MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 7

A curved section of a rod 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 $v$ than speeds lower than $v$
0%
car cannot remain in static equilibrium on the curved section
0%
car will not slip when moving with speed $v$
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None of the above

Explanation

as the diagram shows if there is no friction,

$mg\sin { \theta \le m\dfrac { { v }^{ 2 } }{ R } \cos { \theta } }$

$\Rightarrow v\ge \sqrt { rg\tan { \theta } }$

so at this velocity the car will not slip.

A particle of mass

$M$ is moving in a horizontal circle of radius

$R$ with uniform speed

$v$ . When the particle moves from one point to a dramatically opposite point, its

0%
Momentum does not change
0%
Momentum changes by $2Mv$
0%
Kinetic energy changes by $\cfrac { M{ v }^{ 2 } }{ 4 }$
0%
Kinetic energy changes by $M{ v }^{ 2 }$

Explanation

Velocity of the particle at one point and at diametrical opposite point in

$v$ and

$-v$ , respectively.
Change in momentum

$\Delta P = P_f-P_i$

$=Mv-(-Mv)$

$\Delta P=2Mv$

A point moves along a circle with speed

$v=at$ . The total acceleration of the point at a time when it has traced

$\dfrac{1}{8}^{th}$ of the circumference is

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$8a$
0%
$2a\sqrt { 4+{ \pi }^{ 2 } }$
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$a$
0%
$\cfrac { a }{ 2 } \sqrt { 4+{ \pi }^{ 2 } }$

Explanation

V = at

$1/8^{th}$ of circumference

$\Rightarrow \dfrac{2\pi R}{8} = \dfrac{\pi R}{4}$

$S = \dfrac{1}{2} at^2$

$(\because u =0$ at

$t = 0)$

$\dfrac{\pi R}{4} = \dfrac{1}{2} at^2$

$\Rightarrow t_{0} = \sqrt{\dfrac{\pi R}{2a}}$

At time

$t = t_{0}, v_{0} = at \Rightarrow v_0 = \sqrt{\dfrac{\pi R_0}{2}}$

$a_c = \dfrac{v_{0}^{2}}{R} = \dfrac{\pi R a}{2R}$

$a_{t} = 0$

$\Rightarrow a_{c} = \dfrac{\pi}{2}a$

$a_t = a$

$a_{nf} \sqrt{a_{i}^{2} + a_{t}^{2}} = a\sqrt{1+\dfrac{\pi^2}{4}} \Rightarrow a_{net}\dfrac{a}{2} \sqrt{4+\pi^2}$

1179552_694443_ans_03e4381917384b93b6af62d6e37577b1.jpeg

When a ceiling fan is switched off, its angular velocity fall to half while it makes

$36$ rotations . How many more rotations will be make before coming to rest ? (Assume uniform angular retardations ).

0%
$36$
0%
$24$
0%
$18$
0%
$12$

Explanation

Using

$\omega_2^{2}-\omega_1^{2}=2\,\alpha \theta$ ,

We get,

$(\dfrac{\omega}{2})^{2}-\omega ^{2}=2 \alpha (36 \times 2\pi)$ ........(i)

Similarly ,

$(0)^{2}-(\dfrac{\omega}{2})^{2}=2 \alpha(\pi \times 2 \pi)$ .......(ii)

Dividing Eq.(i) by (ii), we get

$\dfrac{-\dfrac{3}{4} \omega^{2}}{-\dfrac{\omega^{2}}{4}}=\dfrac{36}{n}$

$n=12$

A particle

$A$ moves along a circle of radius

$R = 50\ cm$ so that its radius vector

$r$ relative to the point

$O$ (figure) rotates with the constant angular velocity

$\omega = 0.40\ rad/s$ . Then magnitude of the velocity of the particle, and the magnitude of its total acceleration will be

0%
$v = 0.4\ m/s, a = 0.4\ m/s^{2}$
0%
$v = 0.32\ m/s, a = 0.32\ m/s^{2}$
0%
$v = 0.32\ m/s, a = 0.4\ m/s^{2}$
0%
$v = 0.4\ m/s, a = 0.32\ m/s^{2}$

Explanation

Given,

$R=0.50m$

$\omega=0.4 rad/s$

$\theta=\dfrac{\theta '}{2}$

$\omega '=2\omega$

$v'=2\omega R=2\times 0.40\times 0.50=\underline{0.4 m/s}$

$a=\omega ^2R$

$a'=\omega ' ^2R=(2\omega)^2\times 0.50=\underline{0.32m/s^2}$

A cricket ball is thrown over a triangle from one end of a horizontal base falls on the other end of the base after grazing the vertex. If

$\theta_1$ and

$\theta_2$ are the base angles of projection, then.

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$\tan\theta_1 -\tan\theta_2 =\tan\alpha$
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$\sin\theta_1 +\sin\theta_2 =\sin\alpha$
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$\tan\theta_1 +\tan\theta_2 =\tan\alpha$
0%
$\cos\theta_1 +\cos\theta_2 =\cos \alpha$

Explanation

The equation of trajectory is given by
So,

$y=x\tan\alpha -\displaystyle\frac{x^2\sin\alpha}{\left(\displaystyle\frac{2u^2\sin\alpha \cos\alpha}{g}\right)\cos\alpha}$

$\Rightarrow \displaystyle y=xtan\alpha\left(\displaystyle 1-\frac{x}{r}\right)$ ..........(I)
Coordinates of B are

$(h\cot\theta_1, h)$ and the range AC

$=$ AO

$+$ OC

$=h\cot\theta_1+h\cot\theta_2$
Substituting the coordinate values of B in Eq. (i)

$h=h\cot\theta_1\tan\alpha\left(\displaystyle 1-\frac{h\cot\theta_1}{h(\cot\theta_1+\cot\theta_2)}\right)$

$\Rightarrow 1=\cot\theta_1\cdot \tan\alpha\left[\displaystyle\frac{\cot\theta_1+\cot\theta_2-\cot\theta_1}{\cot\theta_1+\cot\theta_2}\right]$

$\Rightarrow \displaystyle\frac{\cot\theta_1+\cot\theta_2}{\cot\theta_1\cdot \cot\theta_2}=\tan x$

$\Rightarrow \displaystyle \tan\theta_2+\tan\theta_1=\tan\alpha$

$\therefore \displaystyle\tan\theta_1+\tan\theta_2=\tan\alpha$ .

Consider the motion of a particle described by

$x=a\cos { t } ,y=a\sin { t }$ and

$z=t$ . The trajectory traced by the particle as a function of time is

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Helix
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Circular
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Elliptical
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Straight line

Explanation

path is circular in x-y plane

$\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { a }^{ 2 } } =\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1$

$\implies \ x^2+y^2 =a^2$ which is the equation of circle.
In one revolution, the particle moves

$A$ distance of

$1$ unit along z-axis

$\therefore \cfrac { dz }{ dt } =1$
Thus the particle moves along z direction with constant speed tracing a circular path in xy plane.
Hence the overall path is helix.

The position vectors of A, B are a, 6 respectively. The position vector of C is

$\dfrac {5\bar{a}}{3} -\bar{b}$ . Then 3

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C is inside the $\Delta OAB$
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C is outside the $\Delta OAB$ but inside the angle OAB
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C is outside the $\Delta OAB$ but inside the angle OBA
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None of these

Explanation

We have given that

The position vector of

$A$ and

$B$ are

$a$ and

$6$ respectively.

$\therefore$

Now ,

The position vector of

$C$ is

$\dfrac{{5\overline a }}{3} - \overline b$ ,

Then, the position vector at

$C$ os inside the

$\Delta OAB$

Hence, the option

$(A)$ is correct.

A particle is moving in a circle of radius

$r$ with constant speed

$v$ . The change in velocity in moving from

$P$ to

$Q$ is

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$2v\cos 20^{\circ}$
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$2v\sin 20^{\circ}$
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$2v\cos 40^{\circ}$
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$2v\sin 40^{\circ}$

Explanation

$\bf{Step 1: Resultant \ of \ two\ vectors \ being\ subtracted}$

Given,

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

Change in velocity,

$\Delta\vec v = \vec v_q -\vec v_p$

$\Rightarrow \Delta\vec v = \vec v_q +(-\vec v_p)$

Therefore, the Resultant of two vectors being subtracted is given by

$|\Delta \vec {v}|^{2} = v^{2}_q + v^{2}_p - 2v_p v_q \cos \theta$

where

$\theta$ is the angle between

$v_p$ and

$v_q$

From figure, on shifting the vector

$v_p$ at point Q, we see that:

The angle between

$v_p$ and

$v_q$ is

$40^o$

$\bf{Step 2: Putting\ values}$

The change in velocity between point

$P$ and

$Q$ is given by

$|\Delta \vec {v}|^{2} = v^{2} + v^{2} - 2v^{2} \cos (40)^{\circ}$

$= 2v^{2} [1 - (1 - 2\sin^{2} 20^{\circ})]$

$= 4v^{2} \sin^{2} 20^{\circ}$

Therefore,

$|\Delta \vec {v}| = 2v\sin 20^{\circ}$

Hence correct option is B.

$\textbf{Alternate solution:}$

Write initial and final velocities in vector form by taking their components along x-y axes.

$\therefore \ v_p = 0 \hat i+v \hat j = v \hat j$

$\& \ \ v_q = -v\ sin40^{\circ} \hat i +v\ cos40^{\circ} \hat j$

Then find the change in velocity by subtracting the two vectors as shown below:

$\Delta\vec v = \vec v_q -\vec v_p$

2101557_695206_ans_7400b74ceff64d88a22897949ee26df0.png

For which of the following physical quantities, it is necessary to indicate direction along with its magnitude?

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Speed
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Path length
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Displacement
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Temperature

A particle is going with constant speed along a uniform helical and spiral path separately as shown in figure then

[Assume that the vertical acceleration of the particle is negligible in case (a)]

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The velocity of the particle is constant in both cases
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The magnitude of acceleration of the particle is constant in both cases
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The magnitude of acceleration is constant in (a) and decreasing in (b)
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The magnitude of acceleration is decreasing continuously in both the cases

Explanation

Since acceleration,

$a=\cfrac{v^2}{r}$

and velocity is uniform and radius is constant in figure

$(a)$

Therefore, acceleration is constant.

But in figure

$(b)$ , although the velocity is constant but the radius is increasing.

Therefore, acceleration is decreasing.

Two particles

$P$ and

$Q$ are moving on a circle. At a certain instant of time both the particles are diametrically opposite and

$P$ has tangential acceleration

$8\ m/s^{2}$ and centripetal centripetal

$5\ m/s^{2}$ whereas

$Q$ has only centripetal acceleration of

$1 m/s^{2}$ . At that instant acceleration (in

$m/s^{2})$ of

$P$ with respect to

$Q$ is:

0%
$14$
0%
$\sqrt {80}$
0%
$10$
0%
$12$

Explanation

According to the question, we can draw the following diagram.
From the diagram,

$a_{p} = 5\hat {i} + 8\hat {j}$

and

$a_{Q} = - \hat {i}$

Now,

$a_{rel} = a_{P} - a_{Q}$

$a_{rel} = 5\hat {i} + 8\hat {j} - (-\hat {i})$

$a_{rel} = 6\hat {i} + 8\hat {j}$

$|a_{rel}| = \sqrt {(6)^{2} + (8)^{2}} = \sqrt {36 + 64}$

$= \sqrt {100}$

$= 10\ m/s^{2}$ .

A small object of mass of

$100\ gm$ moves in a circular path. At a given instant velocity of the object is

$10\hat {i} m/s$ and acceleration is

$(20\hat {i} + 10\hat {j})m/s^{2}$ . At this instant of time, rate of change of kinetic energy of the object is:

0%
$200\ kgm^{2}/s^{3}$
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$300\ kgm^{2}/s^{3}$
0%
$10000\ kgm^{2}/s^{3}$
0%
$20\ kgm^{2}/s^{3}$

Explanation

Given, mass of object

$(m) = 100\ g$

$= 100\times 10^{-3} kg$
Velocity of object

$(v) = 10\hat {i} m/s$
Acceleration of object

$(a) = (20\hat {i} + 10\hat {j}) m/s^{2}$
We know that,

$\dfrac {d(KE)}{dt} = F\cdot v = ma\cdot v \ \ \ \ \ \ [\because K\cdot E = 1/2 mv^{2}]$

$= (100 \times 10^{-3})(20\hat {i} + 10\hat {j}).(10\hat {i})$

$= 100\times 10^{-3} \times 200$

$= \dfrac {100\times 200}{1000} = 20\ kgm^{2}/s^{3}$ .

A vector having magnitude

$30$ unit makes equal angles with each of

$x,y$ and

$z$ axes. The component along each of

$x,y$ and

$z$ -axis are

0%
$10\sqrt {3}$
0%
$20\sqrt {3}$
0%
$30\sqrt {3}$
0%
$18\sqrt {3}$

Explanation

Vector makes equal angle with each of

$x$ ,

$y$ and

$z$ so components are also equal.

Let vector is

$\overrightarrow A= a\hat i + a \hat j + a\hat k$ ,

$|\overrightarrow A|=\sqrt{a^2 +a^2 +a^2} = \sqrt3a =30$

$\Rightarrow a= 10\sqrt3$

A particle is projected up inclined plane such that its component of velocity along the incline is 10m/s. Time of flight is 2 sec and maximum height above the incline is 5m. Then velocity of projection will be:

0%
$10m/s$
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$10\sqrt{2}m/s$
0%
$5\sqrt{2}m/s$
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None

A wheel is at rest. Its angular velocity increases uniformly and become 80 radians second after 5 seconds. The total angular displacement is:

0%
800 rad
0%
400 rad
0%
200 rad
0%
100 rad

Explanation

Initial angular speed,

$\omega_0=0$

${rad/s}$

Final angular speed,

$\omega=80$

${rad/s}$

Time taken,

$t=5$

$s$

From kinematics:

$\omega-\omega_0=\alpha t$

$\implies 80=\alpha\times 5$

$\implies \alpha=\cfrac{80}{5}=16$

${rad/s^2}$

where,

$\alpha$ is angular acceleration.

Also,

$\theta=\omega_0t+\cfrac{1}{2}\alpha t^2$

$\implies \theta=0+\cfrac{1}{2}\times 16\times 25$

$\implies \theta=200$

$rad$

A particle is projected at an angle

$\theta$ from ground with speed

$u(g=10m/s^2)$ , then which of the following is true?

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If $u=10m/s$ and $\theta = 30^o$ , then time of flight will be 1 sec
0%
If $u=10\sqrt{3}m/s$ and $\theta=60^o$ , then time of flight will be 3 sec
0%
If $u=10\sqrt{3}m/s$ and $\theta=60^o$ , then after 2 sec velocity becomes perpendicular to initial velocity
0%
If $u=10m/s$ and $\theta=30^o$ , then velocity never becomes perpendicular to intial velocity during its flight

Explanation

Using the formula,

$T = \dfrac{2uSin\theta}{g}$

For OPTION A, we have,

$u = 10 \ m/s$

$\theta = 30{^{\circ}}$

$T = \dfrac{2 \times 10 \times Sin30}{10}$

$T = 1 \ sec$

For OPTION B, we have,

$u = 10\sqrt{3} \ m/s$

$\theta = 60{^{\circ}}$

$T = \dfrac{2 \times 10 \sqrt{3} \times Sin60}{10}$

$T = 3\ sec$

Time taken by velocity to become perpendicular to the initial velocity is given by:

$t = \dfrac{u}{gSin\theta}$

For OPTION C, we have,

$u = 10\sqrt{3} \ m/s$

$\theta = 60{^{\circ}}$

$t = \dfrac{10 \sqrt{3}}{10 \times Sin60}$

$t = 2 \ sec$

For OPTION D, we have,

$u = 10 \ m/s$

$\theta = 30{^{\circ}}$

$t = \dfrac{10}{10\times Sin30}$

$t = 2 \ sec$

Since,

$T = 1\ sec \leq t = 2 \ sec$

$\therefore$ The velocity never becomes perpendicular to the initial velocity during its flight as time of flight is less than time when velocity becomes perpendicular to the initial velocity.

Hence, OPTIONS A, B, C, D are correct.

A particle is moving along a circle such that it completes one revolution in

$40$ seconds. In

$2$ minutes

$20$ seconds, the ratio

$\cfrac { \left| displacement \right| }{ distance }$ is

0%
$0$
0%
$\cfrac{1}{7}$
0%
$\cfrac{2}{7}$
0%
$\cfrac{7}{11}$

Explanation

Particle completes revolution in

$40$ seconds.

$20$ minutes

$20$ second

$=140$ second

$=40\times 3+20$ second

Therefore, the player would have completes

$3.5$ revolutions.

$\therefore$ Distance

$=3\times 2\pi r\times \pi r=7\pi r$

where as, Displacement

$=2r$ (the other end of the diameter)

$\therefore \cfrac{|Displacement|}{Distance}=\cfrac{2r}{7\pi r}=\cfrac{2}{7\pi}$

A ball is thrown horizontally with velocity

$30m/s$ from top of a

$60m$ high tower, simultaneously, another ball is thrown vertically upward with velocity

$40m/s$ from the bottom of same tower. The shortest distance between the two balls is

$\left( g=10m/{ s }^{ 2 } \right)$

0%
$36m$
0%
$48m$
0%
$60m$
0%
$45m$

A motorboat is racing towards the north at 25

$kmh^{-1}$ and the water current in that region is 10

$kmh^{-1}$ in the direction of 60

$^{\circ}$ east of south. The resultant velocity of the boat is:

0%
11 $kmh^{-1}$
0%
22 $kmh^{-1}$
0%
33 $kmh^{-1}$
0%
44 $kmh^{-1}$

Explanation

Given,

velocity of the water current ${{v}_{c}}=10\ km/s$

velocity of the motorboat ${{v}_{b}}=25\,km/h$

angle between north and south east is ${{120}^{o}}$

Resultant velocity $v_{R}= \sqrt{v_{b}^{2} +v_{c}^{2} + 2v_{b}v_{c}cos120^{\circ}}$

$\sqrt{{{\mathrm{(25)}}^{\mathrm{2}}}\mathrm{+(10}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+2(25)(10)}\left( \dfrac{\mathrm{-1}}{\mathrm{2}} \right)}$ $=22\,km/{{h}^{-1}}$

$22\,km/{{h}^{-1}}$

Which of the following is not a scalar quantity?

0%
Temperature
0%
Coefficient of friction
0%
Charge
0%
Impulse

Explanation

Impulse

$(I)$

$Force\propto\dfrac{1}{Time \ interval}$

$Force=I\dfrac{I}{Time \ interval(t)}$

$I=Force\times Time \ interval$

$I-F.I$

Since force is a vector quantity and impulse is the product of the sudden force applied and the time interval to bring equivalent the system. So Impulse also a vector quantity which have same direction to the force applied.

A bird flies from (-3m,4m,-3m) to (7m,-2 m,-3m) in the xyz- coordinates. The bird's displacement vector is given by

0%
(4 $\hat{i}+2\hat{j}-6\hat{k}$ )
0%
$(10\hat{i}-6\hat{j})$
0%
( $\hat{i}-2\hat{j}$ )
0%
$(10\hat{i}+6\hat{j}-6\hat{k})$

Explanation

Bird's displacement is

$\Delta \vec{r} = \vec{r}_{2}- \vec{r}_{1}$

$\therefore$

$\Delta\vec{r} = (7 \hat{i} - 2\hat{j} - 3 \hat{k}) - (-3 \hat{i} + 4\hat{j} - 3\hat{k})$

$= 10 \hat{i}-6\hat{j}$

Which of the following is not a property of a null vector?

0%
$\vec{A}$ + $\vec{0}$ = $\vec{A}$
0%
$\lambda$ $\vec{0}$ = $\vec{0}$ where $\lambda$ is a scalar
0%
0 $\vec{A}$ = $\vec{A}$
0%
$\vec{A}$ - $\vec{A}$ = $\vec{0}$

Explanation

Multiplying any vector by zero will result in a null vector.
So, the property of null vector is wrongly mention in

$option(C)$
Hence, the correct option is

$(C)$ .

A disc rotating about its axis, from rest it acquires a angular speed

$100rev/s$ in

$4$ second. The angle rotated by it during these four seconds (in radian) is

0%
$100\pi$
0%
$200\pi$
0%
$300\pi$
0%
$400\pi$

Explanation

Initial angular speed,

$\omega_o=0$

${rad/s}$

Final angular speed,

$\omega=100$

$rps$

$\implies \omega=100\times 2\pi$

${rad/s}$

$\implies \omega=200\pi$

${rad/s}$

Time,

$t=4$

$s$

From kinematics:

$\omega=\omega_o+\alpha t$

$\implies 200\pi=4\alpha$

$\implies \alpha=\cfrac{200\pi}{4}$

${rad/s^2}$

where,

$\alpha$ is angular acceleration.

Also,

$\theta =\omega_ot+\cfrac{1}{2}\alpha t^2$

$\implies \theta =0+\cfrac{1}{2}\times \cfrac{200\pi}{4}\times 16$

$\implies \theta=400\pi$

The (x,y,z) coordinates of two points A and B are given respectively as (0, 4, -2) and (-2, 8, -4). The displacement vector from A to B is:

0%
-2 $\hat{i}$ +4 $\hat{j}$ -2 $\hat{k}$
0%
2 $\hat{i}$ -4 $\hat{j}$ +2 $\hat{k}$
0%
2 $\hat{i}$ +4 $\hat{j}$ -2 $\hat{k}$
0%
-2 $\hat{i}$ -4 $\hat{j}$ -2 $\hat{k}$

Explanation

Here,

$\vec{r}_{A} =0\hat{i}+4\hat{j}-2\hat{k}, \vec{r}_B = -2\hat{i}+8\hat{j}-4\hat{k}$ .

Displacement vector from A to B is given by

$\vec{r}= \vec{r}_B - \vec{r}_{A}= (-2\hat{i}+8\hat{j}-4\hat{k})-(0\hat{i}+4\hat{j}-2\hat{k})=-2 \hat{i}+4\hat{j}-2\hat{k}$

The circular motion of a particle with constant speed is

0%
periodic and simple harmonic.
0%
simple harmonic but not periodic.
0%
neither periodic nor simple harmonic.
0%
periodic but not simple harmonic.

Explanation

The correct answer is option (D).

Hint: Periodic motion has a constant time period.

Step 1: Check for periodic motion.

During the circular motion with constant speed, the particle repeats its motion after a regular interval of time.

Let the radius of the circular orbit be

$r$ .

Time period of motion can be calculated as:

$t=\dfrac{2\pi r}{\omega}$ (where

$\omega$ is the angular velocity of the particle)

Radius of circle is constant.

According to the question, speed of particle is constant.

Hence, both

$r$ and

$\omega$ are constant.

So, time period is also constant.

Thus, the motion of particle is both repeated and constant. So, it is a periodic motion.

Step 2: Check whether the motion is simple harmonic or not.

For the motion to be simple harmonic,

$a\propto -x$ (where,

$a$ is the acceleration,

$x$ is the displacement of the particle)

Thus, the acceleration of the particle should be antiparallel to the displacement of the particle.

However, in a circular motion, the displacement and acceleration are perpendicular to each other. They form an angle of

$90^o$

So, the motion is not simple harmonic.

Hence, in a circular motion with constant speed, the motion of a particle is periodic but not simple harmonic.

The correct answer is option (D).

A car goes around uniform circular track of radius

$R$ at a uniform speed

$v$ once in every

$T$ seconds. The magnitude of the centripetal acceleration is

$a_C$ . If the car now goes uniformly around a larger circular track of radius

$2R$ and experiences a centripetal acceleration of magnitude

$8a_c$ , then its time period is

0%
$2T$
0%
$3T$
0%
$T/2$
0%
$3/2 T$

Explanation

$a_C = mR \omega^2 = mR * 4\pi^2 / T^2$

$8a_C = m 2R * 4\pi^2 / T' ^2$

Dividing,

$8 = (2/ T'^2 )* (T^2/1)= 2T^2 / T'^2$

$T'= \sqrt {T^2/ 4} = T/ 2$

So option C is correct.

Which one of the following statements is true?

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A scalar quantity is the one that is conserved in a process.
0%
A scalar quantity is the one that can never take negative values.
0%
A scalar quantity is the one that does not vary from one point to another in space.
0%
A scalar quantity has the same value for observers with different orientations of the axes.

The equations of motion of a projectile are given by

$x = 36 \ t$ m and

$2y = 96t - 9.8 t^{2}$ m. The angle of projection is

0%
sin $^{-1}$ (4/5)
0%
sin $^{-1}$ (3/5)
0%
sin $^{-1}$ (4/3)
0%
sin $^{-1}$ (3/4)

Explanation

Given:

$x - 36t, 2y = 96t - 9.8t^{2}$

$y = 48t-4.9t^{2}$

Let the initial velocity of projectile be u and angle of projection is

$\theta$ . Then,

Initial horizontal component of velocity,

$u_{x}= u cos\theta= \dfrac{dx}{dt}_{t=0}= 36 \ \ \ or\ \ u cos\theta=36$ ---(i)

Initial vertical component of velocity,

$u_{y}= u sin\theta= \dfrac{dy}{dt}_{t=0}= 48 \ \ or \ \ u sin\theta=48$ ---(ii)

Dividing (ii) by (i), we get

$tan\theta=\dfrac{48}{36}=4/3$

$\therefore sin\theta =4/5$

$\theta$ =

$sin^{-1}(4/5)$

A projectile is thrown with an initial velocity

$(a\hat{i}+b\hat{j})ms^{-1}$ , where

$\hat{i}$ and

$\hat{j}$ are the unit vectors along horizontal and vertical directions respectively. If the range of projectile is twice the maximum height reached by it, then

0%
$b=\frac{a}{2}$
0%
$b=a$
0%
$b = 2a$
0%
$b= 4a$

Explanation

Range

$=\dfrac{2 u_{x}u_{y}}{g}=\dfrac{2ab}{g}$

Maximum Height

$(H)=\dfrac{(u_{y})^{2}}{2g}=\dfrac{b^{2}}{2g}$

Given,

$R=2H$

$\Rightarrow \dfrac{2ab}{g}=2\left(\dfrac{b^{2}}{2g}\right)$

$\Rightarrow b=2a$

1441549_950196_ans_5892723610e345168b24ede1239fb579.png

If a body placed at the origin is acted upon by a force

$\overline{F}=(\hat{i}+\hat{j}+\sqrt2\hat{k})$ , then which of the following statements are correct?
1.Magnitude of

$\overline{F}$ is

$(2+sqrt2)$
2.Magnitude of

$\overline {F}$ is 2
3.

$\overline {F}$ makes an angle of

$45^0$ with the Z-axis.
4.

$\overline {F}$ makes an angle of

$30^0$ with the Z-axis.
Select the correct answer using the codes given below.

0%
1 and 3
0%
2 and 3
0%
1 and 4
0%
2 and 4

Explanation

$\vec{F}=\hat{i}+\hat{j}+\sqrt{2}\hat{k}$

$|\vec{F}|=\sqrt{(1)^{2}+(1)^{2}+(\sqrt{2})^{2}}$

$=\sqrt{1+1+2}$

$=\sqrt{4}=2\Rightarrow 2$

Now

$\hat{F}=\dfrac{\vec{F}}{|\vec{F}|}=\dfrac{\hat{i}+\hat{j}+\sqrt{2}\hat{k}}{2}=\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\dfrac{\sqrt{2}}{2}\hat{k} ... (1)$

Let

$\alpha, \beta, \gamma$ be angles by

$F$ from

$x, y$ and

$z$ axis respectively

So,

$\hat{F}=\cos \alpha\hat{i}+\cos\beta\hat{j}+\cos\gamma \hat{k} .... (2)$

From

$(1)$ and

$(2)$

$\cos\alpha=\dfrac{1}{2}, \cos\beta=\dfrac{1}{2}$ and

$\cos\gamma=\dfrac{1}{\sqrt{2}}$

$\Rightarrow \gamma =45^{o}\Rightarrow 3$

Ans

$(B) 2$ and

$3$

The speed of a projectile at its maximum height is

$\dfrac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is P times the maximum height attained by it, then P equals

0%
$4/3$
0%
$2$ $\sqrt{3}$
0%
$4$ $\sqrt{3}$
0%
$3/4$

Explanation

Given

$\dfrac{\sqrt{3}}{2}u=u\cos \theta$

speed at maximum height or

$\cos\theta =\dfrac{\sqrt{3}}{2}$ or

$\theta=30^o$

Given that

$PH_{max}=R$

We know

$H_{max}=\dfrac{R\tan \theta}{4}$

$P=\dfrac{R}{H_{max}}=\dfrac{4}{\tan \theta}$

$=\dfrac{4}{\tan 30^o}=4\sqrt{3}$

If a body placed at the origin is acted upon by a force

$\vec{F}$ =

$(\hat{i} + \hat{j} + \sqrt 2 \hat{k})$ , then the following statements are correct?
Magnitude of

$\vec{F}$ is

$(2 + \sqrt2)$ .
Magnitude of

$\vec{F}$ is 2.

$\vec{F}$ makes an angle of

$45^0$ with the Z-axis.

$\vec{F}$ makes an angle of

$30^0$ with the Z-axis.
Select the correct answer using the codes given below.

0%
1 and 3
0%
2 and 3
0%
1 and 4
0%
2 and 4

Explanation

Here,

${\vec F}= \hat{i}+\hat{j}+ \sqrt{2}\hat{k}\\$

$|\vec F|= \sqrt{1^{2}+1^{2}+\sqrt{2}^{2}}\\$

$|\vec F| = 2$

Thus

$\hat{F} = \dfrac{\underset{F}{\rightarrow}} {| \vec F| }$

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

Let

$\alpha, \beta and \gamma$ are angled by force F from x,y and z-axis respectively. Then

$\hat{F} = cos\alpha \hat{i}+ cos \beta \hat{j}+ cos \gamma \hat{k}\\$ -- (2)

Comparing equations (1) and (2) , we get

$cos \alpha = \dfrac{1}{2} , so \alpha = 60^{0}\\$

$cos \beta = \dfrac{1}{2} , so \beta = 60^{0}\\$

$cos \gamma = \dfrac{1}{\sqrt{2}} , so \gamma = 45^{0}\\$

Thus statement 2 and 3 are correct.

Correct option is C.

A cyclist starts from centre O of a circular park of radius 1 km and moves along the path OPRQO as shown in figure. If he maintains constant speed of 10

$ms^{-1}$ , what is his acceleration at point R ?

0%
10 $ms^{-2}$
0%
0.1 $ms^{-2}$
0%
0.01 $ms^{-2}$
0%
1 $ms^{-2}$

Explanation

The acceleration at point

$R$ is towards the center of the circle.
Here,

$v = 10 ms^{-1}$ ,

$r = 1 km = 1000 m$
Centripetal acceleration,

$a_{c}=\dfrac{v^{2}}{r}$

$=\dfrac{(10 ms^{-1})^{2}}{1000 m} = 0.1 ms^{-2}$

A bomb is released by a horizontal flying aeroplane. The trajectory of the bomb is

0%
a parabola
0%
a straight line
0%
a circle
0%
a hyperbola

Explanation

The correct answer is option (A).

Hint: Apply the concept of inertia.

Explanation:

Let us consider that the horizontal flying aeroplane possesses a velocity of

$u$ . At the instant when the bomb is released from the plane, it also possesses the initial velocity

$u$ by virtue of its inertia of motion and tends to move in the same direction as that of the plane.

However, due to frictional force, it eventually loses this velocity. Gravitational force acting on it produces a vertical acceleration on it and it starts to fall towards the ground.

The trajectory of the bomb thus forms a parabola as shown below:

A bomb is released by a horizontal flying aeroplane class 11 physics CBSE

Hence, the trajectory of the bomb is a parabola.

The correct answer is option (A).

In uniform circular motion:

0%
both the angular velocity and the angular momentum vary
0%
the velocity varies but the momentum remains constant
0%
magnitude of both the velocity and the momentum stay constant
0%
the momentum varies but the velocity remains constant

Explanation

In the case of uniform circular motion, though direction of the velocity vector changes at every point but the magnitude of the velocity or speed used to be constant.

and we know,
Momentum,

$P=mv$
and magnitude of momentum,

$|P|=|mv|$

As, magnitude of the momentum used to be dependent on magnitude of the velocity.

Hence, the magnitude of both the velocity and the momentum remains constant during uniform circular motion.

The correct option is C.

A stone tied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 22 s, then the acceleration of the stone is then

0%
16 $ms^{-2}$
0%
4 $ms^{-2}$
0%
12 $ms^{-2}$
0%
8 $ms^{-2}$

Explanation

Here,

$r = 100 cm = 1 m$
Frequency,

$\nu$ =

$\dfrac{14}{22}rps$

$\therefore$

$\omega = 2\pi \nu = 2\times \dfrac{22}{7} \times\dfrac{14}{22}$ =

$4 rad s^{-1}$
The acceleration of the stone is

$a_{c} = \omega^{2}r=(4)^{2}(1) = 16ms^{-2}$

The relative velocity of

$B$ as seen from

$A$ is

0%
$-8\sqrt { 2 } \hat { i } + 6\sqrt { 2 } \hat { j }$
0%
$4\sqrt { 2 } \hat { i } + 3\sqrt { 3 } \hat { j }$
0%
$3\sqrt { 5 } \hat { i } + 2\sqrt { 3 } \hat { j }$
0%
$3\sqrt { 2 } \hat { i } + 4\sqrt { 3 } \hat { j }$

Explanation

$V_{B, A}=V_B-V_A$

$=-14\cos 45^o \hat i +14\sin 45^o \hat j-(2\cos 45^o\hat i)-2\sin 45\hat j$

$=-\dfrac{16}{\sqrt 2}\hat i +\dfrac{12}{\sqrt 2}\hat j=-8\sqrt 2\hat i+6\sqrt 2\hat j$

The minimum number of vectors having different planes which can be added to give zero resultant is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

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%
${R_1} \geqslant {R_2}$

Explanation

The weight Mg of the car acts vertically decreases through the center of gravity G of the car

$\therefore { R }_{ 1 }+{ R }_{ 2 }=Mg\\$

and the horizontal force F provides the centripetal force for motion in a circle

$\therefore { F=\frac { m{ v }^{ 2 } }{ r } }\\$

Taking moments about G, there should be no resultant turning effect about the center of gravity

${ Fh+{ R }_{ 1 }a={ R }_{ 2 }a }\\ \therefore { R }_{ 2 }=\frac { Fh+{ R }_{ 1 }a }{ a } \\ or,\quad { R }_{ 2 }={ R }_{ 1 }+\left( \frac { Fh }{ a } \right) \\ \Rightarrow { R }_{ 2 }>{ R }_{ 1 }\\$

A cube is placed so that one corner is at the origin and three edges are along the

$x-, y-,$ and

$z$ -axes of a coordinate system. Use vectors to compute
a. The angle between the edge along the

$z$ -axis (line ab) and the diagonal from the origin to the opposite corner (line ad).
b. The angle between line

$ac$ (the diagonal of a face) and line

$ad$ .

0%
a. $\cos^{-1}\dfrac{1}{\sqrt{2}}$

b. $\cos^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$
0%
a. $\cos^{-1}\dfrac{1}{\sqrt{3}}$

b. $\cos^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$
0%
a. $\cos^{-1}\dfrac{1}{\sqrt{3}}$

b. $\cos^{-1}\dfrac{\sqrt{3}}{\sqrt{3}}$
0%
a. $\cos^{-1}\dfrac{2}{\sqrt{3}}$

b. $\cos^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$

Explanation

$\cos\theta =\dfrac { a }{ \sqrt { 3 } a }$

$\boxed { \theta ={ \cos }^{ -1 }\dfrac { 1 }{ \sqrt { 3 } } }$

$\cos\theta =\dfrac { { \left( \sqrt { 2 } a \right) }^{ 2 }+{ \left( \sqrt { 3 } a \right) }^{ 2 }-{ a }^{ 2 } }{ 2\times \sqrt { 3 } a\times \sqrt { 2 } a }$

$=\dfrac { { 4a }^{ 2 } }{ 2\sqrt { 3 } \sqrt { 2 } { a }^{ 2 } }$

$=\dfrac { \sqrt { 2 } }{ 3 }$

$\boxed { \theta ={ \cos }^{ -1 }\dfrac { \sqrt { 2 } }{ 3 } }$

1167222_980018_ans_30fc788c10f946269339b0dbd190a917.png

The components of a vector along the

$x$ - and

$y$ -directions are

$(n +1)$ and

$1$ , respectively. If the coordinate system is rotated by an angle

$\theta = 60^o$ , then the components change to

$n$ and

$3$ . The value of

$n$ is

0%
$2$
0%
$\,1+ \sqrt{3}$
0%
$\,1- \sqrt{3}$
0%
$\,1\pm \sqrt{3}$

Explanation

Let the vector is A having magnitude a and making an angle x with x axis.so, the components are

$a\cos x=n+1$

$a\sin x=1$

When coordinates are rotated by 60⁰, the angle can be $x+60\,or\,x-60$ ,the new coordinates are

$a\cos (x\pm 60)\,\,=n$

$a\sin (x\pm 60)=3$

Now, solving four equations we get

$a\cos (x\pm 60)=a(\operatorname{cosx}\cos 60\pm \operatorname{sinx}\sin 60)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n\,\,\,\,\,\,\,\,\,\,\,\,=a(\frac{1}{2}\operatorname{cosx}\,\pm \frac{\sqrt{3}}{2}sinx)$

$put,\,a\cos x\,=\,n+1\,\,and\,\,a\sin x\,=\,1$

$so,\,\,n\,=\,1\pm \sqrt{3}$

n depends on direction of rotation.

An object moves with constant acceleration

$\vec { a }$ . Which of the following expression is/are also constant?

0%
$\dfrac { d\left| \vec { v } \right| }{ dt }$
0%
$\left| \dfrac { d\vec { v } }{ dt } \right|$
0%
$\dfrac { d\left( { v }^{ 2 } \right) }{ dt }$
0%
$\dfrac { d\left( \vec { v } /\left| \vec { v } \right| \right) }{ dt }$

A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds.

0%
$300 \pi$ radians.
0%
$400 \pi .$ radians
0%
$250 \pi$ radians.
0%
$200 \pi$ radians.

Explanation

Hint :-Use equation of circular motion.

Step 1: Find angular acceleration of the wheel

Initial angular velocity is 0 and final angular velocity = 100 X 2 π = 200 π rad/s

From the equation of acceleration,

Angular acceleration

$α = \dfrac{\omega\ -\ \omega_0}{t} = \dfrac{200\ \pi}{4} = 50\ π\ rad/s^2$

Step 2: Find the rotation of the angle

Rotation of the angle

$θ = \omega_0t + \dfrac{1}{2} \alpha{\ t}^2$

$= 0 + \dfrac{1}{2} * 50 * 4^2 = 25 * 15 = 400\ π\ radians$

Hence, option B is the correct answer.

A car moves on a circular road describing equal angles about the centre in equal intervals of time. Which of following statements about the velocity of car are not true?

0%
Velocity is constant
0%
Magnitude of velocity is constant but the direction changes
0%
Both magnitude and direction of velocity change
0%
Velocity is directed towards the center of circle

Explanation

Angular speed is constant. Hence linear speed is constant i.e magnitude of velocity is constant but direction is changing.

Hence,

$(a), (c), (d)$ are not true.

A wheel has moment of inertia

$10^{-2} kg-m^2$ and is making 10 rps. The torque required to stop it in 5 secs is

0%
12.56
0%
9.42
0%
6.28
0%
3.14

Explanation

The initial angular velocity is

$\omega_i=10 rps = 10 \times 2 \pi=20 \pi$ radians/sec and the final angular velocity is

$\omega_f =0$ , since the wheel comes to rest

We know that torque

$\tau=I\dfrac{(\omega_f-\omega_i)}{t}=-(10^{-2})(20 \pi/5)=-4 \times 10^{-2} \pi = -12.56 \times 10^{-2}$

A disc rotates through 10 radians in 4 seconds. The disc experienced uniform acceleration. If the disc starts from rest, what is the angular velocity after four seconds?

0%
2.5 radians/sec
0%
5 radians/sec
0%
7.5 radians/sec
0%
10 radians/sec

Explanation

$\theta ={ \omega }_{ \circ }t+\frac { 1 }{ 2 } { \alpha }{ t }^{ 2 }\\ 10=0+\frac { 1 }{ 2 } { \times \alpha \times }{ 4 }^{ 2 }\\ \therefore \quad \alpha =\frac { 20 }{ 16 } =\frac { 6 }{ 4 } \\ \therefore \quad \omega ={ \omega }_{ \circ }+\alpha t\\ =0+\frac { 5 }{ 4 } { \times 4 }\\ =5\frac { radians }{ sec }$

A pendulum bob of mass

$m = 80 mg$ , carrying a charge of

$q = 2 \times 10^{-8}C$ , is at rest a horizontal uniform electric field of

$E = 20,000 V/m$ . The tension

$T$ in the thread of the pendulum and the angle

$\alpha$ it makes with vertical is (take

$g = 9.8 m/s^2$ )

0%
$\alpha \approx 27^o$
0%
$T \approx 880 \mu N$
0%
$T = 8.8 \mu N$
0%
$\alpha \approx 356o$

Explanation

Force

$= q E = (2 \times 10^{-8} \times 20000) N = 4 \times 10^{-4}N$

$T \cos{\theta} =mg$ -----------(I)

$T \sin{\theta} = 4 \times 10^{-4}$ -------------(II)

So,

$\tan{\theta} = \cfrac{4 \times 10^{-4}}{80 \times 10^{-6} \times 9.8} = \cfrac{40}{8 \times 9.8} \approx \cfrac{1}{2}$

So,

$\theta= 27^{\circ}$

$T=\cfrac{mg}{\cos {\theta} } = \cfrac{80 \times 10^{-6} \times 9.8}{0.89}=880 \mu N$

1107412_1017197_ans_c226ff44ba594eb3825a4e48147335b3.png

A body rotates about a fixed axis with an angular acceleration of

$1$ rad/

$sec^2$ . Through what does it rotate during the time in which its angular velocity increases from

$5$ rad/s to

$15$ rad/s.

0%
100 rad
0%
200 rad
0%
250 rad
0%
150 rad

Explanation

Given :

$w_i = 5\ \ rad/s$

$w_f = 15 \ rad/s$

Angular acceleration

$\alpha = 1\ rad/s^2$

Using

$w_f^2 - w_i^2 = 2\alpha \theta$

$15^2 - 5^2 = 2\times 1\times \theta$

$\implies \ \theta = 100 \ rad$

The angular velocity of a wheel increases from

$100\ m/s$ to

$300\ m/s$ in

$10\ s$ . The revolutions made during that time is:

0%
$600$
0%
$1500$
0%
$1000$
0%
$2000$

Explanation

The correct option is D

The angular acceleration(a) of the wheel is given by

$\alpha$ =

$\frac{change of angular velocity}{Time}$

$\frac{(300-100)}{10}$ =20 revolutions/

$s^2$

[We have retained the angular velocity in rps itself since the number of

revolution is to be found out]

Substituting the known values in the equation,

$\omega^2$ =

$\omega^2_0+2a\theta2$ (which is similar to

$v^2$ =

$v_0^2+2as$ in linear motion),We have

$300^2=100^2+2\times20\times\theta$

[The angular displacement

$\theta$ will be in revolutions for the obvious reason]

from above equation ,

$\theta$ =

$2000 revolutions$

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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