$
0%
$\sqrt { 4 }\ m/{ s }^{ 2 }$
0%
$\sqrt { 5 }\ m/{ s }^{ 2 }$
0%
$\sqrt { 2 }\ m/{ s }^{ 2 }$

A car starts from rest and accelerates uniformly at the rate of

$10 cms^{-2}$ and reaches a maximum velocity of

$72 kmph$ . it travels with this maximum velocity for some time and retards uniformly at the rate of

$20 cms^{-2}$ and comes to a stop, if the total distance covered by the body is

$5 km$ , find the total time of travel of the body in sec

0%
$200$
0%
$250$
0%
$300$
0%
$400$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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