MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 10

A block of mass

$5Kg$ is kept on the floor of an elevator kept at rest. The elevator started going down with an acceleration of

$14{ ms }^{ -2 }$ at time

$t=0$ . The magnitude of displacement of the block during

$t=0$ to

$t=0.1s$ is

$\left[ g=10{ ms }^{ -2 } \right]$

0%
$\dfrac { 1 }{ 20 } m$
0%
$20 /m$
0%
$\dfrac { 7 }{ 20 } m$
0%
$\dfrac { 12 }{ 20 } m$

A body is started from rest with acceleration

$2m/{s}^{2}$ till it attains the maximum velocity then retards to rest with

$3m/{s}^{2}$ . If total time taken is

$10$ second then maximum speed attained is

0%
$12m/s$
0%
$8m/s$
0%
$6m/s$
0%
$4m/s$

A train is moving along a straight path with uniform acceleration its engine passes across a pole with a velocity of

$60km/h$ and the end (guard'svan) passes across same pole with a velocity of

$80km/h$ . The middle point of the train will pass across same pole with a velocity

0%
$90km/h$
0%
$70.7 m/h$
0%
$65km/h$
0%
$75km/h$

Explanation

Let S be the length of the train t be the time taken by the full train to cross the pole and a be uniform acceleration of train from -

$\Rightarrow v^2-u^2=2as$

$\Rightarrow \dfrac{80^2-60^2}{2a}=s$

$\Rightarrow s=\dfrac{6400-3600}{2a}$

$=\dfrac{1400}{2a}\longrightarrow \left( 1\right)$

The middle part of the train covers half of total distance

$\dfrac{s}{2}=\dfrac{700}{a}$ from

$v^2-u^2=2as,$

$\Rightarrow v^2-60^2=2a\times \dfrac{70}{a}=1400$

$v^2=1400+3600=\sqrt{5000}=70.7km/hr$

Hence, the answer is

$70.7km/hr.$

Find the odd one out and give the reason : speed, distance , mass, velocity

0%
Mass
0%
Speed
0%
Distance
0%
Velocity

A ball dropped from some height coves half of its total during the last second of its free fall. Find

0%
time of flight
0%
height of its fall
0%
speed with which it strikes the ground.
0%
none of these

A body released from the top of alls through a height of

$5 \mathrm { m }$ during the first second of its fall and

$35 \mathrm { m }$ during the last second of its fall. The height of the tower is

0%
$80 \mathrm { m }$
0%
$60 \mathrm { m }$
0%
$40 \mathrm { m }$
0%
$20 \mathrm { m }$

A man of mass 40 kg is standing on a uniform plank of mass 60 kg lying on horizontal frictionless ice. The man walks from one end to the other end of the plank . the distance walked by the man relative to ice is (given length of plank=5m)

0%
2 m
0%
3 m
0%
5 m
0%
4 m

Explanation

d =

$\dfrac{mL}{m_1 + m_2}$

$\dfrac{40\times 5m}{100}$

= 2m

A body of mass 5kg starts from the origin with an initial velocity

$\overrightarrow u = 30 \hat i + 40 \hat j ms^{-1}$ if a constant force

$\overrightarrow F= - ( \hat i + 5 \hat j) N$ acts on the body , the time in which the y-component of the velocity becomes zero is:

0%
$5 seconds$
0%
$20 seconds$
0%
$40 seconds$
0%
$80 seconds$

Explanation

correct option

$(C)$

1967240_1354450_ans_51c54762e0e14cb0bf4e5f7bfdc905f5.jpg

A stone is dropped into the water from a bridge

$44.1 m$ above the water. another stone is thrown vertically downward

$1$ second later. both strike the water simultaneously . then initial speed of the second stone is

0%
$24.5 ms^{-1}$
0%
$4.9 ms^{-1}$
0%
$9.8 ms^{-1}$
0%
$12.25 ms^{-1}$

A bullet moving at 20 m/sec. it strike a wooden plank and pentrates 4 cm before coming to stop. the time taken to stop is

0%
0.08 sec
0%
0.16 sec
0%
0.004 sec
0%
0.02 sec

Explanation

Given,

$u=20m/s$

$s=4cm=0.04m$

$v=0m/s$

From 3rd equation of motion,

$2as=v^2-u^2$

$2\times a\times 0.04=0^2-20\times 20$

$a=-\dfrac{20\times 20}{2\times 0.04}$

$a=-5000m/s^2$

From 1st equation of motion,

$v=u+at$

$0=20-5000t$

$t=\dfrac{20}{5000}=0.004sec$

A stone is thrown vertically up from a bridge with velocity

$3 ms^{-1}$ if it strikes the water under the bridge after 2 s, the bridge is at a height of

$( g= 10m/s^2)$

0%
$26 m$
0%
$16 m$
0%
$14 m$
0%
$20 m$

The graph predicts the condition of

0%
Body is undergoing positive acceleration
0%
Body is undergoing negative acceleration
0%
Uniform velocity
0%
Uniform speed

A car starts from rest and has an acceleration

$a = 1 \mathrm { m } / \mathrm { s } ^ { 2 }$ .A truck is moving with a uniform velocity of

${ 6 } \mathrm { m } / \mathrm { s }$ . At what distance will the car overtake the truck? direction (at

$t = 0$ both start their motion in the same direction from the same position)

0%
36 m
0%
8 m
0%
32 m
0%
4 m

Explanation

At the

$moment$ when the car will be

$overtaking$ the truck their

$velocitis$ will be

$same.$

so suppose after

$t$ second from the start of the motion the car overtake the truck then the velocity of the car at that moment will be

$v=0+at=t\times 1m/s^2=t$

$m/s$

and the velocity of the truck at that moment will be

$u=6m/s (constant)$

both should be same i.e

$u=v$ or

$6=t$ or

$t=6$

$second$

so the distance covered by truck

$\text{oe car because both will be at same place ast that point of time }$ in this time

will be

$s=vt=6\times 6=36meter$

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is

$S_1$ and that covered in the first 20 seconds is

$S_2$ , then

0%
$S_2= 3S_1$
0%
$S_2= 4S_1$
0%
$S_2= S_1$
0%
$S_2= 2S_1$

A person travelled a distance of 3 km along a straight line in the North direction .Then he travelled 2 km in west direction and then 5 km in south direction.The magnitude of the displace-ment of this person would be ___________.

0%
$2\sqrt { 2 }$ Km
0%
$3\sqrt { 2 }$ Km
0%
$4\sqrt { 2 }$ Km
0%
10 Km

Three particles start from origin at the same time with a velocity

$2 ms^{-1}$ along positive x-axis the second with a velocity

$6 ms^{-1}$ aling negative y-axis. Find the velocity of the third particle along x=y line so that the three particles may always lie in a straight line

0%
$-3 \sqrt 3$
0%
$3 \sqrt 2$
0%
$-3 \sqrt 2$
0%
$2 \sqrt 2$

A ball is thrown up under gravity

$\left( g = 10 \mathrm { m } / \mathrm { sec } ^ { 2 } \right) .$ Find its velocity after

$1.0 sec$ at a height of 10

$\mathrm { m }$

0%
$5$ $\mathrm { m } / \mathrm { sec } ^ { 2 }$
0%
$5$ $\mathrm { m } / \mathrm { sec }$
0%
$10$ $\mathrm { m } / \mathrm { sec }$
0%
$15$ $\mathrm { m } / \mathrm { sec }$

A stone drop from height 'h' on Earth surface fall in 1 sec. If the same stone taken to Moon and drop freely then it will reaches from the surface of the Moon in the time (The 'g' of Moon is 1

$/ 6$ times ofEarth):-

0%
$\sqrt { 6 }$ second
0%
9 second
0%
$\sqrt { 3 }$ second
0%
6 second

Two trains A and B of length 400 m each are moving on two parallel tracks with uniform speed of

20 m/s in the same direction. with A ahead of B, driver of B accelerates at

$1m/{ s }^{ 2 }$ in order to overtrake. if after 50 sec, the guard of B just brushes past driver of A, what was original distance between drivers of the trains?

0%
$450 m$
0%
$1250 m$
0%
$850 m$
0%
$1650 m$

Explanation

The relative speed of

$B$ wrt

$A$ will be

$V_{BA}=V_B-V_A=20-20=0$

their relative acceleration will be

$a_{BA}=a_B-a_A=1-0=1m/s^2$

the distance traveled will be in

$s=ut+\dfrac{1}{2}at^2=0+\dfrac{1}{2}1\times (50)^2=1250meter$

if initial separation was

$x$ then

$x+400m=s$ or

$x=1250-400=850m$

because now the drivers are side by side of each other so the length of train A was added.

The figure represents the velocity-time graph of body moving in a straight line. How much distance does it travel during the last 10 seconds?

0%
20 m
0%
100 m
0%
50 m
0%
200 m

Explanation

Distance travelled in last

$10$ sec, is

$d = velocity \times time$ , which is nothing but area under the curve between the limit

$t = 10s$ to

$t = 20s$

So, area of triangle is

$= \dfrac {base \times height}{2} = \dfrac {(20 - 10)\times 20}{2} = 100\ m$

Hence, correct option is (B).

A particle is dropped from a tower. It is found that it travels 55m in the last second of its journey. Then height of the tower is (

g= 10 m/s2g= 10 m/s2) ?

0%
$125 m$
0%
$180 m$
0%
$100 m$
0%
$55 m$

A ball is dropped from a roof to the ground 8.0 m below. A rock is thrown down from the roof 0.6 s later. If they both hit the ground at the same time, what was the intitial speed of the rock?

0%
4.43 m/s
0%
8.43 m/s
0%
16.43 m/s
0%
12.43 m/s

If initial velocity of an object is 'u' and acceleration is 'a' then find the distance travelled in nth second.

0%
$S_n = un + \dfrac {an^2} {2}$
0%
$S_n = un + an^{2}$
0%
$S_n = u + \dfrac {a} {2} (2n - 1)$
0%
$S_n = (u + \dfrac a 2)n^2$

A particle moving with constant acceleration describes in the last second of its motion

$\dfrac { 5 }{ 9 }$ of the total distance. If it stats from rest it is in motion for a time.

0%
1s
0%
2s
0%
3s
0%
4s

A body dropped from the top of a tower covers

$7/16$ of the total height in the last second of its fall. The time of fall is

0%
$2$ sec
0%
$4$ sec
0%
$1$ sec
0%
$(\cfrac{50}{7})$ sec

What is the angle between instantaneous displacement and acceleration during the retarded motion

0%
$Zero$
0%
$\pi$
0%
$\frac{\pi}{2}$
0%
$\frac{\pi}{4}$

In case of a freely falling body, the ratio of kinetic energy at the end of the third second increase in kinetic energy in the next three seconds is

0%
$1:1$
0%
$1:2$
0%
$1:3$
0%
$1:9$

A body moving with uniform acceleration covers 48m in the

${ 10 }^{ th }$ second and 56m in the

${ 12 }^{ th }$ . Distance moved in the

${ 11 }^{ th }$ second is

0%
50m
0%
51m
0%
52m
0%
53m

How long will a train, running at a speed of 45 kmph cross a standing man, given the length of the train is 450m?

0%
100 sec
0%
150 sec
0%
50 sec
0%
36 sec

Water drugs full from a tap on the floor 5 m below at regular intervals of time, The first drop beings to fall. The height at which the third drop will be from ground (at the instant when the first drop strikes the ground) will be

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

0%
1.25 m
0%
2.15 m
0%
2.75 m
0%
3.75 m

A stone falls from a balloon that is descending at a uniform rate of 12m/s. the displacement of the stone from the point of release 10 sec is

0%
490 m
0%
510 m
0%
610 m
0%
725 m

Rain is falling vertically with a speed of 30 m/s. A women rides a bicycle with a speed of 10m/s in the north to south direction. What is the direction in which she should hold her umbrella?

0%
$37^o$ with vertical towards west.
0%
$37^o$ with vertical towards east.
0%
$71^o$ with vertical towards south east.
0%
$60^o$ with vertical towards east.

Explanation

Since

$tan^{-1} 3=71.4$

When a body is accelerated: (i) its velocity always changes (ii) its speed always changes (iii) its direction always changes (iv) its speed may or may not change Which of the following is correct?

0%
(i) and (ii)
0%
(i) and (iv)
0%
(i) only
0%
(ii) and (iii)

An object falls freely from rest in a vacuum. The graph shows the variation with time t of the
velocity

$v$ of the object.
Which graph, using the same scales, represents the object falling in air?

Explanation

In free fall in air, the only force acting on object is gravity.

$\dfrac{dv}{dt}= g$

Since

$g$ is constant.

$\implies \dfrac{dv}{dt}=$ constant

This implies that slop of

$v-t$ graph is constant.

So, option A is correct.

An object has an initial velocity

$u$ and an acceleration

$a$ . The object moves in a straight line
through a displacement s and has final velocity

$v$ .
The above quantities are related by the equation shown.

$v^{2} = u^{2} + 2as$
Which condition must be satisfied in order for this equation to apply to the motion of the object?

0%
The direction of $a$ is constant and the direction of $a$ is the same as the direction of $s$
0%
The direction of $a$ is constant and the direction of $a$ is the same as the direction of $u$
0%
The magnitude of $a$ is constant and the direction of $a$ is constant
0%
The magnitude of $a$ is constant and the direction of $a$ is the same as the direction of $v$

Explanation

$v^{2} = u^{2} + 2as$

The above equation of motion is valid only for constant acceleration.

So, for constant acceleration , the magnitude and direction of acceleration must be constant.

The following figure gives the movement of an object. Select the correct statement from the given choices.

0%
The total distance travelled by the object is $975\ m$
0%
The maximum acceleration of the object is $2\ ms^{-2}$
0%
The maximum deceleration happened between $25th$ and $85th$ seconds
0%
The object was at rest between $10th$ and $15th$ seconds
0%
At $40th$ second, the speed of object was decelerating

A stone is thrown vertically upward with a velocity of 34.3 m/s from the top of a cliff 147 m high . after it reaches the foot of the cliff, the mine elapsed will be :

0%
3 sec
0%
13 sec
0%
7 sec
0%
10 sec

Explanation

Correct option is

$(D)$

As we know;

$s=u t+\frac{1}{2} a t^{2}$

$\begin{array}{l}-147=(34.3) t+\frac{1}{2}(-9.8) t^{2} \\ -147=(34.3) t-(4.9) t^{2} \\ \text { Hence, }(4.9) t^{2}-(34.3) t-147=0 \\ \text { By Solving this we get; } \\ t=10\end{array}$

A particle starts from rest with a uniform acceleration, It travels a distance

$x$ in first two second & a distance

$y$ in the next two seconds. Then,

0%
Y = x
0%
Y = 2x
0%
Y = 3x
0%
Y = 4 x

Two particles A and B move from rest along a straight line with constant accelerations f and h respectively. If A takes m seconds more than B and describes n units more than that of B acquiring the same speed, then?

0%
$(f+h)m^2=fhn$
0%
$(f-fh)m^2=fhn$
0%
$(h-f)n=\dfrac{1}{2}fhm^2$
0%
$\dfrac{1}{2}(f+h)n=fhm^2$

Explanation

$S+n=\dfrac{1}{2}f(t+m)^2$ and

$S=\dfrac{1}{2}ht^2, V=ht$

$\therefore \dfrac{1}{2}ht^2+n=\dfrac{1}{2}f(t+m)^2$ ……………

$(1)$
Also

$V=0+ht=0+f(t+m)$

$\Rightarrow t+m=\dfrac{ht}{f}$
From equation

$(1)$ ,

$\dfrac{1}{2}ht^2+n=\dfrac{1}{2}f\left(\dfrac{ht}{f}\right)^2$

$\Rightarrow t^2=\dfrac{2nf}{h(h-f)}$
Also,

$ht=f(t+m)$

$\Rightarrow t^2=\dfrac{m^2f^2}{(h-f)^2}$

$\therefore \dfrac{2nf}{h(h-f)}=\dfrac{m^2f^2}{(h-f)^2}$

$\Rightarrow 2n=\dfrac{m^2fh}{h-f}$

$\Rightarrow n(h-f)=\dfrac{1}{2}fhm^2$ .

A boy throws a ball in a train moving at a speed of

$18 km/ph$ , with a velocity of

$15\sqrt{2} m/s$ at an angle of

$45^0$ to the horizontal in the direction of motion of train. The angle with which it strikes the plane of projection w.r.t. an observer watching from the road near by to the train is

0%
$45^0$
0%
$60^0$
0%
$37^0$
0%
None of these.

$x= a ( cos\theta+\theta sin\theta)$ and

$y= a( sin\theta -\theta cos \theta)$ and

$\theta$ increases at uniform rate

$\omega$ . The velocity of particle is

0%
$a\omega$
0%
$\frac{a^2\theta}{\omega}$
0%
$\frac{a\theta}{\omega}$
0%
$a \theta\omega$

A body starts from rest and is uniformly accelerated for

$30s$ . The distance travelled in the first

$10s$ is

$x_{ 1 }$ next

$10s$ is

$x_{ 2 }$ and the last

$10s$ is

$x_{ 3 }$ .Then

$x_{ 1 }:x_{ 2 }:x_{ 3 }$ is the same as :-

0%
$1:2:4$
0%
$!:2:5$
0%
$1:3:5$
0%
$1:3:9$

Explanation

$u = 0$

$s = ut + \dfrac{1}{2}at^2 \\$

$s = o*t + \dfrac{1}{2}at^2$

$s = \dfrac{1}{2}at^2$

In first 10 sec, travelled distance

$x_1 = \dfrac{1}{2}a(10)^2 \\ x_1 = 50a$
After next 10 sec, travelled distance

$x_2 = \dfrac{1}{2}a(10+10)^2 - x_1\\ x_2 = \dfrac{1}{2}a*(400) - 50 a \\ x_2 = 150a$

In last 10 sec, travelled distance

$x_3 = \dfrac{1}{2}a(10+10+10)^2 - x_1 - x_2 \\ x_3 = 450a - 50a - 150a \\x_3 = 250a \\$

$x_1:x_2:x_3 = 50a : 150a: 250a$

$x_1:x_2:x_3 = 1 : 3: 5$

In the two dimensional motions:

0%
$x-t$ graph gives actual path of the particle
0%
$y-t$ graph gives actual path of the particle
0%
$\sqrt{x^2+y^2}$ versus t graph gives the actual path of the particle
0%
$y-x$ graph gives actual path of particle

Mark the correct statement(s)

0%
A particle can have zero displacement and non zero average velocity
0%
A particle can have zero displacement and non zero velocity
0%
A particle can have zero acceleration and non zero velocity
0%
A particle can have zero velocity and non zero acceleration

A person moves 30 m north and then 20 m towards east and finally

$30\sqrt { 2 }$ m in south-west direction. The displacement of the person from the origin will be

0%
10 m along north
0%
10 m along south
0%
10 m along west
0%
Zero

Explanation

$AB=30 m, BC=20 m$ ,

$CD=30\sqrt { 2 } m$

$DBCE$ is isosceles (Fig.), so

$BE=BC=20m$

$AE=AB-BE=30-20=10 m$

$EC=20\sqrt { 2 } m$

$ED=CD-EC=30\sqrt { 2 } -20\sqrt { 2 } =10\sqrt { 2 } m$

$DADE$ is isosceles, so

$AD=AE=10 m$

so the displacement is

$10m$ due west

1572151_1734108_ans_78391e510fe64fa4a4d7d835a8e54d83.png

A parachutist drops first freely from an aeroplane for 10 s and then his parachutist opens out. Now he descends with a net retardation of

$2.5ms^{-2}$ . If he bails out of the plane at a height of 2495 m and

$g = 10 ms^{-2}$ , his velocity on reaching the ground will be

0%
$5 ms^{-1}$
0%
$10 ms^{-1}$
0%
$15 ms^{-1}$
0%
$20 ms^{-1}$

Explanation

The velocity v acquired by the parachutist after 10 s:

$v=u+gt=0+10\times 10=100ms^{-1}$
Then,

$s_1=ut+\dfrac{1}{2}gt^2=0+\dfrac{1}{2}\times 10\times 10^2=500m$
The distance travelled by the parachutist under retardation is

$s_2=2495-500=1995m$
Let

$v_g$ be the velocity on reaching the ground. Then

$v_{ g }^{ 2 }-v^{ 2 }=2as_{ 2 }$
or

$v_{ g }^{ 2 }-(100)^2=2\times(-2.5)\times1995$ or

$v_g=5ms^{-1}$

The velocity-time graph of a body is shown in Fig.4.The displacement of the body in 8 s is

0%
9 $m$
0%
12 $m$
0%
10 $m$
0%
28 $m$

Explanation

Displacement = area under graph

$Area=\quad 2\times 2+\dfrac { 1 }{ 2 } \left( 2+6 \right) \times 1+\dfrac { 1 }{ 2 } \times 1\times 6-\dfrac { 1 }{ 2 } +1\times 6-1\times 6+2\times 4$

$=4+4+3-3-6+8=10m$

$Area=10m=displacement \ in \ 8sec.$

One of the rectangular components of a velocity of

$60$ m/s is

$30$ m/s, the other rectangular component is?

0%
$30$ m/s
0%
$30\sqrt{3}$ m/s
0%
$30\sqrt{2}$ m/s
0%
None of these

A train 100 m long travelling at 40

$m/s^{-1}$ starts overtaking another train 200 m long travelling at 30

$m/s^{-1}$ . The time taken by the first train to pass the second train completely is

0%
30 s
0%
40 s
0%
50 s
0%
60 s

Explanation

Relative velocity of overtaking

$= 40 - 30 = 10 ms^{-1}$ .

Total relative distance covered with this relative velocity during overtaking

$= 100 + 200 = 300 m$

So time taken

$= \dfrac{300}{10 }= 30 s$

A point moves with uniform acceleration and

$v_1,v_2,$ and

$v_3$ denote the average velocities in the three successive intervals of time

$t_1,t_2,$ and

$t_3$ . Which of the following relations is correct?

0%
$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2+t_3)$
0%
$(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$
0%
$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_1-t_3)$
0%
$(v_1-v_2):(v_2-v_3)=(t_1-t_2):(t_2-t_3)$

Explanation

Suppose u be the initial velocity.
Velocity after time

$t_1:v_{11}=u+at_1$
Velocity after time

$t_1+t_2:v_{22}=u+a(t_1+t_2)$
Velocity after time

$t_1+t_2+t_3:$

$v_33=u+a(t_1+t_2+t_3)$
Now

$v_1=\dfrac { u+v_{ 11 } }{ 2 } =\dfrac { u+u+at_{ 1 } }{ 2 } =u+\dfrac { 1 }{ 2 } at_{ 1 }$

$v_2=\dfrac{v_{11}+v_{22}}{2}=u+at_1+\dfrac{1}{2}at^2$

$v_3=\dfrac{v_{22}+v_{33}}{2}=u+at_1+at_2\dfrac{1}{2}at^3$
So

$v_1-v_2=-\dfrac{1}{2}a(t_1+t_2)$

$v_2-v_3=--\dfrac{1}{2}a(t_2+t_3)$

$(v_1-v_2):(v_2-v_3)=(t_1+t_2):(t_2+t_3)$

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.