Match the following : List - I List - II a) Current strength e) Vectors b) Distance,Work done f) Scalars c) Force, torque g) Possesses direction but not vector d) Component of vector, smaller value of angular displacement h) Scalar and vector

$$\mathrm{a}\rightarrow \mathrm{g};\mathrm{b}\rightarrow \mathrm{f};\mathrm{c}\rightarrow \mathrm{e};\mathrm{d}\rightarrow \mathrm{h}$$

$$\mathrm{a}\rightarrow \mathrm{e};\mathrm{b}\rightarrow \mathrm{f};\mathrm{c}\rightarrow \mathrm{g};\mathrm{d}\rightarrow \mathrm{h}$$

$$\mathrm{a}\rightarrow \mathrm{h};\mathrm{b}\rightarrow \mathrm{g};\mathrm{c}\rightarrow \mathrm{f};\mathrm{d}\rightarrow \mathrm{e}$$

$$\mathrm{a}\rightarrow \mathrm{g};\mathrm{b}\rightarrow \mathrm{e};\mathrm{c}\rightarrow \mathrm{f};\mathrm{d}\rightarrow \mathrm{h}$$

MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 4

$\overline{\mathrm{R}}$ is the resultant of vectors

$\vec{A}$ and

$\vec{B}$ . lf

$\overline R=\displaystyle \frac{\vec B}{\sqrt{2}}$ the value of

$\theta$ is :

0%
30 $^{0}$
0%
45 $^{0}$
0%
60 $^{0}$
0%
75 $^{0}$

Explanation

$sin\theta = \vec R/\vec B = 1/\sqrt { 2 } =>\theta = 45^0$

A projectile is given an initial velocity of

$(\hat{i}+2\hat{j}) \ m/s$ , where

$\hat{i}$ is along the ground and

$\hat{j}$ is along the vertical. If

$g=10 \ m/s^{2}$ , the equation of its trajectory is:

0%
$y=2x-5x^{2}$
0%
$4y=2x-5x^{2}$
0%
$4y=2x-25x^{2}$
0%
$y=x-5x^{2}$

Explanation

$\vec{u} = \hat{i} + 2\hat{j};$

$\therefore u \cos \theta = 1; u \sin \theta =2$

$x=u \cos \theta t =t$

$y=u \sin \theta t - \dfrac{1}{2}gt^2 = 2t-5t^{2}$

Substituting

$t=x$
Equation of trajectory is

$y=2x-5x^{2}$

The maximum resultant of two concurrent forces is

$10N$ and their minimum resultant is

$4N$ . The magnitude of the larger force is:

0%
$5N$
0%
$7N$
0%
$3N$
0%
$14N$

Explanation

$\textbf{Step 1: Resultant of two forces}$

Resultant of two forces say

$F_1$ ,

$F_2$ is given by :

$R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2F_{1} F_{2} \cos \theta}$

$\textbf{Step 2: Calculation of}\ R_{\text{max}}\ \textbf{and}\ R_{\text{min}}$

$\text{For R}_\text{max}:$

$R$ is maximum when,

$\cos \theta= 1$

$\Rightarrow\theta = 0^{\circ}$

i.e. The forces are parallel

So,

$R_{\text{max}}= \sqrt{F_1^2+ F_2^2 +2F_1F_2}$

$\Rightarrow\ \ 10\ N= F_{1} + F_{2}$

$....(1)$

$\text{For R}_\text{min}:$

$R$ is minimum when,

$\cos \theta= -1$

$\Rightarrow\theta = 180^{\circ}$

i.e. The forces are anti-parallel

So,

$R_{\text{min}}= \sqrt{F_1^2+ F_2^2 -2F_1F_2}$

$\Rightarrow\ \ 4\ N= F_{1} - F_{2}$

$....(2)$

$\textbf{Step 3: Solving equations}$

Adding eq.

$(1)$ and

$(2)\Rightarrow$

$2F_{1} = 14N$

$\Rightarrow$

$F_{1} = 7N$

So, Equation

$(1)\Rightarrow$

$F_{2} = 3 N$

The larger force is

$7N$ .

Hence, Option(B) is correct.

$\textbf{Alternate Solution:}$

We can solve directly by remembering the result that:

$F_{max} = F_1+F_2\ \ \ \ \text{&}$

$F_{min} = F_1-F_2$

Match the following :

List - I	List - II
a) Current strength	e) Vectors
b) Distance,Work done	f) Scalars
c) Force, torque	g) Possesses direction but not vector
d) Component of vector, smaller value of angular displacement	h) Scalar and vector

0%
$\mathrm{a}\rightarrow \mathrm{e};\mathrm{b}\rightarrow \mathrm{f};\mathrm{c}\rightarrow \mathrm{g};\mathrm{d}\rightarrow \mathrm{h}$
0%
$\mathrm{a}\rightarrow \mathrm{h};\mathrm{b}\rightarrow \mathrm{g};\mathrm{c}\rightarrow \mathrm{f};\mathrm{d}\rightarrow \mathrm{e}$
0%
$\mathrm{a}\rightarrow \mathrm{g};\mathrm{b}\rightarrow \mathrm{e};\mathrm{c}\rightarrow \mathrm{f};\mathrm{d}\rightarrow \mathrm{h}$
0%
$\mathrm{a}\rightarrow \mathrm{g};\mathrm{b}\rightarrow \mathrm{f};\mathrm{c}\rightarrow \mathrm{e};\mathrm{d}\rightarrow \mathrm{h}$

Let

$\vec{A}=2\hat{i}+7\hat{j},\vec{B}=\hat{i}+2\hat{j}+4\hat{k},\ \displaystyle \vec{C}=\dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5}$ .

The ratio in which

$\vec{C}$ divides

$\vec{AB}$ internally is?

0%
$1:4$
0%
$2:3$
0%
$3:2$
0%
$5:1$

Explanation

As shown in the figure let

$\vec{C}$ divide

$\vec{AB}$ in a ratio

$K:1$

$C= \dfrac{KB+A}{K+1}$

$\implies \dfrac{9\hat{i}+30\hat{j}+4\hat{k}}{5} = \dfrac{K(\hat{i}+2\hat{j}+4\hat{k})+(2\hat{i}+7\hat{j})}{K+1}$

$\implies 9K\hat{i}+30K\hat{j}+4K\hat{k}+9\hat{i}+30\hat{j}+4\hat{k} = {5K\hat{i}+10K\hat{j}+20K\hat{k}+10\hat{i}+35\hat{j}}$

$\implies 4K\hat{i}+20K\hat{j}-16K\hat{k}=\hat{i}+5\hat{j}-4\hat{k}$

$\implies 4K(\hat{i}+5\hat{j}-4\hat{k})=\hat{i}+5\hat{j}-4\hat{k}$

$\implies K=\dfrac{1}{4}$
Hence,

$\vec{C}$ divide

$\vec{AB}$ in a ratio

$1:4$

Speed

$v$ of a particle moving along a straight line, when it is at a distance

$x$ from a fixed point on the line is given by

$v^{2} = 108 - 9x^{2}$ (assuming mean position to have zero phase constant)
(all quantities in are in

$cgs$ unit):

0%
The motion is uniformly accelerated along the straight line
0%
The magnitude of the acceleration at a distance $3\ cm$ from the fixed point is $27\ cm/s^{2}$
0%
The motion is simple harmonic about $x = \sqrt{12}\ m$ .
0%
The maximum displacement from the fixed point is $4\ cm$ .

Explanation

$v^2 = 108-9x^2$ -----

$(i)$

Differentiating

$(i)$ w.r.t

$x$

$2 \times v \dfrac{dv}{dx} = -18x$ -----

$(ii)$

We know that acceleration of particle is also given by,

$a$ =

$v \dfrac{dv}{dx}$

From

$(ii)$ we have ,

$2a = -18x$

hence ,

$a = -9x$

This acceleration is clearly not uniform.

$x=3$

$a = -9 \times 3$

$a = -2$

Hence the magnitude of acceleration at a distance of

$3 cm$ is

$27 cm / s^2 .$

Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of

$10\ ms^{-1}$ . It implies that the boy is :

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at rest
0%
moving with no acceleration
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in accelerated motion
0%
moving with uniform velocity

A mass is performing vertical circular motion(see figure).If The average velocity of theparticle is increased, then at which point thestring will break :

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

A wheel, starting from rest, rotates with a uniform angular acceleration of

$2 rad s^{-2}$ . The number of rotations it performs in the tenth second is

0%
3
0%
6
0%
9
0%
12

Explanation

$\omega_0=0$

$\theta_n=\omega_0 +\dfrac {1}{2}\alpha (2n-1)$

$\theta_{10}=\dfrac {\alpha}{2}(2n-1)=\dfrac {2}{2}(2\times 10-1)=19 rad$
Number of revolutions:

$\dfrac {19}{2\pi}=3$

A smooth flat horizontal turntable 4.0 m in diameter is rotating at 0.050 revs per second. A student at the centre of the turntable, and rotating with it, Places a smooth flat puck on the turntable 0.50 m from the edge. Which of the following figures describes the motion of the puck as seen by a stationary observer who is standing at the side of the turntable and above the turntable?

A particle moves on a circular path of radius r. Find the ratio of the magnitude of displacement and the distance covered by the particle when the body moves from one diametric end to the other.

0%
1 : 1
0%
$\pi :2$
0%
$2:\pi$
0%
$\pi :1$

A particle is moving in a circular path of radius

$r$ . The magnitude of displacement after half a circle would be :

0%
Zero
0%
$\pi r$
0%
$2 r$
0%
$2 \pi r$

In a projectile motion from a point of horizontal surface to another point on the same surface (Take

$\vec {a}=$ acceleration and

$\vec {v}=$ instantaneous velocity)

0%
$\vec{a}\cdot\vec{v}=0$ at maximum height
0%
$\vec{a}\cdot \vec{v}=0$ only if angle of projection is $90^{\circ}$
0%
$\vec{a} \times \vec{v}=$ constant every where in air
0%
None of these

Explanation

At highest point

$\vec{a}=-g \widehat{j}$

$\vec{v}=u \cos \theta \hat{i} \Rightarrow \vec{a}\cdot \vec{v}=0$

Option A is correct.

$\vec{v}= u \cos \theta \widehat{i} + u \sin \theta \widehat{j}-gt \widehat{j}$

$\vec {a}=-g\widehat{j}$

$\vec{a}\times \vec{v}=u \cos \theta g\hat{k} = constant$

Option C is correct.

A shell fired from the base of a mountain just clears it. If

$\alpha$ is the angle of projection, then the angular elevation of the summit

$\beta$ is

0%
$\dfrac {1 \alpha}{2}$
0%
$\tan^{-1}\left (\dfrac {1}{2}\right )$
0%
$\tan^{-1}\left (\dfrac {1 \tan \alpha}{2}\right )$
0%
$\tan^{-1}(2 \tan \alpha)$

Explanation

Angular elevation of

ββ

$=\dfrac{\left ( 2u^{2}\sin ^{2}\alpha /2g \right )}{\left ( 2u^{2}\sin\alpha \cos \alpha \right )/g} =\dfrac{\tan \alpha }{2}$

$\therefore \beta = \tan^{-1}\left ( \dfrac{\tan \alpha }{2} \right )$

A particle is moving in a circular path of radius r. Its displacement after moving through half the circle would be :

0%
Zero
0%
$r$
0%
$2r$
0%
$\dfrac{2}{r}$

Explanation

Displacement is the minimum distance between the initial position and final position.
After moving half circle (A to B) displacement will be AB
AB is diameter i.e. $2r$
Option C is correct.

If air resistance is considered, then the maximum height achieved by the projectile

0%
decreases
0%
increases
0%
remains unchanged
0%
very difficult to answer as no data provided

A small body is thrown at an angle to the horizontal with the initial velocity

$\vec{v}_0$ . Neglecting the air drag, find the displacement of the body as a function of time

$r(t)$ .

0%
$\displaystyle\vec{r}=\vec{v}_0t+\dfrac{\vec{g}t^2}{2}$
0%
$\displaystyle\vec{r}=\vec{v}_0t-\dfrac{\vec{g}t^2}{2}$
0%
$\displaystyle\vec{r}=2\vec{v}_0t-\dfrac{\vec{g}t^2}{2}$
0%
$\displaystyle\vec{r}=2\vec{v}_0t+\dfrac{\vec{g}t^2}{2}$

Explanation

$\vec{v}_{o}=u \cos \theta\hat{i}+u \sin \theta \hat{j}$

Using,

$y = u \sin \theta t +\dfrac { 1 }{ 2 } \vec{ g } { t }^{ 2 }$

$x=u \cos \theta t$

so,

$\vec { r } =x\hat{i}+y\hat{j}=\vec { { v }_{ o } } t+\dfrac { 1 }{ 2 } \vec{ g } { t }^{ 2 }$

Find the total acceleration of the point as a function of velocity and the distance covered.

0%
$\sqrt{2}\dfrac{v^2}{R}$
0%
$\sqrt{2}\dfrac{2v^2}{R}$
0%
$\sqrt{3}\dfrac{v^2}{R}$
0%
$\sqrt{3}\dfrac{2v^2}{R}$

Explanation

The total acceleration is

$\sqrt{a_t^2+a_{radial}^2}$

As,

$a_t=a_{radial}=v^2/R$

so the total acceleration is

$\sqrt 2v^2/R$

Two projectiles A and B are thrown with the same speed such that A makes angle

$\theta$ with the horizontal and B makes angle

$\theta$ with the vertical, then

0%
Both must have same time of flight
0%
Both must achieve same maximum height
0%
A must have more horizontal range than B
0%
Both may have same time of flight

Explanation

Depending on the value of

$\theta$ one can have same or different time of flight, range or maximum height.
Therefore must statement is wrong .
They may be equal or may not be.
This is only possible for

$\theta =45^0$

Suppose a player hits several baseballs. Which baseball will be in the air for the longest time?

0%
The one with the farthest range.
0%
The one which reaches maximum height.
0%
The one with the greatest initial velocity.
0%
The one leaving the bat at $45^0$ with respect to the ground.

Explanation

$\textbf{Setp 1: Formula of Maximum Height and Time of flight}$

For a projectile thrown in air at an angle of projection

$\theta$ and speed

$u$

Time of flight is given by

$T = \cfrac{2usin\theta}{g}$

$\Rightarrow\ \ \ u\ sin\theta=\cfrac{gT}{2}$

$....(1)$

Maximum Height is given by

$H = \cfrac{(usin\theta)^2}{2g}$

$....(2)$

$\textbf{Step 2: Relation of T and H}$
Putting Value of

$usin\theta$ from equation

$(1)$ in equation

$(2)$ , we get:

$\Rightarrow\ \ H = \cfrac{(gT/2)^2}{2g}$

$\Rightarrow\ \ \ T= 2\sqrt{\dfrac{2H}{g}}$

$T \propto \sqrt H$

Therefore, the ball which reaches maximum height, will be in the air for longest period of Time.

Hence, option B is correct.

Consider the motion of the tip of the minute hand of a clock. In one hour

0%
the displacement is zero.
0%
the distance covered is zero.
0%
the average speed is zero.
0%
the average velocity is zero.

Explanation

Hint:

Displacement depends on only the initial and final position of the body.

Step 1: Explanation

$\bullet$ In a trip of one hour of a tip of the minute hand, its initial and final positions are the same. Displacement is the distance between the initial and final position Thus, its displacement in this trip is zero. Option A is correct.

$\bullet$ But it has covered some path in this trip, thus distance traveled by the tip of the hand is not zero. Thus, the Average speed that is defined as the ratio of total distance travelled with the total time taken is not zero. Options B and C are incorrect.

$\bullet$ Also, the average velocity is defined as the ratio of total displacement with the total time taken, as displacement is zero, the average velocity of the tip also remains zero. Option D is correct.

Thus, Option A and D are correct.

A ball is thrown from a height of

$12.5m$ from the ground level in the horizontal direction. It falls at a horizontal distance of

$200m$ . The initial velocity of the ball (approximately) is

0%
$140m/s$
0%
$80m/s$
0%
$126m/s$
0%
$120m/s$

Explanation

$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$
consider this equation in vertical direction.

$\Rightarrow 12.5=\dfrac { 1 }{ 2 } (10){ t }^{ 2 }\\ \Rightarrow t=\sqrt { 2.5 } s$
Thus, it requires

$\sqrt { 2.5 } s$ to reach the bottom.
Now, in the horizontal direction,

$velocity=\dfrac { displacement }{ time } =\dfrac { 200 }{ \sqrt { 2.5 } } =126.49$

In uniform circular motion

0%
acceleration is variable.
0%
acceleration is uniform.
0%
the direction and magnitude of acceleration both vary.
0%
if force applied is doubled in circular motion, then angular velocity becomes double.

Explanation

Since the direction of acceleration is different at every instant, hence the acceleration vector is variable. This can also be proved by double differentiating the position vectors x and y, giving accelerations function of time.

If the applied force which is responsible for circular motion is doubled,

$\omega ^2$ doubles.

$\Rightarrow \omega ' = \sqrt{2} \omega$

Find the angle between the vectors of the angular velocity and the angular acceleration at

$t=10.0\:s$

0%
$15^\circ$
0%
$14^\circ$
0%
$17^\circ$
0%
$13^\circ$

Explanation

angular acceleration

$\vec { \alpha } =\dot { \vec { \omega } } \\ \vec { \alpha } =\quad a\hat { i } +\quad 2bt\hat { j } \\ cos\theta \quad =\quad \dfrac { \vec { \alpha } \cdot \vec { \omega } }{ \left| \vec { \alpha } \right| \left| \vec { \omega } \right| } \\ cos\theta \quad =\quad \dfrac { { a }^{ 2 }t+2{ b }^{ 2 }{ t }^{ 3 } }{ \sqrt { { a }^{ 2 }+4{ b }^{ 2 }{ t }^{ 2 } } \sqrt { { a }^{ 2 }{ t }^{ 2 }+{ b }^{ 2 }{ t }^{ 4 } } } \\ cos{ \theta }_{ (t=10) }\quad =\quad 0.955\\ \theta \quad =\quad 17$

Two vectors $\overline{A}$ and $\overline{B}$ lie in a plane. Another vector $\overline{C}$ lies outside this plane, then the resultant of the three vectors $(\overline{A}+\overline{B}+\overline{C})$ :

0%
can be of zero magnitude
0%
cannot be of zero magnitude
0%
lies in the plane containing $(\overline{A}+\overline{B})$
0%
lies in the plane containing $(\overline{A}-\overline{B})$

Explanation

The resultant is always going to contain atleast one component which is non-zero because of

$\vec { C }$ as it lies in another plane
Thus, resultant cannot have a zero magnitude.

If the resultant of three vectors is zero then:

0%
the magnitude of one vector must be equal to the sum of magnitudes of the other two
0%
the magnitude of each vector must be greater than the sum of magnitudes of the other two vectors
0%
the magnitude of each vector must be less than the sum of magnitudes of the other two vectors
0%
all the three vectors must be of equal magnitude

Two particles A and B are moving on different concentric circles with different velocities

$v_A$ and

$v_B$ then angular velocity of B relative to A as observed by A is given by :

0%
$\displaystyle \frac{v_B - v_A}{r_B - r_A}$
0%
$\displaystyle \frac{v_A}{r_A}$
0%
$\displaystyle \frac{v_A - v_B}{r_A - r_B}$
0%
$\displaystyle \frac{v_B + v_A}{r_B + r_A}$

Explanation

Assuming the particles to be the closest, relative velocity

$v_r = \left |\overline {v_B} - \overline {v_A} \right | = v_B - v_A$
and the relative position vector,

$r_r = \left |\overline {r_B} - \overline {r_A} \right | = r_B - r_A$
Using

$\displaystyle \omega = \frac{v}{r}$ , we get relative angular velocity

$\displaystyle \omega = \frac{v_r}{r_r} =,\frac{v_B - v_A}{r_B - r_A}$

A body is moving with uniform speed

$v$ on a horizontal circle from A as shown in the figure. Change in the velocity in the first quarter revolution is:

0%
$v^2$
0%
$\sqrt{2} v$ southwest
0%
$\sqrt{2}$ northwest
0%
$2v$ west

Explanation

For quarter revolution

$\triangle \overline{v} = v_2(-\hat{i})-v_1 \hat{j}$
or

$\triangle \overline{v} = -v\hat{i} - v\hat{j}$

On calculating the magnitude

$\triangle v = \sqrt{v^2+v^2} = \sqrt{2} v$
Also

$\displaystyle \alpha = \tan^{-1} \frac{v}{v} = 45^{\circ}$

$\triangle \overline{v} = \sqrt{2} v$ southwest.

A particle moves in a circle of radius 4 cm clockwise at constant speed of 2 cm

$s^{-1}$ . If

$\hat{x}$ and

$\hat{y}$ are unit acceleration vectors along x and y axes, respectively, the acceleration of the particle at the instant half way between PQ is given by

0%
$-4(\hat{x} - \hat{y})$
0%
$4(\hat{x} + \hat{y})$
0%
$\displaystyle \frac {-(\hat{x} + \hat{y})}{\sqrt{2}}$
0%
$\displaystyle \frac {\hat{x} - \hat{y}}{4}$

Explanation

Mid point acceleration,

$\displaystyle a= \frac {v^2}{r} =\frac{4}{4} = 1ms^{-1}$ at

$45^{\circ}$ with x-axis

$\displaystyle \overrightarrow{a}_x = -a \cos 45^{\circ} \hat{x}$

$= -1 \times \frac{1}{\sqrt{2}} \hat{x} = -\frac{1}{\sqrt{2}} \hat{x}$

$\displaystyle \overrightarrow{a}_y = -a \sin 45^{\circ} \hat{y}$

$= -1 \times \frac{1}{\sqrt{2}} \hat{y} = -\frac{1}{\sqrt{2}} \hat{y}$
Thus

$\hat{a} = \overrightarrow{a_x} + \overrightarrow{a_y} = - \frac{1}{\sqrt{2}} (\hat{x} +\hat{y})$

The minimum number of coplanar vectors having different non-zero magnitudes can be added to give zero resultant is :

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

$\vec {A}=\vec {B}+\vec {C}$ and the magnitude of

$\vec {A}, \vec {B}$ and

$\vec {C}$ are

$5,4$ and

$3$ units, respectively, then the angle between

$\vec {A}$ and

$\vec {C}$ is:

0%
$\displaystyle \frac{\pi }{2}$
0%
$\sin ^{-1}(3/4)$
0%
$\cos^{-1}(3/5)$
0%
$\cos^{-1}(4/5)$

Explanation

Magnitude of

$\overrightarrow { A } =AC=5,\overrightarrow { B } =AB=4\: \&\: \overrightarrow { C } =BC=3\quad$

which forms a Pythagoras triplet
Let

$\theta$ be angle between

$\overrightarrow { A } \& \overrightarrow { C }$

$\cos { \theta =\dfrac { 3 }{ 5 } }$ so,

$\theta =\cos ^{ -1 }{ \dfrac { 3 }{ 5 } }$

When a ceiling fan is switched off, its velocity falls to half in 36 rotations. How many more rotations it will make?

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

If a particle goes from point A to point B in

$1$ s moving in a semicircle(See figure). The magnitude of the average velocity is:

0%
3.14 $ms^{-1}$
0%
2 $ms^{-1}$
0%
1 $ms^{-1}$
0%
zero

Explanation

Here average velocity

$\displaystyle = \dfrac{displacement\quad AB}{time}$

$\displaystyle = \dfrac{2}{1} = 2\ ms^{-1}$

If the magnitude of vectors

$\overrightarrow{A}$ ,

$\overrightarrow{B}$ and

$\overrightarrow{C}$ are 12, 5 and 13, respectively and

$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ , then the angle between the vectors

$\overrightarrow{A}$ and

$\overrightarrow{B}$ is

0%
$\pi/4$
0%
$\pi/3$
0%
$\pi/2$
0%
zero

Explanation

Given:

$\vec C=\vec A+\vec B$

$\vec C.\vec C=(\vec A+\vec B).(\vec A+\vec B)$

$C^2=A^2+B^2+2AB\cos\theta$

$13^2=12^2+5^2+2(12)(5)\cos\theta$

$\cos\theta=0 \Rightarrow \theta=\pi/2$

The second equation of motion in rotatory motion is

0%
$\displaystyle{S=ut+\frac{at^2}{2}}$
0%
$\displaystyle{\theta = \omega_1t+\frac{\alpha t^2}{2}}$
0%
$\displaystyle{\omega^2_2 = \omega^2_1 +2\alpha \theta}$
0%
$\displaystyle{\omega_2 = \omega_1 +\alpha t}$

Explanation

$\displaystyle{S=ut+\frac{at^2}{2}}$ is the second equation of linear motion.

Substituting analogous physical quantities of rotatory/circular motion in this we get

$\theta=\omega_1 t+\frac{1}{2}\alpha t^2$

This is the second equation of circular motion

$G$ and

$G'$ be the centroids of the triangles

$ABC$ and

$A'B'C'$ respectively, then

$\overrightarrow { AA' } +\overrightarrow { BB' } +\overrightarrow { CC' } =$

0%
$\displaystyle \dfrac { 2 }{ 3 } \overrightarrow { GG' }$
0%
$\overrightarrow { GG' }$
0%
$2\overrightarrow { GG' }$
0%
$3\overrightarrow { GG' }$

Explanation

Let P.V. of

$A\left( \overrightarrow { a } \right) ,B\left( \overrightarrow { b } \right) ,C\left( \overrightarrow { c } \right) ,A'\left( \overrightarrow { { a }_{ 1 } } \right) ,B'\left( \overrightarrow { { b }_{ 1 } } \right) ,C'\left( \overrightarrow { { c }_{ 1 } } \right)$

Now P.V. of

$\displaystyle G=\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 }$ ;

Similarly

$\displaystyle G'=\dfrac { \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } }{ 3 }$

Now

$\overrightarrow { AA' } =\overrightarrow { { a }_{ 1 } } -\overrightarrow { a } ,\overrightarrow { BB' } =\overrightarrow { { b }_{ 1 } } -\overrightarrow { b } ,\overrightarrow { CC' } =\overrightarrow { { c }_{ 1 } } -\overrightarrow { c }$

$\overrightarrow { AA' } +\overrightarrow { BB' } +\overrightarrow { CC' } =\left( \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } \right) -\left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } \right)$

$\displaystyle =3\left( \dfrac { \overrightarrow { { a }_{ 1 } } +\overrightarrow { { b }_{ 1 } } +\overrightarrow { { c }_{ 1 } } }{ 3 } -\dfrac { \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } }{ 3 } \right) =3\overrightarrow { GG' }$

Two wheels are constructed, as shown in Figure, with four spokes. The wheels are mounted one behind the other so that an observer normally sees a total of eight spokes but only four spokes are seen when they happen to align with one another. If one wheel spins at 6 rev/min, while other spins at 8 rev/min in same sense, how often does the observer see only four spokes?

0%
4 times a minute
0%
6 times a minute
0%
8 times a minute
0%
Once in a minute

Explanation

The observer will see only four spokes if the difference of angular

displacements of wheel 1 \& 2 is

$n\dfrac { 2\pi }{ 4 } =n\dfrac { \pi }{ 2 } \\$ Difference in angular displacement = (8rev/min - 6rev/min)\times t
$= 2\times 2\pi \times t=4\pi \times t (t in minutes)\\$ The least time after which observer will see only four spokes
is given by

$4\pi \times t = (1)\frac { \pi }{ 2 } \\$ \therefore t= 1/8 min Therefore no of times observer will see only four spokes in a min= 8
We can also think of this in following way: No of times observer will see only four spokes in a time period will be the value of n as it will tell us the number of $(\dfrac { \pi }{ 2 } )$ degrees rotations that the wheels have

taken relative to each other. And everytime they takes a $\frac { \pi }{ 2 }$ degree rotation relative to each other

observer will see only four spokes once.

In 1 min $2\times 2\pi =4\pi =8\frac { \pi }{ 2 }$ . Therefore 8 times a min.

$P$ is a point on the line through the point

$A$ whose position vector is

$\overrightarrow{a}$ and the line is parallel to the vector

$\overrightarrow{b}$ . If

$PA=6$ , the position vector of

$P$ is

0%
$\overrightarrow{a}+6\overrightarrow{b}$
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$\displaystyle \overrightarrow{a}+\dfrac{6}{\left |\overrightarrow{b} \right |}\overrightarrow{b}$
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$\overrightarrow{a}-6\overrightarrow{b}$
0%
$\displaystyle \overrightarrow{b}+\dfrac{6}{\left |\overrightarrow{a} \right |}\overrightarrow{a}$

Explanation

Line is parallel to vector

$\overrightarrow{b}$
The required vector is at a distance of 6 units from

$\overrightarrow{a}$
So, the required position vector will be

$\overrightarrow{a} + \dfrac{6}{|\overrightarrow{b}|} \overrightarrow{b}$ since we need a unit vector in the direction of

$\overrightarrow{b}$

A straight line

$r=a+\lambda b$ meets the plane

$r.n=0$ in

$P$ . The position vector of

$P$ is

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$\displaystyle a+\frac { a.n }{ b.n } b$
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$\displaystyle a+\frac { b.n }{ a.n } b$
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$\displaystyle a-\frac { a.n }{ b.n } b$
0%
None of these

Explanation

A straight line

$r=a+\lambda b$ meets the plane

$r.n=0$ in

$P$ for which

$\lambda$ is given by

$\displaystyle \left( a+\lambda b \right) .n=0\Rightarrow \lambda =-\frac { a.n }{ b.n }$

Thus, the position vector of

$P$ is

$\displaystyle r=a-\frac { a.n }{ b.n } b$

$[$ putting the value of

$\lambda$ in

$r=a+\lambda b]$

What is the work done in

$t$ seconds by a body of mass

$m$ moving in a circular path of radius

$r$ with a constant angular acceleration? Initially the body was at rest.

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$m\alpha^2R^2t^2$
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$2\pi mv^2t$
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$\dfrac{1}{2} m\alpha^2R^2t^2$
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zero

Explanation

The angular velocity

$\omega$ of mass

$m$ after time

$t$ is given by

$\omega=\omega_o+\alpha t=0+\alpha t=\alpha t$
According to work energy theorem, the work done on an object is equal to change in its kinetic energy i.e.

$W=\Delta K.E.$

$\Rightarrow W=K.E_f-K.E_i \\ \Rightarrow W=\dfrac{1}{2}mv^2-0 \\ \Rightarrow W=\dfrac{1}{2}m(\omega R)^2=\dfrac{1}{2}m\omega^2R^2 \\ \Rightarrow W=\dfrac{1}{2}m(\alpha t)^2R^2=\dfrac{1}{2}m\alpha^2R^2t^2$

A car is moving with a speed of

$72\, kmh^{-1}$ . The radius of its wheel its

$50\, cm$ . If its wheels come to rest after

$20$ rotations as a result of application of brakes, then the angular retardation produced in the car will be

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$23.5\,rads^{-2}$
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$0.25\,rads^{-2}$
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$6.35\,rads^{-2}$
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$zero$

Explanation

Given,

$v=72kmph=$

$72*\dfrac{5}{18}ms^{-1}$

$=20$

$ms^{-1}$ .
Now , angular velocity of wheel intially=

$\dfrac{v}{r}$
where ,

$r$ is radius of wheel.

$r=50cm$ {given}

$=0.5 m$
angular velocity intially=

$40 rad$

$s^{-1}$
and also given, That the wheel is rotated by

$20 revolutions$ .
angle of rotation say (

$\theta$ ) =

$20\times2$

$\pi rads^{-1}$

$\theta =40\pi rads^{-1}$
As we know that,

$\omega_{final}^{2}-\omega_{intial}^{2} =2\times \alpha\times \theta$ {let

$\alpha$ be angular accerlation}
By substituting the known values,

$\Rightarrow 0^{2}-40^{2}=2\times \alpha \times 40\pi$

$\Rightarrow \alpha=-6.3661 rads^{-1}$ {- sign indicates retardation}
Therefore answer is Option C.

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first two seconds it rotates through angle

$\theta_1$ . In the next two seconds it rotates through angle

$\theta_2$ . What is the ratio

$\theta_2 / \theta_1$ ?

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$4$
0%
$3$
0%
$2$
0%
$1$

A particle is moving along a circular path of radius 5 m and uniform speed 5

$ms^{-1}$ . What will be the average acceleration when the particle completes half revolution?

0%
$10\pi ms^{-2}$
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$10/\pi ms^{-2}$
0%
$10 ms^{-2}$
0%
zero

Explanation

average acceleration

$=\vec { a } =\dfrac { \Delta v }{ t } =\dfrac { { v }_{ 2 }-{ v }_{ 1 } }{ t } =\dfrac { (-5\hat { i } )-(5\hat { i } ) }{ t } =\dfrac { -10\hat { i } }{ t }$

and

$t=\dfrac { 5\pi }{ 5 } =\pi$

$\Rightarrow \vec { a } =\dfrac { -10\hat { i } }{ \pi }m{ s }^{ -2 } \\ \Rightarrow \left| a \right| =\dfrac { 10 }{ \pi }m{ s }^{ -2 }$

A bullet is fired horizontally with a speed of

$1500\ m/s$ in order to hit a target

$100\ m$ away. If

$g=10\ m/s^2$ . The gun should be aimed

0%
$15\ cm$ above the target
0%
$10\ cm$ above the target
0%
$2.2\ cm$ above the target
0%
directly towards the target

Explanation

The bullet performs a horizontal journey of

$100 cm$ with constant velocity of

$1500 m/s$ . The bullet also performs a vertical journey of h with zero initial velocity and downward acceleration

$g$
Therefore for horizontal journey, time

${t}=\dfrac{Distance}{velocity}$

$t=\dfrac{100}{1500}=\dfrac{1}{15} sec ........(1)$

The bullet performs vertical journey for this time
For vertical journey ,

$h=ut+\dfrac{1}{2}gt^2$

$h=0+\dfrac{1}{2}\times 10\times (\dfrac{1}{15})^2$

$h =\dfrac{10}{2\times 15\times 15}m =\dfrac{10\times 100}{2\times 15\times 15}cm$

$h=2.2 cm$
The gun should be aimed

$\dfrac{20}{9}$ cm above the target

When a lead storage battery is discharged, then:

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SO $_2$ is evolved
0%
lead is formed
0%
lead sulphate is consumed
0%
sulphuric acid is consumed

When a body moves with a constant speed along a circle:

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no work is done on it
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no acceleration is produced in the body
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no force acts on the body
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its velocity remains constant

Two identical particles are located at

$\overrightarrow{x}$ and

$\overrightarrow{y}$ with reference to the origin of three dimensional co-ordinate system. The position vector of centre of mass of the system is given by

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$\overrightarrow { x } - \overrightarrow { y }$
0%
$\displaystyle \frac {\overrightarrow { x } + \overrightarrow { y }} {{2}}$
0%
$(\overrightarrow { x } - \overrightarrow { y })$
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$\displaystyle \frac{\overrightarrow { x } - \overrightarrow { y }} {{2}}$

A wheel initially at rest, is rotated with a uniform angular acceleration. The wheel rotates through an angle

$\theta_1$ in first one second and through an additional angle

$\theta_2$ in the next one second. The ratio

$\theta_2/\theta_1$ is:

0%
$4$
0%
$2$
0%
$3$
0%
$1$

Explanation

Angle revolved in 1 sec and 2 sec are

$\theta_1$ and

$\theta_1 + \theta_2$ respectively.

Given:

$w_1=0$ using

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

$\theta_1= \dfrac{1}{2}\alpha (1)^2$

$\theta_1+\theta_2= \dfrac{1}{2}\alpha (2)^2$

Solving we get,

$\theta_2=\dfrac{3}{2}\alpha$

Thus

$\dfrac{\theta_2}{\theta_1}= 3$

A gun is aimed at a horizontal target. It takes

$\dfrac{1}{2} s$ for the bullet to reach the target. The bullet hits the target

$x$ meter below the aim. Then

$x$ is equal to (Take

$g=9.8\ m/s^2$ )

0%
$\dfrac{9.8}{4} m$
0%
$\dfrac{9.8}{8} m$
0%
$9.8 m$
0%
$19.6 m$

Explanation

$x=\dfrac{1}{2} gt^2$

$x=\dfrac{1}{2}\times 9.8 \times \dfrac{1}{2}\times\dfrac{1}{2} = \dfrac{9.8}{8} m$

A body is revolving with a constant speed along a circle .If its direction of motion is reversed but the speed remains the same. Then,

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the centripetal force will not suffer any change in magnitude
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the centripetal force will have its direction reversed
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the centripetal force will not suffer any change in direction
0%
the centripetal force would be doubled

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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