MCQ Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 9

A train is at rest. It accelerates for a time

$t_1$ at a uniform rate

$\alpha$ and the comes to rest under a uniform retardation rate

$\beta$ in time.

$t_2$ .

$\dfrac{t_1}{t_2}$ is equal to

0%
$\dfrac{\beta}{\alpha}$
0%
$\dfrac{\alpha +\beta}{\alpha}$
0%
$\dfrac{\alpha + \beta}{\beta}$
0%
$\alpha \beta$

Explanation

Initially train is at rest.

From the 1st equation of motion,

$v=u+at$

$v=0+at$

$v=at$

From the above equation,

$\alpha t_1=\beta t_2$

$\dfrac{t_1}{t_2}=\dfrac{\beta}{\alpha}$

The correct option is A.

1530342_1082226_ans_c6e1a78d3bad439584c32f2eb1edc46f.jpeg

A car accelerates from rest at a constant rate

$\alpha$ for some time, after which it decelerates at a constant rate

$\beta$ and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by car is

0%
$(\dfrac{\alpha^2+\beta^2}{\alpha \beta})t$
0%
$(\dfrac{\alpha^2-\beta^2}{\alpha \beta})t$
0%
None of these
0%
$(\dfrac{\alpha \beta r}{\alpha \beta})$

Explanation

Lets consider,

$v=$ maximum velocity acquired by car

From 1st equation of motion,

$v=0+\alpha t_1$

$t_1=\dfrac{v}{\alpha}$

Similarly,

$0=v-\beta t_2$

$t_2=\dfrac{v}{\beta}$

$t_1+t_2=t=\dfrac{v}{\alpha}+\dfrac{v}{\beta}$

$t=v(\dfrac{\alpha +\beta}{\alpha \beta})$

$v=\dfrac{\alpha \beta}{\alpha +\beta}t$

Velocity at the top of vertical journey under gravity when a body is projected upward with velocity

$1000m/s$ is

0%
Zero
0%
$10 m/s$
0%
$100 m/s$
0%
$1000 m/s$

Explanation

As body goes upwards, due to gravitational force the body will stop at the maximum height and fall towards the ground.

Hence, the velocity at the maximum height is zero.

$v_m=0m/s$

The correct option is A.

A particle starts with an initial velocity of

$20 m/s$ and moves with a constant acceleration on a straight line. If its final velocity is

$60 m/s$ , then velocity of midpoint is

0%
$40 m/s$
0%
$22\sqrt{5} m/s$
0%
$20\sqrt{5} m/s$
0%
$24\sqrt{5} m/s$

Explanation

The following velocity at the mid point is give by the formula,

$V_{mid \quad point}=\sqrt{\dfrac{(V_{initial})^2+(V_{final})^2}{2}}$

$V_{mid \quad point}=\sqrt{\dfrac{20^2+60^2}{2}}=\sqrt{\dfrac{400+3600}{2}}$

$=20\sqrt5 m/s$

If a body starts from rest,the time in which it covers a particular displacement with uniform acceleration

0%
inversely proportional to the square root of the displacement
0%
inversely proportional to the displacement
0%
diretily proportional to the displacement
0%
directly proportional to the square root of the displacement

Explanation

Apply equation of kinematic

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

${{t}^{2}}+\dfrac{2ut}{a}-\dfrac{2s}{a}=0$

${{t}^{2}}+pt-ks=0$

Where, $p=\dfrac{2u}{a}\,\,and\,\,k=\dfrac{2}{a}$

$t=\dfrac{-p\pm \sqrt{{{p}^{2}}+ks}}{2}$

Hence, time is directly proportional to root of displacement.

A particle travels

$10$ m in first

$5$ seconds

$10$ cm in next

$3$ seconds. Assuming constant acceleration what is the distance travelled in next

$2$ second ?

0%
$8.3$ m
0%
$9.3$ m
0%
$10.3$ m
0%
None of these

Explanation

Let assume initial velocity is $u$ and constant acceleration $a$ . using $s$ = $ut$ + $(0.5)at^2$ ,

for first $5$ second

$10$ = $5u$ + $(0.5)a(25)$

for next 3 seconds, $t = 8$ and $s = 20$

$20 = 8u + (0.5)a(64)$

Solving,

$a = {\dfrac{1}{3}}$ and $u = {\dfrac{7}{6}}$

Now for next $2$ seconds, $t = 10$ and $S'$ :

$S' = (10){\dfrac{7}{6} +(0.5){\dfrac{1}{3}}(100)}$

$S'$ = ${\dfrac{170}{6}}$

Now distance travelled in $2$ sec = ${\dfrac{170}{6}} - {20} = {\dfrac{50}{6}}$ meter

A body moving with uniform retardation covers

$3\ km$ before its speed is reduced to half of its initial value. It comes to rest in another distance of :

0%
$1\ km$
0%
$2\ km$
0%
$3\ km$
0%
$\dfrac {1}{2} km$

Explanation

Lets

$u=$ initial velocity

$s=3km=3000m$

$v=\dfrac{u}{2}$

3rd equation of motion,

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

$(\dfrac{u}{2})^2-u^2=2a\times 3000$

$-\dfrac{3}{4}u^2=3000\times 2a$

$u^2=-8000a$ . . . . . .(1)

Now body comes to rest. 3rd equation of motion is

$0-(\dfrac{u}{2})^2=2as'$

$\dfrac{-u^2}{4}=2as'$

$-\dfrac{-8000a}{4}=2as'$

$s'=1km$

The correct option is A.

A force of

$20N$ is applied on a body of mass

$5 kg$ resting on a horizontal plane. The body gains a kinetic energy of

$10 joule$ after it moves a distance of

$2m$ . The frictional force is:

0%
$10 N$
0%
$15 N$
0%
$20 N$
0%
$30 N$

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the

$5th$ sec to that covered in

$5$ sec is :

0%
$9/25$
0%
$3/5$
0%
$25/9$
0%
$1/25$

Explanation

Given,

$u=0m/s$

Distance covered by the

$n^{th}$ sec,

$S=ut+a(n-\dfrac{1}{2})$

For

$5^{th}$ sec

$S_1=0+a(5-0.5)$

$S_1=4.5a$ . . . . . .. . . . .(1)

The distance covered by the body in 5 sec

$S_2=ut+\dfrac{1}{2}at^2$

$S_2=0+\dfrac{1}{2}a\times 5\times 5$

$S_2=12.5a$ . . . . . . . . .(2)

$\dfrac{S_1}{S_2}=\dfrac{4.5a}{12.5a}$

$\dfrac{S_1}{S_2}=\dfrac{9}{25}$

A block can slide on a smooth inclined plane of inclination '

$q$ ' kept on the floor of a lift. When the lift is descending with retardation, '

${a}$ '

$m/s^2$ the acceleration of the block relative to the incline will be:

0%
$(g-a)\sin q$
0%
$(g+a)\sin q$
0%
$g\sin q$
0%
$g-a$

Two cars OF SAME LENGTH move in the same direction along parallel roads. One of them is a

$100$ m long travelling with a velocity of

$7.5$

$ms^{-1}$ . How long will it take for the first car to overtake the second car?

0%
$26$ s
0%
$40$ s
0%
$60$ s
0%
$80$ s

A body is thrown with speed 20 m/s vertically upwards,if will return to thrower's hand after a time

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

0%
2 s
0%
4 s
0%
20 s
0%
Never

Explanation

Given that,

Initial velocity $u=20\,m/s$

Final velocity $v=0$

Now, from equation of motion

$v=u-gt$

$0=20-10t$

${{t}_{1}}=\dfrac{20}{10}$

${{t}_{1}}=2\,s$

Total time

$t=2{{t}_{1}}$

$t=2\times 2$

$t=4\,s$

Hence, the time is $4\ s$

If a body loses half of its initial velocity on permenenting 2 cm in a wooden block, them how much it penetrate more before its velocity reduces to one fourth of its initial velocity [Assume retaradation of body in uniform ]

0%
2 cm
0%
1 cm
0%
0 . 5 cm
0%
4 cm

Explanation

Given,

Initial velocity $=u$

After $2\ cm\,$ penetration, velocity $v=\dfrac{u}{2}$

Apply equation of kinematic

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

${{\left( \dfrac{u}{2} \right)}^{2}}-{{u}^{2}}=2a\times 2$

$a=\dfrac{-3{{u}^{2}}}{16}$

When, Final velocity ${{v}_{1}}=\dfrac{u}{4}$

Apply kinematic equation of motion

$v_{1}^{2}-{{u}^{2}}=2a{{s}_{1}}$

${{\left( \dfrac{u}{4} \right)}^{2}}-{{u}^{2}}=2\left( \dfrac{-3{{u}^{2}}}{16} \right){{s}_{1}}$

${{s}_{1}}=2.5\ cm$

Hence, for final velocity to be one fourth of initial velocity, body penetrate $0.5\ cm$ more in wood.

The splash of sound was heard

$6$ s after dropping a stone into a well

$125$ m deep. The velocity of sound is:

0%
$350$ cm/s
0%
$125$ m/s
0%
$392$ cm/s
0%
$0$ cm/s

Explanation

From equation of motion we have

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

where u=

$0$ m/s;g=acceleration due to gravity=

$10 m/{s}^{2}$ and s= distance traveled=

$125$ m

Putting the above-mentioned values in the equation of motion we get

${t}^{2}=\dfrac{250}{10}$

$t=5 s$

Now as per question the sound of splash is after

$6s$ of dropping, thus the time taken for the sound to cover a distance of

$125$ m is

$1$ second.

Speed of sound

$v=\dfrac{distance}{time}$ =

$125$ m/s

A particle is thrown vertically up with speed u so that distance covered in last second of flight is

$35\ m$ . If then initial speed of throw is

0%
$20 m/s$
0%
$26.45m/s$
0%
$40 m/s$
0%
$50 m/s$

Explanation

We know from equation of motion we get,

$v=u-gt$ -------(1)

as velocity of the particle becomes at the very last second of flight

$v=0$

Thus equation 1 can be written as,

$t=\dfrac{u}{g}$ -------(A)

Again we know,

$s=ut-\dfrac{1}{2}g{t}^{2}$

$35=\dfrac{{u}^{2}}{g}-\dfrac{1}{2}g{(\dfrac{u}{g})}^{2}$

$u=\sqrt{700}=26.45$

A man runs at a speed of

$4.0 \,m/s$ to overtake a standing bus. When he is

$6.0 \,m$ behind the door (at

$t = 0$ ), then bus moves forward and continues with a constant acceleration of

$1.2 \,m/s^2$ . The man shall access the door at time

$t$ equal to:

0%
5.2s
0%
4.3s
0%
2.3s
0%
the man shall never gain the door

Explanation

Suppose the man gets the bus after time

$t$ from the start of the bus.

$\text{Then the man needs to cover the distance BY bus + 6meter to catch the Bus }$

so the distance by man ,

$X_M=4\times t$ should be equal to

$6m +\dfrac{1}{2}at^2$

$4t=6+0.6t^2$ on solving it we get

$t=4.38sec$

Just employ the Shridharchcarya Formula

$t=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Where

$t$ is root of

$at^2+bt+c=0$

A car moving along a straight highway with a speed of 108 km/h, is brought to rest within a distance of 100 m. How long does it take for the car to stop?

0%
$\dfrac{14}{2}s$
0%
$\dfrac{20}{3}s$
0%
$\dfrac{25}{3}s$
0%
$\dfrac{32}{3}s$

Explanation

Given,

$u=108km/h=\dfrac{108\times 1000}{60\times 60}=30m/s$

$v=0m/s$

$S=100m$

From 3rd equation of motion,

$2aS=v^2-u^2$

$2\times 100\times a=0^2-30\times 30$

$a=-\dfrac{30\times 30}{2\times 100}=-4.5m/s^2$

From 1st equation of motion,

$v=u+at$

$0=30-4.5t$

$t=\dfrac{30}{4.5}=\dfrac{20}{3}sec$

The correct option is B.

A man drops a ball downside from the roof of a tower of height 400 meters. At then same time another ball is thrown upside with a velocity 50 m/sec from the base of the tower, then they will meet at which height from the base of the tower:

0%
$100$ meters
0%
$320$ meters
0%
$80$ meters
0%
$240$ meters

Explanation

Let both ball meet at point

$P$ after time

$t$ .

The distance travelled by ball

$A$ ,

$h_1=\dfrac{1}{2}gt^2$

The distance travelled by ball

$B$ ,

$h_2=ut-\dfrac{1}{2}gt^2$

$h_1+h_2=400$

$ut=400\rightarrow \boxed{t=8\,sec}$

$h_1=320m$ and

$h_2=80m$

Ans:

$80m$

2029130_1115905_ans_fde6555dc40140dfa788cf98e7f67b8b.png

A person crossing the a road with a certain velocity due north sees a car moving towards east. The velocity of the car with respect to the person is

$\sqrt 2$ times that of velocity of the person. The angle made by the relative velocity with the east is

0%
${30^ \circ}$
0%
${45^ \circ}$
0%
${60^ \circ}$
0%
${90^ \circ}$

Explanation

$\vec{V}$ person

$=v\hat{j}$

$\vec{V}$ car

$=u\hat{i}$

$\vec{V}cp=\vec{V}car\rightarrow person=u\hat{i}-v\hat{j}$

$\Rightarrow |\vec{V}cp|=\sqrt{u^{2}+v^{2}}$

given that

$|\vec{V}cp|=\sqrt{2} V$

$\Rightarrow \sqrt{u^{2}+v^{2}}=\sqrt{2}x\Rightarrow u^{2}+v^{2}=2v^{2}$

$\Rightarrow v^{2}=u^{2}\Rightarrow v=u$

$\Rightarrow \vec{V}cp=v\hat{i}-v\hat{j}$

$\theta=\tan^{-1}\dfrac{v}{v}\Rightarrow \theta=45^{o}$

1404136_1123338_ans_859db6ffcdca4cc3a35cc74cc686d0a8.png

Just after being hit, a baseball has a horizontal speed of 20 m/s and vertical speed of 25 m/s upward. Ignoring air resistance what are these speeds two seconds later?

0%
20 m/s horizontal and 5 m/s downward
0%
20 m/s horizontal and 5 m/s upward
0%
10 m/s horizontal and 5 m/s upward
0%
20 m/s horizontal and 45 m/s downward

Explanation

Given,

Horizontal velocity, ${{v}_{x}}=20\,m{{s}^{-1}}$

Vertical initial velocity, ${{u}_{y}}=25\,m{{s}^{-1}}$

Acceleration of gravity, ${{a}_{y}}=-10\,m{{s}^{-2}}$

Apply equation of kinematic

${{v}_{y}}={{u}_{y}}+{{a}_{y}}t$

${{v}_{y}}=25-10\times 2=5\,m{{s}^{-1}}$

Hence, velocity $20\,m{{s}^{-1}}$ horizontal and $5\,m{{s}^{-1}}$ upward

A body of mass 2 Kg starts from rest and moves with uniform acceleration. It acquires a velocity 20 m/in 4s. The power exerted on the body in 2s in watts is:

0%
50 watt
0%
100 watt
0%
150 watt
0%
200 watt

Explanation

Given that,

Mass $m=2\,kg$

Initial velocity $u=0\,m/s$

Final velocity $v=20\,m/s$

Time $t=4\,s$

Now, the acceleration is

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{20}{4}$

$a=5\,m/{{s}^{2}}$

Now, the force is

$F=ma$

$F=2\times 5$

$F=10\,N$

Now, the displacement is

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

$s=0+\dfrac{1}{2}\times 5\times 16$

$s=40\,m$

Now, the work done is

$W=F\centerdot s$

$W=10\times 40$

$W=400\,J$

Now, the power is

$P=\dfrac{W}{t}$

$P=\dfrac{400}{2}$

$P=200\,watt$

Hence, the power is $200\ watt$

A bullet moving with a speed of

$100$

$ms^{-1}$ can just penetrate two planks of equal thickness. Then the number of such planks penetrated by the same bullet when the speed is doubled will be:

0%
$4$
0%
$8$
0%
$6$
0%
$10$

Explanation

equation of motion can be be applied:

$v^2 = u^2 + 2as$ , where acceleration is -ve.

Hence

$v^2 = u^2 – 2as$ .

v(final velocity) = 0.

Therefore,

$u^2 = 2as$ .

i.e. s is proportional to

$u^2 = {\dfrac{s_1}{ s_2}} = {\dfrac{u_1^2} {u_2^2}}$

Therefore,

$s_2 = {\dfrac{u_2^2} {u_1^2}} \times s_1 = {\dfrac{200^2}{100^2}} \times 2 (planks) = 8 planks$

A body is started from rest with acceleration

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

$3 m/s^2$ . If total time taken is 10 second then maximum speed attained is

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

Explanation

Given that,

Acceleration $a=2\,m/{{s}^{2}}$

Retardation $a=-3\,m/{{s}^{2}}$

Time $t=10\,s$

Now, from equation of motion

The maximum velocity attained be v at t

$v=u+at$

$v=0+2\times t$

$v=2t\,m/s.....(I)$

Now, again it reaches velocity 0 after 10 s with retardation

$v=u+at$

$0=v-3\left( 10-t \right)$

Now from equation (I)

$2t-3\left( 10-t \right)=0$

$2t-30+3t=0$

$5t=30$

$t=6\,s$

Now, put the value of t in equation (I)

$v=2t$

$v=2\times 6$

$v=12\,m/s$

Hence, the maximum speed is $12\ m/s$

A juggler maintains four balls in the air with air with throwing speed

$20$ m/s upwards in regular time intervals. When one ball is about to leave his hand the height of balls in air from the ground will be:

0%
$60m, 80m, 60m, 0m$
0%
$30m, 40m, 30m, 0m$
0%
$15m, 20m, 15m, 0m$
0%
$10m, 20m, 10m, 0m$

Explanation

Initial velocity $u=20\,m/s$

Final velocity $v=0\,m/s$

Acceleration $a=-10\,m/{{s}^{2}}$

Now, the time by ball reach its maximum height

From equation of motion

$v=u+at$

$0=20-10t$

$t=2\,s$

So, the time taken by the same ball to return to the hands of the juggler is

$=\dfrac{2u}{g}$

$=\dfrac{2\times 20}{10}$

$=4\,s$

So, he is throwing the balls after 1 sec each,

Let at some instant he throws ball number 4.

Now, Before 1 s of throwing it, he throws ball 3.

So, the height of ball 3 is

${{h}_{3}}=20\times 1-\frac{1}{2}\times 10\times {{\left( 1 \right)}^{2}}$

${{h}_{3}}=15\,m$

Before 2s, he throws ball 2, so the height of ball 2 is

${{h}_{2}}=20\times 2-\frac{1}{2}\times 10\times 4$

${{h}_{2}}=20\,m$

Before 3s, he throws ball 1, so the height of ball 1 is

${{h}_{1}}=20\times 3-\frac{1}{2}\times 10\times 3\times 3$

${{h}_{3}}=15\,m$

$15\ m$ , $20\ m$ , $15\ m$ and $0\ m$

A particle travels

$10\ m$ in the first

$5\ sec$ and

$10\ m$ in the next

$3\ sec$ . Assuming constant acceleration, what is the distance travelled in the next

$2\ sec$ ?

0%
$8.3\ m$
0%
$9.3\ m$
0%
$20\ m$
0%
$16.6\ m$

A ship is moving due east with a velocity of

$12\ m/sec$ . A truck is moving across on the ship with a velocity of 4m/sec.A monkey is climbing a vertical pole mounted on the truck, with a velocity of

$3m/sec$ . Find the velocity of the monkey, as observed by a man on the shore.

$(m/sec)$

0%
$10$
0%
$15$
0%
$13$
0%
$16$

An aeroplane is in a level flight at

$144km/hr$ at an altitude of

$1000m$ . How far from a given target a bomb be released from it to hit the target ?(take

$g=10m/s^2$ )

0%
$566m$
0%
$671.43m$
0%
$471.34m$
0%
$371.34m$

Explanation

as time to fall

$S=\dfrac{1}{2}\times g\times t^2$

$\dfrac{100\times 2}{10}=t^2$

$t=\sqrt{200}=10\sqrt{2}$

$=14.14s$
so Speed =

$144\times \dfrac{1000}{3600}=40$
so,

$x=40\times t=40\times 14.14$

$=565.6m$

The two ends of train moving with constant acceleration pass a certain point with velocities u and 2 u. The velocity with which, the middle point of the train passes the same points is.

0%
$\sqrt { 2.5 } u$
0%
$\sqrt { 1.5 } u$
0%
$\sqrt { 5 } u$
0%
u

Explanation

Hint: Apply kinematical equations.

Correct Option: Option A

Explanation for correct answer:

Step 1: Find the length of train in terms of velocity and acceleration.

The displacement of the train is given by:

$S = \dfrac{{{v^2} - {u^2}}}{{2a}}$ , where

$v$ and

$u$ are the initial and final velocities of the train and

$a$ is the acceleration of the train

If $L$ is the length of the train, then:

$L = \dfrac{{{{(2u)}^2} + {u^2}}}{{2a}} = \dfrac{{3{u^2}}}{{2a}}$ ------(1)

Step 2: Find the velocity of midpoint.

As, midpoint is at distance $\dfrac{L}{2}$ ,

$\dfrac{L}{2} = \dfrac{{{v^2} - {u^2}}}{{2a}}$

$\Rightarrow \dfrac{{3{u^2}}}{{4a}} = \dfrac{{{v^2} - {u^2}}}{{2a}}$

$\Rightarrow v = \sqrt {2.5} u$ $m/s$

So, the velocity with which, the middle point of the train passes the same point is $\sqrt {2.5} u$ $m/s$ .

Acceleration of a body is given by

$\frac { dv }{ dt } =4-2v$ .The body started from rest t=Then

0%
The acceleration of the body is initially + ve but after some time it becomes negative
0%
The body will ultimately acquire a constant velocity after a long time
0%
The body will finally come rest
0%
The acceleration will finally attain a constant non-zero value

Explanation

Given that,

Acceleration, $a=\frac{dv}{dt}=4-2t$

At t = 0 s , $a=4m/{{s}^{2}}$

At t = 1 s, $a=2\,m/{{s}^{2}}$

At t = 2 s, $a=0$

At t = 3 s, $a=4-6=-2\,m/{{s}^{2}}$

Hence, the acceleration of the body is initially positive but after some time it becomes negative.

An elevator is going down with a constant acceleration. A coin dropped from a point

$1.8m$ above the elevator floor takes one second to reach the floor. The magnitude of the acceleration of the the elevator in

$ms^{-2}$ is

Given:

$g=10 \ ms^{-2}$

0%
$3.6$
0%
$5$
0%
$7.2$
0%
$6.4$

Water drops fall at regular intervals from a tap 5 m above ground. The 3 rd drop is leaving tap when first drop reaches ground. The distance of 2 nd drop at the instant is

0%
2.5 m
0%
3.75 m
0%
4 m
0%
1.25 m

Explanation

Given that,

Height $h=5\,m$

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

Now, for first drop

${{s}_{1}}=ut+\dfrac{1}{2}g{{t}^{2}}$

$5=0+5{{t}^{2}}$

$t=1\,\sec$

It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every $0.5\ sec$

Now, distance covered by the second drop in $0.5\ s$

${{s}_{2}}=ut+\dfrac{1}{2}g{{t}^{2}}$

$s=0+\dfrac{1}{2}\times 10\times 0.5\times 0.5$

$s=1.25\,m$

Therefore, the distance of the second drop above the ground

$s={{s}_{1}}-{{s}_{2}}$

$s=5-1.25$

$s=3.75\,m$

Hence, this is the required solution

A stone is released with acceleration 'a' from an upwardy moving left. Find out the acceleration and direction of the stone.

0%
a in upward direction
0%
(g+a) in downward direction
0%
(g-a) in upward direction
0%
g in downward direction

Particle is moving in a straight path with an acceleration

$a = bt^n$ , where b is a positive constant

$\left( n\neq -1 \right)$ . The relation between the velocity and the position vector of the particle at time t = 1s [Assuming particle is in rest at origin at t=0]

0%
$V = r$
0%
$V=\left( n+2 \right) r$
0%
$V=\frac { \left( n+3 \right) r }{ 2 }$
0%
$V=\frac { \left( n+1 \right) r }{ 2 }$

Explanation

Given that,

$a=b{{t}^{n}}$

$\dfrac{dv}{dt}=b{{t}^{n}}$

$dv=b{{t}^{n}}dt$

$v=b\dfrac{{{t}^{n+1}}}{n+1}........(1)$

$\dfrac{dr}{dt}=b\dfrac{{{t}^{n+1}}}{n+1}$

$dr=\dfrac{b}{n+1}\int{{{t}^{n+1}}dt}$

$r=\dfrac{b}{n+1}\dfrac{{{t}^{n+2}}}{n+2}$

$r=\dfrac{Vt}{n+2}\,\,(from\,1)$

$at\,\,t=\,1\,s$

$V=(n+2)r$

A particle moves in a straight line with a constant acceleration. It changes its velocity from

$10$ m/s to

$20$ m/s while passing through a distance of

$135$ m in t seconds. The value of t is:-

0%
$12$
0%
$9$
0%
$10$
0%
$1.8$

Explanation

solution

Applying second

$eq^{n}$ of motion, we get

$s = ut +\dfrac{at^{2}}{2} ...(i)$

Also from first

$eq^{n}$ of motion,

we have

$v-u = at$

$\Rightarrow a = \dfrac{v-u}{t}...(ii)$

From (i) and (ii), we get

$s = ut + \dfrac{(v-u)}{2}t$

$\Rightarrow s = (\dfrac{u+v}{2})t$

Here,

$135 = (\dfrac{10+20}{2})t$

$t = 9 s$

option - B is correct.

1143059_1164375_ans_b26142583d0547499673ecef08579b5e.jpg

A machine which is

$75$ % efficient uses

$12$ joules of energy in lifting up a

$1$ kg mass through a certain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is(in

$ms^{-1}$ ):

0%
$\sqrt{24}$
0%
$\sqrt{32}$
0%
$\sqrt{18}$
0%
$\sqrt{9}$

Explanation

$100$ joules is given in the machine and

$75$ joules is output

So, If

$12$ joules is given in the machine and

$\dfrac { 75 }{ 100 } \times 12=9$ joules is output

$w=f\times d$

$9=1\times 9.8\times d$

$\dfrac { 9 }{ 9.8 } =d$

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

$2as=2\times 9.8\times \dfrac { 9 }{ 9.8 } =18$

${ v }^{ 2 }=18$

$v=\sqrt { 18 } \dfrac { m }{ s }$

A body travels 200 cm in the first two seconds and 220 cm in the next 4 seconds with same acceleration. The velocity of the body at the end of the 7th second is:

0%
10 cm/s
0%
5 cm/s
0%
12 cm/s
0%
2 cm/s

A body of mass

$2\ kg$ fall vertically, passing through two points

$A$ and

$B$ . The speeds of the body as it passes

$A$ and

$B$ are

$1\ m/s$ and

$4\ m/s$ respectively. The resistance against which the body falls is

$9.6\ N$ . What is the distance

$AB$ ?

0%
$2\ m$
0%
$3\ m$
0%
$6\ m$
0%
$1.5\ m$

Explanation

$u=1\dfrac { m }{ s }$

$v=4\dfrac { m }{ s }$

$a=\dfrac { F }{ m }$

$a=\dfrac { 9.6 }{ 2 } =4.8\dfrac { m }{ { s }^{ 2 } }$

using

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

${ 4 }^{ 2 }={ 1 }^{ 2 }+2\times 4.8\times S$

$S=1.5625m$

A ball, thrown by a boy, is caught by another boy in 2s. What is the maximum height attained by the ball?

0%
10 m
0%
5 m
0%
20 m
0%
7.5 m

Explanation

The formula for time of flight is

$T=\dfrac{2u\sin \theta }{g}$

It is given that time is 2 seconds. So,

$2=\dfrac{2u\sin \theta }{g}$

$g=u\sin \theta$

Squaring both sides

${{g}^{2}}={{u}^{2}}{{\sin }^{2}}\theta ........(1)$

Height travelled is

$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$

$H=\dfrac{{{g}^{2}}}{2g}\,\,\,\,(from\,\,1)$

$H=\dfrac{g}{2}=5\,m$

A bus begins to move with an acceleration of

$\dfrac{1}{2} \,m/s^2$ . A person who is

$40 \,m$ behind the bus, runs at a rate of

$7 \,m/s$ . Find the time taken by the person to catch the bus.

0%
$10$
0%
$20$
0%
$30$
0%
$40$

Explanation

Distance covered by mat

$S = vt$

$S_1 = 7t$ ....(1)

Distance covered by bus

$S = \dfrac{1}{2} \,at^2$

$S = \dfrac{1}{2} \times \dfrac{1}{2} \times t^2$

$S_2 = \dfrac{1}{4} t^2$

$S_2 + 40 = S_1$

$\dfrac{1}{4} t^2 + 40 = 7t$

$t^2 + 160 = 28t$

$t^2 - 20t - 8t + 60 = 0$

$t(t - 20) - 8(t - 20) = 0$

Relative disp

$= 40$

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

$40 = 7t + \dfrac{1}{2} \left(\dfrac{1}{2}\right) t^2$

$t^2 - 28t + 160 = 0$

$t^2 - 20t - 8t + 160 = 0$

$(t - 20)(t - 8)$

$t = 20$ sec.

A ball is thrown upwards from the ground with an initial velocity u. The ball is at a height of

$80$ m at two times. The time interval being

$6s$ . Take

$g=10m/s^2$ , the value of u is

0%
$100$ m/s
0%
$80$ m/s
0%
$50$ m/s
0%
$30$ m/s

Explanation

From situation

$B$ to

$C$

$t=6\ s$

$\implies \cfrac{2v}{g}=6$

$\implies v=\cfrac{6\times 10}{2}=30\ m/s$

Now from

$A$ to

$B$

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

$\implies 30^2=u^2+2(-10)(80)$

$\implies u=50\ m/s$

1221058_1212321_ans_56043aadbc6d4427bbbc1121b2213bd6.PNG

A laser gun of power

$3 \,mW$ and mass

$50 \,gm$ emits photons of wavelength

$500 \,nm$ . It is gravity frequency space and emits for one hour. Find the distance covered by gun due to recoil in this one hour (approx).

0%
$1.3 \,mm$
0%
$2.1 \,\mu m$
0%
$1.1 \,cm$
0%
$8.5 \,mm$

Explanation

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

$t = 1 \,hr = 3600 \,sec$
Let us find force on the gun

$=$ no. of photon emitted

$\times$ change in momentum

$= \left(\dfrac{Total \,energy \,emitted}{Energy \,of \,photon}\right) \times \left(\dfrac{h}{\lambda}\right)$

$= \dfrac{3 \times 10^{-3} Joules}{\dfrac{hc}{\lambda}} \times \dfrac{h}{\lambda}$

$= \dfrac{3 \times 10^{-3}}{3 \times 10^8} = 10^{-11} N$

$\therefore a = \dfrac{Force}{Mass} = \dfrac{10^{11}}{\left(\dfrac{50}{1000}\right)} = 20 \times 10^{-11} \,m/s^2$

$\therefore S = \dfrac{1}{2} \times 20 \times 10^{-11} \times (3600)^2$

$= 10^{-10} \times 12.96 \times 10 \,m$

$= 1.296 \,mm$

$= 1.3 \,mm$

a ship of mass

$3\times 10^{7}$ kg initially at rest is pulled by a force of

$5\times 10^{4}$ N through a distance of 3 m. Assuming that the resistance due to water is negligible, what will be the speed of the ship ?

0%
0.1 m/s
0%
1.5 m/s
0%
5 m/s
0%
0.2 m/s

Explanation

Given,

Mass of ship, $M=3\times {{10}^{7}}\,kg$

Force, $F=5\times {{10}^{4}}\,N$

Distance, $s=3\,m$

Acceleration, $a=\dfrac{F}{M}=\dfrac{5\times {{10}^{4}}}{3\times {{10}^{7}}}=\dfrac{1}{600\,}\,m{{s}^{-2}}$

Apply kinematic equation of motion

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

$v=\sqrt{2as}=\sqrt{2\times \dfrac{1}{600}\times 3}=0.1\,m{{s}^{-1}}$

Hence, velocity of ship $0.1\,m{{s}^{-1}}$

A parachutist after bailing out falls

$50 m$ without friction. When parachute opens, it decelerate at

$2 m/s^2$ . He reaches the ground with a speed of

$3 m/s$ . At what height, did he bail out nearly.

0%
$298 m$
0%
$111 m$
0%
$91 m$
0%
$182 m$

A body travelling with uniform acceleration crosses two points

$A$ and

$B$ with velocities

$20\ m/sec$ and

$30\ m/sec$ respectively then the speed of the body at mid point of

$A$ and

$B$ is

0%
$25\ m/sec$
0%
$25.5\ m/sec$
0%
$24\ m/sec$
0%
$10\sqrt { 6 }\ m/sec$

Explanation

Let the acceleration of the car is

$= a$ and the distance between

$A$ and

$B$ is

$=d$

$v^2-u^2=2ad$

Given,

$v=30\,m/s$ and

$u=20\,m/s$

$2ad=30^2-20^2$

$ad=\dfrac{(900-400)}{2}=250$

when the car is at the mid point of

$AB$ then let the speed of the car is

$v_1$

$v_1^2-20^2=2a(\dfrac d2)$

$v_1^2=ad+400=250+400=650$

therefore

$v_1$ is

$= 25.5\,m/s$

If you decide to launch a ball vertically to a friend located 45 m above you who can catch it. What is the minimum launch speed you can use?

0%
4.5 m/s
0%
1.50 m/s
0%
45 m/s
0%
29.7 m/s
0%
90 m/s

Explanation

Given that,

Distance $s=45\,m$

Initial velocity $u=0$

We know that,

By using equation of motion

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

${{v}^{2}}=0+2\times 9.8\times 45$

${{v}^{2}}=882\,m/s$

$v=29.7\,m/s$

Hence, the minimum speed is $29.7\ m/s$

A body is through vertically upwards with an initial velocity 'u' reaches a maximum height in 6s. The ratio of the distance travelled by the body in the first second to the seventh second is then

0%
1:1
0%
11:1
0%
1:2
0%
1:11

Explanation

It is given that, the body reaches the maximum height in 6s. Let u be the initial velocity. At maximum height the velocity is zero. Using the equation of motion as

$v=u+at$

$0=u-(10)(6)$

$u=60\,m/s$

Distance travelled in n^th second is given by:

${{S}_{n}}=u+\left( \dfrac{a}{2} \right)\left( 2n-1 \right)$

Displacement in 1^st second is ${{S}_{1}}=60-\left( \dfrac{10}{2} \right)(2-1)=55\,m$

Displacement in 7^th second is ${{S}_{2}}=60-\left( \dfrac{10}{2} \right)(14-1)=-5\,m$

So, $\dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{55}{5}=\dfrac{11}{1}$

The ratio is $11:1$ .

A particle falling from rest under gravity covers a distance

$x$ in

$4\ s$ . If it continues falling then next

$2\ x$ distance will be covered in approximately

0%
$1.41\ s$
0%
$1.73\ s$
0%
$2.05\ s$
0%
$2.92\ s$

Explanation

$h = \frac{1}{2} gt^{2}$

given

$x = \frac{1}{2} g (4)^{2} = 89$ ....(1)

Next 2x means total x+2x=3x from rest

so just use

$h=\frac{1}{2}gt^{2}$

& put h = 3x

we get

$3x = \frac{gt^{2}}{2}$

$\Rightarrow t^{2} = \frac{6x}{9}$

$\Rightarrow t = \sqrt{\frac{6x}{9}} = \sqrt{\frac{6}{9} \times 89}$ ; from equation

$\Rightarrow t = \sqrt{48} = 4\sqrt{3} = 4 \times 1.732 = 6.928 sec$

So time for distance b/w x & 3x

will be 6.928 - 4 = 2.928 sec

1179997_1295271_ans_768223e65b4b452aba747f63a025b5a1.jpg

A body is moving with velocity

$1$ m/s at a height

$3$ m from the ground. What will be its speed at height

$2$ m from the ground?

0%
4.54 m/s
0%
1 m/s
0%
6 m/s
0%
5.32 m/s

A stone is allowed to fall freely from rest. The ratio of the time taken to fall through the first meter and the second meter distance is :

0%
$\sqrt { 2 } - 1$
0%
$\sqrt { 2 } + 1$
0%
$\sqrt { 2 }$
0%
None of these

If a car at rest accelerates uniformly to a speed of 144 km/h in 20s , it covers a distance of

0%
400 m
0%
1440 m
0%
2880 m
0%
none of these

CBSE Questions for Class 11 Engineering Physics Motion In A Straight Line Quiz 9 - 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 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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