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

$\overrightarrow A = 3\widehat i + 2\widehat j\;and\;\overrightarrow B = 7\widehat i + 24\widehat j,$ find a vector having the same magnitude as

$\overrightarrow B$ and parallel to

$\overrightarrow A$ .

0%
$9\widehat i + 6\widehat j$
0%
$6\widehat i + 10\widehat j$
0%
$15\widehat i + 20\widehat j$
0%
None of these

For a particle in uniform circular motion, the acceleration a at a point

$P(R,9)$ on the circle of radius

$R$ is (Here 0 is measured from the x- axis)

0%
$\frac { v ^ { 2 } } { R } \hat { i } + \frac { v ^ { 2 } } { R } \hat { j }$
0%
$- \frac { v ^ { 2 } } { R } \cos \theta \hat { i } + \frac { v ^ { 2 } } { 2R } \sin \theta \hat { j }$
0%
$- \frac { v ^ { 2 } } {2 R } \cos \theta \hat { i } + \frac { v ^ { 2 } } { R } \sin \theta \hat { j }$
0%
$- \frac { v ^ { 2 } } { R } \cos \theta \hat { i } - \frac {2 v ^ { 2 } } { R } \sin \theta \hat { j }$

A ring takes time

${t}_{1}$ in slipping down an inclined plane of length

$L$ and takes time

${t}_{2}$ in rolling down the same plane. The ratio

$\frac{{t}_{1}}{{t}_{2}}$ is

0%
$\sqrt{2}:1$
0%
$1:\sqrt{2}$
0%
$1:2$
0%
$2:1$

An object moving in a circular path at constant speed has constant

0%
Energy
0%
Velocity
0%
Acceleration
0%
Displacement

Explanation

An object undergoing uniform circular motion is moving with a constant speed nonetheless

$,$ it is accelerating due to its change in direction

$.$ The direction of the acceleration is inwards

$.$ An object moving in uniform circular motion will cover the same linear distance in each second of time

$($ constant speed

$)$

Hence

$,$ option

$(D)$ is correct

$.$

There are m points on a straight line

$AB$ and

$n$ points on the line

$AC$ none of them being the point

$A$ . Triangles are formed with these points as vertices, when:

(i)

$A$ is excluded.

(ii)

$A$ is included. The ratio of number of triangles in the two

cases is:

0%
$\dfrac { m + n - 2 } { m + n }$
0%
$\dfrac { m + n - 1 } { m + n }$
0%
$\dfrac { m + n - 2 } { m + n + 2 }$
0%
$\dfrac { m ( n - 1 ) } { ( m + 1 ) ( n + 1 ) }$

Explanation

Case i) Triangles without vertex A

We must choose 2 vertices from line AB and one from line AC.

or two lines AC and one vertice from line AB.

vertices

$= ^{m}C_{2}\times ^{n}C_{1}+^{n}C_{2}\times ^{m}C_{1}$

$= \frac{m(m-1)}{2}\times n+\frac{n(n-1)}{2}m$

$= \frac{mn}{2}[m-1+n-1]$

$= \frac{mn}{2}(m+n-2)$

Case ii) when A is included

(a) we have already 1 vertex (A)

Therefore, we must choose 1 vertex from line AB

and 1 vertex from line AC.

$= 1\times ^{m}C_{1}\times ^{n}C_{1}$

$= mn$

(b) If point A is not selected, then

selection is same as

$1^{st}$ case, so total no.

of ways would be

$\frac{mn(m+n-2)}{2}$

So total possibility =

$mn+\frac{mn(m+n-2)}{2}$

$= mn[1+\frac{m+n-2}{2}]=\frac{mn(m+n)}{2}$

Taking Ratio =

$\frac{\frac{mn(m+n-2)}{2}}{\frac{mn(m+n)}{2}}= \frac{m+n-2}{m+n}$

1202639_1249485_ans_64219ef650d944edbac50d0faf71a2f6.jpg

A force of

$2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +4\overset { \wedge }{ k }$ acts on a body for 4 seconds and produces a displacement of

$3\overset { \wedge }{ i } +4\overset { \wedge }{ j } +5\overset { \wedge }{ k } m.$ The power used is

0%
9.5 W
0%
7 W
0%
6.5 W
0%
4.5 W

The component of a vector is

0%
always less than its magnitude
0%
always greater than its magnitude
0%
always equal to its magnitude
0%
none of these

Explanation

Correct Answer: Option D

Hint: Each part of a two-dimensional vector is known as it's components.

Explanation of Incorrect Options:

Since the magnitude of component can be less than the magnitude of vector as in fig.(i) or the magnitude of component can be equal to the magnitude of vector as in fig.(ii) ,Options A,B and C are incorrect

Explanation of Correct Option:

In both the figures, we can see a vector

$A$ and its component

From figure

$(i)$ , it is clear that magnitude of component of

$A$ is less than the magnitude of vector

$A$ .

Consider fig.

$(ii)$ , When component is taken in the same direction as that of the vector, magnitude of the component is maximum, i.e, magnitude of the component is equal to the magnitude of the vector.

Since the magnitude of component can be less than or equal to the magnitude of vector, Option D is correct

2148701_1266522_ans_92eed059a076484489e4d84b73577dd2.png

A particle moves in the xy plane with velocity

${u}_{x}=8t-2$ and

${u}_{y}=2$ . If it passes through the point

$x=14$ and

$y=4$ at

$t=2s$ , the equation of the path is

0%
$x={y}^{2}-y+2$
0%
$x={y}^{2}-2$
0%
$x={y}^{2}+y-6$
0%
None of these

A particle moves with constant speed

$v$ along a circular path of radius

$r$ and completes the circle in time

$T$ . The acceleration of the particle is

0%
$2\pi v/T$
0%
$2\pi r/T$
0%
$2\pi {r}^{2}/T$
0%
$2\pi {v}^{2}/T$

Explanation

We know that

$\vec{\Delta s}$

$=\vec{\Delta Q}\times$

$\vec{\Delta r}$

$\dfrac{\vec{\Delta s}}{\Delta t}$

$=\dfrac{\vec{\Delta Q}}{\Delta t}\times$

$\vec{\Delta r}$

$\vec{\Delta v}$

$=\vec{\Delta w}\times$

$\vec{\Delta r}\Rightarrow |v|=wr(\because \theta=90^0)$

$\vec{a}=\dfrac{d\vec{v}}{dt}(\vec{w}\times \dfrac{d\vec{r}}{dt})+(\dfrac{d\vec{w}}{dt}\times \vec{r})$

$w=$ then

$\dfrac{dw}{dt}=0$

$\vec{a}=\vec{w}\times \vec{v}$ or

$a=wv$

$a=\dfrac{v}{r},v=\dfrac{v^2}{r},a=rw^2$

$a=rw^2$

$=r\left[\dfrac{2\pi}{T}\right]^2$

$=r.\dfrac{4\pi^2}{T^2}$

$a=vw$

$v.\dfrac{2\pi}{T}$

Hence, option

$A$ is correct
2030500_1287482_ans_d718bd47011341fc9fbb430afb35f7a3.png

A rigid body rotates about a fixed axis with variable angular velocity equal to

$\alpha - \beta t$ , at the time

$t$ , where

$\alpha$ ,

$\beta$ are constants. The angular through which it rotates before its stops

0%
$\frac{{{\alpha ^2}}}{{2\beta }}$
0%
$\frac{{{\alpha ^2} - {\beta ^2}}}{{2\alpha }}$
0%
$\frac{{{\alpha ^2} - {\beta ^2}}}{{2\beta }}$
0%
$\frac{{\left( {\alpha - \beta } \right)\alpha }}{2}$

$\vec a,\vec b, \vec c$ are three coplanar vectors, then

$v\left[ 2\vec { a } +3\vec { b } ,2\vec { b- } 5\vec { c } ,2\vec { c } +3\vec { a } \right]$ is

0%
$0$
0%
$1$
0%
$-\sqrt 3$
0%
$\sqrt 3$

$u,\ v,\ w$ are non-coplanar vector and

$p,\ q$ are real numbers, then the equality

$[3u\ pv\ pw]-[pv\ w\ qw]-[2w\ qv\ qu]=0$ holds for

0%
Exactly two values of $(p,\ q)$
0%
More than but not all values of $(p,\ q)$
0%
All values of $(p,\ q)$
0%
Exactly one values of $(p,\ q)$

Explanation

$\begin{array}{l} Since, \\ \left[ { 3u\, \, pv\, \, pw } \right] -\left[ { pv\, \, wq\, \, qv } \right] =0 \\ \therefore 3{ p^{ 2 } }\left[ { u.\left( { v\times w } \right) } \right] -pq\, \left[ { v.\left( { w\times u } \right) } \right] \\ -2{ q^{ 2 } }\left[ { w.\left( { v\times u } \right) } \right] =0 \\ \Rightarrow \left( { 3{ p^{ 2 } }-pq+2{ q^{ 2 } } } \right) \left[ { u.\left( { v\times w } \right) } \right] =0 \\ But\, \, \left[ { u\, \, \, v\, \, \, w } \right] \ne 0 \\ \Rightarrow 3{ p^{ 2 } }-pq+2{ q^{ 2 } }=0 \\ \therefore p=q=0 \end{array}$

A body moves in a circle covers equal distance in equal intervals of time. Which of the following remains constant

0%
Velocity
0%
Acceleration
0%
Speed
0%
Displacement

Which of the following statements best describes an object undergoing uniforms circular motion?

0%
The velocity and acceleration vectors are always tangential to each other
0%
Objects in uniform circular motion tend to stay in uniform circular motion
0%
The object moves with a constant velocity
0%
The object would travel in a straight line if the acceleration vector goes to zero

The equation of a projectile is

$y=ax-b{x}^{2}$ . Its horizontal range is

0%
$\cfrac{a}{b}$
0%
$\cfrac{b}{a}$
0%
$a+b$
0%
$b-a$

A ball is thrown horizontally from the top of a tower with an initial velocity of 20 meters per second, what is the horizontal velocity of the ball as it hits the ground ?

0%
9.81 m/s
0%
34.3 m/s
0%
20.0 m/s
0%
68.6 m/s

Explanation

Since, there is no acceleration of ball in horizontal direction, its velocity in x-direction remain same

$i.e\; 20\;m/s$

Therefore, C is correct option.

A vector of magnitude 10 has its rectangular component as as 8 and 6 along x axis and y axis respectively. Find the angles it make with the x and y axis?

0%
$36.87^o with\ x -axis, 53.13^o with y -axis$
0%
$53.14^o with\ x-axis, 36.86^o$ with y -axis
0%
$47.16^o with\ x- axis 23.92^o with y - axis$
0%
$23.92^o with\ x-axis, 47.16^o with y-axis$

Explanation

Angle with the

$X-axis$

$\Rightarrow tan \theta = \dfrac{P}{h} = \dfrac{6}{8}$

$\Rightarrow \theta = tan^{-1}(3/4) = 36.86^o$

Angle with the

$Y-axis$

$\Rightarrow \theta = tan^{-1}(3/4) = 36.86^o$

$\Rightarrow \alpha= 90^o - 36.86^o = 53.14^o$

2068621_1312032_ans_c2f03a3243654c5baa13471386623b16.png

The path of projectile in the absence of air drag is shown in the figure by dotted line. If the air resistance is not ignored then which one of the path shown in the figure is appropriate for the projectile

0%
B
0%
A
0%
D
0%
C

A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1m, by means of a string at an initial speed of 10 R.P.M. Keeping the radius constant,the tension in the string is reduced to one quarter of its initial value. the new speed is

0%
5 r.p.m
0%
10 r.p.m
0%
20 r.p.m
0%
14 r.p.m

A ball is projected vertically down with an initial velocity from a height of

$15$ m on to a horizontal floor. During the impact it loses

$25$ % of its energy and rebounds to the same height, the initial velocity of its projections is

0%
$20\,m{s^{ - 1}}$
0%
$15\,m{s^{ - 1}}$
0%
$10\,m{s^{ - 1}}$
0%
$5\,m{s^{ - 1}}$

Explanation

Let ball is projected vertically downward with velocity v from height h
Total energy at point

$A = \dfrac {1}{2} mv^2 +mgh$
During collision loss of energy is 50% and the ball rises up to same height. It means it possess only potential energy at same level.

$50$ %

$( \dfrac {1}{2} mv62 + mgh ) = mgh$

$\dfrac {1}{2} ( \dfrac {1}{2} mv^2 + mgh) = mgh$

$v = \sqrt { 2gh} = \sqrt { 2 \times 10 \times 20 }$

$\therefore v = 20 m/s$
1703460_1301612_ans_8576dda35e9045c5901fa772a5e4c1f9.png

Abody projected up with certain velocity along an inclined plane of coefficient of friction

$0.6$ and slope

$\frac{3}{4}$ travels for 't' seconds and comes to rest Its time of decent from that position is

0%
$t$
0%
$9t$
0%
$3t$
0%
$5t$

The length of the second hand of a clock is

$4\ cm$ the speed of the tip of the second hand is:

0%
$0.24\ cm/s$
0%
$0.32\ cm/s$
0%
$0.42\ cm/s$
0%
$0.50\ cm/s$

Explanation

$0.42cm/s$

circumference

$=2\pi r$

$=2 \times \dfrac{22}{7} \times 4$

$=25.1cm$

time taken to complete one round

$=60s$

distance = speed×time

$s=\dfrac{25.1}{60}=0.419\approx 0.42cm/s$

A body is projected with a velocity

$50m{s}^{-1}$ . Distance traveled in 6th second is
(

$g=10m{s}^{-2}$ )

0%
$5m$
0%
$10m$
0%
$15m$
0%
$20m$

A particle is projected from ground with initial speed

$u$ at angle

$\theta$ with horizontal. If air friction is absent, then the average velocity of complete motion will be:

0%
$u\cos{\theta}$
0%
$u\sin{\theta}$
0%
$2u\cos{\theta}$
0%
$2u\sin{\theta}$

Explanation

To know the average time traveled from being one point of height to other which is same from the ground determines the parabolic shapes of projectile.

Hence given as speed u at angle \theta and time as

${t_1}$ and

${t_1}$ . Horizontal component of the velocity be u cos \theta and the total displacement in travelling from

${t_1}$ to

${t_2}$ be

$,$

$u\cos \theta \times \left( {{t_1} - {t_2}} \right)$

=> Total time taken be

$(t_1-t_2)$

Therefore the average velocity be

$,$

${V_a} = \dfrac{{Total\,\,displacement}}{{Total\,\,time}}$

$={V_a} = u\cos \theta \times \dfrac{{\left( {{t_1} - {t_2}} \right)}}{{\left( {{t_1} - {t_2}} \right)}} =u \cos \theta$

Hence,

option

$(A)$ is correct answer.

If the vectors

$\bar { AB } =3\hat { i } +4\hat { k }$ and

$\bar { AC } =5\hat { i } -2\hat j+4\hat k$ are the sides of a triangle ABC, then the length of the median through A is:

0%
$\sqrt { 18 }$
0%
$\sqrt { 72 }$
0%
$\sqrt { 33 }$
0%
$\sqrt { 45 }$

Explanation

We have,

$AB=3\overrightarrow{i}+4\overrightarrow{k}$

$AC=5\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k}$

So, the coordinate of

$B\,is\,\left( 3,0,4 \right)$

$C\,is\,\left( 5,-2,4 \right)$

The mid point of $BC$ can be easily found by midpoint formula of two points as

$$\begin{align}

$D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( \dfrac{3+5}{2},\dfrac{0-2}{2},\dfrac{4+4}{2} \right)$

$D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 5,-2,4 \right)$

So, the length of median through A is

$AD=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( -1-0 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}}$

$AD=\sqrt{33}$

Hence, this is the answer

The maximum range of a gun or horizontal range is 16 km. If g = 10

${ ms }^{ -1 }$ . The muzzle velocity of the shell must be:-

0%
$1600 \mathrm { ms } ^ { - 1 }$
0%
$400 \mathrm { ms } ^ { - 1 }$
0%
$200 \sqrt { 2 } \mathrm { ms } ^ { - 1 }$
0%
$160 \sqrt { 10 } \mathrm { ms } ^ { - 1 }$

Explanation

Horizontal Range

$R = \frac{{{u^2}\sin 2\theta }}{g}$

For maximum horizontal range

$\theta = {45^0}$

$,$

${R^{\max }} = \frac{{{u^2}}}{g};$ where

$u$ is the muzzle velocity of the shell

$\therefore 1600\,m = \frac{{{u^2}}}{{10\,m/{s^2}}}$

$\therefore u = 400m/s$

Hence,

option

$(B)$ is correct answer.

A body rotates at

$300$ rotation per minute. The value in radian of the angle described in

$1$ sec is

0%
$5$
0%
$5\pi$
0%
$10$
0%
$10 \pi$

Explanation

Given,

$\omega=300 \ rev/min$

i.e, In

$1\ min, 300\times 2\pi=600\pi\ rad$

$1\ sec=\dfrac{600\pi}{60}=10\pi\ rad$

Hence the answer is

$10\pi\ rad$

Option

$\textbf D$ is the correct answer

A projectile is projected with a kinetic energy

$K$ . If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be

0%
$K/4$
0%
$K/2$
0%
$3K/4$
0%
$K$

Explanation

$\textbf{Step 1: Maximum range of the projectile }$

Let

$u$ be the initial velocity with which the projectile is projected.

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

$R_{max}= \dfrac{u^2}{g}$ ,when

$\theta=45^0$

$\textbf{Step 2: Velocity at the highest point }$

$u_y= 0$

$u_x=u\cos \theta$

Put

$\theta = 45^o$

$u_x = u \cos 45^o$

$\therefore u_x = \dfrac{u}{\sqrt{2}}$

$....(1)$

$\textbf{Step 3: Kinetic Energy of the projectile }$

Kinetic energy at the ground,

$K = \dfrac{1}{2}mu^2$

Let Kinetic Energy at the highest point be

$K^{'}$

$K' = \dfrac{1}{2}mu_x^2$

$(1)\Rightarrow$

$K' = \dfrac{1}{2}m (\dfrac{u}{\sqrt{2}})^2$

$\Rightarrow K' = \dfrac{1}{4}mu^2 = \dfrac{K}{2}$

Hence, Option(B) is correct

The potential gradient is a_______

0%
vector quantity
0%
scalar quantity
0%
conversion factor
0%
constant

If a particle of mass

$m$ is moving in a horizontal circle of radius

$r$ with a centripetal force

$( - \frac{k}{{{r^2}}})$ , the total energy is

0%
$- \frac{3k}{{2r}}$
0%
$- \frac{k}{r}$
0%
$- \frac{{2k}}{r}$
0%
$- \frac{{4k}}{r}$

For a particle in uniform circular motion:

0%
both velocity and acceleration are constant
0%
acceleration and speed are constant but velocity changes
0%
both acceleration and velocity changes
0%
Speed is constant

Two particles having position vectors

$\vec { r } = ( 3 \vec { i } + 5 j ) m$ and

$\vec { r } _ { 2 } = ( - 5 i + 3 j ) m$ are moving with velocities

$\vec { V } _ { 1 } = ( 4 \hat { i } - 4 \hat { j } ) m s ^ { - 1 }$ and

$\vec { V } _ { 2 } = ( a \hat { i } - 3 \hat { j } ) m s ^ { - 1 }$ . If they collide after

$2$ seconds , the value of

$a$ is

0%
- $2$
0%
- $4$
0%
- $6$
0%
- $8$

Explanation

Let us consider the values as following:

$\begin{array}{l} { r_{ 1 } }=\left( { 3i+5j } \right) m \\ { r_{ 2 } }=\left( { -5i+3j } \right) m \\ { v_{ 1 } }=\left( { 4i-4j } \right) \, m/s \\ and\, \, let\, \, { v_{ 2 } }\, \, be=\left( { -ai-3j } \right) m/s \\ t=2\, \, \sec \end{array}$

If two particles collides after time

$"t"$ they travel equal distance.

$\begin{array}{l} { r_{ 1 } }+{ v_{ 1 } }t={ r_{ 2 } }+{ v_{ 2 } }t \\ \left( { 3i+5j } \right) +\left( { 4i-4j } \right) \times 2=-5i+3j+\left( { -ai-3j } \right) 2 \\ on\, \, solving\, \, we\, \, get, \\ 3i+8i+5j-8j=-5i+3j-2ai-6i \\ 11i+5i-3j=-2ai-3j \\ 16i=-2ai \\ a=\dfrac { { -16 } }{ 2 } \\ =-8\, m/{ s^{ 2 } } \end{array}$

Physical quantities are mainly classified into

0%
scalars
0%
vectors
0%
scalars and vectors
0%
scalars, vectors and tensors

Which of the following is not a vector quantity

0%
speed
0%
velocity
0%
torque
0%
displacement

Time of flights for a particle thrown for equal range at different angles are

$4 s$ and

$3 s$ . Speed of projection is

0%
$10 m/s$
0%
$20 m/s$
0%
$25 m/s$
0%
$50 m/s$

A particle move in a circle of radius 5 cm with constant speed and time period 0.2

$\pi$ s . The acceleration of the particle is :

0%
15 $m/s^2$
0%
25 $m/s^2$
0%
36 $m/s^2$
0%
5 $m/s^2$

Explanation

Given,

Radius,

$r=5\,cm=5\times 10^{-2}\,m$

Time period,

$T=0.2\pi s$

We know that acceleration,

$a=r\omega^2$

$=\dfrac{4\pi^2}{T^2}r$

$\dfrac{4\times \pi^2\times 5\times 10^{-2}}{(0.2\pi)^2}=5\,ms^{-2}$

Position of a particle in a rectangular co - or dinate system is (3 , 2, 5,).Then its position vector will be

0%
$3 \hat i + 5 \hat j + 2 \hat k$
0%
$3 \hat i + 2 \hat j + 5 \hat k$
0%
$5 \hat i + 3 \hat j + 2 \hat k$
0%
None of these

The position vector of a particle is determined by the expression

$\vec { r } = 3 t ^ { 2 } \hat { i } + 4 t ^ { 2 } \hat { j } + 7 \hat { k }$ , The distance traversed in first

$10$ sec is

0%
$500 m$
0%
$300 m$
0%
$150 m$
0%
$100 m$

Explanation

$r(i)$ =Initial position vector

$r(f)$ =Final position vector

Since r is the fiction to time.

$\begin{array}{l} r\left( i \right) =r\left( 0 \right) =7\widehat { k } \\ r\left( f \right) =r\left( 0 \right) =3\times { 10^{ 2 } }\widehat { i } +4\times { \left( { 10 } \right) ^{ 2 } }\widehat { j } +7\widehat { k } \end{array}$

Now,

Displacement vector

$=r(f)-r(i)$

$\begin{array}{l} D=\left| { 300\widehat { i } +400\widehat { j } +\left( { 7-7 } \right) \widehat { k } } \right| \\ =\sqrt { { { 300 }^{ 2 } }+{ { 400 }^{ 2 } } } \\ =500 \end{array}$

Hence, Option

$A$ is correct.

A particle moves along a semicircle of radius

$10\ m$ in

$5$ seconds. The average velocity of the particle is

0%
$2\pi\ ms^{-1}$
0%
$4\pi\ ms^{-1}$
0%
$2\ ms^{-1}$
0%
$4\ ms^{-1}$

A mass m rotates in a vertical circle, of radius R and has a circular speed

$v_{c}$ at the top, . If the radius of the circle is increased by a factor of 4, circular speed at the top will be.

0%
decreased by a factor of 2
0%
decreased by a factor of 4
0%
increased by a factor of 2
0%
increased by a factor of 4

Explanation

Circular speed at the top will be decreased by a factor of

$2.$

Hence,

option

$(A)$ is correct answer.

A cricket player throws a ball at an angle of

$53^o$ with the horizontal.What is the velocity of the projection of the ball if it is caught at the same level by a fielder after 2 seconds?

$[take g = 10 m/s^2 ]$

0%
$6.25 m/s$
0%
$25 m/s$
0%
$17.5 m/s$
0%
$3.53 m/s$

If two vectors

$\vec { { A }_{ 1 } } =\hat { i } -3\hat { i } +5\hat { k }$ and

$\vec { { A }_{ 2 } } =\hat { i } -3\hat { j } +a\hat { k }$ are equal, then the value of a is:

0%
+5
0%
-5
0%
-3
0%
2

A vector

$\vec{A}$ has magnitude A and

$\hat{A}$ is unit vector in the direction of

$\vec{A}$ , then which of the following are correct

0%
$\vec{A}. \hat{A} = A$
0%
$\hat{A} = \dfrac{\vec{A}}{A}$
0%
$\vec{A}. \vec{A} = A^2$
0%
All of the above

Explanation

$\textbf{Step 1: Analysing option A}$

$\vec{A}\rightarrow$ vector;

$|\vec {A}|=A \rightarrow$ magnitude;

$\hat {A} \rightarrow$ Unit vector;

We know,

$\hat {A} =\dfrac{\vec{A}}{ |\vec {A}|}$

So,

$\vec{A} . \hat {A} = \vec{A}. \dfrac{\vec{A}}{{A}} = \dfrac{A^2}{A} = A$

Hence option

$A$ is correct.

$\textbf{Step 2: Analysing option B}$

$\hat{A} = \dfrac {\vec{A}}{A}$ [Unit Vector]

Hence option

$B$ is also correct.

$\textbf{Step 3: Analysing option C}$

$\vec{A}.\vec{A} = A^2$

Hence option

$C$ is also correct.

All the options are correct.

$\therefore$ Option

$D$ is the correct answer.

In case of uniform circular motion which of the following physical quantity do not remain constant

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A.Speed
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B. Momentum
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C.Kinetic energy
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D.Mass

A stone is thrown vertically upward with an initial velocity

$V_0$ . The distance traveled in time

$4v_0 / 3g$ is

0%
$\dfrac{2v_0^2}{g}$
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$\dfrac{v_0^2}{2g}$
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$\dfrac{4v_0^2}{9g}$
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$\dfrac{5v_0^2}{9g}$

Explanation

Lets consider,

$v_0=$ initial velocity of the stone

$g=$ acceleration due to gravity

$t=\dfrac{4v_0}{3g}$

2nd equation of motion,

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

$s=v_0\dfrac{4v_0}{3g}-\dfrac{1}{2}g (\dfrac{4v_0}{3g})^2$

$s=\dfrac{4v_0^2}{3g}-\dfrac{8v_0^2}{9g}$

$s=\dfrac{4v_0^2}{9g}$

The correct option is C.

A particle is projected with velocity 'u' at an angle $\theta$ with an inclined plane of inclination $\theta <45$ with the horizontal.The time taken when velocity of projectile becomes parallel to the plane

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$\cfrac { u \sin \theta } { g }$
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$\cfrac { u \cot \theta } { g }$
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$\cfrac { 2 u \tan \theta } { g }$
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$\cfrac { u \tan \theta } { g }$

A particle is projected vertically upward with speed

$u=10m/s$ . During its journey air applies a force of

$-.02{v}^{2}$ on the particle. What is the maximum height attained by the particle (

$m=2kg$ )?

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$5\ln{(2)}$
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$3\ln{(2)}$
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$2\ln{(2)}$
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$10\ln{(2)}$

Vector $\overrightarrow {\text{A}}$ lies in xy plane and makes an angle will positive y direction. The x components of $\overrightarrow {\text{A}}$ is

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$A\cos \theta$
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$A\sin \theta$
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$A\tan \theta$
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$A\sec \theta$

Explanation

$\left| \vec A\right|=A$

Hence, the answer is

$A\;cos\;\theta.$

1380587_1451556_ans_e85ae96b12194bcbb407ee6df5dc3529.png

A particle moves so that its position vector is given by

$\overrightarrow { r } =\cos ^{ }{ \omega t } \hat { x } +\sin { \omega t } \hat { y }$ , where

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

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Velocity is perpendicular to $\overrightarrow { r }$ and acceleration is directed away from the origin
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Velocity and acceleration both are perpendicular to $\overrightarrow { r }$
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Velocity and acceleration both are parallel to $\overrightarrow { r }$
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Velocity is perpendicular to $\overrightarrow { r }$ and acceleration is directed towards the origin

An object of mass

$m$ moves with constant speed in a circular path of radius

$R$ under the action of a force constatn magnitude

$F$ . The kinetic energy of object is

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$\cfrac{1}{2}FR$
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$FR$
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$2FR$
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$\cfrac{1}{4}FR$

Explanation

Force F is providing the centripetal force keeping the object moving in a circular path.

$F = \dfrac{mv^2}{R}$

$FR = mv^2$

Hence,

Kinetic Energy(KE)

$= \dfrac{1}{2}mv² = \dfrac{1}{2}FR$

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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