The following table represents the velocity of a moving body at different instants of time. Times (s) 0 5 10 15 20 25 30 Velocity $$(m\,s^{-1})$$ 10 15 20 20 30 15 0 The velocity-time graph is as shown in figure. For which interval of time the body has a uniform motion ? Find the velocity in this time interval ?

from t = 10 s to t = 15 s; 20 m/s

from t = 0 s to t = 5 s; 10 m/s

from t = 5 s to t = 10 s; 20 m/s

from t = 15 s to t = 25 s; 30 m/s

If $$\vec{a}$$ is a vector and $$x$$ is a non-zero scalar, then

$$\;x\vec{a}$$ is a vector collinear to $$\vec{a}$$

$$\;x\vec{a}$$ is a vector in the direction perpendicular to $$\vec{a}$$

$$\;x\vec{a}$$ and $$\vec{a}$$ have independent directions

MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 5

A wheel rotates with the constant acceleration of

$2.0\ rad/s^2$ , if the wheel starts from rest the number of revolution it makes in first ten seconds will be approximately

0%
32
0%
24
0%
16
0%
8

Explanation

For circular angular motion, the formula for angular displacement

$\theta$ and angular acceleration

$\alpha$ is

$\theta=\omega t+\dfrac{1}{2}\alpha t^2 or \theta =0+\dfrac{1}{2}\alpha t^2$ where

$\omega =$ initial velocity
or

$\theta=\dfrac{1}{2}\times (2)(10)^2$

$\theta=100\ radians$

$2\pi$ radian are covered in 1 revolution.

$\therefore$ 1 radian is covered is covered in

$\dfrac{1}{2\pi}$ revolution
or 100 radian are covered in

$\dfrac{100}{2\pi}$ revolution.

$\therefore$ Number of revolution

$= \dfrac{50}{3.14}=16$

$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B},$ then

0%
$|\overrightarrow{C}|$ is always greater than $|\overrightarrow{A}|$
0%
It is possible to have $|\overrightarrow{C}| < |\overrightarrow{A}|$ and $|\overrightarrow{C}|<|\overrightarrow{B}|$
0%
$|\overrightarrow{C}|$ is always equal to $|\overrightarrow{A}|+|\overrightarrow{B}|$
0%
$|\overrightarrow{C}|$ is never equal to $|\overrightarrow{A}+\overrightarrow{B}|$

Explanation

A: It's not necessary, eg. if

$\vec { A } = \hat { i } , \vec { B } =-\hat { i }$ and hence

$\vec { C } = 0$

B: Again for the same example, statement is true.

$\vec { A } = \hat { i } , \vec { B } =-\hat { i }$ and hence

$\vec { C } = 0$

When

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

$|\vec{C}| =|\vec{A} + \vec{B}|$

C & D: it is not necessary

Which one of the following is not the vector quantity?

0%
Torque
0%
Displacement
0%
Velocity
0%
Speed

Match column I with column II and select the correct answer using the codes given below:
Column I Column II
(A) F = constant

$\neq$ 0 (p) Directional change
(B) F = 0 (q) Variable acceleration
(C) F

$\neq$ constant (r) No change in momentum
(D) F

$\neq$ 0 (s) If a body moves uniformly along a circle

0%
A - (p); B - (s); C - (q); D - (r)
0%
A - (q); B - (p); C - (r); D - (s)
0%
A - (p); B - (r); C - (q); D - (s)
0%
A - (s); B - (q); C - (p); D - (r)

Which one of the following is a scalar quantity?

0%
Displacement
0%
Electric field
0%
Acceleration
0%
Work

The following table represents the velocity of a moving body at different instants of time.

Times (s)	0	5	10	15	20	25	30
Velocity $(m\,s^{-1})$	10	15	20	20	30	15	0

The velocity-time graph is as shown in figure. For which interval of time the body has a uniform motion ? Find the velocity in this time interval ?

0%
from t = 0 s to t = 5 s; 10 m/s
0%
from t = 5 s to t = 10 s; 20 m/s
0%
from t = 10 s to t = 15 s; 20 m/s
0%
from t = 15 s to t = 25 s; 30 m/s

Explanation

According to above graph we conclude that the body has uniform motion from t

$=$ 10 s to t

$=$ 15 s in part 'bc'. Velocity is constant and is equal to

$20m{s}^{-1}$ during this interval.

Which one is a vector quantity?

0%
Time
0%
Temperature
0%
Flux density
0%
Magnetic field intensity

A disc is rotating in a room. A boy standing near the rim of the disc of radius

$R$ finds the water droplet falling from the ceiling is always falling on his head. As one drop hits his head, other one starts from the ceiling. If height of the roof above his head is

$H$ , then angular velocity of the disc is

0%
$\displaystyle \pi \sqrt{\frac{2g R}{H^{2}}}$
0%
$\displaystyle \pi \sqrt{\frac{2g H}{R^{2}}}$
0%
$\displaystyle \pi \sqrt{\frac{2g}{H}}$
0%
None of these

Explanation

Using equation of motion for drop,

$H= 0+ \dfrac{1}{2}gt^2$

$t= \sqrt{\dfrac{2H}{g}}$ which is equal to time period of the disc.

Let w be the angular velocity of disc, thus

$t= \dfrac{2\pi}{w}$

$\sqrt{\dfrac{2H}{g}}= \dfrac{2\pi}{w}$

$w= \pi\sqrt{\dfrac{2g}{H}}$

A point moves along a circle having a radius

$20 cm$ with a constant tangential acceleration

$5 cm$

$\displaystyle s^{-2}.$ How much time is needed after motion begins for the normal acceleration of the point to be equal to tangential acceleration?

0%
1 s
0%
2 s
0%
3 s
0%
4 s

Explanation

$a_t= a_c \implies 5= \dfrac{v^2}{r}$

$\dfrac{v^2}{20}= 5$ gives

$v= 10 cm/s$

Using,

$v= u+ a_tt$ , where u=0

$10= 0+ 5t$

$t= 2 s$

Three particles A, B and C move in a circle in anticlockwise direction with speeds 1

$\displaystyle ms^{-1}, 2.5 ms^{-1}\:and\:2 ms^{-1}$ respectively. The initial positions of A, B and C are as shown in figure. The ratio of distance travelled by B and C by the instant A, B and C meet for the first time is

0%
3:2
0%
5:4
0%
3:5
0%
data insufficient

Explanation

Let the particles meet at point P after time t.

Thus using

$\theta'=wt$ where

$w=\dfrac{v}{r}$

For A:

$\theta= \dfrac{1}{r}t$ ..................(1)

For B:

$\theta + \dfrac{\pi}{2}= \dfrac{2.5}{r}t$ ................(2)

For C:

$\theta + \pi= \dfrac{2}{r}t$ .....................(3)

Dividing (2) by (3), we get

$\theta= - 3\pi$

Thus, distance moved by B,

$\theta_B = -3\pi + \dfrac{\pi}{2}= -2.5 \pi$

Distance moved by C,

$\theta_C = -3\pi + \pi= -2 \pi$

$\theta_B : \theta_C= 5:4$

A shaft initially rotating at

$1725$ rpm is brought to rest uniformly in

$20s.$ The number of revolutions that the shaft will make during this time is

0%
$1680$
0%
$575$
0%
$287$
0%
$627$

Explanation

$0= \omega_{0}-\alpha t$

$\displaystyle \therefore \alpha= \frac{\omega_{0}}{t}=\frac{(2\pi)(1725/60)}{20}$

$=9 rad/s^{2}$

$\displaystyle \theta =\frac{1}{2} \alpha t^{2}=\frac{1}{2} \times 9 \times (20)^{2} =1800\: rad$

$\displaystyle N=\frac {\theta}{2\pi} =286.4 =287$

Which of the following quantities is not a vector?

0%
Mass
0%
Displacement
0%
Weight
0%
Acceleration

A particle rotates in a circular path of radius

$54 m$ with varying speed

$\displaystyle v= 4t^{2}.$ Here

$v$ is in m/s and

$t$ in second. Find angle between velocity and acceleration at

$t=3 s$ .

0%
$\displaystyle 45^{\circ}$
0%
$\displaystyle 75^{\circ}$
0%
$\displaystyle 90^{\circ}$
0%
$\displaystyle 15^{\circ}$

Explanation

Given :

$r =54$ m

$t = 3$ s

The velocity is given as

$v = 4t^2$ m/s

Radial acceleration

$a_r = \dfrac{v^2}{r} =\dfrac{16t^4}{54}$

$\therefore$

$a_r\bigg|_{t =3} = \dfrac{16 \times 3^4}{54} = 24$

$m/s^2$

Tangential acceleration

$a_t = \dfrac{dv}{dt} = 8t$

$\therefore$

$a_t\bigg|_{t =3} = 8\times 3 = 24$

$m/s^2$

Thus angle between

$a$ and

$v$ ,

$tan\theta = \dfrac{a_r}{a_t} = \dfrac{24}{24} =1$

$\implies$

$\theta =45^o$

In a golf tournament, if we designate the hole as the origin, and the putter (like a bat) is at -5.0 m at time 0 s, the velocity-time graph for the golf ball will be

0%
a straight line inclined to the time axis
0%
a straight line parallel to the time axis
0%
a concave curve
0%
a convex curve

Explanation

Displacement = change in position

From the given data,

$s = t$

$v=\dfrac{s_2-s_1}{t_2-t_1}$

$=\dfrac{t_2-t_1}{t_2-t_1} = 1\ m/s$

What is the change in velocity in the question?

0%
$\displaystyle u\sin \alpha$ (downwards)
0%
$\displaystyle u\cos \alpha$ (downwards)
0%
$\displaystyle 2u\sin \alpha$ (downwards)
0%
$\displaystyle 2u\cos \alpha$ (downwards)

Explanation

From figure,

$\vec{u_x} = u cos\alpha \hat{i}$

$\vec{u_y} =u sin\alpha \hat{j}$

x direction :

$a_x = 0$

$\implies \vec{V_x} = \vec{u_x} = u cos\alpha \hat{i}$

y direction :

$V_y^2 - u_y^2 = 2a_y S_y$

$\therefore$

$V_y^2 - u_y^2 = 2 (g) (0)$

$\implies \vec{V_y} = - \vec{u_y} = - u sin\alpha \hat{j}$

$\therefore$ Change in velocity

$\Delta\vec{V} = [u cos\alpha \hat{i}-usin\alpha \hat{j}] -[ ucos\alpha \hat{i} + u sin\alpha \hat{j}] = -(2u) sin\alpha \hat{j}$

$\implies$ Change in velocity

$= 2u$

$sin\alpha$ (downwards)

A particle is projected from horizontal making an angle

$\displaystyle 60^{\circ}$ with initial velocity

$40\ ms^{-1}$ . Find the time taken by the particle to make angle

$\displaystyle 45^{\circ}$ from horizontal.

0%
$1.5\ s$
0%
$2.5\ s$
0%
$3.5\ s$
0%
$4.5\ s$

Explanation

At $\displaystyle 45^{\circ}, v_{x}= v_{y}$

or $\displaystyle u_{x}= u_{y}-gt$

$\displaystyle \therefore t= \dfrac{u_{y}-u_{x}}{g}$

$\displaystyle = \dfrac{40\left ( \sin 60^{\circ}-\sin 30^{\circ} \right )}{9.8}=1.5 \ s$

Following sets of three forces act on a body. Whose resultant cannot be zero?

0%
$\;10,10,10$
0%
$\;10,10,20$
0%
$\;10,20,20$
0%
$\;10,20,40$

Explanation

$\textbf{Step 1: Conclusion from triangle law of addition}$

From the triangle law of vector addition,

The resultant must be zero if the sum of two forces

$\geq$ remaining force

$\textbf{Step 2: Analysing options}$

In all the options except D, the sum of two forces is greater or equal to the remaining force

However, For option D,

$10 + 20 = 30$

$30 < 40$

Hence, the resultant can not be zero for option D.

$\therefore$ Option

$D$ is the correct answer.

The distance between the point

$P(x,y,z)$ and plane

$xz$ is :

0%
$x$
0%
$y$
0%
$z$
0%
$xy$

Explanation

The coordinate of any point in xz plane is

$Q(x,0,y)$

Thus, the distance between point P and Q is

$=\sqrt{(x-x)^2+(y-0)^2+(z-z)^2}=y$

$A, B, C$ are fixed points and

$P$ is a variable point such that the resultant of forces at

$P$ represented by

$\displaystyle \vec{PA}$ and

$\displaystyle \vec{PB}$ always passes through

$C$ and

$D$ is the mid-point of

$AB$ , then locus of

$P$ is

0%
circle passing through $A, B, C$ and $D$
0%
straight line passing through $C$ and $D$
0%
straight line passing through $A$ and $B$
0%
none of these

Explanation

Resultant of forces represented by

$\bar {PA}$ and

$\bar{PB}$ i.e. by

$2\bar{PD}$ where

$D$ is the middle point of

$AB$ . Since

$A$ and

$B$ are fixed. Thus

$D$ is also fixed point. But the resultant also passed through

$C$ , another fixed point. Hence the locus of

$P$ is a straight line joining

$C$ and

$D$

On a bright sunny day, a boy standing on a sea shore is flying a kite. In what direction will his kite fly ?

0%
Towards the sea
0%
upward the sea
0%
downward the sea
0%
none of these

Figure shows three vectors

$\vec{a},\vec{b}$ and

$\vec{c}$ , where

$R$ is the midpoint of

$PQ$ . Then which of the following relations is correct?

0%
$\;\vec{a}+\vec{b}=2\vec{c}$
0%
$\;\vec{a}+\vec{b}=\vec{c}$
0%
$\;\vec{a}-\vec{b}=2\vec{c}$
0%
$\;\vec{a}-\vec{b}=\vec{c}$

Explanation

$\textbf{Step 1: Applying triangle law of vector addition [Refer Fig.]}$

From triangle law of vector addition,

$\Delta OPR$

$\vec {a} = \vec {c} + \vec {RP}$

$.... (1)$

and In

$\Delta ORQ$

$\vec {b} = \vec {c} + \vec {RQ}$

$.... (2)$

$\textbf{Step 2: Equation solving}$

Adding

$eq^{n} (1)$ and

$(2)$

$\vec {a} + \vec {b} = 2\vec {c} + \vec {RP} + \vec {RQ}$

Since R is midpoint of PQ , therefore

$\vec {RP} = -\vec {RQ}$

$\Rightarrow \vec {a} + \vec {b} = 2\vec {c} + \vec {RP} - \vec {RP}$

$\Rightarrow \vec {a} + \vec {b} = 2\vec {c}$

Hence option

$A$ is correct.

2111265_272943_ans_1f26daa765444758a07dc2f150c00a5c.png

How many minimum number of coplanar vectors which represent same physical quantity having different magnitudes can be added to give zero resultant.

0%
(A) $\;2$
0%
(B) $\;3$
0%
(C) $\;4$
0%
(D) $\;5$

Explanation

As magnitude of the vectors are not equal so two vectors can not give zero resultant. According to triangle law of vector addition, minimum three vectors are needed to get zero resultant. From figure,

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

So,

$\vec A+\vec B-\vec C=0$

A physical quantity which has a direction:-

0%
must be a vector
0%
may be a vector
0%
must be a scalar
0%
must be zero.

$\vec{a}$ is a vector and

$x$ is a non-zero scalar, then

0%
$\;x\vec{a}$ is a vector in the direction perpendicular to $\vec{a}$
0%
$\;x\vec{a}$ is a vector collinear to $\vec{a}$
0%
$\;x\vec{a}$ and $\vec{a}$ have independent directions
0%
$x\vec a$ is a scalar

Explanation

By multiplying

$x$ which is a non zero scalar to

$\vec{a}$ we get

$x\vec{a}$ its magnitude increases by

$x$ .

Now it is a different vector which makes an angle of zero degree with

$\vec{a}$ therefore it is collinear to

$\vec{a}$ .

$\displaystyle \vec{a},\vec{b},\vec{c},\vec{d},\vec{e},\vec{f}$ are position vectors of 6 points A, B, C, D, E & F respectively such that

$\displaystyle 3\vec{a}+4\vec{b}=6\vec{c}+\vec{d}=4\vec{e}+3\vec{f}=\vec{x}$ then

0%
$\displaystyle \overline{AB}$ is parallel to $\displaystyle \overline{CD}$
0%
line AB, CD and EF are concurrent
0%
$\displaystyle \frac{\vec{x}}{7}$ is position vector of the point dividing CD in ratio 1 : 6
0%
A, B, C, D, E & F are coplanar

Explanation

Given condition

$3\vec{a}+4\vec{b}=6\vec{c}+\vec{d}=4\vec{e}+3\vec{f}=\vec{x}$

It is clearly seen that sum of coefficient

$7=7=7=\vec{x}$

So lines are concurrent

position vector of c and d

$position\ vector=\dfrac{6c+d}{7}$

from above condition

$position\ vector=\dfrac{x}{7}$

When a ceilling fan is switched off its angular velocity reduces to

$50\%$ while it makes

$36$ rotations. How many more rotation will it make before coming to rest (Assume uniform angular retardation):

0%
$\;18$
0%
$\;12$
0%
$\;36$
0%
$\;48$

Explanation

Let the initial angular velocity of the fan be

$w$

$\therefore$ Its angular velocity when the fan makes 36 rotation

$w_1 = 0.5w$

Using

$w_2^2 - w_1^2 = 2\alpha S$ where

$S = 2\pi (36)$ radian

$\therefore$

$(0.5w)^2 - w^2 =2 \alpha \times 2\pi (36)$

$\implies 4\pi \alpha =\dfrac{-0.75w^2}{36}$

Let the fan makes

$n$ more rotation before coming to rest.

Using

$w_3^2 - w_2^2 = 2\alpha S$ where

$S =2\pi n$ radian and

$w_3 = 0$

$\therefore$

$0 -(0.5w)^2 = 2 \alpha (2\pi n)$

$- 0.25 w^2 = \dfrac{-0.75 w^2}{36} \times n$

$\implies n = 12$

Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first-

0%
The faster one
0%
The slower one
0%
Both will reach simultaneously
0%
Depends on the masses

An object is projected with a velocity of

$20 m/s$ making an angle of

$45$ with horizontal. The equation for the trajectory is

$h = Ax - Bx^{2}$ , where

$h$ is height,

$x$ is horizontal distance,

$A$ and

$B$ are constants. The ratio

$A : B$ is-

$($

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

$)$

0%
$1 : 5$
0%
$5 : 1$
0%
$1 : 40$
0%
$40 : 1$

Explanation

horizontal and vertical component of velocity are same and equal to $20 cos 45^o = 10\sqrt{2}$

hence horizontal distance is $x = 10\sqrt{2} t$

and vertical distance is $s = ut + \dfrac{1}{2}at^2 = 10\sqrt{2}t - \dfrac{1}{2}gt^2$

therefore $h = 10\sqrt{2}t - 5t^2 = A \times 10\sqrt{2}t -B\times 200t^2$

$10\sqrt{2}A = 10\sqrt{2}$

$A = 1$

$200B = 5$

$B = \dfrac{1}{40}$

therefore $A:B = \dfrac{1}{\dfrac{1}{40}} = 40:1$

Alternatively, $y = x( A-Bx ) =0$ when particle reaches the ground.

Hence, $R = \dfrac{A}{B}$

But, $R = \dfrac{u^2 \times \sin( 2 \theta)}{g} = \dfrac{20^2 \sin 90^0}{10}= 40$

$\Rightarrow \dfrac{A}{B} = 40: 1$

Which of the following is scalar ?

0%
Current
0%
Velocity
0%
Force
0%
Acceleration

Explanation

Although the current has both magnitude and direction but it is a scalar quantity because it does not follow the vector addition rule. Mathematically it is defined as

$I=\int \vec{J}. \vec{dA}$ .

An athlete completes half a round of a circular track of radius,

$R$ . Then, the displacement and distance covered by the athlete are

0%
$2R\;and\;\pi R$
0%
$\pi R\;and\;2 R$
0%
$R\;and\;2\pi R$
0%
$2\pi R\;and\;R$

Explanation

For completing the half round of circle, the distance covered will be half of circumference of the circle i.e.,

$d=\pi R$ .
Displacement is the shortest distance between initial and final position, so the displacement will be just the diameter of circle.

$S= 2R$

Which of the following pairs of quantities have same dimensional formula of which one is a vector and other is scalar ?

0%
Work and energy
0%
Torque and work
0%
Impulse and momentum
0%
Power and pressure

Explanation

Work is equal to dot product of force and distance travelled =

$\vec{F}.\vec{r}$ and hence is a scalar. Torque is equal to

$\vec{r}\times \vec {F}$ and hence is a vector, though having same dimensions.

An athlete completes one round of a circular track of radius

$R$ in

$40$ seconds. What will be his displacement at the end of

$2$ minute

$20$ seconds?

0%
$0$
0%
$2R$
0%
$2\pi R$
0%
$7\pi R$

Position vector of a particle is given by

$\vec{r}=a cos\omega t\hat {i}+a sin\omega t\hat {j}$ . Which of the following is/are true?

0%
velocity vector is parallel to position vector
0%
velocity vector is perpendicular to position vector
0%
acceleration vector is directed towards the origin
0%
acceleration vector is directed away from the origin

Explanation

$\vec{r}=acos\omega t\hat {i}+asin \omega t\hat {j}$

$velocity=\vec{v}=\displaystyle\frac{d\vec{r}}{dt}=-a\omega sin \omega t\hat {i}+a\omega cos\omega t\hat {j}$

$Acceleration=\displaystyle\frac{d^2\vec{r}}{dt^2}=-a\omega^2cos\omega t\hat {i}-a\omega^2sin\omega t\hat {j}=-\omega^2\vec{r}$

Forces proportional to

$AB,BC$ and

$2CA$ act along the sides of triangle

$ABC$ in order. Their resultant represented in magnitude and direction as

0%
$\;\overrightarrow{CA}$
0%
$\;\overrightarrow{AC}$
0%
$\;\overrightarrow{BC}$
0%
$\;\overrightarrow{CB}$

Explanation

For triangle $ABC\,\colon\,\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}$

Now $\overrightarrow{AB}+\overrightarrow{BC}+2\overrightarrow{CA}$

$=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}+\overrightarrow{CA}=\overrightarrow{CA}$

A car runs at constant speed on a circular track of radius 100 m taking 62.8 s on each lap. What is the average speed and average velocity on each complete lap?

$(\pi\, =\, 3.14)$

0%
Velocity 10 m/s, speed 10 m/s
0%
Velocity zero, speed 10 m/s
0%
Velocity zero, speed zero
0%
Velocity 10 m/s, speed zero

Explanation

Radius of lap = 100m
Circumference

$= 2\pi R = 200 \pi = 628m$
time taken to cover one lap = 62.8s
Hence average speed

$= \dfrac{628}{62.8} = 10ms^{-1}$
Average velocity = 0 as the inital and final positions are same (after completing a lap) (displacement = 0)
Hence option B is correct.

A ball is thrown upwards. It returns to ground describing a parabolic path.Which of the following remains constant.

0%
Speed of the ball
0%
Kinetic energy of the ball
0%
Vertical component of velocity
0%
Horizontal component of velocity.

Explanation

$\textbf{Explanation}$

When the ball is thrown upwards , acceleration due to gravity is acting downwards and so it is represented as $-g$ .
So, this means that the vertical component of the velocity is changing.
The speed in vertical direction decreases, that is the reason ball has zero velocity in vertical direction ultimately comes to rest at highest point of path.
Now, the acceleration along the horizontal component is zero. So, the horizontal component of velocity is constant.

Since the speed is changing, this means that the kinetic energy of the ball also changes.

Hence, the horizontal component of velocity is constant.

The correct answer is option (D).

A body is projected with a velocity of

$98\;ms^{-1}$ . If acceleration due to gravity be

$9.8\;m/s^{2}$ , then:-

0%
It will go upto the maximum height of $980 m$ , after $10second$
0%
It will go upto the height of $490 m$ , is $5 second$
0%
It will come back to earth in $10 seconds$
0%
Its speed reduced to $49\;ms^{-1}$ after $5 second$

Explanation

$v = u+at$

$v = 98 - 9.8 \times 5 = 98-49 = 49 m/s$

Vector sum of two forces of 10N and 6N cannot be :-

0%
4N
0%
8N
0%
12N
0%
2N

Explanation

Magnitude of

$\vec A + \vec B$ has to lie between

$A-B$ and

$A+B$ , i.e

$10-6$ and

$10+6$ , or

$4$ and

$16$ .

$2$ is out of this range, hence it's the correct answer.

Figure shows the three vectors

$\overrightarrow a , \overrightarrow b$ and

$\overrightarrow c$ , where R is the midpoint of PQ. Then which of the following relations is correct ?

0%
$\overrightarrow a + \overrightarrow b = 2 \overrightarrow c$
0%
$\overrightarrow a + \overrightarrow b = \overrightarrow c$
0%
$\overrightarrow a - \overrightarrow b = 2 \overrightarrow c$
0%
$\overrightarrow a - \overrightarrow b = \overrightarrow c$

Explanation

Position vector of P is

$\vec a - \vec 0 = \vec a$
Similarly, Position vector of Q is

$\vec b - \vec 0 = \vec b$
Thus, position vector of the mid-point of PQ should be

$\frac{\vec a + \vec b}{2}$
Hence, Position vector of R is

$\vec c - \vec 0 = \vec c$ , which is equal to

$\frac{\vec a + \vec b}{2}$
Thus,

$\vec a + \vec b = 2\vec c$

A car runs at constant speed on a circular track of radius

$10\ m$ taking

$6.28\ s$ on each lap. The average speed and average velocity on each complete lap is

0%
Velocity $10\ m/s$ , speed $10\ m/s$
0%
Velocity $0$ , speed $10\ m/s$
0%
Velocity $0$ , speed $0$
0%
Velocity $10\ m/s$ , speed $0$

Explanation

Since after each lap the position of the car is same as initial so the average velocity will be zero.
Total Distance

$= 2\pi r =2\times 3.14 \times 10 = 62.8\ m$
Total time

$= 6.28\ s$
Average speed

$= \dfrac dt =\dfrac{62.8}{6.28}= 10\ m/s$

The equation of the path of the projectile is

$y=0.5x-0.04x^{2}$ The initials speed of the projectile is

0%
$10\;m/s$
0%
$15\;m/s$
0%
$12.5\;m/s$
0%
None

Explanation

$y=0.5x-0.04x^{2}= x( 0.5-0.04x)$

When $y=0, x=0\: or\: R$

$\therefore R = \dfrac{0.5}{0.04}=12.5$

$\displaystyle y=x\;tan\theta\left[1-\dfrac{x}{R}\right]\Rightarrow tan\theta =\dfrac{1}{2};R=12.5m$

But $\displaystyle R=\dfrac{4u^{2}sin\theta\;cos\theta}{g}$

$\displaystyle \Rightarrow u=\sqrt{\dfrac{Rg}{2\;sin\theta\;cos\theta}}=\sqrt{\dfrac{12.5\times 10}{2\times \dfrac{1}{\sqrt{5}}\times \dfrac{2}{\sqrt{5}}}}=12.5m/s$

Following sets of three forces act on a body. Whose resultant cannot be zero ?

0%
10, 10, 10
0%
10, 10, 20
0%
10, 20, 20
0%
10, 20, 40

Explanation

Let, the three forces be

$\vec { P } , \vec { Q }, \ \vec { R }$ .

And

$\vec { R }$ has the maximum magnitude.

$\angle \vec { P } \vec { Q } = \theta \\ \vec { P } + \vec { Q } + \vec { R } = 0\\ \Rightarrow \vec { R } = -\left( \vec { P } \quad +\quad \vec { Q } \right) \\ R= \sqrt { { P }^{ 2 }\quad +\quad { Q }^{ 2 }\quad +\quad 2PQ\cos { \theta } } \\ \cos { \theta } \ \in \ \left[ -1,\ 1 \right]$

Checking all the four options, only option D is not possible as:

$40> \sqrt { 500+ 400 } \qquad \left( substituting\ \cos { \theta } =1 \right)$

A body moves along a curved path of a quarter circle.Calculate the ratio of distance to displacement :

0%
$11:7$
0%
$7:11$
0%
$11:7\sqrt{2}$
0%
$7:11\sqrt{2}$

Explanation

Perimeter of circle is

$2\pi R$

Hence, distance on the quarter circle is

$d$ =

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

Displacement is

$D$ =

$R\sqrt{2}$

$\therefore$

$d:D = \dfrac{\pi R}{2} :R\sqrt{2}$

$\pi = \dfrac{22}{7}$

$d:D = 11:\sqrt{2}\times 7$

Statement-1: Electric curent is a scalar quantity
and
Statement -2: Electric current does not obey law of vector addition

0%
Statement -1 is True Statement -2 is True;

Statement -2 is a correct explanation for Statement -1
0%
Statement -1 is True Statement -2 is True;

Statement -2 is NOT a correct explanation for Statement -1
0%
Statement -1 is True Statement -2 is False;
0%
Statement -1 is False Statement -2 is True;

Velocity - Time graph of a particle is given. Which of the following statement(s) is/are correct.

0%
net displacement is zero
0%
total distance travelled is 80 metre
0%
acceleration of the particle is constant
0%
direction of velocity is constant

A body is projected horizontally from the top of a tower with initial velocity

$18\ ms^{-1}$ . It hits the ground at angle

$45^{\circ}$ . What is the vertical component of velocity when strikes the ground?

0%
$18 \sqrt{2}\, ms^{-1}$
0%
$18 ms^{-1}$
0%
$9 \sqrt{2}\, ms^{-1}$
0%
$9 ms^{-1}$

Explanation

The horizontal component of velocity remains a constant.
i.e. Horizontal component is always equal to

$18ms^{-1}$
When the body strikes the ground, (strikes at

$45^{\circ}$ )
Both horizontal and vertical components are equal.
Hence vertical component

$= 18ms^{-1}$
Option

$B$ is correct.

For four particles A, B, C & D, the velocities of one with respect to other are given as

$\bar{V}_{DC}$ is 20 m/s towards north,

$\bar{V}_{BC}$ is 20 m/s towards east and

$\bar{V}_{BA}$ is 20 m/s towards south. Then

$\bar{V}_{DA}$ is:

0%
20 m/s towards north
0%
20 m/s towards south
0%
20 m/s towards east
0%
20 m/s towards west.

Explanation

It is given that:

${V}_{DC} = 20 \hat {j}$ m/s

${V}_{BC} = 20 \hat {i}$ m/s

${V}_{BA} = -20 \hat {j}$ m/s

${V}_{DA}$ can be calculated by using the given relative velocities.

${V}_{DA}$

$={V}_{DC}$ +

${V}_{BA}$ -

${V}_{BC}$

$=20\hat { j }$ + (-

$20\hat {j}$ ) -

$20\hat { i }$

$=-20\hat { i}\ m/s$

The direction of

${V}_{DA}$ is west and the magnitude is

$20\ m/s$ .

The maximum and minimum magnitudes of the resultant of two forces are 17 newton and 7 newton respectively, if these two forces are acting at right angle to each other, then the value of resultant of these forces in newton will be:

0%
10
0%
12
0%
13
0%
24

Explanation

$|F_{max}|=17\ N\Rightarrow F_1+F_2=17$

$|F_{min}|=7\ N \Rightarrow F_1-F_2=7$

$\Rightarrow F_1=12N$ and

$F_2=5N$

When,

$\theta =90^0$

$F_{resultant}=\sqrt{F_1^2+F_2^2+2F_1F_2Cos\theta}$

$= \sqrt{12^2+5^2+0} \\=13N$

A particle is dropped from height 100 m and another particle is projected vertically up with velocity 50 m/s from the ground along the same line. Find out the height where two particle will meet? (take g = 10

$m/s^2$ )

0%
100 m
0%
80 m
0%
120 m
0%
40 m

Explanation

Relative displacement

$= 100\ m$

Relative acceleration

$= 0\ m/s^2$ because both have acceleration

$g$

Relative initial velocity

$= -50\ m/s$

Using relative motion,

$s = ut + (1/2)at^2$

$100 = 50 t + (1/2)\times 0 \times t^2$

$t = 2\ s$

For particle projected from ground,

$s= ut + (1/2)at^2$

$s = 50\times 2 - (1/2) \times 10 \times 2^2$

$= 80\ m$

A fan is running at

$3000\ rpm$ . It is switched off. It comes to rest by uniformly decreasing its angular speed in

$10\ seconds$ . The total number of revolutions in this period.

0%
$150$
0%
$250$
0%
$350$
0%
$300$

Explanation

$\omega_f = \omega_i + \alpha t$

$0 = 3000 / 60 + \alpha (10)$

$\alpha = -5 \ revolutions/s^2$

$\omega_f^2 = \omega_i^2 + 2 \alpha \Delta \theta$

$0 = (3000/60)^2 + 2 \times (-5) \times \Delta \theta$

$\Delta \theta = 250\ revolutions$

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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