Match List $$I$$ with List $$II$$ for a projectile List$$ - I$$ List $$- II$$ $$(a)$$ For two angles $$2\theta $$ and $$2(90 - \theta )$$ with same magnitude of velocity of projection $$(e)\ \dfrac{P\hat {i}\cdot P\hat{i}}{g}$$ $$(b)$$ Equation of parabola of a projectile $$y = PX -QX^{2}$$ $$(f)$$ Maximum height$$ = 25\%$$ of $$\dfrac{P^{2}}{Q}$$ $$(c)$$ Radius of curvature of path of a body projected with velocity $$ \left[P\hat{i}+Q\hat{j}\right] ms^{-1}$$ at highest point $$(g)$$ Range $$=$$ Maximum height $$(d)$$ Angle of projection $$\theta =\tan ^{-1} (4)$$ $$(h)$$ Range is same The correct match is :

$$a\rightarrow h,b\rightarrow f, c\rightarrow e,d\rightarrow g $$ .

$$ a\rightarrow f,b\rightarrow h, c\rightarrow g,d\rightarrow e.$$

$$a\rightarrow e,b\rightarrow g, c\rightarrow f,d\rightarrow h $$.

$$ a\rightarrow e,b\rightarrow g, c\rightarrow h,d\rightarrow f $$.

Angle between velocity and acceleration vectors in the following cases are given below. Match the correct pairs. List I List II a) Vertically projected body e) $$90^{0}$$ b) For freely dropped body f) changes from point to point c) For projectile g) zero d) In uniform circular motion h) $$180^{0}$$

$$a\rightarrow h,b\rightarrow g,c\rightarrow f,d\rightarrow e.$$

$$a\rightarrow f,b\rightarrow g,c\rightarrow h,d\rightarrow e$$.

$$a\rightarrow e,b\rightarrow f,c\rightarrow h,d\rightarrow g$$.

$$a\rightarrow g,b\rightarrow h,c\rightarrow e,d\rightarrow f.$$

Study the following List - I List - II a) Horizontal motion of a projectile e) zero velocity b) Freely falling body f) retarded motion from a small height c) Parachutist g) uniform descending down acceleration. d) Maximum height of a body thrown vertically up h) uniform velocity The correct match is

$$a\rightarrow h,b\rightarrow g,c\rightarrow f,d\rightarrow e,g$$.

$$a\rightarrow g,b\rightarrow f,c\rightarrow h,d\rightarrow e,g$$.

$$a\rightarrow e,b\rightarrow h,c\rightarrow f,d\rightarrow g,c$$.

$$a\rightarrow f,b\rightarrow e,c\rightarrow g,d\rightarrow h$$.

MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 3

A fly wheel rotating at 600 rev/min is brought under uniform deceleration and stopped after 2 minutes, then what is angular deceleration in

$rad/sec^2$ ?

0%
$\frac{\pi}{6}$
0%
$10 \pi$
0%
$\frac{1}{12}$
0%
300

AT the point of the trjectory of a projectile, the acceleration is then

0%
Maximum.
0%
Minimum
0%
Zero
0%
g

Explanation

as per equation

$h=ut-1/2gt^2$

at the point of trajectory acceleration is maximum

What is the rage of a projectile thrown with velocity

$98 \,m/sec$ with angle

$30^o$ from horizontal?

0%
$490\sqrt{3}$ meter
0%
$9800\sqrt{3}$ meter
0%
$245\sqrt{3}$ meter
0%
$100\sqrt{3}$ meter

Explanation

$R = \dfrac{u^2 \sin(2\theta)}{g} = \dfrac{98 \times 98 \times \sin(60)}{9.8} = 490\sqrt{3}$ meter

A curved section of road is banked for a speed v. if there is no friction between road and tyres of the car , then :

0%
car is more likely to slip at speeds higher than V
0%
car cannot remain in static equilibrium on the curved section
0%
car will not slip when moving with speed v
0%
none of the above

Explanation

As for the given speed

$v$ , the car will not slip but since there is no friction then the car is likely to slip when the speed is higher than

$v$ .

$\overrightarrow{A}$ and

$\overrightarrow{B}$ are two unit vectors and the angle between them is

$90^o$ . Then

$\dfrac{1-\overrightarrow{A}. \overrightarrow{B}}{1+ \overrightarrow{A}. \overrightarrow{B}}is$

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

Explanation

$\bar A \cdot \bar B = \left| A \right|\left| B \right|1\cos {90^ \circ }$

$= 0$

So , ${{1 - \bar A \cdot \bar B} \over {1 + \bar A \cdot \bar B}} = {{1 - 0} \over {1 + 0}}$

$= 1$

$\vec{r} = \vec{x}\hat{i}+\vec{y}\hat{j}$ is the equation of:

0%
$yoz$ plane
0%
a straight line joining the points $\vec{i}$ and $\vec{j}$
0%
$zox$ plane
0%
$xoy$ plane

Explanation

A vector is defined by its magnitude and its orientation with respect to a set of coordinates. It is often useful in analyzing vectors to break them into their component parts. For two-dimensional vectors, these components are horizontal and vertical.

Vector can always be represented in two dimension and three dimension. In two dimension, vector is represented in the form of

$x\hat{i}+y\hat{j}$ . In three dimension, it is represented by

$x\hat{i} + y\hat{j} + z\hat{k}$ .

Thus,

$\overrightarrow{r}=x\hat{i}+y\hat{j}$ is vector in

$xy$ plane.

A ball is thrown across a flat field.
Which statement describes the motion of the ball, when the effects of air resistance are ignored?

0%
The ball lands with the same velocity at which it is thrown
0%
The horizontal component of acceleration is constant throughout the motion
0%
The horizontal and vertical components of acceleration are both zero at the highest point of
the motion
0%
The horizontal and vertical components of velocity are both zero at the highest point of the
motion

$\vec{a}$ be the position vector whose tip is (5,-3), find the coordinates of a point B such that

$\vec{AB} = \vec{a},$ the coordinates of A being (4,-1).

0%
(9, -4)
0%
(-9, -4)
0%
(9, 4)
0%
none of these

Explanation

$\vec{a}$ is position vector of

$(5, -3)$

$\vec{a}=5\hat{i}-3\hat{j}$

Coordinates of

$A=(4, -1)$

$\vec{OA}=4\hat{i}-\hat{j}$

$\vec{AB}=\vec{a}=5\hat{i}-3\hat{j}$

$\vec{OB}-\vec{OA}=5\hat{i}-3\hat{j}$

$\Rightarrow \vec{OB}=5\hat{i}-3\hat{j}+\vec{OA}$

$=5\hat{i}-3\hat{j}+4\hat{i}-3\hat{j}$

$=a\hat{i}-4\hat{j}$

$\vec{OB}=9\hat{i}-4\hat{j}$

Coordinates of point B

$=(9, -4)$ .

find the coordinate of the tip of the position vector which is equivalent to

$\vec{AB}$ , where the coordinates of A and B are (-1, 3) and (-2, 1) respectively.

0%
(+1,+2)
0%
(+1,-2)
0%
(-1,+2)
0%
(-1,-2)

Explanation

$A(-1, 3)$

$B(-2, 1)$

'O' be the origin

$\vec{OA}=-\hat{i}+3\hat{j}$ ;

$\vec{OB}=-2\hat{i}+\hat{j}$

$\vec{AB}=\vec{OB}-\vec{OA}=(-2\hat{i}+\hat{j})-(-\hat{i}+3\hat{j})$

$=-2\hat{i}+\hat{j}+\hat{i}-3\hat{j}$

$=-\hat{i}-2\hat{j}$

$\therefore$ Coordinates

$=(-1, -2)$ .

If the position vectors of the points

$A(3,4),B(5, -6)$ and

$C(4,-1)$ are

$\vec{a}, \vec{b}, \vec{c}$ respectively, compute

$\vec{a}+2\vec{b}-3\vec{c}.$

0%
$-5\hat{i}-1\hat{j}$
0%
$\hat{i}-5\hat{j}$
0%
$\hat{i}+5\hat{j}$
0%
none of these

Explanation

Position vectors of points

$A(3, 4), B(5, -6), C(-4, -1)$ are

$\vec{a}, \vec{b}, \vec{c}$

$\vec{a}=3\hat{i}+4\hat{j}$

$\vec{b}=5\hat{i}-6\hat{j}$

$\vec{c}=4\hat{i}-\hat{j}$

$\vec{a}+2\vec{b}-3\vec{c}=3\hat{i}+4\hat{j}+2(5\hat{i}-6\hat{j})-3(4\hat{i}-\hat{j})$

$=3\hat{i}+4\hat{j}+10\hat{i}-12\hat{j}-12\hat{i}+3\hat{j}$

$=\hat{i}-5\hat{j}$

$\vec{a}+2\vec{b}-3\vec{c}=\hat{i}-5\hat{j}$ .

A particle of mass M is moving in a circle with constant speed v , between two position at the ends of a diameter, the particle's

0%
Momentum changes by mv
0%
kinetic energy changes by $1/2 mv^2$
0%
Momentum changes by 2 mv
0%
Kinetic energy changes by $mv^2$

Explanation

On the diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore, change in momentum is

$MV-(- MV) = 2MV$

A ball falls from a table top with initial horizontal speed

$V_o$ . In the absence of air resistance, which of the following statement is correct.

0%
The vertical component of the acceleration changes with time
0%
The horizontal component of the velocity does not changes with time
0%
The horizontal component to the acceleration is no zero and finite
0%
The time taken by the ball to touch the ground depends on $V_0$
0%
The vertical component of the acceleration varies with time

Explanation

When the ball falls from a table-top with initial speed

$v_o$ , its horizontal component of the velocity will remain unchanged with time because there is no air resistance so no acceleration in the horizontal direction.

Surface area is

0%
Scalar
0%
Vector
0%
Neither scalar not vector
0%
Both scalar and vector

Which of the following is a scalar quantity

0%
Displacement
0%
Electric field
0%
Acceleration
0%
Work

Assertion :A physical quantity cannot be called as a vector if its magnitude is zero
Reason :A vector has both, magnitude and direction.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect
0%
assertion is false but reason is true

Three forces are acted on a body. Their magnitudes are

$3\ N, 4\ N$ and

$5\ N$ . Then

0%
The acceleration of body must be zero
0%
The acceleration of body may be zero
0%
The acceleration of the body must not be zero
0%
None of the above

Write true or false for the following statement:
A quantity which can be completely specified by magnitude as well as direction is called a scalar quantity

0%
True
0%
False

Which of the following physical quantity is scalar?

0%
Mass
0%
Force
0%
Impulse of force
0%
Momentum

$a = 0.72$ and

$r = 0.24$ , then the value of

$t_r$ is

0%
$0.02$
0%
$0.04$
0%
$0.4$
0%
$0.2$

Explanation

$t_r=1-(a+r)$

$t_r=1-0.96$

$t_r=0.04$

A particle is moving in a circle in a radius

$r$ with a constant speed

$v$ . The change in velocity after the particle has traveled a distance equal to

$(\dfrac{1}{8})^{th}$ of the circumference of the circle is:

0%
zero
0%
$0.5\; V$
0%
$0.765 \;V$
0%
$0.125 \;V$

Explanation

Let initial position of particle be A velocity be

$\vec{v}_A = v \hat{i} + 0 \hat{j}$
After it has transversed

$\dfrac {1}{8}^{th}$ of circle

$\Rightarrow (\dfrac {360^0}{8} = 45^0)$
So,

$\theta = 45^0$
So, now velocity is,

$\vec{v}_B = v cos 45^0 \hat{i} - v sin 45^0 \hat{j}$
So, change in velocity in x direction

$= v cos 45^0 - v$

$= - 0.292v$
Change in velocity in y direction

$= - v sin 45^0 - 0$

$= - 0.707 v$
Net change in velocity

$= -0.292 v \hat {i} - 0.707v \hat {j}$

$= \sqrt {(0.292)^2 + (0.707)^2} = 0.765 \ V$

If a particle moves in a circle, describing equal angles in equal intervals of time, the velocity vector:

0%
Remains constant.
0%
Changes in magnitude.
0%
Changes in direction.
0%
Changes both in magnitude and direction.

Explanation

Hint: Angular Speed is the change in the angle of the object per unit of time.

${\textbf{Explanation:}}$

$\bullet$ The figure clearly describes the motion of the particles and it is evident that the velocity vector changes direction continuously during the motion.

$\bullet$ As the particle is covering equal angles in equal time, the magnitude of the velocity (speed) remains constant while the direction changes.

Match List

$I$ with List

$II$ for a projectile

List

$- I$ List

$- II$

$(a)$ For two angles $2\theta$ and $2(90 - \theta )$ with same magnitude of velocity of projection	$(e)\ \dfrac{P\hat {i}\cdot P\hat{i}}{g}$
$(b)$ Equation of parabola of a projectile $y = PX -QX^{2}$	$(f)$ Maximum height $= 25\%$ of $\dfrac{P^{2}}{Q}$
$(c)$ Radius of curvature of path of a body projected with velocity $\left[P\hat{i}+Q\hat{j}\right] ms^{-1}$ at highest point	$(g)$ Range $=$ Maximum height
$(d)$ Angle of projection $\theta =\tan ^{-1} (4)$	$(h)$ Range is same

The correct match is :

0%
$a\rightarrow f,b\rightarrow h, c\rightarrow g,d\rightarrow e.$
0%
$a\rightarrow h,b\rightarrow f, c\rightarrow e,d\rightarrow g$ .
0%
$a\rightarrow e,b\rightarrow g, c\rightarrow f,d\rightarrow h$ .
0%
$a\rightarrow e,b\rightarrow g, c\rightarrow h,d\rightarrow f$ .

Explanation

$(a)$

$\sin2(90-\theta )\\ \Rightarrow \sin(180-2\theta )=\sin2\theta$
Thus, Range will be same since,

$\sin2\theta =\sin2(90-\theta )$

$(b)$

$y=Px-Q{ x }^{ 2 }\\ \dfrac { dy }{ dx } =P-2Qx=0\\ \Rightarrow x=\dfrac { P }{ 2Q }$

$\Rightarrow y=\dfrac { { P }^{ 2 } }{ 2Q } -\dfrac { { P }^{ 2 } }{ 4Q } =0.25\dfrac { { P }^{ 2 } }{ Q }$

$(c)$

$\dfrac { \overrightarrow { P} .\overrightarrow { P} }{ g }$

$(d)$

$\tan\theta =4\\ \dfrac { { H }_{ max } }{ R } =\dfrac { tan\theta }{ 4 } =1\\ \Rightarrow { H }_{ max }=R$

For a projectile, the physical quantities which remain constant are:

0%
vertical component of velocity and kinetic energy.
0%
potential energy and kinetic energy.
0%
acceleration and horizontal component of velocity
0%
potential energy and acceleration.

Explanation

$\textbf{Step 1: Variable Quantities}$

$\textbf{[Ref. Fig.]}$

Here speed is changing at every instant, and height is also changing during motion.

Kinetic Energy

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

Potential Energy

$= m \, gh$

$v$ and

$h$ are changing so kinetic and potential energy will change.

$\textbf{Step 2: Constant Quantities}$

$x$ direction there is no acceleration

$(a_x)$ so horizontal component of velocity will not change (constant).

$y$ direction acceleration is constant i.e.

$g$ . Therefore total acceleration is also constant.

Hence option

$C$ is correct.

Angle between velocity and acceleration vectors in the following cases are given below. Match the correct pairs.

List I List II

a) Vertically projected body	e) $90^{0}$
b) For freely dropped body	f) changes from point to point
c) For projectile	g) zero
d) In uniform circular motion	h) $180^{0}$

0%
$a\rightarrow h,b\rightarrow g,c\rightarrow f,d\rightarrow e.$
0%
$a\rightarrow f,b\rightarrow g,c\rightarrow h,d\rightarrow e$ .
0%
$a\rightarrow e,b\rightarrow f,c\rightarrow h,d\rightarrow g$ .
0%
$a\rightarrow g,b\rightarrow h,c\rightarrow e,d\rightarrow f.$

Explanation

(a) The direction of motion gives velocity which is in upward direction and the acceleration due to gravity acting in downward direction always hence angle is

$180^0$ .

(b) Body is moving in downward direction hence velocity is in downward direction and acceleration is acting downwards hence angle is zero.

(c) In projectile motion, particle has velocity changing from point to point but acceleration acting in downward direction always.

(d) In uniform circular motion velocity is always tangential to the circular path at each point and centripetal acceleration acting always inwards towards center hence angle is

$90^0$ .

A ball is thrown with a velocity

$u$ making an angle '

$\theta$ ' with the horizontal. Its velocity vector is normal to initial velocity vector

$u$ after a time interval of

0%
$\dfrac{u \sin \theta} {g}$
0%
$\dfrac {u} {g \cos \theta}$
0%
$\dfrac{u} {g \sin \theta}$
0%
$\dfrac{4 \cos \theta}{g}$

Explanation

Initial velocity, u

$u = \underbrace{u\cos \theta \hat{i}} + u \sin \theta \hat{j}$

$u_{x} =$ constant

$u_{x}(t)= u\cos \theta$

$u_{y}(t) = u\sin \theta - gt$
So,

$u(t) = u\cos \theta \hat{i} +(u \sin \theta -gt ) \hat{j}$

Now, for

$u(t)$ to be perpendicular to

$u$ , dot product should be

$0$ .

$\vec{u(t)}\cdot \vec{u} = 0$

$(u\cos \theta \hat{i}+(u \sin \theta -gt ) \hat{j}) \cdot (u\cos \theta \hat{i} + u \sin \theta \hat{j}) = 0$

$\Rightarrow u^{2} \cos^{2} \theta + u^{2} \sin^{2} \theta - u \sin \theta gt = 0$

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

At the maximum height of a projectile, the velocity and acceleration are

0%
parallel to each other
0%
antiparallel to each other
0%
perpendicular to each other
0%
inclined to each other at $45^{0}$

Explanation

At maximum height ;

$v_{x} = v_{x} \cos \theta$ in horizontal direction and

$v_{y} = 0$

$a_{x} = 0 , a_{y} = -g$
Clearly angle between them is

$90^{\circ}.$

A particle moves in a plane with a constant acceleration in a direction different from the initial velocity. The path of the particle is:

0%
a straight line
0%
an arc of circle
0%
a parabola
0%
an ellipse

Explanation

A particle moves with constant acceleration in a direction different from initial velocity.
Let the velocity be

$v$ , acceleration be

$a$ , angle between them is

$\theta\ [ \theta <90^{0} ]$

Component of Velocity in the Direction of

$a$ is

$vcos\theta$ and in a direction perpendicular to

$a$ is

$vsin \theta$

From equations of Kinematics,
In the direction of acceleration,

$y = vcos \theta t + \frac {1}{2}a t^{2} ------ (1)$

In the direction perpendicular to acceleration,

$x = v sin \theta t$

$t=\dfrac{x}{vsin \theta} -------- (2)$

Substituting 2 in 1, we get:

$y = vcos \theta ( \dfrac{x}{vsin \theta}) + \dfrac{1}{2}a(\dfrac{x^2}{sin^2 \theta})$

$y = \dfrac {x}{tan \theta} + \dfrac{1}{2}a\dfrac{x^2}{sin^2 \theta}$

This is in the form of a parabola:

$y = Ax^2+Bx+C$

Study the following

List - I List - II

a) Horizontal motion of a projectile	e) zero velocity
b) Freely falling body	f) retarded motion from a small height
c) Parachutist	g) uniform descending down acceleration.
d) Maximum height of a body thrown vertically up	h) uniform velocity

The correct match is

0%
$a\rightarrow g,b\rightarrow f,c\rightarrow h,d\rightarrow e,g$ .
0%
$a\rightarrow h,b\rightarrow g,c\rightarrow f,d\rightarrow e,g$ .
0%
$a\rightarrow e,b\rightarrow h,c\rightarrow f,d\rightarrow g,c$ .
0%
$a\rightarrow f,b\rightarrow e,c\rightarrow g,d\rightarrow h$ .

Explanation

a) Horizontal motion of a projectile
The only force acting on the body during projectile motion is gravity which is acting in downward direction (vertically). No force is acting horizontally and therefore the velocity is uniform. (Option h)

b) Freely falling body
The force acting will be gravity in downward direction with a constant value

$mg$ where m is mass of the body and g is acceleration due to gravity. So the motion of the body will be uniform descending downward with constant acceleration. (Option g)

c) Parachutist
The motion is almost like free falling but due to presence of parachute it is retarded as air resistance increases. (Option f)

d) Maximum height of a body thrown vertically up
When the body reaches maximum height then for a fraction second it stops and then starts going down. Therefore. at maximum height the velocity becomes zero. And after that motion becomes free falling. (Option e,g)

A ball is thrown with a velocity of

$8\ ms^{-1}$ making an angle of

$60^{o}$ with the horizontal. It's velocity will be perpendicular to the direction of initial velocity of projection after a time of (

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

0%
$\dfrac{1.6}{\sqrt{3}}s$
0%
$\dfrac{4}{\sqrt{3}}s$
0%
$4\sqrt{3} s$
0%
$0.8\sqrt{3} s$

Explanation

$u(t)= v\cos 60^{\circ}\hat{i} + (v\sin 60^{\circ}- gt)\hat{j}$

$u(o)= v\cos 60^{\circ}\hat{i} + v\sin 60^{\circ}\hat{j}$

At time

$t$ :

$u(t)\cdot u(0)= 0$ for these to be

$\perp$

$\Rightarrow (v\cos 60^{\circ})^{2}+ (v\sin 60^{\circ})^{2}-v\sin 60^{\circ}gt = 0$

$\Rightarrow v^{2}= v\sin 60^{\circ}gt$

$\Rightarrow \displaystyle t=\dfrac{v}{g}\frac{2}{\sqrt{3}} = \dfrac{8\times 2}{g\sqrt{3}} = \dfrac{16}{10\times 3}\sqrt{3}$

$= \dfrac{1.6}{\sqrt{3}}$ s

A body is projected with a velocity

$u$ at an angle of

$60^{o}$ to the horizontal. The time after which it will be moving in a direction of

$30^{o}$ to the horizontal is

0%
$\dfrac{1}{\sqrt3} \dfrac{u}{g}$
0%
$\dfrac{\sqrt{3}u}{g}$
0%
$\dfrac{\sqrt{3}u}{2g}$
0%
$\dfrac{2u}{\sqrt{3}g}$

Explanation

$u(t)= u_{x}\hat{i} + u_{y}\hat{j}$

$u(t)= (u\cos \theta)\hat{i} + (u\sin \theta-gt)\hat{j} \ \ (\theta = 60 ^{0})$
Now, for this to make

$30 ^{0}$ with horizontal.

$\tan 30 ^{0} = \dfrac{u_{y}}{u_{x}} = \dfrac{u\sin 60 ^{0}-gt}{u\cos 60 ^{0}}$

$\Rightarrow \dfrac{1}{\sqrt{3}}= \dfrac{\dfrac{\sqrt{3}}{2}u -gt}{\dfrac{4}{2}} \Rightarrow u(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}})= gt$

$\Rightarrow \left | t \right | = \left | \dfrac{u}{g}(+\frac{1}{\sqrt{3}}) \right |$

$\Rightarrow t = \dfrac{u}{g}\dfrac{1}{\sqrt{3}}$

Assertion (A): In the motion of projectile the horizontal component of velocity remains constant

Reason (R) : The force on the projectile is the gravitational force which acts only in the vertically downward direction

0%
A and R are correct and R is the correct explanation of A
0%
A and R are correct and R is not the correct explanation of A
0%
A is true and R is false
0%
A is false and R is true

The distances covered by a particle thrown in a vertical plane, in horizontal and vertical directions at any instant of time 't' are x

$=$ 3t and

$y =4t - 5t^{2}$ .The acceleration due to gravity is

0%
$8 m/s^{2}$
0%
$9 m/s^{2}$
0%
$10 m/s^{2}$
0%
$16 m/s^{2}$

Explanation

Range,

$x=3t$
vertical height,

$y=4t-5t^{2}=4t-\dfrac{1}{2}\left ( 10 \right ) t^{2}$

equivalent formula,

$s=u t+\dfrac{1}{2}gt^{2}$

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

A body is projected with a velocity of

$20 m/s$ making an angle

$45^{o}$ with the horizontal. Its path (in m) is represented by

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

0%
$y=x-\dfrac{x^{2}}{20}$
0%
$y=x-\dfrac{x^{2}}{40}$
0%
$y = \sqrt{3}x-\dfrac{x^{2}}{40}$
0%
$y = \dfrac{x}{\sqrt{3}}-\dfrac{x^{2}}{40}$

Explanation

Motion in horizontal direction:

$x = 20 \cos 45 ^{o}t$

$t = \dfrac{x}{20\cos 45}$ ...(1)

Motion in vertical direction:

$y = 20 (\sin 45 ^{o})t - \dfrac{1}{2} gt^{2}$ -(2)

putting value of

$t$ in equation (2) from (1) we get

$y = x - \dfrac{x^{2}}{40} (m)$

A circular disc is rotating about its own axis at constant angular acceleration. If its angular velocity increases from

$210\ rpm$ to

$420\ rpm$ during

$21$ rotations then the angular acceleration of disc is :

0%
$5.5\ rad/s^{2}$
0%
$11\ rad/s^{2}$
0%
$16.5\ rad/s^{2}$
0%
$22\ rad/s^{2}$

Explanation

Hint:

Third Equation of Uniform Motion is,

${\left( {{\omega _f}} \right)^2}\, - \,{\left( {{\omega _i}} \right)^2} = 2\alpha \theta$

Where,

${\omega _f}$ is final angular velocity,

${\omega _i}$ is initial angular velocity,

$\alpha$ is angular acceleration,

$\theta$ is angular displacement.

Note that, $1rpm=\dfrac{2\pi }{60}rad{{s}^{-1}}$

Step 1: Note all the given values.

Initial Angular velocity is, ${{\omega }_{i}}=210rpm=\dfrac{210\times 2\pi }{60}rad{{s}^{-1}}$

Final Angular velocity, ${{\omega }_{f}}=420rpm=\dfrac{420\times 2\pi }{60}rad{{s}^{-1}}$

$No. of rotations = 21$

Therefore, angular displacement $\theta =21\times 2\pi \,=42\pi \,rad$

Step 2: Calculate what is required.

Using Third Equation of motion,

${{\left( {{\omega }_{f}} \right)}^{2}}\,-\,{{\left( {{\omega }_{i}} \right)}^{2}}=2\alpha \theta$

$\Rightarrow {{\left( 14\pi \right)}^{2}}\,-\,{{\left( 7\pi \right)}^{2}}=2\alpha 42\pi$

$\Rightarrow 196{{\pi }^{2}}\,-\,49{{\pi }^{2}}=84\pi \alpha$

$\Rightarrow 147\pi \,=84\alpha$

$\Rightarrow \alpha =5.5\,rad{{s}^{-2}}$

Thus, acceleration of disc is, $\alpha =5.5\,rad{{s}^{-2}}$

The value is closest to option A, thus Option A is correct.

A stationary wheel starts rotating about its own axis at a constant angular acceleration. If the wheel completes

$50$ rotations in the first

$2\ s$ , then the number of rotations made by it in the next

$2\ s$ is:

0%
$75$
0%
$100$
0%
$125$
0%
$150$

Explanation

$\omega _{i}=0$

$t=2x$

$0=50(2\pi )=100\pi rad$

$\therefore \alpha =\dfrac{20}{t^{2}}=\dfrac{22\pi }{4}=50\pi rads^{-2}$

$'0'$ in 4 sec

$0=\dfrac{1}{2}at^{2}=\dfrac{1}{2}(50\pi )(16)$

$400\pi rad$

in the laser 2 sec,

$0=400\pi -100\pi =300\pi rad$

$\therefore$ no of rotations =

$\dfrac{0}{2\pi }=\dfrac{300\pi }{2\pi }=150$

If the angle of projection is

$60^{o}$ , the height of the projectile when it has travelled a distance

$\dfrac{3R}{4}$ is (

$R$ is the range)

0%
$\dfrac{3\sqrt{3}R}{16}$
0%
$\dfrac{2}{3}R$
0%
$\dfrac{\sqrt{3}R}{16}$
0%
$\dfrac{4}{5}R$

Explanation

Motion in horizontal direction :

Range

$R= \dfrac{u^{2}\sin 2 \theta}{g} = \dfrac{u^{2}\sin 120^0}{g} = \dfrac{u^{2}\sqrt{3}}{2g} -(1)$

time taken in traveling

$\dfrac{3R}{4}$ distance

$\dfrac{3R}{4} = v \cos 60^{o}t$

$t = \dfrac{3R}{2V}$

Motion in vertical direction :

So,

$H = u \sin 60 ^{o}t -\dfrac{1}{2} gt^{2}$

$= (4\dfrac{\sqrt{3}}{2}\times \dfrac{3R}{2V}) - \dfrac{1}{2} \times g (\dfrac{3R}{2V})^{2}$

$= \dfrac{3\sqrt{3}R}{4} - \dfrac{g}{2} \times \dfrac{9R^{2}}{8RG}\sqrt{3}$

$= \dfrac{R}{4}(3\sqrt{3}-\dfrac{9\sqrt{3}}{4}) = \dfrac{3\sqrt{3}R}{16}$

For a projectile, the ratio of maximum height reached to the square of the time of flight is

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

0%
$5:4$
0%
$5:2$
0%
$5:1$
0%
$10:1$

Explanation

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

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

$\displaystyle\dfrac{H_{max}}{T^{2}}= \dfrac{(u^{2}\sin^{2} \theta)}{2g} \frac{g^{2}}{4u^{2}\sin^{2} \theta}= \dfrac{g}{8} = \dfrac{10}{8} = \dfrac{5}{4}$

A stationary wheel starts rotating about its own axis with an angular acceleration of

$5.5\ rad / s^{2}$ . To acquire an angular velocity

$420$ revolutions per minute, the number of rotations made by the wheel is:

0%
$14$
0%
$21$
0%
$28$
0%
$35$

Explanation

Given, angular acceleration

$\alpha =5.5rad/s^2$

angular velocity

$=420\: rpm=420\times \dfrac{2\pi }{60}$

$=14\pi\:rad/s$

$\omega_1^2- \omega_2^2=2\alpha \theta$

or,

$(14\pi)^2=2\times 5.5\times \theta$

or,

$\dfrac{98\pi ^2}{5.5}=\theta$

$\therefore$ no of revolutions

$=\dfrac{\theta }{2\pi }=\dfrac{49\pi }{5.5}=28$

A motor car is moving in a circular path with uniform speed,

$V$ . Suddenly the car rotates through an angle

$\theta$ . Then, the magnitude of change in its velocity is

0%
$2V\cos\dfrac{\theta }{2}$
0%
$2V\sin\dfrac{\theta }{2}$
0%
$2V\tan\dfrac{\theta }{2}$
0%
$2V\sec\dfrac{\theta }{2}$

Explanation

The Magnitude of change in velocity

$=\sqrt { { v }^{ 2 }+{ u }^{ 2 }-2uv\cos { \theta } }$

$=\sqrt { { v }^{ 2 }+{ v }^{ 2 }-2{ v }^{ 2 }\cos { \theta } }$

$\left( \because v=u \right)$

$=v\sqrt { 2 } \times \sqrt { 1-\cos { \theta } }$

$=v\sqrt { 2 } \times \sqrt { 2\sin ^{ 2 }{ \left( \dfrac { \theta }{ 2 } \right) } }$

$=2v\sin { \left( \dfrac { \theta }{ 2 } \right) }$

$=2V\sin { \left( \dfrac { \theta }{ 2 } \right) }$

$\left( \because v=V \right)$

A:The sum and difference of two vectors will be equal in magnitude when two vectors are perpendicular to each other.
B:The sum and difference of two vectors will have the same direction, when the vectors have unequal magnitudes but are in the same direction.

0%
Both A and B are true
0%
A is true but B is false
0%
B is true but A false
0%
Both A and B are false

Explanation

As we know the resultant of two vector can be written as

$V_{Resultant}= \sqrt{V_1^2+V_2^2 + V_1V_2 cos \theta}$

Now

$\theta= 90^0$

$cos90=0$

Hence A is correct.

The sub and difference of two vectors will not have the same direction, when the two vectors have unequal magnitudes but in the same direction. Because the resultant of of sum of two vectors will be the opposite of the resultant of subtraction of two vectors. Hence B is wrong.

A wheel has a speed of

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

$4\ radians/s^{2}$ . The number of revolutions it makes before coming to rest is:

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

Explanation

Given angular retardation

$=4\ rad/s^2$
Angular velocity

$= 1200\times \dfrac{2\pi }{60}rad/s$

$=40\pi\ rad/s$

$\therefore \omega^2=2\alpha \theta$

$\Rightarrow \theta =\dfrac{\omega ^2}{2\alpha }=\dfrac{40\pi \times 40\pi }{2\times 4}=200\pi ^2$

$Number\; of\: revolutions =\dfrac{\theta }{2\pi }=100\pi =314$

lf the initial velocity of a body is

$\mathrm{V}_{1}$ and final velocity is

$\mathrm{V}_{2}$ and the angle between

$\mathrm{V}_{1}$ and

$\mathrm{V}_{2}$ is

$\phi$ , then the change in velocity is:

0%
$\mathrm{V}_{2}-\mathrm{V}_{1}$
0%
$\sqrt{\mathrm{V}_{2}^{2}+\mathrm{V}_{1}^{2}}$
0%
$\sqrt{\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}-2\mathrm{V}_{1}\mathrm{V}_{2}\cos\phi}$
0%
$\sqrt{\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}+2\mathrm{V}_{1}\mathrm{V}_{2}\cos\phi}$

Explanation

Change in velocity will be vector quantity ${V}_2 - {V}_1$ , whereas all other options are scalar quantities.

In circular motion if

$\bar{v}$ is velocity vector,

$\bar{a}$ is acceleration vector,

$\bar{r}$ is instantaneous position vector,

$\bar{p}$ is momentum vector and

$\bar{\omega}$ is angular velocity of particle, then:

0%
$\bar{v}$ , $\bar{\omega}$ , $\bar{r}$ are mutually perpendicular
0%
$\bar{p}$ , $\bar{v}$ , $\bar{\omega}$ are mutually perpendicular
0%
$\bar{r} \times \bar{v} = 0$ and $\bar{r} \times \bar{\omega} = 0$
0%
$\bar{r}.\bar{v} = 0$ and $\bar{r}.\bar{\omega} = 0$

Explanation

$p$ in the direction of velocity

$\therefore \vec{v},\vec{w},\vec{r}$ are mutually tan

$\vec{r}.\vec{v} = 0; \vec{r}.\vec{w}=0$

$\vec{v}\times\vec{p}=0$

A force:

${F}=(6\hat{i} - 8\hat{j} + 10\hat{k})\ N$ produces an acceleration of

$1\ m/s^{2}$ in a body. The mass of body would be:

0%
$200\ Kg$
0%
$20\ Kg$
0%
$10\sqrt{2}\ Kg$
0%
$6\sqrt{2}\ Kg$

Explanation

Magnitude of the force is given as:

$F = \sqrt{6^2 +(- 8^2) + 10^2}$ N

$= 10\sqrt{2}$ N

Force = Mass $\times$ Acceleration

$10\sqrt{2}$ = Mass

$\times$

$1$
Mass =

$10\sqrt{2} Kg$

Out of the following options, the resultant of which pair cannot be 4N?

0%
2N and 2N
0%
3N and 8N
0%
2N and 6N
0%
4N and 8N

Explanation

In the options given above only the second option i.e. 3N & 8N will give

$F_{max}= 8+3 =11N$ &

$F_{min}=8 -3 =5N$ . The resultant force always lies between maximum & minimum value & 4 is outside the range.
So

$3N$ and

$8N$ cannot give resultant as

$4N$ .

Assertion(A): Electric current and velocity of light both have direction as well as the magnitude but still they are not considered as vectors.
Reason(R): Electric current and velocity of light do not follow laws of vector addition.

0%
Both A and R are true and the R is correct explanation of the A
0%
Both A and R are true, but R is not correct explanation of the A
0%
A is true, but the R is false
0%
A is false, but the R is true

A wheel having a radius of

$10\ cm$ is coupled by a belt to another wheel of radius

$30\ cm$ . 1st wheel increases its angular speed from rest at a uniform rate of 1.57

$rad/s^2$ . The time for the 2nd wheel to reach a rotational speed of

$100\ rev/min$ is (assume that the belt does not slip):

0%
$20 \ s$
0%
$10\ s$
0%
$1.5\ s$
0%
$15 \ s$

Explanation

$\alpha_1 = 1.57 rad/s^2$

$\omega_2 = 100 \ rev/m$

$= 100 \times \dfrac {2\pi}{60}$

$= \dfrac {10\pi}{3} rad/s$
linear velocity of belt is equal

$\therefore \omega_{1} r_{1} = \omega_{2} r_{2}$ and

$\alpha_{1}r_{1} = \alpha_{2}r_{2}$

$\therefore \alpha_2 = (1.57)(\dfrac {10}{30})$

$= \dfrac {1.57}{3} rad/s^2$

$\omega = \omega_0 + \alpha t$

$\dfrac {10\pi}{3} = 0 + \dfrac {1.57}{3} \times t$

$t = 20 \ s$

A circular disc is rotating about its own axis at uniform angular velocity

$\omega$ . The disc is subjected to uniform angular retardation by which its angular velocity is decreased to

$\dfrac{\omega}{2}$ during

$120$ rotations. The number of rotations further made by it before coming to rest is:

0%
$120$
0%
$60$
0%
$40$
0%
$20$

Explanation

$\omega_1^2-\omega_2^2=2\alpha \theta$

$\Rightarrow (\omega )^2-\left ( \dfrac{\omega }{2} \right )^2=2\alpha(120)$

$\Rightarrow \dfrac{3\omega ^2}{4}=240\alpha$
No. of rotations further made

$=n$

$\Rightarrow \dfrac{\omega ^2}{4}=2\alpha(n)$

$3=\dfrac{240}{2n}$

$n=40$

The length of a seconds hand in a watch is

$1$ cm. The change in its velocity in

$15$ s is

0%
$0\ cm/s$
0%
$\displaystyle \dfrac{\pi}{30\sqrt{2}}\ cm /s$
0%
$\dfrac{\pi}{30}\ cm / s$
0%
$\displaystyle \dfrac{\pi}{30}\sqrt{2}\ cm / s$

Explanation

$\omega = \pi/30$
Hence v =

$\pi/30$ cm/s
Change in velocity by 90 degrees in 15 seconds =

$v\sqrt{2} = \pi/30\sqrt{2}$ cm/s

Two projectiles of same mass and with same velocity are thrown at an angle 60 & 30 with the horizontal, then which quantity will remain same :

0%
Time of flight
0%
Horizontal range of projectile
0%
Max height acquired
0%
All of them

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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