MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 1

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is

$v$ , the total area around the fountain that gets wet is :

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$\displaystyle \pi\dfrac{\mathrm{v}^{2}}{\mathrm{g}}$
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$\displaystyle \pi\dfrac{\mathrm{v}^{4}}{\mathrm{g}^{2}}$
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$\displaystyle \dfrac{\pi \mathrm{v}^{4}}{2\mathrm{g}^{2}}$
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$\displaystyle \pi\dfrac{\mathrm{v}^{2}}{\mathrm{g}^{2}}$ .

Explanation

As we know that

$\displaystyle \mathrm{A}=\pi \mathrm{R}_{\mathrm{m}\mathrm{a}}^{2}$ .....(1)
Maximum Range=

$\displaystyle \mathrm{R}_{\mathrm{m}\mathrm{a}}^{}$ =

$\displaystyle \frac{\mathrm{v}^{2}}{\mathrm{g}}$ ............(2)
From (1) and (2) we get,

$\displaystyle \mathrm{A}=\frac{\pi \mathrm{v}^{4}}{\mathrm{g}^{2}}$ .

A particle starts from the origin at

$t=0$ with an initial velocity of

$3.0\ \hat i\ m/s$ and moves in the

$x-y$ plane with a constant acceleration

$(6.0\ \hat i +4.0\ \hat j) m/s^2$ . The

$x-$ coordinates of the particle at the instant when its

$y-$ coordinates is

$32\ m$ is

$D$ meters. The value of

$D$ is:

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$50$
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$60$
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$40$
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$32$

Explanation

Here acceleration is constant so we can use equations of motion.

Now taking motion along y axis

displacement along

$y (s_y) = 32 m$

acceleration along

$y (a_y) = 4 ms^{-2}$

initial velocity along y axis

$(u_y) = 0$

using second equation of motion

$s_y = u_y \times t + \dfrac{1}{\alpha} a_y \ t^2$

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

$\Rightarrow 32 = 0 \times t + \dfrac{1}{2} \times 4 \times t^2$

$\Rightarrow t^2 = 16$

$\Rightarrow t = 4$ ___(i)

Now taking motion along x axis.

displacement along x axis

$(s_x) = ?$

acceleration along x axis

$(ax) = 6$

initial velocity along x axis

$(u_n) = 3$

So,

$S_x = u_x t + \dfrac{1}{2} a_x t^2$

$= 3 \times 4 + \dfrac{1}{2} \times 16$

$= 12 + 48$

$= 60 m$

For a particle in uniform circular motion the acceleration

$\vec{\mathrm{a}}$ at a point

$\mathrm{P}(\mathrm{R}, \theta)$ on the circle of radius

$\mathrm{R}$ is (here 'v' is the speed of the particle and

$\theta$ is measured from the x-axis)

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$-\displaystyle \frac{\mathrm{v}^{2}}{\mathrm{R}}\cos\theta\hat{\mathrm{i}}+\frac{\mathrm{v}^{2}}{\mathrm{R}}\sin\theta\hat{j}$
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$-\displaystyle \frac{\mathrm{v}^{2}}{\mathrm{R}}\sin\theta\hat{\mathrm{i}}+\frac{\mathrm{v}^{2}}{\mathrm{R}}\cos\theta\hat{j}$
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$-\displaystyle \frac{\mathrm{v}^{2}}{\mathrm{R}}\cos\theta\hat{\mathrm{i}}-\frac{\mathrm{v}^{2}}{\mathrm{R}}\sin\theta\hat{j}$
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$\displaystyle \frac{\mathrm{v}^{2}}{\mathrm{R}}\hat{\mathrm{i}}+\frac{\mathrm{v}^{2}}{\mathrm{R}}\hat{j}$

Explanation

For a particle in uniform circular motion,

$\displaystyle \vec{\mathrm{a}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}$ towards centre of circle

$\therefore \vec{\mathrm{a}}=\dfrac{\mathrm{v}^{2}}{\mathrm{R}}(-\cos\theta\hat{\mathrm{i}}-\sin\Theta\hat{j})$
or

$\displaystyle \vec{\mathrm{a}}=-\frac{\mathrm{v}^{2}}{\mathrm{R}}\cos\theta\hat{\mathrm{i}}-\frac{\mathrm{v}^{2}}{\mathrm{R}}\sin\theta\hat{j}$

A uniform circular disc of radius

$50\ cm$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of

$2.0\ rad {s}^{-2}$ . Its net acceleration in

${ms}^{-2}$ at the end of

$2.0 \ s$ is approximately:

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$8.0$
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$7.0$
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$6.0$
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$3.0$

Explanation

Given:

Radius,

$R=0.5 m$

Angular acceleration,

$\alpha=2\ {rad/s}^2$

Time,

$t=2 s$

Assumption:

Acceleration asked is for a point on the rim.

After 2 s, Angular velocity is given by:

$\omega=\omega_o+\alpha t$

$\omega=0+2 \times 2$

$\omega=4\ rad/s$

Centripetal Acceleration is given by,

$a_c=\omega^2 R=4^2 \times 0.5=8\ m/s^2$

Tangential Acceleration is given by,

$a_t=R\alpha=0.5 \times 2=1\ m/s^2$

Net acceleration,

$a_{net}=\sqrt {a_c^2+a_t^2}$

$a_{net}=\sqrt{8^2+1^2} \approx 8 m/s^2$

The position vector of a particle

$\overrightarrow { R}$ as a function of time is given by:

$\overrightarrow { R} = 4\, sin(2\pi t)\hat{i}+4 cos (2\pi t)\hat{j}$ ,
where

$R$ is in meters,

$t$ is in seconds and

$\hat{i}$ and

$\hat{j}$ denote unit vectors along x-and y- directions, respectively. Which one of the following statements is wrong for the motion of particle?

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Path of the particle is a circle of radius $4\ m$
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Acceleration vector is along $-\overrightarrow { R}$ .
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Magnitude of acceleration vector is $\displaystyle \frac{v^2}{R}$ , where $v$ is the velocity of particle.
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Magnitude of the velocity of particle is $8\ m/s$

Explanation

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

$\vec {R} = 4\sin (2\pi t)\hat {i} + 4(\cos 2\pi t) \hat {j}$

$.... (1)$

Velocity is given by

$\vec {V} = \dfrac {d\vec {R}}{dt}$

$= 8\pi \cos (2\pi t)\hat {i} - 8\pi \sin (2\pi t)\hat {j}$

$.... (2)$

Acceleration is given by

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

$= -16\pi^{2} \sin (2\pi t)\hat {i} - 16\pi^{2} \cos (2\pi t)\hat {j}$

$.... (3)$

From equation

$(1)$ and

$(3)$

$\vec {a} = -(2\pi)^{2}\vec {R}$

$\therefore$ Acceleration is along

$-\vec {R}$

Hence, Option (B) is correct.

$\textbf{Step 2: Equation solving}$

Magnitude of velocity

$|\vec {V}| = \sqrt {[8\pi \cos (2\pi t)]^{2} + [-8\pi \sin (2\pi t)]^{2}}$

$= 8\pi \sqrt {[\cos (2\pi t)]^{2} + [\sin (2\pi t)]^{2}}$

$= 8\pi$

Magnitude of acceleration

$|\vec {a}| = |- (2\pi)^{2}\vec {R}| = 4\pi^{2} |\vec {R}|$

From (1)

$|\vec {R}| = \sqrt {4^{2}} \sqrt {\sin^{2} (2\pi t) + \cos^{2} 2\pi t}$

$= 4$ (Radius of circle)

$\Rightarrow |\vec {a}| = 16\pi^{2} = \dfrac {|\vec {V}|^{2}}{|\vec {R}|}$

Hence option (A) and (C) are also correct.

The magnitude of velocity is coming

$8\pi$

Hence, Option(D) is incorrect.

The angular displacement of a particle is given by

$\theta =t^{3}+t^{2}+t+1$ then, its angular velocity at

$t = 2$ sec is _____

$rads^{-1}$ :

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$27$
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$17$
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$15$
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$6$

Explanation

$\omega= \dfrac {d\theta}{dt}$

$=3t^2 + 2t +1$
at,

$t = 2 sec,$

$\omega = 12 + 4 + 1$

$= 17\ rad/s$

The pair containing a scalar and a vector quantity is:

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impulse and angular momentum
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work and frequency
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electromotive force and force
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current density and current

Which of the following is a scalar quantity?

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Displacement
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Electric field
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Acceleration
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Work

(1) In uniform circular motion, tangential acceleration is zero.

(2) In uniform circular motion, velocity is constant.

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Both 1 and 2 are true and 2 is correct explanation of 1
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Both 1 and 2 are true and 2 is not correct explanation of 1
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1 is true and 2 is false
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1 is false and 2 is true

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Assertion is incorrect but Reason is correct

Starting from rest the fly wheel of a motor attains an angular velocity of

$60 rad/s$ in

$5\ s$ , the angular acceleration of the fly wheel is:

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$6\ rad/s^{2}$
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$12\ rad/s^{2}$
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$300\ rad/s^{2}$
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$150\ rad/s^{2}$

Explanation

Given,

$\omega_0=0$ ,

$\omega = 60\ rad/s$ ,

$t=5\ s$

Using the equations of motion,

$\omega = \omega_0 + \alpha t$

or,

$60 = 0 + \alpha 5$

or,

$\alpha = 12\ rad/s^2$

(1) : In uniform circular motion the kinetic energy of the body is constant.
(2) : In uniform circular motion the tangential force is zero.

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Both 1 and 2 are true and 2 is correct explanation of 1
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Both 1 and 2 are true and 2 is not correct explanation of 1
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1 is true and 2 is false
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1 is false and 2 is true

Explanation

In uniform circular motion , tangential force is zero. This implies that the angular velocity is constant.

$\therefore v=\omega r$

$\Rightarrow v$ is constant

$(\because \omega$ is constant).

$KE=\dfrac{1}{2} mv^{2}$

$\therefore KE$ constant

$(\because$ v is constant.)

A stationary wheel starts rotating about its own axis at uniform angular acceleration

$8\ rad / s^{2}$ . The time taken by it to complete

$77$ rotations is:

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$5.5\ s$
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$7\ s$
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$11\ s$
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$14\ s$

Explanation

From the equations of circular motion, we get:

$\theta =\omega_0 t+\frac { 1 }{ 2 } \alpha {t }^{ 2 }$

From the given conditions,

$\theta =77\times 2\pi\ rad$ ,

$\alpha = 8\ rad/s^2$ and

$\omega_0=0$ as the wheels starts rotating from rest
So, we get:

$77\times 2\pi =\dfrac { 1}{ 2} \times 8\times { t}^{ 2 }$
or,

$t=11\ s$

The horizontal component of weight:

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is always zero
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is always equal to the weight
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is less than the weight, but is not zero
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depends on the angle of longitude

The set containing only vector quantities is:

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Thermal Capacity, Magnetic susceptibility and Electric charge
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Magnetic moment, Electric field intensity and Torque
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Magnetic flux, Electric potential and Force
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Magnetic induction, Electric capacitance and Impulse

Which of the following units could be associated with a scalar quantity?

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$Nm^{-1}$
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$Nms^{-1}$
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$kgms^{-2}$
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$Ns$

Explanation

$Nm^{-1}$ is unit of surface tension which is a scalar quantity.

$Nms^{-1}$ is the unit of power which is a scalar quantity.

Which one is the pseudo vector in the given options ?

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Linear momentum
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Angular momentum
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Force
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All the above

Explanation

Angular Momentum is used in rotational mechanics and it defined as (momentum about a point)

$\overrightarrow{L} = \overrightarrow{r} \times m\overrightarrow{V}$
Angular momentum is always directed along the axis of rotation for a rigid body which can only rotate about an axis. Such vectors are also known as Angular Vectors.
Pseudo vectors are nothing but angular or axial vector only.

Which of the following is a vector?

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Displacement
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Density
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Volume
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Temperature

The set containing only scalar quantities is:

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velocity, specific heat and latent heat
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electric intensity, electric potential and electric capacity
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pole strength, permeability and permittivity
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torque, angular acceleration and linear momentum

Explanation

Gradient of any scalar quantity is a Vector.
Pole strength

$= \dfrac{W}{i}$ .
Permeability is a Scalar and it is material property.
Permittivity is the measure of the resistance that is encountered when forming an electric field in a medium.
All these three are scalar quantities.

Which of the following units could be associated with a scalar quantity?

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Nm $^{-2}$
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ms $^{-1}$
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kg ms $^{-2}$
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rad s $^{-2}$

Explanation

Pressure is force per unit area applied in a direction perpendicular to the surface of an object.
Pressure is a scalar physical quantity.

$P = \dfrac{F}{A}$

Which one is polar vector in the given options ?

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Acceleration
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Moment of inertia
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Angular momentum
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Velocity

Which of the following is not a vector ?

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Linear momentum
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Electric field intensity
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Kinetic energy
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Acceleration

Angular momentum is:

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a scalar
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a polar vector
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an axial vector
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not a physical quantity

Which one of the following is a scalar quantity?

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Momentum
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Electric charge
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Shearing force
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Displacement

In the following V-t graphs, identify the graph that represents a body moving with uniform velocity and then with uniform retardation until it stops.

Identify the scalar quantity in the given options:

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work
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impulse
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force
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acceleration

Explanation

Impulse, force and acceleration are vector quantities since they specify the direction while work done is a scalar quantity. There is no direction in which the work is done.
Work done

$W=\vec F \cdot \vec s$ .
Work is a scalar product of force vector

$\vec F$ and displacement

$\vec s$ and hence work is a scalar quantity.

Which of the following quantities is a scalar?

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magnetic moment
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acceleration due to gravity
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electric field
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electrostatic potential

A body moving along a circular path will have :

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a constant speed
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a constant velocity
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no tangential velocity
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a radial acceleration

Is it possible to have an accelerated motion with a constant speed? Name such type of motion.

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Yes, Uniform Circular Motion.
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Yes, Non-Uniform Circular Motion.
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Yes, Uniform Linear Motion
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No

I started walking down a road to day-break facing the sun. After walking for some time, I turned to my left then I turned to the right once again. In which direction was I going?

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northeast
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south
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east
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northwest

If the position vectors of the points

$A, B, C, D$ are

$(0,2, 1)$ ,

$(3,1,1),$

$(-5,3,2)$ ,

$(2,4,1)$ respectively and if

$PA+PB+PC+PD=0$ then the position vector of P is

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$(0,\displaystyle \dfrac{5}{2},\dfrac{5}{4})$
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$(\dfrac{5}{2},\dfrac{5}{2},\dfrac{5}{4})$
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$(\displaystyle \dfrac{5}{2},0,\dfrac{5}{4})$
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$(\displaystyle \dfrac{5}{2},\dfrac{5}{4},0)$

Explanation

Taking

$P$ common from given condition

$4\vec{P}-(\vec{A}+\vec{B}+\vec{C}+\vec{D})= 0$

$\vec{P}= \dfrac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}$

$\vec{P}=(0, \dfrac{10}{4}, \dfrac{5}{4})$

Which of the following is not a vector quantity?

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Retardation
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Acceleration due to gravity
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Average speed
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Displacement

Explanation

Since speed

$=$ rate of change of distance with time and distance is not a vector quantity. So speed is also not a vector quantity.

Which of the following changes when a particle is moving with uniform velocity?

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Speed
0%
Velocity
0%
Acceleration
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Position vector

Explanation

Acceleration

$=0$ , Velocity

$=$ Speed

$=$ constant
As the position of particle changes with time so position vector is changing.

SI unit of the angular velocity is

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m/s
0%
rad
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rad/s
0%
m/rad

Explanation

Since angular velocity

$= \dfrac{ \text{angular displacement}}{\text{time taken}}$
SI units of angular displacement is radians and time is seconds.
Therefore SI units of angular velocity is radians/s

Which of the following is not a vector quantity?

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retardation
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acceleration due to gravity
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average speed
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displacement

A wheel is rotating at

$900\space r.p.m,$ about its axis. When power is cut off it comes to rest in

$1\space minute$ . The angular retardation in

$rad/s^2$ is:

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$\pi/2$
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$\pi/4$
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$\pi/6$
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$\pi/8$

Explanation

Initially

$\nu=900 r.p.m= \dfrac{900}{60}= 15\ s^{-1}$

Thus

$\omega_1= 2\pi \nu= 30\pi\ rad/s$ and

$\omega_2= 0$

Using,

$\omega_2=\omega_1 + \alpha t$ , where

$t= 60\ s$ , we get:

$0=30\pi + \alpha (60)$

or,

$\alpha= -\dfrac{\pi}{2}\ rad/s^2$ (-ve sign shows angular retarding acceleration)

Pressure is a

$\underline{ }$ quantity

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scalar
0%
vector
0%
tensor
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cant say

Projectile motion is a 3-dimensional motion

Projectile motion (at low speeds) is uniformly accelerated motion.

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True
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False

Explanation

Let a stone is projected with velocity

$u$ making an angle

$\theta$ with the horizontal.

Acceleration of the stone in x direction

$a_x =0$

Acceleration of the stone in y direction

$a_y = -g$

Net acceleration

$a = -g = constant$ (in -ve y direction)

Hence projectile motion is an uniformly accelerated motion.

State whether given statement is True or False.

Projectile motion is a 3-dimensional motion.

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True
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False

Under what conditions the formulae of range, time of flight and maximum height can be applied directly in case of a projectile motion?

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Between two points lying on the same horizontal line.
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Between two points lying on the same vertical line.
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Can be applied for any case
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None of the above

Which of the following quantity remains constant in a uniform circular motion?

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velocity
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speed
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acceleration
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both velocity and speed

At the top of the trajectory of a projectile, the acceleration is :

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Maximum
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Minimum
0%
Zero
0%
$g$

If air resistance is not considered in projectiles, the horizontal motion takes place with :

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Constant velocity
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Constant acceleration
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Constant retardation
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Variable velocity

At the uppermost point of a projectile its velocity and acceleration are at an angle of :

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$180^o$
0%
$90^o$
0%
$60^o$
0%
$45^o$

Explanation

At the uppermost point in the trajectory the vertical velocity is zero so velocity direction is completely horizontal but the acceleration is always downwards due to earth's gravity so the angle is

$\pi /2$

Name a physical quantity that remains constant in a uniform circular motion.

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Velocity
0%
Acceleration
0%
Momentum
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Angular speed

A marble is dropped vertically, Another identical marble

$B$ is projected horizontally from the same point at the same instant :

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$A$ will reach the ground earlier than $B$
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$B$ will reach the ground earlier than $A$
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Both $A$ and $B$ will reach the ground at the same instant
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None of the above

State whether given statement is True or False.

Scalar has both magnitude and direction.

0%
True
0%
False

A ball is projected upwards. Its acceleration at the highest point is :

0%
zero
0%
directed upwards
0%
directed downwards
0%
such as cannot be predicted

Explanation

Acceleration of the ball during any stage is constant in g and direction ,it is

$g = 9.8 m/s^2$ and its downwards.

If in a wrench either the force or the couple of the wrench is zero, then wrench is called

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null wrench
0%
degenerate wrench
0%
non-degenerate wrench
0%
central wrench

During projectile motion, acceleration of a particle at the highest point of its trajectory is:

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g
0%
zero
0%
less than g
0%
dependent upon projection velocity

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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