MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 8

If the magnitude of the cross product of two vector is

$\sqrt { 3 }$ times to the magnitude of their scalar product the angle between two vector will be :

0%
$\pi$
0%
$\frac { \pi }{ 2 }$
0%
$\frac { \pi }{ 3 }$
0%
$\frac { \pi }{ 6 }$

Explanation

$\textbf{Step 1: Simplify the given equation}$

Let the two vectors be

$\vec{A}$ and

$\vec{B}$

Magnitude of Cross product of vectors=

$\sqrt{3}\times$ Magnitude of Dot product of vectors

$\Rightarrow|\vec{A}\times \vec{B}|=\sqrt{3}|\vec{A}.\vec{B}|$

Squaring on both the sides

$|\vec{A}\times \vec{B}|^2=3|\vec{A}.\vec{B}|^2$

$\Rightarrow\ \ (\vec{A}\times \vec{B})(\vec{A}\times \vec{B})=3(\vec{A}.\vec{B})(\vec{A}.\vec{B})$

$\Rightarrow\ \ (|\vec{A}||\vec{B}|\sin\theta)(|\vec {A}||\vec{B}|\sin\theta)$

$=(|\vec{A}||\vec{B}|\cos\theta)(|\vec {A}|\vec{B}|\cos\theta)$

$....(1)$

$\textbf{Step 2: Angle between the Vectors}$

$(1)$

$\Rightarrow\ \sin^2\theta=3 \cos^2\theta$

$\Rightarrow\ \ tan^2\theta=3$

$\Rightarrow\ \ \theta=\tan^{-1}(\sqrt{3})$

$=\dfrac{\pi}{3}$

Hence, Option C is correct.

If a particle is rotating with an angular velocity

$\omega$ and angular acceleration

$\alpha$ , then ,

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the particle is slowing down, if the directions of both angular velocity and angular acceleration are shown in the same directions
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the particle is speeding up, if the directions of both angular velocity and angular acceleration are shown in the same directions
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the particle is slowing down, if both angular velocity and angular acceleration are shown in the opposite directions
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the particle is speeding up, if both angular velocity and angular acceleration are shown in the opposite directions

A plate rotates about a fixed perpendicular axis such that the angle changes with time as

$\theta = 2t^2$ . Find the angular acceleration in

$rad/s^2$ .

0%
16
0%
12
0%
4
0%
0

Explanation

Angular velocity is

$\omega=d\theta/dt = 4t$ and Angular acceleration is

$\omega/dt=4 rads/s^2$

The correct option is (c)

The rotating rod starts from rest and acquires a rotational speed

$\omega =600 rev/min$ in

$2$ seconds with constant angular acceleration. The angular acceleration of rod is?

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$20 \pi rad/s$
0%
$10 \pi rad/s$
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$30 \pi rad/s$
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$5 \pi rad/s$

Explanation

Initial angular velocity

$w_i = 0$

Final angular velocity

$w_f = 600\ rev/min = 600\times \dfrac{2\pi}{60} = 20\pi \ rad/s$

Time

$t = 2 \ s$

Using

$w_f = w_i +\alpha t$

$20\pi = 0+\alpha \times 2$

$\implies \ \alpha = 10\pi \ rad/s$

The centripetal acceleration of a particle varies inversely with the square of the radius r of the circular path. The KE of this particle varies directly as:

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r
0%
$r^2$
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$r^-2$
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$r^{-1}$

Explanation

The correct option is C.

We have,

Acceleration of a particle varies inversely with the square of the radius r

Let the centripetal acceleration be a,

The radius of the circular path is R.

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

Where c is the proportionality constant.

By using the definition a:

$a=\dfrac{dv}{dR}$

Where vis the velocity.

$\dfrac{dv}{dR}=\dfrac{c}{R^2}$

$dv=\dfrac{c}{R^2}dR$

Integrating both sides the we get,

$\int dv=\int \dfrac{c}{R^2}dR$

$V+C=-\dfrac{c}{R}+C$

Now the KE of the particle of mass is given as:

$KE=\dfrac{mv^2}{2}=\dfrac{m}{2R^2}$

Thus, KE is inversely proportional to the square of the radius.

WHere,

$\dfrac{m}{2}$ is constant.

$\overrightarrow { A } +\overrightarrow { B }$ is a unit vector along x-axis and

$\overrightarrow { A } =\hat { i } -\hat { j } +\hat { k }$ then what is

$\overrightarrow { B }$ ?

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

In a Rutherford scattering experiment when a projectile of charge

$Z_{1}$ and mass

$M_{1}$ approaches a target nucleus of charge

$Z_{2}$ and mass

$M_{2}$ the distance to closest approached is

$r_{0}$ . The energy of the projectile is

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directly proportional to $M_{1} \times M_{2}$
0%
directly proportional to $Z_{1} Z_{2}$
0%
directly proportional to to $Z_{1}$
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directly proportional to mass $M_{1}$

Explanation

The particle of mass

$M_1$ and charge

$Z_1$ possess initial velocity

$u$ when it at a large distance from the nucleus of an atom having atomic number

$Z_2$ . At the distance of closest approach, the kinetic energy of the particle is completely converted to potential energy.

$\dfrac{1}{2}M_1u^2=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Z_1Z_2}{r_0}$

So the energy of the particle is directly proportional to

$Z_1Z_2$

A force has magnitude 20N. Its one rectangular component is 12N, the other rectangular component must be:

0%
8 N
0%
14 N
0%
16 N
0%
32 N

Explanation

By rectangular component,

the two forces

$F_1(12N)$ and

$F_2$ are perpendicular forces with resultant as

$20N$

Using Pythagorus theorem

$R^2=F^2_1+F_2^2\\20^2=12^2+F_2^2\\400=144+F^2_2\\F_2^2=400-144\\F_2^2=-256\\F_2=16N$

A particles is projected from ground in vertically upward direction such that the distance travelled by it in

$5th$ and

$8th$ second are equal. The time of flight of the particle is:

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$10\ s$
0%
$12\ s$
0%
$13\ s$
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$16\ s$

The minimum number of vectors of unequal magnitude required to produce a zero resultant is :

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$2$
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$3$
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$4$
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$more\ than\ 4$

The linear and angular acceleration of a particle are

$10\ m/sec^2$ and

$5\ rad/sec^2$ respectively, it will be at a distance of ___ m from the axis of rotation

0%
$50\ m$
0%
$1/2\ m$
0%
$1\ m$
0%
$2\ m$

Explanation

Given :

$a_t = 10 \ m/s^2$

$\alpha = 5 \ rad/s^2$

Using

$a_t = r\alpha$

$10 = r\times 5$

$\implies \ r = 2 \ m$

A particle starts travelling on a circle with constant tangential acceleration. The angle between velocity vector and acceleration vector, at the moment when particle complete half the circular track, is:

0%
$\tan ^{ -1 }{ (2\pi ) }$
0%
$\tan ^{ -1 }{ (\pi ) }$
0%
$\tan ^{ -1 }{ (3\pi ) }$
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$\tan^{-1} (2)$

The direction of a vector

$\overrightarrow { A\\ }$ is reversed. Find the value of

$\triangle \overrightarrow { A }$ and

$\triangle \overrightarrow { \left| A \right| }$ ?

0%
- $\overrightarrow { A\\ }$ ,0
0%
$2\overrightarrow { A\\ }$ ,0
0%
$\frac { \overrightarrow { A } }{ 2 }$ ,0
0%
$\overrightarrow { A\\ }$ ,0

If s=a

$\sin { \omega t } \hat { i }+b\cos { \omega t } \hat { j }$ , the equation of path of particle is

0%
${ { x }^{ 2 }+y }^{ 2 }=\sqrt { { a }^{ 2 }+{ b }^{ 2 } }$
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$\dfrac { { x }^{ 2 } }{ { b }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { a }^{ 2 } } =1$
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$\dfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\dfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
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$none\ of\ these$

three vectors

$\overrightarrow a ,\overrightarrow b \,{\text{and}}\,\overrightarrow c$ are such that

$\left| {\overrightarrow a + \overrightarrow b \, + \,\overrightarrow c } \right| = 0$ a + b + c = t unit. Maximum value of

${(\overrightarrow a - \overrightarrow b)}{(\overrightarrow b - \overrightarrow c)}$ is.

0%
$\dfrac{{t^2 }} {{\sqrt 3 }}$
0%
$\dfrac{{\sqrt 3 t^2 }} {4}$
0%
$\dfrac{{t^2 }} {{12\sqrt 3 }}$
0%
$\dfrac{{t^2 }} {{2\sqrt 3 }}$

Calculate the angle between two vectors 2F and $\sqrt 2 \,{\rm{F}}$ so that the resultant force is ${\rm{F}}\sqrt {10} .$

0%
120 degrees
0%
90 degrees
0%
60 degrees
0%
45 degrees

Explanation

Let

$\theta$ be the angle between

$2F$ and

$\sqrt F$ resultant:-

$\sqrt{(2F)^2+(\sqrt{2F})^2+2(2F)(\sqrt2 F\cos\theta)}=F\sqrt{10}\\ \sqrt{4F^2+2F^2+4F^2\sqrt{2\cos\theta}}=F\sqrt{10}\\\sqrt{6F^2+4F^2\sqrt{2\cos\theta}}=(\sqrt{10}F)^2\\6F^2+4F^2\sqrt2\cos\theta=10F^2\\3+2\sqrt2\cos\theta=5\\ \cos\theta=\cfrac{5-3}{2\sqrt2}\\ \cos\theta=\cfrac{2}{2\sqrt2}=\cfrac{1}{\sqrt2}\\ \theta=\cos^{-1}\cfrac{1}{\sqrt2}=45°$

A body of mass

$5\ kg$ under the action of constant force

$\vec F={F}_{x}\hat i+{F}_{y}\hat j$ has velocity at

$t=0\ s$ as

$\vec v=(6\hat i-2\hat j)m/s$ and at

$t=10s$ as

$\vec v=+6\hat j\ m/s$ . The force

$\vec F$ is:

0%
$(-3\hat i+4\hat j)N$
0%
$(-\dfrac {3}{5}\hat i+\dfrac {4}{5}\hat j)N$
0%
$(3\hat i-4\hat j)N$
0%
$(\dfrac {3}{5}\hat i-\dfrac {4}{5}\hat j)N$

Explanation

Given,

$\vec v_i=(6\hat i-2\hat j)m/s$ at

$t=0sec$

$\vec v_f=6\hat j m/s$ at

$t=10sec$

$m=5kg$

Constant force,

$F=\dfrac{dP}{dt}$

$F=m\dfrac{dv}{dt}$

$F=5\dfrac{\vec v_f-\vec v_i}{t_f-t_i}$

$F=5\dfrac{(6\hat j)-(6\hat i-2\hat j)}{10}$

$F=\dfrac{-6\hat i+8\hat j }{2}$

$F=(-3\hat i +4\hat j)$

The correct option is A.

A vector

$\vec A$ when added to the vector

$\vec B = 3 \hat i + 4 \hat j$ yields a resultant vector that is an in the position y -direction and has magnitude equal that of

$\vec B$ find the magnitude

$\vec A$

0%
$\sqrt {10}$
0%
$10$
0%
$5$
0%
$\sqrt {15}$

Explanation

$\begin{array}{l} \overrightarrow { A } +\left( { 3\widehat { i } +4\widehat { j } } \right) =5\widehat { j } \\ \overrightarrow { A } =-3\widehat { i } +\widehat { j } \\ \left| { \overrightarrow { A } } \right| =\sqrt { 10 } \\ \therefore \, option\, \, A\, \, is\, \, correct. \end{array}$

A fire extinguishing hose pipe disposes watch at a speed of 10 m/s to put off fire on a building. Assuming safe distance from the building on ground is 5 m, What is the maximum height at which water strikes building?

0%
1.75 m
0%
2.5 m
0%
3.75 m
0%
4.75 m

Explanation

Range

$R=5\times 2=10 m$

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

$u=10 m/s$

$10=\dfrac{100\times \sin2\theta}{10}$

$\sin 2\theta=1$

$\theta=45^\circ$

$H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{100 \times (1/2)}{2\times 10}$

$=2.5 m$

$\vec P+\vec Q=\vec P-\vec Q$ , then

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$\vec P+\vec 0$
0%
$\vec Q+\vec 0$
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$|\vec P|=1$
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$|\vec Q|=1$

The initial position of an object at rest is given by

$3\hat i-8\hat j$ it moves with constant acceleration and reaches to the position

$2\hat i+4\hat j$ after

$4\ s$ . What is its acceleration?

0%
$-\dfrac{ 1 }{ 8 } \hat { i } +\dfrac{ 3 }{ 2 } \hat { j }$
0%
$2\hat { j } -\dfrac{ 1 }{ 8 } \hat { j }$
0%
$-\dfrac{ 1 }{ 2 } \hat { i } +8\hat { j }$
0%
$8\hat { i } -\dfrac{ 3 }{ 2 } \hat { j }$

A tree trunk of diameter

$20cm$ lies in the horizontal field. A lazy grasshopper wants to jump over the trunk. Find the minimum takeoff speed of the grasshopper that will suffice. (No air drag)

0%
$1.1m/sec$
0%
$2.2m/sec$
0%
$3.3m/sec$
0%
$4.4m/sec$

Explanation

$\begin{array}{l}u = \sqrt {2gr(1 + \sqrt 2 )} \\ = \sqrt {2 \times 10 \times 0.1(1 + \sqrt 2 )} \\ = 2.2m/s\end{array}$

The speed of a projectile at the highest point becomes $\dfrac{1}{\sqrt{2}}$ times its initial speed. The horizontal range of the projectile will be

0%
$(a) \frac{u^2}{g}$
0%
$(b) \frac{u^2}{2g}$
0%
$(c) \frac{u^2}{3g}$
0%
$(d) \frac{u^2}{4g}$

Explanation

At heighest point speed

$=ucos\theta$

given that :

$ucos\theta=\dfrac{u}{\sqrt2}$

$cos\theta=\dfrac{1}{\sqrt2}$

$\theta =45^o$

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

$=\dfrac{u^2sin90^o}{g}$

$=\dfrac{u^2}{g}$

If two particles are moving on same circle with different angular velocities

${\omega}_{1}$ and

${\omega}_{2}$ and different time period

${T}_{1}$ and

${T}_{2}$ , then the time taken by

$2$ to complete one revolution w.r.t particle 1 is

0%
$T=\cfrac { { T }_{ 1 }{ T }_{ 2 } }{ { T }_{ 2 }-{ T }_{ 1 } }$
0%
$T=\cfrac { { T }_{ 1 }+{ T }_{ 2 } }{ 2 }$
0%
$T=\cfrac { { T }_{ 1 }{ T }_{ 2 } }{ { T }_{ 2 }+{ T }_{ 1 } }$
0%
${T}_{2}-{T}_{1}$

The resultant of two forces, one double the other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is:

0%
$120^0$
0%
$135^0$
0%
$90^0$
0%
$150^0$

Explanation

Hint:

Resultant of two vectors $\vec a$ and $\vec b$ is given as,

$\vec r = \vec a + \vec b$

Step 1: Note al the values given.

It is given that two Forces are acting at some angle, let angle is

$x$ . Let, two forces are,

$\vec A$ and

$\vec B$ . It is given that magnitude of one is twice the other. So let,

$| \vec B | = 2 | \vec A |$

Step 2: Write expression for their Resultant.

Let, their Resultant is given by,

$\vec R = \vec A + \vec B$

By dot product with

$\vec A$ both the sides, we have

$\vec R . \vec A = \vec A . \vec A + \vec B . \vec A$

It is given that the resultant is perpendicular to the smaller Force, that is

$\vec A$ .

Therefore,

$\vec R . \vec A = 0$

Thus, by eq.1 we have,

$\Rightarrow 0 = | \vec A |^2 + 2 | \vec A|^2 cosx$

$\Rightarrow 0 = 1 + 2cosx$

$\Rightarrow 2cosx = -1$

$\Rightarrow cosx = \dfrac{-1}{2}$

$\Rightarrow x = 120^0$

Thus, angle between two Forces is $120^0$ .

Option A is correct.

A particle is fired horizontally with a velocity $98m{s^{ - 1}}$ from the top of a tower 490m high. The time taken by the projectile to hit the ground is: $:\left( {g = 9.8m/{s^2}} \right)$

0%
2 s
0%
5 s
0%
10 s
0%
20 s

Explanation

$h=490 m , u_x=98m/s , u_y=0m/s$

using 2nd equation of motion in Y direction

$s_y=u_yt+\dfrac{gt^2}{2}$

$h=\dfrac{gt^2}{2}$

$t=\sqrt{\dfrac{2h}{g}}$

$t=10s$

An object has a displacement from position vector $\vec{r_1} = (2\hat{i}+ 3\hat{j} )m$
to $\vec{r_2} = (4\hat{i}+ 6\hat{j} )m$ under a force $\vec{F} = (3x^2 \hat{i} + 2y \hat{j} )N,$ then work done by the force is:

0%
$24J$
0%
$33J$
0%
$83J$
0%
$45J$

Explanation

$\begin{array}{l} w=\int _{ \overrightarrow { { r_{ 1 } } } }^{ \overrightarrow { { r_{ 2 } } } }{ \overrightarrow { F } \left( { dx\widehat { i } +dy\widehat { j } } \right) } \\ =\int _{ \overrightarrow { { r_{ 1 } } } }^{ \overrightarrow { { r_{ 2 } } } }{ \left( { 3{ x^{ 2 } }\widehat { i } +2y\widehat { j } } \right) }\left( { dx\widehat { i } +dy\widehat { j } } \right) \\ =\int _{ 2 }^{ 4 }{ 3{ x^{ 2 } }dx+ }\int _{ 3 }^{ 6 }{ 3y\, dy } \\ =\left[ { { x^{ 3 } } } \right] _{ 2 }^{ 4 }+\left[ { { y^{ 2 } } } \right] _{ 3 }^{ 6 } \\ =\left( { 64-8 } \right) +\left( { 36-9 } \right) \\ =56+27 \\ =83\, J \end{array}$

$\therefore$ Option

$C$ is correct.

The horizontal component of a projectile velocity is 2 m/s. The equation of the projectile is

$y = 16 x - (\dfrac{5} {4}) x^2.$ If

$g = 10 m/s^2,$ the horizontal range is

0%
16 m
0%
8 m
0%
3.2 m
0%
12.8 m

Explanation

$\begin{array}{l} Given, \\ y=16x-\left( { \dfrac { 5 }{ 4 } } \right) { x^{ 2 } } \\ y=16x\left[ { 1-\dfrac { x }{ { \left( { \dfrac { { 64 } }{ 5 } } \right) } } } \right] \\ \therefore R=\dfrac { { 64 } }{ 5 } =12.8 \\ \therefore \, Option\, \, D\, \, is\, \, correct. \end{array}$

The

$P.V.'s$ of the vertices of a

$\triangle ABC$ are

$\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+5\bar {j}+\bar {k}$ . The

$P.V.$ of the circumcentre of

$\triangle ABC$ is

0%
$\dfrac{5}{2}\bar {i}+3\bar {j}+\bar {k}$
0%
$5\bar {i}+\dfrac{3}{2}\bar {j}+\bar {k}$
0%
$5\bar {i}+3\bar {j}+\dfrac{1}{2}\bar {k}$
0%
$\bar {i}+\bar {j}+\bar {k}$

Explanation

We have,

$A(1,1,1)$

$B(4,1,1)$

$C(4,5,1)$

$AB=3$

$BC=4$

$CA=5$

Hence,

Triangle is right angle at

$B$ .

$\therefore$ CIrcumcentre is mid point of

$AC = \left( {\frac{5}{2},3,1} \right)$

$= \left( {\frac{5}{2}\overline i ,3\overline j ,\overline k } \right)$

Then,

Option

$A$ is correct answer.

Resultant of which of a following may be equal to zero?

0%
10N, 10 N, 10 N
0%
10N, 10 N, 25 N
0%
10N, 10 N, 35 N
0%
None of these

Explanation

If any three vectors can make a triangle than the resultant of these vectors will be zero when the sum two smaller sides of the triangle is greater than the third side.

From given options, it is possible only in

$A$ option.

A flywheel is initially rotating at

$20 rad/s$ and has a constant angular accelerations After

$9.0s$ it has rotated through

$450rad$ .Its angular acceleration is

0%
$3.3 rad/s$
0%
$4.4 rad/s$
0%
$5.6 rad/s$
0%
$6.7 rad/s$

The positive vector of a particle is determined by the expression

$\vec{r} = 3 t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k}$ . The distance traversed in first 10 sec is:

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

Explanation

$\vec r=3t^2 \hat i+4t^2\hat j+7\hat k$

$t=10 sec$

$\vec r=300\hat i+400\hat j+7\hat k$

Its magnitude is

$r=\sqrt{(300)^2+(400)^2+(7)^2}=500m$

The correct answer is A.

A wheel initially has an angular velocity of

$18rad/s$ .IT has a constant acceleration of

$2.0 rad/s^2$ and is slowing at first.What time elapses before its angular velocity is

$18rad/s$ in the direction opposite to its initial angular velocity?

0%
$3.0s$
0%
$6.0s$
0%
$9.0s$
0%
$18s$

Explanation

Given initial angular velocity

${\omega}_{i}=18$ rad/s

final angular velocity

${\omega}_{f}=-18$ rad/s

And we know acceleration

$\alpha=\dfrac{{\omega}_{f}-{\omega}_{i}}{t}=\dfrac{-18-18}{t}$

$t=\dfrac{-36}{2}=-18=18$ s

$\hat {i},\hat {j},\hat {k}$ are positive vectors of

$A,B,C$ and

$\vec {AB}=\vec {CX}$ , then positive vector of

$X$ is

0%
$-\hat {i}+\hat {j}+\hat {K}$
0%
$\hat {i}-\hat {j}+\hat {K}$
0%
$\hat {i}+\hat {j}-\hat {K}$
0%
$\hat {i}+\hat {j}+\hat {K}$

A projectile is thrown at an angle of

$60^\circ$ with the horizontal with an initial speed of 2 m/sec, with H being highest point of its trajectory. Another particle P is now forced to move along its speed is continuously increasing. When the particle P is at H,

$\vert \vec{V}_p\vert = 20 m/sec, \vert \vec{a}_p\vert = 50 m/sec^2$ , then acceleration vector

$\vec a_p$ at H equals (take

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

0%
$50 \hat{i} m/s^2$
0%
$- 50 \hat{i} m/s^2$
0%
$(20 \sqrt{6} \hat i - 10 \hat{j}) m/s^2$
0%
$(30 \hat i - 40 \hat{j}) m/s^2$

Explanation

Given,

$\theta=60°\\u=2m/s\\|\overrightarrow{V_p}|=20m/s\\|a_p|=50m/s^2$

In this context, the radius of the curvature at point

$H$ be

$R$

Now,

$R=\cfrac{V^2}{g}=\cfrac{20\cos60°}{10}\\ \quad=2\times\cfrac{1}{2}=1m$

Radius acceleration at H

$=\cfrac{(V_p)^2}{R}\\=\cfrac{(20)^2}{1}\\=400m/s^2$

Tangential acceleration at

$H=\sqrt{(9)^2-(40)^2}=30m/s^2$

Therefore, in terms of vectors net acceleration at

$H=(30\hat{i}-40\hat{j})m/s^2$

1231333_1056636_ans_ebb4ce97c6b04a35b1369bb31fcf8d97.png

Given in the diagram OB = BA. The time taken to cover OB is

$T_1$ and the time take to cover BA is

$T_2$ , then the ratio

$T_1 /T_2$ is

0%
= 1
0%
> 1
0%
< 1
0%
cannot be determined

Explanation

By the helps of given figure

Consider a distance

$K$ as shown from projectile path to

$B$ .

Consider the inclined place be absent.

Then

$\begin{array}{l} OP=OB\cos { 30^{ \circ } } -K\cos { 60^{ \circ } } \\ OP'=2OB\cos { 60^{ \circ } } \\ PP'=OB\cos { 30^{ \circ } } +K\cos { 60^{ \circ } } \\ \end{array}$

Since the horizontal component of velocity remains constant. time taken to cover layer distance is larger.

since

$PP'>OP$

Time taken to travel

$OB<Time\,\,taken\,\,to\,\,cover\,\,BA$ along the inclined plane.

so,

$>1$

Hence, the option

$B$ is the correct answer.

A particle is moving on a circular path of

$10 m$ radius. At any instant of time its speed is

$5 m/s$ and the speed is increasing at a rate of

$2 m/s^2$ . At this instant the magnitude of the net acceleration will be nearly.

0%
$3.2 m/s^2$
0%
$2 m/s^2$
0%
$2.5 m/s^2$
0%
$4.3 m/s^2$

Explanation

Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration thus find both of them as follows

$:$

$A_c-$ be centripetal acceleration

$.$

$A_t-$ be tangential acceleration

${A_{net}} = \sqrt {\left( {A_c^2 + A_t^2} \right)}$

Now then

$,$

$A_c$ is

$\dfrac{{{V^2}}}{R} = \dfrac{{25}}{{10}} = 2.5\,m/{s^2}$

$A_t$ is given as

$2.\,m/{s^2},$ thus

${A_{net}} = \sqrt {\left( {2.5^2 + 2^2} \right)}$

$= 3.20\,m/{s^2}.$

Hence,

option

$(A)$ is correct answer.

The position vector of a particle is given by

$\vec { r } ={ \vec { r } }_{ 0 }(1-at)t$ , where

$t$ is the time and

$a$ as well as

${\vec { r } }_{ 0 }$ are constant. After what time the particle returns to the starting point?

0%
$a$
0%
$\dfrac{1}{a}$
0%
$A^2$
0%
$\dfrac{a}{a^2}$

Explanation

Given,

$\vec{r}={{\vec{r}}_{o}}(1-at)t$

Initially at $t=0,$ $\vec{r}=0$

After completing one rotation

Position vector, $\vec{r}=0$

$\vec{r}={{{\vec{r}}}_{o}}(1-at)t$

${{{\vec{r}}}_{o}}(1-at)t=0$

$t=\,\,0\,,\,\,\dfrac{1}{a}$

Hence, at time $t=\dfrac{1}{a}$ particle return to starting point.

A wheel starts from the rest and attains an angular velocity of 20 radian/s after being uniformly accelerated for 10 s.The total angle in radian through which it has turned in 10 second is

0%
$20 \pi$
0%
$40 \pi$
0%
$100$
0%
$100 \pi$

Explanation

From kinematics laws for rotational motion:

$\omega=\omega_0+\alpha t$ ..(1)

$\theta=\theta_0+\omega_0 t+\dfrac{1}{2} \alpha t^2$ ....(2)

$\omega^2-\omega_0^2=2\alpha \theta$ .....(3) where,

$\omega$ is final angular velocity

$\omega_0$ is initial angular velocity

$\theta$ is angular displacement

$\alpha$ is angular acceleration and

$t$ is time

From(1) ,

$20=0+\alpha \times 10$

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

From(3) ,

$(20)^2-0^2=2\times 2 \times \theta$

$400=4\theta$

$\theta=100 rad$

From a point on the ground a particle is projected with initial velocity

$u$ , such that its horizontal range is maximum. The magnitude of average velocity during its ascent.

0%
$\dfrac{\sqrt 5u}{2\sqrt 2}$
0%
$\dfrac{5u}{4}$
0%
$\dfrac{\sqrt 3}{2 \sqrt2}$
0%
$none$

Two forces $\widehat {\text{i}}{\text{ + }}\widehat {\text{j}}{\text{ + }}\widehat {\text{k}}\;{\text{N}}\;{\text{and}}\;\widehat {\text{i}}{\text{ + 2}}\widehat {\text{j}}{\text{ + 3}}\widehat {\text{k}}\;{\text{N}}$ act on a particle and displace it from (2,3,4) to point (5,4,3). Displacement is in m. Work done is:

0%
$5 J$
0%
$4 J$
0%
$3 J$
0%
None of these

Explanation

Given,

$\vec F_1=\hat i+\hat j+\hat k \,, \vec F_2=\hat i+2\hat j+3\hat k$

$F_n=F_1+F_2=2\hat i+3\hat j+4\hat k$

Position vector between two points

$(2,3,4) \, to \, (5,4,3)$ is

$\vec r=(5-2)\hat i+(4-3)\hat j+(3-4)\hat k$

$\vec r=3\hat i+\hat j-\hat k$

Work done,

$W=\vec F_n.\vec r$

$W=(2\hat i+3\hat j+4\hat k).(3\hat i+\hat j-\hat k)$

$W=2\times 3+3\times 1-4\times 1$

$W=6+3-4=9-4=5J$

The correct option is A.

A particle is attached at one end of massless rod whose other end is fixed at

$O$ as shown in figure. A particle is given minimum velocity at lower most point to complete vertical circular motion about

$O$ . Find net force on the particle when it is at position

$P$ . Length of rod is

$\ell$ .

0%
$\dfrac{18mg}{5}$
0%
$\dfrac{23mg}{5}$
0%
$\dfrac{\sqrt{333}mg}{5}$
0%
$None\ of\ above$

The height

$y$ and horizontal distance

$x$ covered by a projectile in a time

$t$ seconds are given by the equations

$y = 8t - 5t^2$ and

$x = 6t$ . If

$x$ and

$y$ are measured in meters, the velocity of projection is:-

0%
$10 \ ms^{-1}$
0%
$6 \ ms^{-1}$
0%
$8 \ ms^{-1}$
0%
$14 \ ms^{-1}$

Explanation

$v_x=\dfrac{dx}{dt}=6$

$v_y=\dfrac{dy}{dt}=8-10t$

$\vec v=v_x\hat{i}+v_y\hat{j}$

$\vec u=\vec v_{t=0}$

$\vec u=6\hat{i}+8\hat{j}$

$u=\sqrt{6^2+8^2}$

$u=10m/s$

A ball is thrown at angle

$\theta$ and another ball is thrown at an angle

$(90-\theta)$ with the horizontal direction from the same point with the same speed

$40 ms^{-1}$ . The second ball reaches

$50m$ higher than the first ball. Find their individual heights.

0%
$20m , 70m$
0%
$25m , 75m$
0%
$15m , 65m$
0%
$10m , 60m$

Explanation

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

$H = \cfrac{u^2 (\sin(90- \theta))^2 }{2g}$

$H-h=50$

$\cfrac{u^2}{2g}[\cos^2 \theta -\sin^2 \theta]=50$

$\cos 2\theta = \cfrac{50 \times 2g}{u^2}$

$\cos 2\theta = \cfrac{50 \times 2 \times 10}{40^2}$

or,

$\cos 2 \theta = \cfrac{5}{8}$

$\therefore h= \cfrac{u^2}{2g}(\cfrac{1-\cos 2 \theta}{2})$

$= \cfrac{40^2}{2 \times 10}(\cfrac{1-\cfrac{5}{8}}{2})$

$= 15m$

$H = 50+h = 50+15 =65m$

978389_1074250_ans_5e4bccb108b54ac3aea80d5f7ef55cf9.PNG

For two particles A and B, given that

$\overrightarrow {r}_A = 2\hat{i} + 3\hat{j},$

$\overrightarrow{r}_B = 6\hat{i} + 7\hat{j},$

$\overrightarrow {V}_A = 3\hat{i} - \hat{j}$ and

$\overrightarrow{v}_B = x\hat{i} - 5\hat{j}.$ what is the value of x if they collide.

0%
1
0%
7
0%
2
0%
-2

A ball has been thrown vertically up such that distance covered by it in

$4th$ and

$6th$ second is same. If

$g=10m/s^2$ , the initial speed of the ball is

0%
$45m/s$
0%
$25m/s$
0%
$50m/s$
0%
$10m/s$

Explanation

When it travel

$s$ distance from ground.

$\Rightarrow t=4s$

Now when it reaches the same height from the ground

$t=6s$

$S=ut-\cfrac{1}{2}gt^2$

Where

$u=$ Initial speed.

$S=$ Distance traveled.

$S=ut-5t^2$

$t=4s$

$S=4u-80$

$\therefore 4u-S=80\longrightarrow(i)$

and, at

$t=6s$

$S=6u-180$

$6u-S=180 \longrightarrow(ii)$

Subtract

$(i)$ From

$(ii)$

$2u=100$

$u=50m/s$

A body starts rotating from rest and completes 10 revolutions in 4 sec. Find its angular acceleration

0%
$2.5\pi \,\,{\text{rad/}}{{\text{s}}^{\text{2}}}$
0%
$5\pi \,\,{\text{rad/}}{{\text{s}}^{\text{2}}}$
0%
$7.5\pi \,\,{\text{rad/}}{{\text{s}}^{\text{2}}}$
0%
$10\pi \,\,{\text{rad/}}{{\text{s}}^{\text{2}}}$

Explanation

Given,

N0. of revolution,

$n=10$ ,

$1 rev=2\pi$

$\omega_0=0 rad/sec$

$\theta=20\pi$ rad

$t=4sec$

2nd equation of motion in rotational motion,

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

$20\pi=0\times 4+\dfrac{1}{2}\alpha (4)^2$

$\alpha =2.5 \pi rad/s^2$

The correct option is A.

A body is projected vertically upwards with a velocity of

$19.6 \ m/s$ . The total time for which the body will remain in the air is (Take

$g = 9.8 m/s^2$ )

0%
$4 \ s$
0%
$6 \ s$
0%
$9 \ s$
0%
$12 \ s$

Explanation

$u=19.6 , a=-9.8$

using :

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

when the body reaches ground displacement ,

$s=0$

$ut-\dfrac{gt^2}{2}=0$

$t(2u-gt)=0$

we get

$t=0, \dfrac{2u}{g}$

since at

$t=0$ body is projected so Time of flight

$=\dfrac{2u}{g}$

$T=\dfrac{2\times 19.6}{9.8}$

$T=4s$

If the direction of cosines of a vector are

$\dfrac{3}{5 \sqrt 2}, \dfrac{4}{5 \sqrt 2}$ and

$\dfrac{1}{\sqrt 2}$ respectively, then the vector is:

0%
$3 \hat{i} + 4 \hat{j} + \hat{k}$
0%
$3 \hat{i} + 4 \hat{j} + \sqrt 5 \hat{k}$
0%
$3 \sqrt{2} \hat{i} + 4 \sqrt{2} \hat{j} + \hat{k}$
0%
$3 \hat{i} + 4 \hat{j} + 5 \hat{k}$

Explanation

Given, the direction cosines,

$(\dfrac{a}{|a|},\dfrac{b}{|b|},\dfrac{c}{|c|})=(\dfrac{3}{5\sqrt 2},\dfrac{4}{5\sqrt 2},\dfrac{1}{\sqrt 2})$

Then the direction ratios,

$(a,b,c)=(3,4,1)$

We know equation of a vector,

$\vec{v}=a\hat{i}+b\hat{j}+c\hat{k}$

Then the vector with given direction ratios will be

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

A ball is thrown from a point on ground at some angle of projection. At the time a bird starts from a point directly above this point of projection at a height

$h$ horizontally with speed

$u$ . Given that in its flight ball just touches the bird at one point. Find the distance on ground where ball strikes:

0%
$2 u \sqrt{\dfrac{h}{g}}$
0%
$u \sqrt{\dfrac{2h}{g}}$
0%
$2 u \sqrt{\dfrac{2h}{g}}$
0%
$u \sqrt{\dfrac{h}{g}}$

Explanation

As at one point the ball touches the bird it simply means that

$\text{Horizontal velocity of the ball which is constant over time is EQUAL with that of the bird and as the ball }$

$\text{JUST TOUCHES one time it simply menas that it will be at its top point during that moment so that}$

$\text{ it never become at that height again}$

So if

$vcos \theta$ is the horizontal velocity of the ball then

$u=v cos \theta$ and

$h=\dfrac{(vSin \theta)^2}{2g}$ or

$vsin \theta =\sqrt[2]{2gh}$

Where

$v$ is projection velocity and

$\theta$ is projection angle.

Now the question ask the horizontal range

$R=\dfrac{2vsin \theta}{g}\times vcos \theta=\dfrac{2\times \sqrt[2]{2gh}}{g}\times u=2u\sqrt[2]{\dfrac{2h}{g}}$

Option C is correct.

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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