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CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 11 - MCQExams.com
CBSE
Class 11 Engineering Physics
Motion In A Plane
Quiz 11
A radio-controlled toy car travels along a straight line for a time of $$15s$$. The variation with time t of the velocity v of the car is shown.
What is the average velocity of the toy car for the journey shown by the graph?
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$$-1.5 ms^{-1}$$
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$$0.0 ms^{-1}$$
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$$4.0 ms^{-1}$$
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$$4.5 ms^{-1}$$
Explanation
Total Displacement $$=3\times10-6\times5=0 m$$
Total time takes$$=15s$$
Average Velocity $$=\dfrac {displacment}{time}=\dfrac 0{15}=0 m/s$$
The vector sum of the forces of 10 newton and 6 newton can be:
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2N
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8N
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18N
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20N
The height y and the distance X along the horizontal for a body projected in the vertical plane given by $$y=8t-{ 5t }^{ 2 }$$ and $$x=6t.$$ Then initial velocity of the projected body is $$\left( g=10{ ms }^{ -2 } \right) .$$
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$${ 8ms }^{ - 1}$$
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$${ 4ms }^{ -1 }$$
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$${ 10ms }^{ -1 }$$
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$${ 10/3ms }^{ - 1}$$
The position vector of a particle changes with time according to the relation $$\vec{r}(t)=15t^2\hat{i}+(4-20t^2)\hat{j}$$. What is the magnitude of the acceleration at $$t=1$$?
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$$40$$
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$$100$$
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$$25$$
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$$50$$
Explanation
$$\overrightarrow{r}(t)=15t^2 \hat{i}+(4-20t^2)\hat{j}$$
$$\overrightarrow{r}(t)=15(2 t^2)\hat{i}+(4\hat{i}-20(2t))$$
$$\overrightarrow{r}(t)=30t \hat{i}+(0-40\hat{j})$$
$$\overrightarrow{r}(t)=30t\hat{i}-40t\hat{j}$$
$$|\overrightarrow{r}(t)|=\sqrt{(30)^2+(-40)^2}$$
$$\Rightarrow \sqrt{900+1600}$$
$$\Rightarrow \sqrt{2500}$$
$$\Rightarrow 50$$
The position vector of a point P is $$\overrightarrow r=x \overrightarrow i + y \overrightarrow j+x \overrightarrow k$$, Where $$x,y,z,\epsilon N$$ and $$\overrightarrow a= \overrightarrow i+ \overrightarrow j+\overrightarrow k$$. If $$\overrightarrow r. \overrightarrow a=10$$, then the number of possible positions of P is ___________.
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$$30$$
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$$72$$
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$$66$$
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$$36$$
Explanation
The number of positive integral solution of the equation $$x_1+x_2+x_3+...x_r=n$$ is $$^{n-1}C_{r-1}$$
We have $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
Also,$$\vec{r}.\vec{a}=10$$
$$\Rightarrow\,x+y+z=10$$
Thus number of positive integral solution is $$= ^{10-1}C_{3-1}= ^{9}C_{2}=\dfrac{9!}{7!2!}=\dfrac{9\times 8}{2}=36$$
Option$$(d)$$ is correct.
A particle starts from points A with constant speed V on a circle of radius R find magnitude of average velocity after half revolution
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$$ \frac {2v}{\pi} $$
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$$ \frac {3v}{4\pi} $$
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$$ 3 \sqrt {3v} $$
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$$ \frac { 3\sqrt { 3v } }{ 2\pi } $$
Explanation
Diameter=$$2\times radius $$ $$= 2R$$
Distance covered $$=$$ Half the circumference $$= πR$$
Speed $$=$$Distance $$/$$ Time
Time $$=$$ Distance $$/$$ Speed $$= πR / V$$
Average Velocity $$=$$ Total Displacement $$/$$ Total Time
$$= 2R \times V / πR$$
$$= 2V / π$$
Hence,
option $$(A)$$ is correct answer.
A projectile thrown with an initial velocity of $$10 ms^{-1}$$ at an angle $$\alpha$$ with the horizontal, has a range of $$5 m.$$ Taking $$g = 10 ms^{-2}$$ and neglecting air resistance, what will be the estimated value of $$\alpha$$ ?
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$$15^o$$
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$$30^o$$
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$$45^o$$
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$$75^o$$
If the length of second's hand of a clock is $$10\, cm$$, the speed of its tip (in $$cm\, s^{-1}$$) is nearly
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$$2$$
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$$0.5$$
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$$1.5$$
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$$1$$
Explanation
$$\,\,r = wR = 10 \times \dfrac{{2\pi }}{{60}}$$
$$ = \dfrac{\pi }{3}cm/s \approx 1\,cm/s$$
Ans. (D)
The real force $$F$$ acting on a particle of mass $$m$$ performing circular motion acts along the radius of circle $$r$$ and is directed towards the center of circle. The square root of magnitude of such force is? ($$T =$$ Time Period)
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$$\dfrac{2\pi}{T}\sqrt{mr}$$
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$$\dfrac{Tmr}{4\pi}$$
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$$\dfrac{2\pi T}{\sqrt{mr}}$$
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$$\dfrac{T^2mr}{4\pi}$$
Explanation
For particle performing circular motion ,
$$Force = \dfrac{m \times v^2}{r}$$
$$v = r \omega$$
$$\therefore Force = m \times r \times {\omega}^2$$
also $$\omega = \dfrac{2 \pi}{T}$$
$$\therefore Force = m \times r \times \dfrac{2 \pi}{T} $$
$$\therefore \sqrt {Force} = \dfrac{2 \pi}{T} \sqrt{m r}$$
If the position vector $$\vec{a}$$ of point $$(12, n) $$ is such that $$\left | \vec{a} \right | = 13$$, then find the value (s) of $$n$$.
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$$\pm 6$$
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$$\pm 4$$
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$$\pm 5$$
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$$\pm 7$$
The minimum magnitude of resultant force is
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$$= 0$$
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$$> 0$$
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$$< 0$$
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$$\leq 0$$
Explanation
The resultant force is the combined effect of the several other forces present in the region. Since it is the combined effect, the resultant force cannot be less than zero. It can either have a positive value or it may be zero in case if all the forces cancel each other.
So option $$A$$ is correct.
A person from a truck, moving with a constant speed of $$60$$ km/h, throws a ball upwards with a speed of $$60$$ km/h. Neglecting the effect of Earth and choose the correct answer from the given choice.
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The person cannot catch the ball when it comes down since the truck is moving
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The person can catch the ball when it comes down, if the truck is stopped immediately after throwing the ball
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The person can catch the ball when it comes down, if the truck moves with speed less than $$60$$ km/h but does not stop
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The person can catch the ball when it comes down, if the truck moves with speed more than $$60$$ km/h
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The person can catch the ball when it comes down, if the truck continues to move with a constant speed of $$60$$ km/h
Explanation
When the person throws a ball in upward direction from moving truck, ball also acquire a speed of $$60$$ km/h in horizontal direction. So, if truck continues to move with same speed, ball comes down again in hands of person.
If electric current is assumed as vector quantity, then :
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Charge conservation principle fails
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Charge conservation principle does not fail
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Coulomb's law fails
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None of the above
A particle has a velocity $$u$$ towards east at $$t=0$$. Its accelerations towards west and is constant. Let $$x_{A}$$ and $$x_{B}$$ be the magnitude of displacement in the first $$10$$ seconds and the next $$10$$ seconds
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$$x_{A} < x_{B}$$
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$$x_{A}=x_{B}$$
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$$x_{A}>x_{B}$$
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the information is insufficient to decide the relation of $$x_{A}$$ with $$x_{B}$$
Explanation
the magnitude of displacement depends upon initial velocity, time duration, and acceleration. In this case, only time is known while magnitude of initial velocity and acceleration is unknown. These two displacements could only be comparable if direction of initial velocity and acceleration had been the same.since in this case directions are opposite, so there is not sufficient information.
Option D is correct.
A situation may be described by using different sets of coordinate axes having different orientations. Which of the following do not depend on the orientation of the axes?
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The value of a scalar
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Component of a vector
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A vector
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The magnitude of a vector
A stone is projected upwards and it returns to ground on a parabolic path. Which of the following remains constant?
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Speed of the ball
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Horizontal component of velocity
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Vertical component of velocity
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None of the above
Explanation
(A) Net acceleration is not equal to zero and hence speed of the ball changes.
(B) Acceleration along horizontal = 0, hence horizontal component of velocity does not change.
(C) Acceleration along vertical component $$=-g$$
(Negative sign indicates downward direction). Hence vertical component of velocity changes.
Mark correct option or options:
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Radial acceleration is equal to time derivative of radial velocity.
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Radial acceleration is not equal to time derivative of radial velocity
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Transverse acceleration is time derivative of transverse velocity
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Both $$(b)$$ and $$(c)$$ are correct
What is the $$\nabla \Phi$$ at the point $$(0, 1, 0)$$ of a scalar function $$\Phi$$, if $$\Phi = 2x^{2} + y^{2} + 3z^{2}$$?
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$$2\hat {j}$$
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$$3\hat {j}$$
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$$4\hat {i} + 2\hat {j}$$
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$$3\hat {i} + 3\hat {j}$$
If a number of particles are projected from the same point in the same plane so as to describe equal parabolas, then the vertices of their paths lie on a:
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Parabola
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Circle
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Square
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Rectangle
A particle is projected at an angle $$60^o$$ with the horizontal with a speed $$10$$ m/sec. Then latus rectum is: (Take $$g=10m/s^2$$)
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$$5$$ m
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$$15$$ m
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$$10$$ m
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$$0$$
Two buses A and B are moving around concentric circular paths of radii $$r_A$$ and $$r_B$$. If the two buses complete the circular paths in the same time, the ratio of their linear speeds is?
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$$1$$
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$$\dfrac{r_A}{r_B}$$
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$$\dfrac{r_B}{r_A}$$
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None of these
In figure the angle of inclination of the inclined plane is $$30^o$$. Find the horizontal velocity $$V_0$$ so that the particle hits the inclined plane perpendicularly.
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$$V_0 =\sqrt {\dfrac {2gH}{5}}$$
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$$V_0 =\sqrt {\dfrac {2gH}{7}}$$
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$$V_0 =\sqrt {\dfrac {gH}{5}}$$
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$$V_0 =\sqrt {\dfrac {gH}{7}}$$
Explanation
In a direction along the inclined plane.
$$0=V_0 \cos 30^o -g \sin 30^o t\ \Rightarrow t =\sqrt 3 V_0 /g$$
In a direction perpendicular to incline,
$$-H\ \cos 30^o =-V \sin 30^o t -\dfrac {1}{2}g \cos 30^o t^2$$
Putting the value of $$t$$ and solving, we get $$V_0 =\sqrt {\dfrac {2gH}{5}}$$
Mark the correct statement.
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$$|\vec{a} + \vec{b}| \geq |\vec{a}| +| \vec{b}|$$
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$$|\vec{a} + \vec{b}| \leq |\vec{a}| +| \vec{b}|$$
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$$|\vec{a} - \vec{b}| \geq |\vec{a}| +| \vec{b}|$$
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All of the above
Which of the following two statements is more appropriate:
(A) Two velocities are added using the triangle rule because velocity is a vector quantity.
(B) V
elocity is a vector quantity because two velocities are added using the triangle rule.
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Statement A
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Statement B
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Both A and B
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None of these
A. stone is thrown horizontally with a velocity of 10 m/ s at t = The radius of curvature of the stone's trajectory at t = 3 s is : $$ [ take g = 10 m/s^2 ] $$
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$$ 10\sqrt {10} m $$
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$$ 100 m $$
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$$ 100 \sqrt {10} m $$
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$$ 1000 m $$
The resultant of $$\underset{A}{\rightarrow}$$ and $$ \underset{B}{\rightarrow} $$ prependicular to $$ \underset{A}{\rightarrow} $$ what is a angle between $$ \underset{A}{\rightarrow} $$ and $$ \underset{B }{\rightarrow} $$
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$$ cos^{-1}\left ( \dfrac{a}{b} \right ) $$
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$$ cos^{-1}\left ( -\dfrac{a}{b} \right ) $$
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$$ sin^{-1}\left ( \dfrac{a}{b} \right ) $$
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$$ sin^{-1}\left ( -\dfrac{a}{b} \right ) $$
A body is undergoing uniform. circular motion then which of the following quantity is constant :
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velocity
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acceleration
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force
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kinetic energy
Which of the following two statement is not more appropriate
(a) Two volecities is added using triangle rule because velocity is vector quantity.
(b) velocity is vector quantity because two velocity are added using triangle rule
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(a)
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(b)
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(a) and (b)
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None of this
A plane is revolving around the earth with a speed if $$100\ km/hr$$ at a constant height from the surface of earth. The change in the velocity as it travels half circle is
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$$200\ km/hr$$
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$$150\ km/hr$$
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$$100\sqrt 2\ km/hr$$
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$$0$$
Explanation
As we know,
$$\Delta v=2v\sin \left( \dfrac{\theta}{2}\right)=2\times v\times \sin 90^o$$
$$=2\times 100=200\ km/hr$$
A particle is moving in a circular path of radius $$r$$. The displacement after half a circle would be:
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Zero
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$$\pi r$$
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$$2r$$
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$$2\pi r$$
Explanation
$$(c)\ 2r$$
Explanation: After half a circle, the particle will be diametrically opposite to its origin Hence, displacement is equal to diameter.
Figure below show a body of mass $$M$$ moving with the uniform speed on a circular path of radius, $$R$$. What is the change in acceleration in ging from $$P_1$$ to $$P_2$$
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Zero
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$$v^2/2R$$
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$$2v^2/R$$
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$$\dfrac{v^2}{R} \times \sqrt 2$$
Explanation
we know,
$$\triangle a=2 a \sin \left(\dfrac{\theta}{2}\right)=2 a \times \sin 45^o = \sqrt 2 a=\sqrt 2 \dfrac{v^2}{R}$$
The position vector of the point which divides the join of points with position vectors $$\vec a +\vec b$$ and $$2\vec a-\vec b$$ in the ratio $$1:2$$ is
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$$\dfrac {3\vec a+2\vec b}{3}$$
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$$\vec a$$
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$$\dfrac {5\vec a-\vec b}{3}$$
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$$\dfrac {4\vec a+\vec b}{3}$$
Explanation
$$(D)$$ is the correct answer. Applying section formula, the position vector of the required point is $$\dfrac {2(\vec a+\vec b)+1(2\vec a-\vec b)}{2+1}=\dfrac {4\vec a+\vec b}{3}$$
Since, the position vector of a point $$R$$ divides the line segment joining the points $$P$$ and $$Q$$, whose position vectors are $$\vec p$$ and $$\vec q$$ in the ration $$m:n$$ internally, is given by $$\dfrac{m\vec q +n\vec p}{m+n}$$
In uniform circular motion
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Both the angular velocity and the angular momentum vary
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The angular velocity varies but the angular momentum remains
constant
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Both the angular velocity and the angular momentum stay
constant
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The angular momentum varies but the angular velocity remains
constant
Explanation
It is known that $$L = I\omega$$
In uniform circular motion $$\omega = constant \therefore L = constant.$$
Which of the following is a vector
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Pressure
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Surface tension
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Momentum of inertia
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None of these
Explanation
Surface tension is a scalar insofar as it is always oriented parallel to the surface.
In higher classes we say moment of inertia is a tensor quantity but in 11th class it is given that moment of inertia is a scalar quantity.
We can put the walls of our container anywhere inside the gas, and the force per unit area (the pressure) is the same. Therefore, pressure is a scalar quantity.
hence option D is correct.
A particle is moving in a circle with uniform speed. Its motion is
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Periodic and simple harmonic
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Periodic but not and simple harmonic
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A periodic
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None of the above
Explanation
In SHM the particle should always have acceleration that is opposite to that of its displacement. In other words,
$$a\propto -x$$
This is clearly not the case when a particle moves in a circle. Hence, it will not be in SHM.
On the other hand, when the speed in uniform, the particle starts from one point in the circle and always returns to that point after a fixed time $$T$$ which is called as the time period. Hence, this motion is periodic.
We can conclude that when a particle is moving in a circle with uniform speed then the motion is periodic but not simple harmonic.
Velocity of a body on reaching the point from which it was projected upwards, is
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$$ \nu = 0 $$
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$$ \nu = 2u $$
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$$ \nu = 0.5 u $$
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$$ \nu = u $$
Explanation
Body reaches the point of projection with same velocity according to law of conservation of potential and kinetic energy .Also the value of acceleration is constant and time of ascent is equal to that of descent . Hence , option (D) is correct .
Which of the following is not a scalar quantity ?
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Time
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Volume
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Displacement
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Work
Explanation
C.Displacement
Let $$\overrightarrow{a}$$ and $$\overrightarrow{b}$$ be unit vectors inclined at an variable angle $$\theta \left( \theta \epsilon \left( 0,\frac{\pi}{2} \right)(\frac{\pi}{2}, \pi) \right).$$
Let $$g(\theta)= {\int}_{-(\overrightarrow{a}.\overrightarrow{b})^2}^{-\lambda} f^2(x)dx+ {\int}_{\lambda}^{|\overrightarrow{a}\times \overrightarrow{b}|^2} f^2(x)dx-\frac{2}{\lambda}, where \lambda > 0 , $$is function satisfying $$ \displaystyle f(x)+f(y)=\frac{x+y}{xy}, x, y\epsilon R-[0] \; and \; h(\theta)=-g(\theta)+|\overrightarrow{a}\times \overrightarrow{b}|^2.(\overrightarrow{a}. \overrightarrow{b}_1)^2, \overrightarrow{b}_1 = 2\overrightarrow{b}$$
If $$|g(\theta)|$$
is attaining its minimum value, then minimum distance between origin and the point of intersection
of lines $$\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{a}\times \overrightarrow{b}$$ and $$\overrightarrow{r}\times \overrightarrow{b} = \overrightarrow{b}\times \overrightarrow{a}$$ is
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$$\sqrt{2-\sqrt{2}}$$
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$$\sqrt{2+\sqrt{2}}$$
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$$\sqrt{\sqrt{2}+1}$$
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$$\sqrt{\sqrt{2}-1}$$
Explanation
$$|g(\theta)|=4 cosec^{2} 2\theta$$ is minimum when $$cosec^{2} 2\theta =1$$
So $$\theta=\dfrac{\pi}{4}, \dfrac{3 \pi}{4}$$
Distance between origin and point of intersection of given lines $$= |\overline{a}+ \overline{b}|=\sqrt{(\overline{a}+ \overline{b})(\overline{a}+ \overline{b})}$$ $$= \sqrt{1+1+2cos\theta}$$
So minimum distance $$=\sqrt{2+2(\dfrac{-1}{\sqrt{2}})}$$
$$=\sqrt{2-\sqrt{2}}$$
How much deep inside a man should go so that his weight becomes one forth of that point which is at height $$R$$ above the surface of the earth.
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$$\mathrm{R}/4$$
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$$15\mathrm{R}/16$$
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$$3\mathrm{R}/4$$
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$$\mathrm{R}/2$$
A particle projected from O and moving freely under gravity strikes the horizontal plane passing through O at a distance R from the starting point O as shown in the figure. Then
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there will be two angles of projection if $$R g < u^{2}$$
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the two possible angles of projection are complementary
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the product of the possible times of flight from O to A is $$\frac{2R}{g}$$
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there will be more than two angles of projection if $$R g = u^{2}$$
Explanation
$$R=\dfrac { u^2 \sin2\theta }{ g } $$
$$\Rightarrow \dfrac { Rg }{ u^2 } =\sin2\theta <1$$
Therefore there should be two values of $$\theta $$,
i.e. $$\theta_1$$ and $${ 90 }^{ o }-\theta_1$$ for same range.
It is obvious that both these angles are complementary.
$$T_1=\dfrac { 2u \sin\theta_1 }{ g } $$ and $$T_2=\dfrac { 2u \sin({ 90 }^{ o }-\theta_1) }{ g } $$
$$\therefore T_1\times T_2=\dfrac { 2u^2 \sin2\theta_1 }{ g^ 2 } $$
$$\Rightarrow \dfrac { 2R }{ g } $$
A particle moves in the $$x-y$$ plane according to the law $$x = kt, y = kt (1 - \alpha t) $$ where $$k$$ and $$\alpha $$ are positive constants and $$t$$ is time. The trajectory of the particle is:
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$$y=kx$$
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$$y=x- \dfrac {\alpha x^{2}}{k}$$
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$$y=-\dfrac{\alpha x^{2}}{k}$$
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$$y=\alpha x$$
Explanation
We have
$$x=kt ; t=\dfrac{x}{k}$$
Also
$$y=kt\left (1-\alpha t \right )$$
We put $$t=\dfrac{x}{k}$$, thus we get
$$y=\dfrac{kx}{k}\left (1-\frac{\alpha x}{k}\right); y=x-\dfrac{\alpha x^{2}}{k}$$
Velocity of a particle varies as $$\vec{V}=y\hat{i}-x\hat{j}$$ under the effect of a single variable force. Then
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force on the particle is conservative
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acceleration of the particle is ( $$x\vec{i}-y \vec{j}$$)
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particle moves with constant speed
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possible path of the particle is only a circle
Explanation
$$\displaystyle \vec{V}= y\vec{i}-x\vec{j}$$
$$ \displaystyle \Rightarrow \dfrac{dx}{dt}=y $$ & $$ \dfrac{dy}{dt}=-x \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y} \implies x^{2}+y^{2} = C(circle)$$
$$\displaystyle speed =\sqrt{x^{2}+y^{2}}= \sqrt{C}= constant$$
$$\displaystyle \vec{a}=\dfrac{dy}{dt}\vec{i}-\dfrac{dx}{dt}\vec{j}$$
$$\displaystyle \vec{a}=-x\vec{i}-y\vec{j} \Rightarrow F = -mx\vec{i}-my\vec{j} (conservative)$$
For a particle performing uniform circular motion, choose the correct
statement(s) from the following:
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Magnitude of particle velocity (speed) remains constant.
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Particle velocity remains directed perpendicular to radius vector.
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Direction of acceleration keeps changing as particle moves.
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Angular momentum is constant in magnitude but directionkeeps changing.
Explanation
Correct option are (A, B, C)
While a particle is in uniform circular motion. Then the following statements are true.
(i) speed will be always constant throughout.
(ii) velocity will be always tangential in the direction of motion at a particular point.
(iii) the centripetal acceleration a = v2/r and its direction will always towards centre of the circular trajectory.
(iv) angular momentum (mvr) is constant in magnitude and direction. And its direction is perpendicular to the plane containing r and v.
Important point: In uniform circular motion, magnitude of linear velocity and centripetal acceleration is constant but direction changes continuously.
Two particles projected from the same point with same speed u at angles of projection $$\alpha$$ and $$\beta$$ strike the horizontal ground at the same point. If $${ h }_{ 1 }$$ and $${ h }_{ 2 }$$ are the maximum heights attained by projectiles, $$R$$ be the range for both and $${ t }_{ 1 }$$ and $${ t }_{ 2 }$$ be their time of flights respectively, then
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$$\alpha +\beta =\dfrac { \pi }{ 2 } $$
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$$R=4\sqrt { { h }_{ 1 }{ h }_{ 2 } } $$
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$$\dfrac { { t }_{ 1 } }{ { t }_{ 2 } } =tan \alpha $$
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$$tan\alpha =\sqrt { \dfrac { { h }_{ 1 } }{ { h }_{ 2 } } } $$
Explanation
$$A$$ $$B$$
$$\theta = \alpha \theta = \beta$$
$$R = R R = R$$
$$H = h_{1} H = h_{2}$$
$$u = u u = u$$
$$t_{1} = \dfrac{2u\sin \alpha}{g} t_{2} = \dfrac{2u\sin \beta}{g}$$
$$\dfrac{t_{1}}{t_{2}}= \dfrac{2u\sin \alpha}{g} \cdot \dfrac{g}{2u\sin \beta} = \dfrac{\sin \alpha}{\sin (\pi/2 -\alpha )}=\tan \alpha $$ (Option C)
$$R_{1} = \dfrac{u^{2}\sin 2\alpha}{g} R_{2} = \dfrac{u^{2}\sin 2\beta}{g}$$
$$\because R_{1} = R_{2} \Rightarrow \sin 2\alpha = \sin 2\beta $$
$$\Rightarrow 2\alpha = \pi - 2 \beta$$
$$\Rightarrow \alpha + \beta= \pi/2$$ (Option A)
$$H_{1}= \dfrac{u^{2}\sin ^{2}\alpha}{2g} \ \ \ \ \ \ H_{2}= \dfrac{u^{2}\sin
^{2}\beta }{2g}$$
$$ \sqrt{\dfrac{H_{1}}{H_{2}}}= (\dfrac{u^{2}\sin^{2}\alpha}{2g} \cdot \dfrac{2g}{u^{2}\sin^{2}\beta})^{1/2}$$
$$ \sqrt{\dfrac{H_{1}}{H_{2}}}= \dfrac{\sin \alpha}{\sin \beta} = \dfrac{\sin \alpha}{\sin (\pi/2 -\alpha)} = \tan \alpha$$ (Option D)
$$\sqrt{H_{1}H_{2}} = \dfrac{u^{2}\sin \alpha \sin \beta }{2g} = \dfrac{2u^{2}}{4g}\sin \alpha \sin (\pi/2 -\alpha)$$
$$\sqrt{H_{1}H_{2}} =\dfrac{u^{2}}{4g}\sin 2\alpha$$
$$4\sqrt{H_{1}H_{2}} = R$$ (Option B)
What is the tangential acceleration?
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$$ 3\alpha$$
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$$ 2\alpha$$
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$$\alpha$$
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$$ 4\alpha$$
Explanation
We have
$$s=\alpha t^{2}$$
So, $$v=\displaystyle \frac{ds}{dt}=2\alpha t$$
and, $$\alpha '$$(tangential acceleration)$$\displaystyle =\frac{dv}{dt}=2\alpha $$
Two particles move on a circular path (one just inside and the other just outside) with angular velocities $$\omega $$ and $$ 5 \omega $$ starting from the same point. Then:
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they cross each other at regular intervals of time $$\displaystyle \frac{\pi }{2 \omega }$$ when their angular velocities are directed opposite to each other
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they cross each other at points on the path subtending an angle of 60$$^o$$ at the centre if their angular velocities are directed opposite to each other
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they cross at intervals of time $$\displaystyle \frac{\pi }{3 \omega }$$ if their angular velocities are directed opposite to each other
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they cross each other at points on the path subtending $$90^{\circ}$$ at the centre if their angular velocities are similar to each other
Explanation
Case 1
When both move in opposite direction
$$\theta_1 = wt$$
$$\theta_2 = 5 wt$$
They meet each other when
$$\theta_2 + \theta_1 = 2 \pi$$
$$\therefore 5 wt + wt = 2 \pi$$
$$\therefore wt = \displaystyle \frac{\pi}{3} = 60^o$$
i.e. the bodies cross each other at points subtending an angle of $$60^{\circ}$$ if their angular velocities are directed opposite to each other.
Case 2
When both move in same direction
$$\theta_1 = wt$$
$$\theta_2 = 5 wt$$
They meet each other when
$$\theta_2 - \theta_1 = 2 \pi$$
$$\therefore 5 wt - wt = 2 \pi$$
$$\therefore wt = \displaystyle \frac{\pi}{2} = 90^o$$
i.e. the bodies cross each other at points subtending an angle of $$90^{\circ}$$ if their angular velocities are similar.
Now, when oppositely directed, beat frequency:
$$= n_2 - n_1$$
$$\displaystyle = \frac{5 w}{2 \pi} - \left ( \frac{-w}{2 \pi} \right )$$
$$= \displaystyle \frac{3w}{\pi}$$
$$\therefore T = \dfrac{\pi}{3 w}$$
A body projected vertically upward with a velocity $$v$$ at $$t=0$$ is found at a height $$h$$ after $$1$$ second and after further $$6$$ seconds, it is found at the same height ($$g=10ms^{-2}$$). Then,
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$$h$$ is $$30 m$$
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$$v$$ is $$40 m/s$$
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maximum height is $$80 m$$
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distance moved in $$5th$$ second is $$5 m$$
Explanation
$$\theta =90\\ \Rightarrow t=\dfrac { 2v }{ g } \quad \& \quad H=\dfrac { { v }^{ 2 } }{ 2g } $$
Total time will be 8 seconds as it takes 1s to reach $$h$$ from initial position and then 6s to return back to $$h$$ and 1s again to reach the initial position.
$$t=8=\dfrac { 2v }{ g } \\ \Rightarrow v=40\quad m/s$$
$$H=\dfrac { { v }^{ 2 } }{ 2g } \\ \Rightarrow H=\dfrac { 40\times 40 }{ 2\times 10 } =80m$$
Now, Velocity will be 0 at top also it will reach there at $$t=4s$$ as total time is 8s
$$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$
$$s=\dfrac { 1 }{ 2 } \times 10\times 1=5m$$
Particles P and Q are undergoing uniform horizontal circular motions along concentric circles of different radii in clockwise sense. P completes each round in 2 minutes while Q does it in 5 minutes. Time required by Q to make one revolution around P is
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3 minutes
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10 minutes
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10/3 minutes
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This is not possible as Q is moving slower than P.
A particle has initial velocity, $$\displaystyle \vec{v}=3\hat{i}+4\hat{j}$$ and a constant force $$\displaystyle \vec{F}=4\hat{i}-3\hat{j}$$ acts on it. The path of the particle can be:
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straight line
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parabolic
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circular
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hyperbolic
Explanation
As it can be seen easily, the Force vector is constant and perpendicular to the velocity vector (dot product is zero). Thus such a path would be circular motion.
A particle moves in a circle of radius $$4cm$$ clockwise at constant speed $$2 cm/s$$. If $$\widehat{x}$$ and $$\widehat{y}$$ are unit acceleration vector along x and y-axis respectively (in $$cm/s^2$$), the acceleration of the particle at the instant half way between P and Q is given by
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$$-4(\widehat{x}+\widehat{y}) $$
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$$4(\widehat{x}+\widehat{y}) $$
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$$-(\widehat{x}+\widehat{y})/\sqrt{2} $$
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$$(\widehat{x}-\widehat{y})/4 $$
Explanation
Radial acceleration $$a_r=\omega^2 r=\dfrac{v^2}{r}=2^2/4=1 cm^2/s$$
Now this acceleration is directed toward origin in third quadrant.
Component of acceleration along negative x axis is $$a_x=-a_rcos45\, \hat x$$
Component of acceleration along negative y axis is $$a_y=-a_rsin45\, \hat y$$
$$\vec a_r=a_x+a_y=\dfrac{-(\hat x+\hat y)}{\sqrt{2}}$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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