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

A cyclist turns a corner with a radius of 50 m at a speed of 20 m/s . What is the magnitude of the cyclist's acceleration?

0%
$8 m/s^2$
0%
$2.5 m/s^2$
0%
$400 m/s^2$
0%
$1000 m/s^2$

Explanation

Centripetal acceleration and tangential acceleration is normal to each other.

Centripetal acceleration, ${{a}_{c}}=\dfrac{{{v}^{2}}}{r}=\dfrac{{{20}^{2}}}{50}=8\,m{{s}^{-2}}$

Tangential acceleration, ${{a}_{t}}=0$

Hence, magnitude of acceleration is $8\,m{{s}^{-1}}$

The horizontal ranges described by two projectiles, projected at angles

$(45^{\circ} - \theta)$ and

$(45^{\circ} + \theta)$ from the same point and same velocity are in the ratio.

0%
$2:1$
0%
$1:1$
0%
$2:3$
0%
$1:2$

Explanation

$The\quad horizontal\quad range\quad of\quad the\quad complementary\quad angle\quad of\quad projection\quad always\quad remains\quad same.\\ And\quad given\quad projected\quad angles\quad =({ 45 }^{ 0 }\quad +\theta )\quad and\quad ({ 45 }^{ 0 }\quad -\quad \theta )\quad which\quad are\quad complementary\quad to\quad each\quad other.\\ Thus\quad ratio\quad of\quad the\quad project\quad angles=\frac { { u }^{ 2 }\quad sin2({ 45 }^{ 0 }\quad -\quad \theta ) }{ { u }^{ 2 }\quad sin2({ 45 }^{ 0 }\quad +\quad \theta ) } =\quad \frac { \quad sin({ 90 }^{ 0 }\quad -\quad \theta ) }{ sin({ 90 }^{ 0 }\quad +\quad \theta ) } =\quad \frac { cos2\theta }{ cos2\theta } =1\\ u\quad and\quad g\quad gets\quad cancel\quad out.\\ \quad \quad$

$A = B$ , then angle between

$(\vec A + \vec B)$ and

$(\vec A - \vec B)$ is?

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$90 ^\circ$
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$30 ^\circ$
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$60 ^\circ$
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$100 ^\circ$

If the line join of ( 6, 1 ) and ( -10, -8 ) subtends a right at P. Then the locus of P is

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$x^{2}+y^{2}+4x-3y+148=0$
0%
$x^{2}+y^{2}+4x-3y-148=0$
0%
$x^{2}+y^{2}-4x-3y+148=0$
0%
$x^{2}+y^{2}-4x-3y-148=0$

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, the magnitude of acceleration is :

0%
$20m{s^{ - 2}}$
0%
$12m/{s^2}$
0%
$27.53m{s^{ - 2}}$
0%
$8m{s^{ - 2}}$

Two particle separated at a horizontal distance

$X$ as shown in figure, they projected at the same line as shown in figure with different initial speed. The timer after which the horizontal distance between them become zero -

0%
$\frac{x}{u}$
0%
$\frac{u}{2x}$
0%
$\frac{2u}{x}$
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None of these

For an object projected from the ground with speed

$u$ . The horizontal range is two times the maximum height attained by it. The horizontal range of the object is:

0%
$\dfrac{2 u^2}{3g}$
0%
$\dfrac{3 u^2}{4g}$
0%
$\dfrac{3 u^2}{2g}$
0%
$\dfrac{4 u^2}{5g}$

Explanation

$\textbf {Hint :}$ Use the maximum height of projectile motionis

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

Horizontal range,

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

$\textbf{Step 1: }$ Recall the formula of maximum height and horIzontal range of projectile motion

The maximum height is

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

Horizontal range,

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

$\textbf{Step 2: }$ According to question,

$R=2H$

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

$2cos\theta=sin\theta$

$tan\theta=2$

$\textbf{Step 3: }$ We know that,

$1+tan^2\theta=sec^2\theta=\dfrac{1}{cos^2\theta}$

$1+2^2=\dfrac{1}{cos^2\theta}$

$cos^2\theta=\dfrac{1}{5}$

$cos\theta=\dfrac{1}{\sqrt{5}}$

And,

$sin^2\theta=1-cos^2\theta$

$sin^2\theta=1-\dfrac{1}{5}=\dfrac{4}{5}$

$sin\theta=\dfrac{2}{\sqrt{5}}$

$\textbf{Step 4: }$ Horizontal range,

$R=\dfrac{2u^2sin\theta .cos\theta}{g}$

$(\because sin2\theta =2sin\theta cos\theta)$

$R=\dfrac{2u^2}{g}\dfrac{2}{\sqrt{5}}\dfrac{1}{\sqrt{5}}$

$R=\dfrac{4u^2}{5g}$

${\textbf{Correct option: A}}$

Figure shows three vectors

$P, q$ and

$r$ where

$C$ is the mid-point of

$AB$ . Then which of the following relation is correct?

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$p + q = 2r$
0%
$p + q = r$
0%
$p - q = 2r$
0%
$p - q = r$

A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the:

0%
$\dfrac{6v}{5l}$
0%
$\dfrac{3v}{5l}$
0%
$\dfrac{6v}{11l}$
0%
$\dfrac{6v}{l}$

Explanation

Moment of inertia of uniform rod of length(l) -

$I=\dfrac { M{ l }^{ 2 } }{ 12 }$

- wherein

About axis passing through its center & perpendicular to its length.

Law of conservation of angular moment -

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

- wherein

If net torque is zero

i.e $\dfrac { d\vec { L } }{ dt } =0,\quad$

$\overrightarrow { L } =$ constant

angular momentum is conserved only when external torque is zero .

Centre of mass from system from 0

$=>\dfrac { 8m\times 0\quad +m(\dfrac { L }{ 3 } )-\quad 2m(\dfrac { L }{ 6 } ) }{ 8m+m+2m } =0$

So,centre of mass is at zero.

From conservation of ang.momentum; ${ L }_{ i }={ L }_{ f }$

${ L }_{ i }=m(2v)\times (\dfrac { L }{ 3 } )+\quad 2mv\times (\dfrac { L }{ 6 } )=mvL$

${ L }_{ f }=[(8m)\times \dfrac { { L }^{ 2 } }{ 12 } +\quad m{ (\dfrac { L }{ 3 } ) }^{ 2 }+2m{ (\dfrac { L }{ 6 } ) }^{ 2 }]\omega$

$=[\dfrac { 2 }{ 3 } m{ L }^{ 2 }+\dfrac { m{ L }^{ 2 } }{ 9 } +\dfrac { m{ L }^{ 2 } }{ 18 } ]\omega$

$=(\dfrac { 12+2+1 }{ 18 } )m{ L }^{ 2 }\omega =\frac { 5 }{ 6 } m{ L }^{ 2 }\omega$

$\dfrac { 5 }{ 6 } m{ L }^{ 2 }\omega =mvl,$

$\therefore \omega =\dfrac { 6v }{ 5L }$

A particle is moving in $xy - plane$ in a circular path with center at the origin. If at an instant the position of the particle is given by $\frac{1}{{\sqrt 2 }}\left( {\hat i + \hat j} \right)$ , then the velocity of the particle is along

0%
$\frac{1}{{\sqrt 2 }}\left( {\hat i - \hat j} \right)$
0%
$\frac{1}{{\sqrt 2 }}\left( {\hat j - \hat i} \right)$
0%
$\frac{1}{{\sqrt 2 }}\left( {\hat i + \hat j} \right)$
0%
Either 1 and 2

Explanation

$\vec{V}=\vec{r}\times \vec{w}$

$[\vec{w}=\hat{k}\ or\ -\hat{k}]$

$=\dfrac{1}{\sqrt{2}}(\hat{i}+\hat{j})\times (\pm \hat{k})$

$=\pm \dfrac{1}{\sqrt{2}}(\hat{i}-\hat{j})$

$\vec{V}=\dfrac{1}{\sqrt{2}}(\hat{i}-j)$ or

$\vec{V}=\dfrac{1}{\sqrt{2}}(\hat{j}-\hat{i})$

A particle moves a distance

$x$ in time t according to equation

$x=(t+5)^{-1}$ . The acceleration of particle is proportional to

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$(Velocity)^{3/2}$
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$(Distance)^2$
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$(Distance)^{-2}$
0%
$(Velocity)^{2/3}$

Explanation

Correct answer: Option A

Hint: Differentiation of displacement gives velocity and differentiation of velocity gives acceleration with respect to time.

Explanation of correct option:

Given: Displacement - Time relation

$x = {(t + 5)^{ - 1}}$ ...................(i)

Step 1:Differentiating both side w.r.t. $'t'$ of equation (i)

We get,

$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}{(t + 5)^{ - 1}}$

$\therefore\dfrac{dx}{dt} = ( - 1){(t + 5)^{ - 2}}(1 + 0)$

$\dfrac{dx}{dt}$ is velocity.

$\Rightarrow v = - {(t + 5)^{ - 2}}$ ...................(ii)

Step 2:Differentiating both side w.r.t. $'t'$ of equation (ii)

We get,

$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}[ - {(t + 5)^{ - 2}}]$

$\therefore \dfrac{dv}{dt}$ = $- ( - 2){(t +5)^{ - 3}}.(1)$

$\dfrac{dv}{dt}$ is acceleration.

Formatting equation (iii) to out the value of equation (ii)

$\Rightarrow a = 2 \times \left\{ { - {{(t + 5)}^{ - 2}}} \right\}\left\{ { - {{(t + 5)}^{ - 2 \times \dfrac{1}{2}}}} \right\}$

$\Rightarrow a = 2v \times {v^{\frac{1}{2}}}$

$\therefore a = 2{v^{\frac{3}{2}}}$

As $2$ is constant so $a$ varies with $v$ such as

$a \propto {v^{\frac{3}{2}}}$ i.e. $(acceleration) \propto (velocity)^{\frac{3}{2}}$

Hence, Acceleration of a particle is proportional to $(velocity)^{\frac{3}{2}}$

If position vector of a point varies with time

$"t"$ as

$\vec {r}=(t+t^{2})(\hat {i}+\hat {j})$ meter then velocity at time

$t=4\ s$ will be`

0%
$(6\hat {i}+6\hat {j})m/s$
0%
$(6\hat {i}+9\hat {j})m/s$
0%
$(9\hat {i}+6\hat {j})m/s$
0%
$(9\hat {i}+9\hat {j})m/s$

Explanation

Given,

$\vec r=(t+t^2)(\hat i+\hat j)m$

Velocity,

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

$\vec v=(1+2t)(\hat i+\hat j)$

$t=4sec$

$\vec =(1+2\times 4)(\hat i+\hat j)$

$\vec v=(9\hat i+9\hat j)m/s$

The correct option is D.

The relation between

$\hat{A} \ and \ \hat{B}$ is

$4(\hat{A} \cdot \hat{B}) = 3$

$|\hat{A} \times \hat{B}|.$
Then

$|\hat{A} - \hat{B}|^2$ is equal to

0%
$A^2 + B^2 - \dfrac{8}{5}AB$
0%
$A^2 + B^2 - \dfrac{6}{5}AB$
0%
$A^2 + B^2 + \dfrac{8}{5}AB$
0%
$A^2 + B^2 + \dfrac{6}{5}AB$

The velocity of a particle at

$t = 0$ is

$(\hat{i} + 3 \hat{j})m/s$ . It has a uniform acceleration of

$2 \hat{i} \ m/s^2$ . The radius of curvature of its path at

$t = 1.5 s$ is nearly

0%
$18.6 \ m$
0%
$16.8 \ m$
0%
$15.6 \ m$
0%
$20.8 \ m$

A partial is projected vertically upwards. Its height

$'h'$ and time

$t'$ are related by

$h = 60t - 16t^2$ . The velocity at which it hits the ground is:

0%
$60$
0%
$30$
0%
$90$
0%
$180$

Explanation

$h = 60t - 16{t^2}$

at ground

$h=0$

$\begin{array}{l} \Rightarrow 60t-16{ t^{ 2 } }=0 \\ t=0 \\ 60-16t=0 \\ t=\frac { { 60 } }{ { 16 } } =\frac { { 15 } }{ 4 } \sec . \\ \frac { { dh } }{ { dt } } =60-32t \\ at\, \, t=\frac { { 15 } }{ 4 } \\ v=60-32\times \frac { { 15 } }{ 4 } \\ =60-120 \\ =60\, m/s \end{array}$

$\therefore$ Option

$A$ is correct.

the position vector of two particles is given is given as

$\acute { { r }_{ 1 } } =\sin { t\hat { i } } +\cos { t\hat { j } } +2t\hat { k } and { r }_{ 2 }=t\hat { k }$ The path followed by particle

$1$ as seen by particles

$2$ is

0%
Straight line
0%
Helical
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Parabolic
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Circular

$\vec{r}$ satisfies

$\vec{r} \times (\bar{i} + 2\bar{j} + \bar{k}) = \bar{i} - \bar{k},$ then for any scalar

$t$ ,

$\vec{r} =$

0%
$\bar{i} + t(\bar{i} + 2\bar{j} + \bar{k})$
0%
$\bar{j} + t(\bar{i} + 2\bar{j} + \bar{k})$
0%
$\bar{k} + t(\bar{i} + 2\bar{j} + \bar{k})$
0%
$\bar{i} - \bar{k} t(\bar{i} + 2\bar{j} + \bar{k})$

A particle is projected at

$20m/s$ horizontally. What is its speed at

$5\sec \,\left( {g = 10m/{s^2}} \right)$ :-

0%
$54\,\,m/s$
0%
$20\,\,m/s$
0%
$50\,\,m/s$
0%
$70\,\,m/s$

Explanation

Horizontal speed will remain

$constant$ but the vertical component will get chnged from

$zero$ to

$gt=10\times 5=50m/s$

So resultant speed will be

$\sqrt[2]{50^2 +20^2}=53.85m/s$ which is nearly

$54m/s$

Option A is correct.

A particle is projected with initial velocity

$\vec {v} = (10\hat i + 15\hat j)m/s$ in

$x-y$ plane. The magnitude of displacement of the particle at time

$t=1\ s$ is
(

$\vec {g} = - 10\hat j\,m/{s^2}$ )

0%
$10\ m$
0%
$10\sqrt 2\ m$
0%
$20\ m$
0%
$20\sqrt 2\ m$

Explanation

Given,

$\vec v=(10\hat i+15\hat j)m/s$

$g=-10\hat j m/s^2$

$t=1s$

From 2nd equation of motion,

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

$\vec s=(10\hat i+15\hat j)\times 1+\dfrac{1}{2}(-10\hat j)(1)^2$

$\vec s=10\hat i+15\hat j-5\hat j$

$\vec s=10\hat i+10\hat j$

Magnitude of the displacement,

$|\vec s|=\sqrt{(10)^2+(10)^2}=\sqrt{200}m$

$|\vec s|=10\sqrt{2}m$

The correct option is B.

A player kicks a foot ball at a speed of

$20 ms^{-1}$ so that its horizontal range is maximum. Another player

$24$ m away in the direction of kick starts running in the same direction at the same instant of hit, if he has to catch the ball just before it reaches to ground, he should run with a velocity equal to (

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

0%
$4ms^{-1}$
0%
$5.65ms^{-1}$
0%
$10ms^{-1}$
0%
$12ms^{-1}$

Explanation

Maximum range of ball

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

Now,

$R= \cfrac {20 \times 20 \times 1}{10}=40m$

Distance travelled by second man=

$40-24=16m$

Time=

$t= \cfrac {40}{u \cos 45^o}=2\sqrt {2}S$

Speed

$v= \cfrac {distance}{time}=\cfrac {16}{2\sqrt {2}}=5.65 m/s$

If vectors

$\vec {A} = 2\hat {i} + 3\hat {j} + 4\hat {k}, \vec {B} = \hat {i} + \hat {j} + 5\hat {k}$ and

$\vec {C}$ form a left-handed system, then

$\vec {C}$ is

0%
$11\hat {i} - 6\hat {j} - \hat {k}$
0%
$-11\hat {i}+ 6\hat {j} + \hat {k}$
0%
$11\hat {i} - 6\hat {j} + \hat {k}$
0%
$-11\hat {i} + 6\hat {j} - \hat {k}$

$a$ and

$b$ are the angles made by a vector, from positive

$x$ and positive

$y$ axes, respectively. Which set of

$a$ and

$b$ is not possible ?

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$45^{o},\ 60^{o}$
0%
$30^{o},\ 60^{o}$
0%
$60^{o},\ 60^{o}$
0%
$30^{o},\ 45^{o}$

A circular wheel of 24 spokes is rotated at 300 rpm in a uniform magnetic field of

$2x{10^{ - 3}}$ tesla.Length of each spoke is 25 cm and magnetic field is along the axis of wheel, then the value of induced emf between rim and center of wheel is

0%
1.96 mv
0%
219.6 v
0%
47.1 mv
0%
19.6 mv

A string of length

$\ell$ has one end fixed and a particle of mass m is attached to the other end. If the particle describe a horizontal circle at an angular speed

$\omega$ , in gravity free space.

0%
The tension in the string = $m\omega^2\ell$
0%
The speed of the particle is $\ell \omega$
0%
The resultant force acting on the particle has no vertical component
0%
None of these

Explanation

When a move in circular motion, centripetal force will act on the body toward center.

When body moves with uniform angular velocity in circular path.

Then,

(i)Centripetal force, ${{F}_{c}}=m\omega {{l}^{2}}$

(ii)Speed of the particle, $v=\omega l$

(iii) Because it is horizontal circle and there is no gravity. Therefore there is no force in vertical direction.

If the angle

$(\theta)$ between velocity vector and the acceleration vector is

$90^o < \theta < 180^o.$ The body is moving on a:

0%
Straight path with retardation
0%
Straight path with acceleration
0%
Rectilinear path with acceleration
0%
Rectilinear path with retardation

A vector is represented by

$3 \hat { i } + \hat { j } + 2 \hat { k }$ , has length in

$XY$ plane

0%
2
0%
$\sqrt { 14 }$
0%
$\sqrt { 10 }$
0%
$\sqrt { 5 }$

Explanation

We know,

The length in

$XY$ plane is due to the

$X$ and

$Y$ component of the line,

Hence, the

$XY$ components are

$3i+j$

And length

$=\sqrt{3^2+1^2}=\sqrt 10$

Option

$\textbf C$ is the correct answer

If the magnitude of difference of two vectors

$\vec A$ and

$\vec B$ of equal magnitude is equal to magnitude of either of the given vectors. The angle between the vectors

$\vec A$ and

$\vec B$ will be

0%
$30^\circ$
0%
$60^\circ$
0%
$120^\circ$
0%
$90^\circ$

The resultant of three vectors of magnitudes

$1,2\ and\ 3$ units respectively and whose direction are those of an equilateral triangle taken in the same order is:

0%
${\sqrt{3}}$
0%
$12$
0%
${\sqrt{14}}$
0%
$6$

A particle is projected in xy planes pointalong vertical, the point of projection is origin.The equation of the path is

$y = \sqrt { 3 } x - \frac { g } { 2 } x ^ { 2 }$ .where y and x are in

$m$ . Then the speed ofprojection in

$m s ^ { - 1 }$ is

0%
$2$
0%
$\sqrt { 3 }+1$
0%
$4$
0%
$\frac { \sqrt { 3 } } { 2 }$

The maximum speed that can be achieved without skidding by a ar on circular unbanked road of radius R and coefficient of static friction

$\mu$ , is

0%
$\mu Rg$
0%
$Rg\sqrt { \mu }$
0%
$\mu \sqrt { Rg }$
0%
$\sqrt { \mu Rg }$

Explanation

${\textbf{Step1: Identify the values that are given.}}$

${\text{Radius = R}}$

${\text{Static friction = }}\mu$

${\text{Now, the centripetal force is providing by frictional force between the road and tyre}}$

${\textbf{Step2: Find v. }}$

$\text{The centripetal force is given by } \dfrac{{m{v^2}}}{R}$

$\text{and the frictional force is } = \mu mg$

$\text{For avoiding car to skid the cetripetal force must be equal to frictional force }$

$\dfrac{{m{v^2}}}{R} = \mu mg$

$v = \sqrt {\mu Rg}$

${\text{Hence,this is the required solution}}$

${\textbf{Hence,}}$ ${\textbf{the final answer is }}$ $\mathbf {v =}$ $\mathbf { { \sqrt {\mu Rg} .}}$

OABC is a current carrying square loop an electron is projected from the centre of loop along its diagonal AC as shown. Unit vector in the direction of initial acceleration will be

0%
$\hat { k }$
0%
$- \left( \cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } } \right)$
0%
$-\hat { k }$
0%
$\cfrac { \hat { i } +\hat { j } }{ \sqrt { 2 } }$

Explanation

current loop is clockwise so megentic field

at center of square loop will be into the

plane of loop i.e

$\vec{B} =B_{0}\hat{k}$

election always AC

$\Rightarrow \vec{V} = \dfrac{v_{0}}{\sqrt{2}} (\hat{i}-\hat{i})$ (velocity of electron)

$\Rightarrow$ forever on it

$\vec{f} = z\vec{V}\times \vec{B} = (-e) \dfrac{V_{0}}{\sqrt{2}} B_{0}(\hat{k}\times \hat{i}-\hat{k}\times \hat{i})$

$\vec{f} = \dfrac{V\sqrt{0}B_{0}e}{\sqrt{2}} (-\hat{i}-\hat{i})$

$\Rightarrow$ acceleration

$\vec{a} = \dfrac{\vec{f}}{m} = \dfrac{V_{0}B_{e}e}{\sqrt{2}m}(-\hat{i}-\hat{i})$

1185146_1165433_ans_90b211a1af3d451cbbc5d5e67ce9f2d6.jpg

The position vectors of the points A,B,C are

$\overline { i } + 2 \overline { j } - \overline { k } , \overline { i } + \overline { j } + \overline { k } , 2 \overline { i } + 3 \overline { j } + 2 \overline { k }$ respectively. If A is chosen as the origin then the position vectors of B and C are

0%
$\overline { i } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$
0%
$\overline { j } + 2 \overline { k } , \vec { i } + \overline { j } + 3 \overline { k }$
0%
$- \overline { j } + 2 \overline { k } , \vec { i } - \overline { j } + 3 \overline { k }$
0%
$- \overline { j } + 2 \overline { k } , \overline { i } + \overline { j } + 3 \overline { k }$

Explanation

$\textbf{Step1-Find position vector from given point}$

$\text{The origin is shifted so the new position vectors of B and C would}$

$\text{ be the difference of the old position vectors and the new origin vector i.e.}$

$(i+j+k) - (i+2j-k) = -j+2k$ (

$\text{ New position vector of point B}$ )

$(2i+3j+2k) - (i+2j-k) = i+j+3k$ (

$\text{ New position vector of point A}$ )

$\textbf{Hence , option D is correct}$

.............. is not a scalar quantity.

0%
Momentum
0%
temperature
0%
length
0%
Time

A particle of mass

$0.2\ kg$ moves along a path given by the relation

$\vec {r}=2\ \cos\omega\ t\hat{i} + 3\ \sin\omega\ t\hat{j}$ . Then torque on the particle about original is

0%
$\sqrt{13}\ \hat{k}\ Nm$
0%
$\sqrt{2}{3}\ \hat{k}\ Nm$
0%
$\sqrt{3}{2}\ \hat{k}\ Nm$
0%
$0\hat{k}$

A particle moving at a circle. If at an instant its linear velocity vector, angular velocity and position vector are

$\overline { v } , \overline { \omega }$ and

$\vec { r }$ respectively then, the correct option is

0%
They are collinear vectors
0%
They are mutually perpendicular vectors
0%
They are coplanar vectors
0%
They are axial vectors

Explanation

We know,

$\overrightarrow A\times\overrightarrow B=\overrightarrow C$

$\overrightarrow A,\overrightarrow B, \overrightarrow C$ are mutually perpendicular vectors.

Also,

$\overrightarrow r\times \overrightarrow v=\overrightarrow{\omega}$

Hence,

$\overrightarrow r,\overrightarrow v,\overrightarrow{\omega}$ are mutually perpendicular.

Option

$\textbf B$ is the correct answer

A motor requires 2 s to go from a speed of 60 rpm to 120 rpm with constant acceleration. Number of revolutions it makes in this time is

0%
$1.50$
0%
$4.5$
0%
$3$
0%
$6.0$

An object is projected at an angle

$\alpha$ with the level ground. At an instant time

$t$ , the angle made by its position vector (from the point of projection) is

$\beta$ . Initial velocity of the projectile is:

0%
$\dfrac{gt\cos(\beta)}{2\sin(\alpha-\beta)}$
0%
$\dfrac{gt\sin(\alpha-\beta)}{2\cot(\beta)}$
0%
$\dfrac{gt\cos(\beta)}{2\sin(\alpha+\beta)}$
0%
$u = \dfrac {gtcos\beta}{sin(\alpha - \beta)}$

A short dipole

$\vec {p}$ is at a distance

$r$ from a point charge

$q$ and oriented so that

$\vec {p}$ makes angle

$\theta$ with position vector

$\vec {r}$ from

$q$ to

$\vec {p}$ . If

$\theta$ is variable, then what is ratio of minimum and maximum magnitudes of force on the dipole.

0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{4}$
0%
$0$
0%
$1$

A car is moving on a circular path and takes a turn. if

${ R }_{ 1 }$ and

${ R }_{ 2 }$ , be the reactions on the inner and outer wheels respectively, then

0%
${ R }_{ 1 }$ = ${ R }_{ 2 }$
0%
${ R }_{ 1 }$ < ${ R }_{ 2 }$
0%
${ R }_{ 1 }$ >{ R }_{ 2 }$$
0%
${ R }_{ 1 }$ $\ge$ ${ R }_{ 2 }$

Explanation

we have,

${R_1}$ is the reaction due to inner surface.

${R_2}$ is the reaction due to outer surface.

${R_1}$ is smaller than

${R_2}$ .

As outer surface covers larger distance than the inner surface.

Hence,

Option

$B$ is correct answer.

A particle in one-dimensional motion:

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with zero speed at an instant may have non-zero acceleration at that instant
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with zero speed may have non-zero velocity
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with constant speed must have zero acceleration
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with positive value of acceleration must be speeding up

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

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$3\overset { \wedge }{ i }+5 \overset { \wedge }{ j+ } \overset { \wedge }{ k }$
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$3\overset { \wedge }{ i }+2 \overset { \wedge }{ j+ } 5\overset { \wedge }{ k }$
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$5\overset { \wedge }{ i }+3 \overset { \wedge }{ j+ } 2\overset { \wedge }{ k }$
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None of tese

Explanation

$\textbf{Step 1: Position vector}$

$\textbf{[Refer Fig.]}$

Position vector is given by

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\textbf{Step 2: Putting values}$

$\vec{r}=3\hat{i}+2\hat{j}+5\hat{k}$

Hence the correct answer is option

$B$ .

2110440_1190412_ans_828216ec57d14d1c9d6d0d13a07b4499.png

If a particle covers half the circle of radius R with constant speed then

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Momentum change is mvr
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change in K.E. is $1/2 mv^2$
0%
change in K.E. is $mv^2$
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change in K.E. is zero

Explanation

$|\vec{v_1}|=|\vec{v_2}|=v$

Since particle performs circular motion and completed half circle with constant speed.

i) change in Momentum'

$\Delta P=P_f-P_i$

$P_i=m\vec{U_i}$

$P_f=m\vec{v_2}=-m\vec{v_1}$ (

$\because \vec{v_1}=-\vec{v_2}$ )

$\Delta P=-m\vec{v_1}-m\vec{v_2}$

$\Delta P=-2m\vec{v_1}$

(ii) Change in kinetic energy

$\Delta KE=\dfrac{1}{2}mv^2_2-\dfrac{1}{2}mv^2_1$

$=0$

So, option Last is correct.

2030499_1229985_ans_f3a174a6a71742e581ef751bfe5ebb4d.png

The moment of the force,

$\vec{F}=4\hat{i}+5\hat{j}-6\hat{k}$ at (2,0,-3) about the point (2,-2,-2), is given by

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$-8\hat{i}-4\hat{j}-7\hat{k}$
0%
$-4\hat{i}-\hat{j}-8\hat{k}$
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$-7\hat{i}-8\hat{j}-4\hat{k}$
0%
$-7\hat{i}-4\hat{j}-8\hat{k}$

Explanation

The force given as

$\bar F=4\vec{j}+5\vec{j}-6\vec{k}$

The force acting the point

$(2, 0, -3)$

The point under construction

$(2, -2, -2)$

Position vector

$(2, 0, -3)$ w.r.to

$(2, -2, -2)$

$\vec r=(2-2)i+(0-(-2))j+(-3-2)\vec{k}=0\vec{i}+2\vec{j}-\vec{k}$

moment of force is

$\vec{M}=\vec{r}\times \vec{f}$

$M=(0i+2\vec{j}-\vec{k})\times (4i+4j-6\vec{k})$

$M=\begin{vmatrix} i & j & k \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix}=-7\vec{i}-4\vec{j}-8\vec{k}$

A particle

$P$ is a moving in a circle of radius

$r$ with a uniform speed

$v.C$ is the centre of the circle, and

$AB$ is a diameter, If

$P$ is at

$B$ , the angular velocities of

$P$ about

$A$ and

$C$ are in the ratio

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$1:1$
0%
$2:1$
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$1:2$
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$4:1$

Explanation

Angular velocity about centre of circle = 2 times about any point on circumference

${W_C} = 2{W_A},\dfrac{{{W_A}}}{{{W_A}}} = \dfrac{1}{2}$

An object is projected upwards with a velocity of

$100\ m/s$ . It will strike (approximately)

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$10\ sec$
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$20\ sec$
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$15\ sec$
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$5\ sec$

Explanation

Given,

$u=100m/s$

$v=0m/s$ ( at the highest point)

$g=-10m/s^2$

From the 1st equation of motion,

$v=u-gt$

$0=100-10t$

$t=10s$

The time taken to reach the ground is

$10s+10s=20s$ .

The correct option is B.

What is the acceleration of a projectile at its heighest point

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maximum
0%
minimum
0%
zero
0%
g

Explanation

The acceleration of a projectile at its highest point is

$9.8m/s^2$ , which is the value of acceleration due to gravity

$(g)$ .

The correct option is D.

If the position of a particle changes from (1,2,3) m to (5,4,2) m, then displacement vector is

0%
$4\overset { \wedge }{ i } +2\overset { \wedge }{ j } -\overset { \wedge }{ k }$
0%
$\overset { \wedge }{ i } +5\overset { \wedge }{ j } +4\overset { \wedge }{ k }$
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$\overset { \wedge }{ i } -5\overset { \wedge }{ j } +4\overset { \wedge }{ k }$
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None of these

$\overset { \rightarrow }{ A } +\overset { \rightarrow }{ B }$ is perpendicular to

$\overset { \rightarrow }{ A } -\overset { \rightarrow }{ B }$ then

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A = B
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A = 2B
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Both A and B
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None of the above

Two balls are projected from a tower such that one is vertically upward with a velocity of 9 m/s and other horizontally with same speed. The relative acceleration of one ball w.r.t. other is

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

0%
10 $m/s^2$
0%
20 $m/s^2$
0%
18 $m/s^2$
0%
$Zero$

Explanation

Acceleration of both the balls

$a_1=a_2 = 10 \ m/s^2$ (downwards)

So, relative acceleration

$a_{rel} = a_1-a_2 = 10-10 = 0 \ m/s^2$

A body is moving in a circular path such that its speed is decreasing with time. Angle between its velocity and acceleration may be

0%
0
0%
90
0%
180
0%
135

Explanation

When a body moves in circular path then its velocity is changing as velocity is a vector quantity and since the direction of the body will change it's velocity will also change continuously. This change implies acceleration. Thus it allows the body to go round the circle having the velocity vector tangent to the circular trajectory and acceleration at

$90°$ to it directed towards the center.

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

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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