MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 6

An old man and a boy are walking towards each other and a bird is flying over them as shown in the figure. Find the velocity of bird as seen by the boy.

0%
$12{\hat{j}}$
0%
$16{\hat{j}}$
0%
$-12{\hat{j}}$
0%
$-16{\hat{j}}$

Explanation

From the figure,

Velocity of Boy with respect to ground is,

$\vec{v_B} = 16\hat{i}$

Velocity of Bird with respect to ground is,

$\vec{v_b} = 16\hat{i}+12\hat{j}$

So, velocity of bird with respect to boy is,

$\vec{v_b} -\vec{v_B} = 16\hat{i}+12\hat{j}-16\hat{i}=12\hat{j}$

So option A is correct.

The position vector of a particle moving in x-y plane is given by

$\displaystyle F= (A\sin \omega t)i+(A\cos \omega t)j$ then motion of the particle is

0%
$SHM$
0%
on a circle
0%
on a stralght line
0%
with constant acceleration

Explanation

let us assume that position vector is F = x i + y j

$x = A sin \omega t \space and \space y= A sin\omega t$

by squaring and adding

we get that

$x^2 + y^2 = A^2$

which is equation of a circle.

So option B is correct.

The angular velocity of rotation of hour hand of a watch is how many times the angular velocity of Earth's rotation about its own axis?

0%
Three
0%
Four
0%
Two
0%
Six

Explanation

Answer is C.

Hour hand completes one rotation in 12 hours while Earth completes one rotation in 24 hours. So, angular velocity of hour hand is double the angular velocity of Earth.
Angular velocity

$\omega =\dfrac { 2\pi }{ T }$ .
Hence, the angular velocity of rotation of hour hand of a watch is two times the angular velocity of Earth's rotation about its own axis.

The rear wheels of a car are turning at an angular speed of

$60$ rad/s. The brakes are applied for

$5$ s, causing a uniform angular retardation of

$8\, rad/s^{-2}$ . The number of revolutions turned by the rear wheels during the braking period is about:

0%
$48$
0%
$96$
0%
$32$
0%
$12$

Explanation

Given :

$w_i = 60$ rad/s

$t = 5$ s

$\alpha =-8$

$rad/s^2$

Using

$w_f = w_i+ \alpha t$

$\therefore$

$w_f =60 + (-8)\times 5$

$\implies w_f = 20$ rad/s

Using

$w_f^2-w_i^2 = 2\alpha \theta$

$\therefore$

$(20)^2 - (60)^2 = 2\times (-8)\theta$

$\implies \theta =200$ rad

Thus number of revolutions:

$N = \dfrac{\theta}{2\pi} \approx 32$ revolutions

Work done upon a body is a vector quantity. State true or false.

0%
True
0%
False

A wheel of radius 1 m rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

0%
$2 \pi m$
0%
$\sqrt2 \pi m$
0%
$\sqrt{\pi^2 +4} m$
0%
$\pi m$

Explanation

r = 1 m
Displacement in half a revolution =

$\pi r$

$AB' = \pi$
Displacement ofthe point initially in contact
=AA'

$=\sqrt{AB'^2 + A'B'^2} = \sqrt{\pi^2 +(2)^2}$
=

$\sqrt{\pi^2 + 4} m$

Which one of the following is not a vector ?

0%
Velocity
0%
Acceleration
0%
Force
0%
Energy

A cyclist taking a turn bends inside because

0%
He feels pleasure in doing so
0%
He increases speed in doing so
0%
He obtains necessary centripetal force
0%
He avoids accidents

How is average velocity found by using velocity-time graphs?

0%
$\text{Average velocity} = \dfrac{\text{Area under v-t graph in the time-interval}}{\text{time interval}}$
0%
$\text{Average velocity} = \dfrac{v(t_2)-v(t_1)}{t_2-t_1}$
0%
$\text{Average velocity} = \dfrac{v(t_2)+v(t_1)}{2}$
0%
None of the above

Explanation

$\text{Average velocity} = \dfrac{\text{Area under v-t graph in the time-interval}}{\text{time interval}}$

The above formula is valid for all v-t graphs.

Option B and option C are valid only for constant acceleration graphs.

Given that

$\vec {P} + \vec {Q} + \vec {R} = \vec {0}$ . Two out of the three vectors are equal in magnitude. The magnitude of the third vector is

$\sqrt {2}$ times that of the other two. Which of the following can be the angles between these vectors?

0%
$90^{\circ}, 135^{\circ}, 135^{\circ}$
0%
$45^{\circ}, 45^{\circ}, 90^{\circ}$
0%
$30^{\circ}, 60^{\circ}, 90^{\circ}$
0%
$45^{\circ}, 90^{\circ}, 135^{\circ}$

Explanation

Let magnitude of vectors

$P$ and

$Q$ be

$A$ i.e

$|\vec{P}| = |\vec{Q}| = A$

$\therefore$

$|\vec{R}| = \sqrt{2}A$

Let the angle between vectors

$P$ and

$Q$ be

$\theta$

Using

$|\vec{R}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2|\vec{P}| |\vec{Q}| cos\theta$

$(\sqrt{2}A)^2 = A^2 + A^2 + 2A^2 cos\theta$

$\implies cos\theta = 0$ OR

$\theta = 90^o$

Also as

$|\vec{P}| = |\vec{R}| cos\theta_1$

$\therefore$

$A = \sqrt{2}A cos\theta_1$

$\implies cos\theta_1 = \dfrac{1}{\sqrt{2}}$ OR

$\theta_1 = 45^o$

Thus the angle between

$\vec{P}$ and

$\vec{Q}$

$= 180^o - 45^o = 135 ^o$

Similarly

$|\vec{Q}| = |\vec{R}| cos\theta_2$

$\therefore$

$A = \sqrt{2}A cos\theta_2$

$\implies cos\theta_2 = \dfrac{1}{\sqrt{2}}$ OR

$\theta_2 = 45^o$

Thus the angle between

$\vec{R}$ and

$\vec{Q}$

$= 180^o - 45^o = 135 ^o$

A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored?

Explanation

For a body going in upward direction $v = u gt$

The slope of the graph, $\displaystyle\frac{dv}{dt} = -g$ (constant).

But when we take into account the effect of resistance it will have sharper slope.

Curve (d) fits into this result.

For a particle in a uniformly accelerated circular motion

0%
Velocity is radial and acceleration has both radial and transverse components
0%
Velocity is transverse and acceleration has both radial and transverse components
0%
Velocity is. radial and acceleration is transverse only
0%
Velocity is transverse and acceleration is radial only

The driver of a car travelling at velocity

$v$ suddenly see a broad wall in front of him at a distance

$d$ . He should :

0%
Brake sharply
0%
Turn sharply
0%
Both (a) and (b)
0%
None of these

Explanation

When driver applies brakes and the car covers distance

$x$ before coming to rest, under the effect of retarding force

$F$ .
Then

$\dfrac { 1 }{ 2 } m{ v }^{ 2 }=Fx$
or

$x=\dfrac { m{ v }^{ 2 } }{ 2F }$
But when he takes turn, then

$\dfrac { m{ v }^{ 2 } }{ r } =F$

$\Rightarrow r=\dfrac { m{ v }^{ 2 } }{ F }$
It is clear that

$x=\dfrac { r }{ 2 }$
i.e., by the same retarding force the car can be stopped in a less distance if the driver apply brakes. This retarding force is actually a friction force.

A particle is executing a two dimensional motion. What is the minimum number of velocity-time graphs required to study the motion of the particle using graphs?

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

A and B travel around a circular path at uniform speeds in opposite direction starting from the diagrammatically opposite point at the same time. They meet each other first after B has traveled 100 meters and meet again 60 meters before A complete one round. What is the circumference of the circular path?

0%
$240m$
0%
$360m$
0%
$480m$
0%
$300m$

A wheel has a speed of 1200 revolutions per minute and is made to slow down at a rate of 4 rad/s

$^2$ . The number of revolutions it makes before coming to rest is:

0%
143
0%
272
0%
314
0%
722

Explanation

Initial frequency

$\nu_i = 1200$ r.p.m

$= \dfrac{1200}{60} = 20$ r.p.s

Initial angular velocity:

$\omega_i = 2\pi \nu_i = 2\pi \times 20 = 40 \pi$

$rad/s$

Final angular velocity:

$\omega_f = 0$

Using,

$\omega_f^2 - \omega_i^2 = 2\alpha S$

$\therefore$

$0 - (40\pi)^2 = 2 (-4) S$

$\implies S = 200\pi^2$ radian

Number of revolutions:

$N = \dfrac{200 \pi^2}{2\pi} =100 \pi \approx 314$

Uniform linear motion is a/an _______ motion while uniform circular motion is a/an _______ motion.

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accelerated, non accelerated
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non accelerated, accelerated
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deviated, retarded
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uniform, retarded

A wheel has a speed of

$1200$ revolution per minute and is made to slow down at a rate of

$4\;rad/s^2$ . The number of revolution it makes before coming to rest is

0%
$143$
0%
$272$
0%
$314$
0%
$722$

Explanation

Initial speed

$\nu_i =1200$

$r.p.m = \dfrac{1200}{60}$

$r.p.s = 20$

$r.p.s$

$\therefore$

$w_i = 2\pi \nu_i = 2\pi \times 20 = 40 \pi$

$rad/s$

Final angular velocity

$w_f= 0$

Using

$w_f^2 - w_i^2 = 2\alpha S$

$\therefore$

$0 - (40\pi)^2 = 2 (-4) S$

$\implies$

$S = 200 \pi^2$

$rad$

Number of revolution

$N = \dfrac{200 \pi^2}{2\pi} = 100 \pi = 314$

A particle is moving with constant speed

$v$ in circle. What is the magnitude of average velocity after half rotation?

0%
$2v$
0%
$2\dfrac{v}{\pi}$
0%
$\cfrac{v}{2}$
0%
$\cfrac{v}{2\pi}$

Explanation

The particle starts from point

$P$ and after half rotation, it reaches at

$Q$ via path

$PMQ$ .

Distance covered by particle to reach at

$Q$ is

$d = \pi R$

Time taken,

$t = \dfrac{d}{v} = \dfrac{\pi R}{v}$

Displacement of the particle,

$s = PQ = 2R$

$\therefore$ Average velocity,

$v_{avg} = \dfrac{s}{t} = \dfrac{2R}{\dfrac{\pi R}{v}} = \dfrac{2}{\pi } v$

When a ceiling fan is switched on, it makes 10 revolution in the first 3 s. Assuming a uniform angular acceleration, how many rotation it will make in the next 3 s?

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

Explanation

Let the angular acceleration be

$\alpha$ .

From the given data:

$\Rightarrow 10\times2\pi=\dfrac{1}{2}\alpha (3)^2$

$\Rightarrow 20\pi=\dfrac{9\alpha}{2}$

Let us take revolution for first six seconds:

$\Rightarrow \theta_o=\dfrac{1}{2}\alpha(6)^2=18\alpha=80\pi$

The no. of revolutions in first six seconds are 40.

The no. of revolutions in the last three seconds

$=40-10=30$ revolutions

A wheel has angular acceleration of

$3.0\ rad/s^2$ and an intial angular speed of

$2.00\ rad/s$ . In a time of

$2s$ it has rotated through an angle (in radians) of:

0%
$6$
0%
$10$
0%
$12$
0%
$4$

Explanation

Given:

$\omega=2$

$\alpha=3$

To find is angular displacement in

$2$ s:

Using equation of motion

$\theta=\omega t+\dfrac{1}{2}\alpha {t}^{2}$

Putting the values we get:

$\theta=2\times 2+\dfrac{1}{2}\times 3\times {2}^{2}$

$\Rightarrow \theta =10$

Certain neutron stars are believed to be rotating at about

$1$ rev/s. If such a star has a radius of

$20$ km, the acceleration of an object on the equator of the star will be.

0%
$20\times 10^8m/s^2$
0%
$8\times 10^5m/s^2$
0%
$120\times 10^5m/s^2$
0%
$4\times 10^8m/s^2$

Explanation

It is known

$a=\omega^2 r=4\pi^2v^2r$

putting values:

$a=4\times \pi^2\times 1^2\times 20\times 10^3$

$a=8\times 10^5m/s^2$

A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of

$15^o$ in time

$t$

$secs$ . The increase in angle through which it rotates in the next

$2t$

$secs$ is:

0%
$90^o$
0%
$120^o$
0%
$30^o$
0%
$45^o$

Explanation

Let the angular acceleration be

$\alpha$

Initial angular velocity of wheel

$\omega _i= 0$

Angle rotated by wheel in

$t$ seconds

$\theta_1 = 15^o$

Using

$\theta_1 = \omega_1t + \dfrac{1}{2}\alpha t^2$

$\therefore$

$15^o = 0 + \dfrac{1}{2}\alpha t^2$

$\implies \alpha t^2 =30^o$

Let angle rotated by wheel in

$t'$ seconds be

$\theta_2$ where

$t'= t + 2t = 3t$

$\therefore$

$\theta_2 = \omega_i t' + \dfrac{1}{2}\alpha (t')^2$

$\theta_2 = 0 + \dfrac{1}{2}\alpha (3t)^2 = \dfrac{9}{2}\alpha t^2$

$\implies \theta_2 = \dfrac{9}{2} \times 30 =135^o$

Increase in angle:

$\Delta \theta = \theta_2 - \theta_1 = 135 - 15 =120^o$

An object is thrown towards the tower which is at a horizontal distance of

$50\ m$ with initial velocity of

$\displaystyle 10{ ms }^{ -1 }$ and making an angle

$\displaystyle { 30 }^{ \circ }$ with the horizontal. The object hits the tower at certain height. The height from the bottom of the tower where the object hit the tower is (

$\displaystyle g=10{ ms }^{ -2 }$ ) :

0%
$\displaystyle \frac { 50 }{ \sqrt { 3 } } \left[ 1-\frac { 10 }{ \sqrt { 3 } } \right] m$
0%
$\displaystyle \frac { 50 }{ 3 } \left[ 1-\frac { 10 }{ \sqrt { 3 } } \right] m$
0%
$\displaystyle \frac { 100 }{ \sqrt { 3 } } \left[ 1-\frac { 10 }{ \sqrt { 3 } } \right] m$
0%
$\displaystyle \frac { 100 }{ 3 } \left[ 1-\frac { 10 }{ \sqrt { 3 } } \right] m$

Explanation

Time in which the body covers the horizontal distance of

$50\ m=\dfrac{50}{v_x}=\dfrac{50}{v cos\theta}=\dfrac{50}{10\times cos30^{\circ}}$

$=\dfrac{10}{\sqrt{3}}s$

Thus, the height that the object reaches in this time =

$s=v_yt-\dfrac{1}{2}gt^2$

$=v sin\theta t-\dfrac{1}{2}gt^2$

$=10\times \dfrac{1}{2}\times \dfrac{10}{\sqrt{3}}-\dfrac{1}{2}\times 10\times (\dfrac{10}{\sqrt{3}})^2$

$=\dfrac{50}{\sqrt{3}}-\dfrac{500}{3}\$

$=\dfrac{50}{\sqrt{3}}[1-\dfrac{10}{\sqrt{3}}]\ m$

A particle moves so that its position vector is given by

$\vec{r}=\cos \omega t \hat{x}+\sin \omega t\hat{y}$ . Where

$\omega$ is a constant. Which of the following is true?

0%
Velocity an acceleration both are perpendicular to $\vec{r}$
0%
Velocity and acceleration both are parallel to $\vec{r}$
0%
Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin
0%
Velocity is perpendicular to $\vec{r}$ and acceleration is directed away from the origin.

Explanation

$\textbf{Step 1: Velocity and acceleration calculation}$

$\vec {r} = \cos \omega t \hat {x} + \sin \omega t \hat {y}$

$.... (1)$

Velocity is given by

$\vec {v} = \dfrac {d\vec {r}}{dt}$

$\Rightarrow$

$\vec {v} = -\omega \sin \omega t \hat {x} + \omega \cos \omega t \hat {y} \ m/s$

$.... (2)$

Acceleration is given by

$\vec {a} = \dfrac {d\vec {v}}{dt}$

$\Rightarrow$

$\vec {a} = -\omega^{2} \cos \omega t \hat {x} - \omega^{2} \sin \omega t \hat {y} \ m/s^{2}$

$.... (3)$

$\textbf{Step 2: Analysing}$

$\vec {r}, \vec {v}$ and

$\vec {a}$

If the dot product between two vectors is zero that means the vectors are perpendicular to each other.

$\vec {v}\cdot \vec {r} = -\omega \sin \omega t \cos \omega t + \omega \sin \omega t \cos \omega t$

$= 0$

$\therefore \vec {v}$ is perpendicular to

$\vec {r}$

$\vec {a} \cdot \vec {r}= -\omega^2 \cos^2\omega t-\omega^2\sin^2\omega t \neq 0$

$\therefore \vec {a}$ is not perpendicular to

$\vec {r}$

Negative sign in

$eq^{n} (3)$ indicates acceleration is towards the origin (Opposite to

$\vec {r}$ )

Hence ,

$\vec{v}$ is perpendicular to

$\vec{r}$ and the acceleration is directed towards the origin

$\therefore$ Option C is the correct answer.

When the tail of vector A is set at the origin of the

$XY$ -axis, the tip of A reaches

$(3,6)$ . When the tail of vector B is set at the origin of the XY -axis, the tip of Breaches

$(-1,5).$ If the tail of vector

$\vec {AB}$ were set at the origin of the

$XY$ - axis, what point would its tip touch?

0%
$(2,11)$
0%
$(2, 1)$
0%
$(-2,7)$
0%
$(-4, -1)$
0%
$(4,11)$

Explanation

Any vector stretching from a point

$X$ to a point

$Y$ can be written as

$\vec{Y}-\vec{X}$

Thus,

$\vec{A} = 3\hat{i}+6\hat{j} - 0\hat{i}-0\hat{j} = 3\hat{i}+6\hat{j}$

Similarly.

$\vec{B} = -1\hat{i}+5\hat{j} - 0\hat{i}-0\hat{j} = -1\hat{i}+5\hat{j}$

Thus,

$\vec{AB} = \vec{B}-\vec{A} = -1\hat{i}+5\hat{j}-3\hat{i}-6\hat{j} = -4\hat{i}-1\hat{j}$

Thus, when

$\vec{AB}$ tail is kept at origin, its tip goes to

$(-4,-1)$

Which of the following are vector quainities?

0%
Force
0%
Velocity
0%
Momentum
0%
All of the above

Which of the following vectors best represents the vector A + B ?

Find out the angular acceleration of a washing machine, starting from rest, accelerates within

$3.14 s$ to a point where it is revolving at a frequency of

$2.00 Hz$ .

0%
$0.100 {rad}/{{s}^{2}}$
0%
$0.637 {rad}/{{s}^{2}}$
0%
$2.00 {rad}/{{s}^{2}}$
0%
$4.00 {rad}/{{s}^{2}}$
0%
$6.28 {rad}/{{s}^{2}}$

Explanation

Given :

$t = 3.14$ s

$w_i = 0$ rad/s

$\nu = 2$ Hz

Final angular velocity

$w_f =2\pi \nu = 2\pi \times 2 =4\pi$ rad/s

Using

$w_f =w_i + \alpha t$

$\therefore$

$4\pi = 0 + \alpha (3.14)$

$\implies \alpha = 4.00$

$rad/s^2$

Two coins are sitting on a horizontal disk, which is rotating at a constant speed.
One coin is a penny located halfway between the center of the disk the edge. The other coin is a dime located at the edge of the disk.
How does the magnitude of the dime's acceleration compare to the magnitude of the penny's acceleration?

0%
The magnitude of the acceleration of the dime is the same as the magnitude of the acceleration of the penny
0%
The magnitude of the acceleration of the dime is twice the magnitude of the acceleration of the penny
0%
The magnitude of the acceleration of the dime is half the magnitude of the acceleration of the penny
0%
The magnitude of the acceleration of the dime is four times the magnitude of the acceleration of the penny
0%
The magnitude of the acceleration of the dime is one-fourth the magnitude of the acceleration of the penny

Explanation

The angular speed of both the coins is same as that of the disc(let

$\omega$ ).

The angular acceleration of the coin at a distance

$r$ from the center of the disc is given by

$\omega^2 r$

Since

$r_2=2r_1$ ,

The ratio of acceleration of dime to that of penny=

$\dfrac{\omega^2r_2}{\omega^2r_1}=2$

Hence correct answer is option B.

The objects pictured above are coins moving around on a record player. The coins are not sliding on the record player surface. Which coins is moving faster and how many times faster it is moving than the other one ?
The space between each vertical line along the horizontal arrow is one-seventh the radius of the circular record player surface.

0%
1-three times faster than slowest
0%
3-five-seventh's faster than slowest
0%
3-five times faster than slowest
0%
1-five-seventh's faster than slowest
0%
2-three-seventh's faster than slowest

Explanation

As all the three coins are on the same record , therefore their angular velocities will be same ,

$\omega$ ,

now we have , linear velocity

$v=\omega r$ ,

where

$r=$ radius ,

according to given figure , radius of first coin

$r_{1}=r/7$ ,

radius of second coin

$r_{2}=\dfrac{r}{7}+\dfrac{r}{7}+\dfrac{r}{7}=3r/7$ ,

similarly , radius of third coin

$r_{3}=\dfrac{5r}{7}$

velocity of first coin

$v_{1}=\omega r_{1}=\omega r/7$ ,

velocity of second coin

$v_{2}=\omega r_{2}=\omega \times3 r/7$ ,

velocity of third coin

$v_{3}=\omega r_{3}=\omega \times5r/7$

it is clear that 3-five times than slowest (1) .

A skater moves over ice in circular path at a constant speed. He later moves over ice in a circular path at a constant speed, but this time with five times as much acceleration as before.
What single difference in the motion of the skater might have caused his acceleration to be five times as great?

0%
He may have been moving five times as fast as before
0%
He may have been moving around a curve with five times the radius of the original circular path.
0%
He may have been moving a little over two times as fast
0%
He may have been moving around a curve with a radius just over twice as much as the original radius.
0%
he may have been wearing more massive skates.

Explanation

for objects moving at constant speed in circular paths, the only two factors that affect acceleration are speed and circular radius. the equation for this "centripetal" or "center-seeking" acceleration is:

${ a }_{ c }=\dfrac { { v }^{ 2 } }{ r }$

We see from this equation that the acceleration for an object moving at a constant speed in a circular path is directly proportional to the square of the object's speed and inversely proportional to the radius of its circular path.

in this case, we want to make the acceleration 5 times as great.

since the acceleration of the skater is directly proportional to the square of the speed, making the speed 5 times as great will make the acceleration 25 times as great.

if we make the speed just over two times as great, we cause the acceleration to be just over "2 squared" times as great, or 5 times as great. (we would need to be more exact in our choice of speed to make the acceleration exactly 5 times as great).

The correct answer is choice C.

A dog walks

$140\ m$ due east and then turns west and runs

$65\ m$ . He then turns north and trots

$40\ m$ . After completing his journey, he is

$85\ m$ northeast of his home. When he hears his master's call, he runs directly home.
Which part of the trip is a negative vector?

0%
The eastward leg
0%
The westward leg
0%
The northward leg
0%
The distance from home
0%
The distance to home

Explanation

When dog goes

$65\ m$ towards west, he goes against

$140\ m$ of east and to get his position now we subtract

$65\ m$ from

$140\ m$ .

It means we add a positive vector (

$140\ m$ due east) to a negative vector (

$65\ m$ due west). That's why westward leg is a negative vector.

Which of the following are vector quantities?

0%
Magnetic moment
0%
Magnetic permeability
0%
Magnetic induction
0%
Magnetic pole strength

Explanation

Magnetic moment, also known as magnetic dipole moment of a magnet is defined as the product of the strength of one of the poles and the distance between the poles. It is directed from the south pole towards the north, inside the magnet. Thus magnetic moment has magnitude as well as direction and is a vector

If two magnetic poles are kept in air or vacuum, then the force of attraction(if the poles are unlike) or repulsion(if they are like) is given by

$F={\cfrac{\mu_0}{4\pi}}\times{\cfrac{m_1m_2}{r^2}}$

Here

$\mu_0$ is the magnetic permeability of free space. It only has magnitude. Hence it is a scalar. If the poles are in a material, then

$\mu_0$ will be replaced by

$\mu$ which is the permeability of that material.

Magnetic induction is nothing but the magnetic field intensity at a point in a magnetic field. It is defined as the force acting on a north pole of unit magnitude, kept in the field. Since force is a vector, magnetic induction is also a vector.

Magnetic pole strength is the strength of one of the poles of a magnet. It has only magnitude and hence is a scalar.

Starting from rest, a fan take five seconds to attain the maximum speed of 400 rpm. Assume constant acceleration, find the time taken by the fan attaining half the maximum speed.

0%
11 s
0%
2.0 s
0%
5.0 s
0%
2.5 s

Explanation

The maximum angular velocity is given,

$w_m=400 rpm=400\times \dfrac{2\pi}{60}=\dfrac{40\pi}{3} rad /sec$

Initial angular velocity is

$w_m=0$

So the angular acceleration,

$\alpha=\dfrac{w_m-w-0}{t}=\dfrac{40\pi/3-0}{5}=\dfrac{8\pi}{3} rad/sec^2$

Let the time

$t_{1/2}$ taken by the fan attaining half the maximum speed.

Now using,

$w=w_0+\alpha t$ we get

$w_m/2=0+\alpha t_{1/2}$

$t_{1/2}=\dfrac{w_m}{2\alpha}=\dfrac{(40\pi/3)}{2(8\pi/3)}=2.5 s$

If the horizontal range is given as 12.8 m, then the equation of parabolic trajectory can be given by:

0%
$16x - \dfrac{5x^2}{4}$
0%
$16x - \dfrac{3x^2}{4}$
0%
$14x - \dfrac{5x^2}{4}$
0%
$12x - \dfrac{5x^2}{4}$

Explanation

Equation of trajectory:

$y = xtan\theta - \dfrac{g^2 x^2}{2u^2 cos^2\theta}$

$\implies$

$y = \dfrac{g}{2 u^2 cos^2\theta}(Rx-x^2)$

where,

$R = \dfrac{u^2sin2\theta}{g} = 12.8$ m

Now,

$y = 16x - \dfrac{5x^2}{4}$

$\implies y =0.8(12.8x-x^2) = 0.8(Rx-x^2)$

Hence, the equation of parabolic trajectory,

$y =16x-\dfrac{5x^2}{4}$

The velocity time graph of a particle moving along a straight line has the form of a parabola

$t^{2}-6t+8$ m/s. Find the velocity (in m/s) when acceleration of particle is zero :

0%
$-1$
0%
$-2$
0%
$-3$
0%
$-4$

Explanation

Here, velocity

$v=t^2-6t+8$

So, acceleration,

$a=\frac{dv}{dt}=2t-6$

When acceleration is zero,

$2t-6=0$ or

$t=3$

and the veclocity will be

$v(t=3)=3^2-6(3)+8=9-18+8=-1 m/s$

If the angle of projection is

$60^0$ , the equation of trajectory can be given by

0%
$\sqrt {2} x- \dfrac{gx^2}{2}$
0%
$\sqrt {3} x- \dfrac{gx^2}{4}$
0%
$\sqrt {3} x- \dfrac{2gx^2}{u^2}$
0%
$\sqrt {3} x- \dfrac{gx^3}{2}$

Explanation

The equation of a projectile motion is a parabola, which is given by:

$y=x\tan\theta-\dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}$

y=xtan⁡θ−(g2u2cos2⁡θ)x2

where, x =x= horizontal distance travelled by object in time

y =y=vertical distance travelled in time

θ =θ= angle of projection

u =u= velocity of projection

Given:

$\theta=60^{o}$

Hence,

$y=x\tan60-\dfrac{gx^{2}}{2u^{2}\cos^{2}60}$

$y=\sqrt3x-\dfrac{gx^{2}}{2u^{2}\times1/4}$

$y=\sqrt3x-\dfrac{2gx^{2}}{u^{2}}$

Equation of oblique projectile can be written as:

0%
$y=xtan\theta(1-\dfrac{x}{T})$
0%
$y=xtan\theta(1-\dfrac{x}{R})$
0%
$y=xsin\theta(1-\dfrac{x}{R})$
0%
$y=xtan\theta(1-\dfrac{x}{H})$

Explanation

The equation of an oblique projectile is given by:

$y=x\tan\theta-\dfrac{gx^{2}}{2u^{2}\cos^{2}\theta}$

$y=x\tan\theta-\dfrac{x^{2}}{(2u^{2}\cos^{2}\theta/g)}$

$y=x\tan\theta-\dfrac{x^{2}\sin\theta}{\cos\theta(2u^{2}\sin\theta\cos\theta)/g}$

$y=x\tan\theta-\dfrac{x^{2}\sin\theta}{\cos\theta(u^{2}\sin2\theta/g)}$

$y=x\tan\theta-\dfrac{x^{2}\tan\theta}{R}$

$y=x\tan\theta(1-\dfrac{x}{R})$

A motor of an engine is rotating about its axis with an angular velocity of

$100$ rev/min . It comes to rest in

$15$ s, after being switched off. Assuming constant angular deceleration, calculate the number of revolutions made by it before coming to rest.

0%
$10.5$
0%
$12.5$
0%
$15.0$
0%
$20.5$

Explanation

Deceleration can be found by using the equation of motion

$\omega=\omega_0+\alpha t$

$\implies \alpha=-\dfrac{\omega_0}{t}$

$=-\dfrac{100}{15\times 60}rev/s^2$

$=-\dfrac{1}{9}rev/s^2$

Now using the equation of motion:

$\omega^2-\omega_0^2=2\alpha \theta$

$\implies \theta=\dfrac{\omega_0^2}{2\alpha}$

$=12.5$

A wheel rotates with a constant acceleration of

$2 \ \ rads^{-2}$ . If the wheel starts from rest, how many revolution will it make in the first

$10$ seconds?

0%
8
0%
12
0%
16
0%
20

Explanation

$\theta$ be the angular displacement, then

$\theta=\dfrac{1}{2}\alpha t^2=0.5(2)(10)^2=100$ radians (using equation :

$S=ut+1/2at^2$ )

Thus, number of revolutions

$=\dfrac{\theta}{2\pi}=\dfrac{100}{2\pi}=15.92 \sim 16$

The ratio between the length and the breadth of a rectangular park is 3 :If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

0%
15360
0%
153600
0%
30720
0%
307200

Explanation

Perimeter = Distance travelled in 8 minutes,

$\Rightarrow perimeter\, = \dfrac{{12000}}{{60}} \times 8 = 1600\,m.\,\,\left[ {because\,\,Dis\tan ce = Speed \times Time} \right]$
As per question length is 3x and width is

$2x$
We know perimeter of rectangle is

$2(L+B)$
So

$, 2(3x+2x) = 1600$

$=> x = 160$
So Length

$= 160 \times 3 = 480 m$
and Width

$= 160 \times 2 = 320 m$
Finally

$,$

$Area = length \times breadth$

$= 480 \times 320 = 153600$

Hence,

option

$(B)$ is correct answer.

Find the change in velocity of the tip of the minute hand (radius =

$10 cm$ ) of a clock in

$45$ minutes. (in cm/min)

0%
$\pi \dfrac{\sqrt{2}}{3}$
0%
$\pi \dfrac{\sqrt{2}}{2}$
0%
$\pi \dfrac{\sqrt{3}}{3}$
0%
$\pi \dfrac{2}{3}$

Explanation

Speed of the tip of the minute will always remain same.

Speed = angular velocity

$\times$ radius

where radius

$=10\ cm$

Time taken by minute hand to rotate

$2\pi$ angle is

$60$ minutes.

Thus its angular velocity

$=\dfrac{2\pi}{60}\:rad/min$

Therefore, speed of the tip of minute hand becomes:

Speed

$= \dfrac{2\pi}{60} \times 10 =\dfrac{\pi}{3}\:cm/min$

Since velocity is a vector, we need to find:

$\vec{v}_{final}-\vec{v}_{initial}$

As seen in the attached diagram, initial velocity is along

$x$ direction and the final velocity is in

$y$ direction, therefore

$\vec{v}_{final}-\vec{v}_{initial}=\dfrac{\pi}{3}(\hat{(j)}-\hat(i) \:)$

Taking the magnitude of

$\vec{v}_{final}-\vec{v}_{initial}$ ,

we get

$|\vec{v}_{final}-\vec{v}_{initial}| =\dfrac{\pi}{3} \times \sqrt{1^2 + 1^2}=\pi \dfrac{\sqrt{2}}{3}$

$cm/min$

A horse runs on a circular track of length

$720$ metres in

$20$ seconds and returns to the starting point. Calculate the average speed

0%
$36\ metre/second$
0%
$0\ metre/second$
0%
$18\ metre/second$
0%
$72\ metre/second$

Explanation

Total distance covered by the horse

$S = 720 m$

Time taken

$t = 20 s$

Thus average speed of the horse

$= \dfrac{S}{t} = \dfrac{720}{20} = 36 m/s$

A particle moves along a circle of radius '

$r$ ' with constant tangential acceleration. If the velocity of the particle is '

$v$ ' at the end of second revolution, after the revolution has started then the tangential acceleration is

0%
$\dfrac{v^2}{8 \pi r}$
0%
$\dfrac{v^2}{6 \pi r}$
0%
$\dfrac{v^2}{4 \pi r}$
0%
$\dfrac{v^2}{2 \pi r}$

Explanation

Initial speed of the particle

$u=0$

Final speed of the particle is

$v$

Distance covered in 2 revolutions

$S = 2(2\pi r) = 4\pi r$

Using

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

where

$a$ is the tangential acceleration

$\therefore$

$v^2 - 0 = 2a(4\pi r)$

$\implies$

$a = \dfrac{v^2}{8\pi r}$

If a vector

$A$ having a magnitude of

$8$ is added to a vector

$B$ which lies along x-axis, then the resultant of two vectors lies along y-axis and has magnitude twice that of

$B$ . The magnitude of

$B$ is

0%
$\cfrac { 6 }{ \sqrt { 5 } }$
0%
$\cfrac { 12 }{ \sqrt { 5 } }$
0%
$\cfrac { 16 }{ \sqrt { 5 } }$
0%
$\cfrac { 8 }{ \sqrt { 5 } }$

Explanation

According to condition

$\vec{A}+B\hat{i}=R\hat{j}$ (Where

$R$ is the resultant vector)
Also

$R=2B$

$\therefore$

$\vec{A}+B\hat{i}=2B\hat{j}$
or

$\vec{A}=2B\hat{j}-B\hat{i}$

$\therefore { A }^{ 2 }=4{ B }^{ 2 }+{ B }^{ 2 }=5{ B }^{ 2 }$
Here,

$A=8$

$\therefore 64=5{ B }^{ 2 }$

$\Rightarrow B=\sqrt { \cfrac { 64 }{ 5 } } =\cfrac { 8 }{ \sqrt { 5 } }$

The locus of a projectile relative to another projectile is a

0%
straight line
0%
circle
0%
ellipse
0%
parabola

Explanation

Equation of path of projectile is

$S=u\cos { \theta } -\cfrac { g{ t }^{ 2 } }{ 2 }$

If two projectiles are projected from horizontal ground with different initial velocity and angle then at any time t

Co ordinates of

$A={ S }_{ 1 }=\left( { u }_{ a }\cos { A } \right) { t }_{ i }-\cfrac { g{ t }^{ 2 } }{ 2 } j$

Co-ordinates of

$B={ S }_{ 2 }=\left( { u }_{ b }\cos { B } \right) { t }_{ i }-\cfrac { g{ t }^{ 2 } }{ 2 } j$

Co ordinates of A with respect to B is

${ S }_{ 2 }-{ S }_{ 1 }=t\left( { u }_{ a }\cos { A } -{ u }_{ b }\cos { B } \right) i+0j$

${ x }_{ i }+{ y }_{ j }=k{ t }_{ i }+0j$

[

${ u }_{ a }\cos { A } -{ u }_{ b }\cos { B }=$ constant

$=k$ ]

i.e.,

$X=kt$

It is a straight line.

In the entire path of a projectile, the quantity that remains unchanged is

0%
Vertical component of velocity
0%
Horizontal component of velocity
0%
Kinetic energy
0%
Potential energy
0%
Linear momentum

When a ceiling fan is switched off, its angular velocity reduces to half its initial value after it completes

$36$ rotations. The number of rotations it will make further before coming to rest is Assuming angular retardation to be uniform

0%
$10$
0%
$20$
0%
$18$
0%
$12$

Explanation

1st case:

We can write the equation of motion for circular motion as:

$\omega^2=\omega_0^2+2a\theta_n$

Now,

$\omega=\dfrac{\omega_0}{2}, \theta=36\times 2\pi$ (given)

So,

$\dfrac{\omega_0^2}{4}=\omega_o^2+2\alpha\times36\times2\pi$

So,

$\alpha=\dfrac{-3\omega_0^2}{4\times144\pi}$

2nd case:

$0 = \dfrac{\omega_0^2}{4}-2\times\dfrac{3\omega_0^2}{4\times 144\pi}\times\theta$

So,

$\theta=24\pi$

So, number of rotations made by the fan before coming to rest

$=\dfrac{24\pi}{2\pi}=12$

If the position vector

$\overrightarrow{a}$ of the point

$(5, n)$ is such that

$|\overrightarrow{a}|=13$ , then the value/values of n be

0%
$\pm 8$
0%
$\pm 12$
0%
8 only
0%
12 only

Explanation

Solution:

Given that:

$|\overrightarrow a|=13$

We have,

$\overrightarrow a=5\hat i+n\hat j$

or,

$|\overrightarrow a|=\sqrt{5^2+n^2}$

or,

$13=\sqrt{25+n^2}$

or,

$n^2=169-25=144$

or,

$n=\pm12$

Hence, B is the correct option.

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 6 - 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 Plane Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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