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CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 13 - MCQExams.com
CBSE
Class 11 Engineering Physics
Motion In A Plane
Quiz 13
If $$\overline {A}=2\hat {i}+3\hat {j}-\hat {k}$$ and $$\overline {B}=-\hat {i}+3\hat {j}+4\hat {k}$$, then projection of $$\overline {A}$$ on $$\overline {B}$$ will be:
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$$\dfrac{3}{\sqrt{13}}$$
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$$\dfrac{3}{\sqrt{26}}$$
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$$\sqrt{\dfrac{3}{26}}$$
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$$\sqrt{\dfrac{3}{13}}$$
Explanation
Given,
$$\vec A=2\hat i+3\hat j-\hat k$$
$$\vec B=-\hat i+3\hat j+4\hat k$$
Projection of $$\vec A$$ on $$\vec B$$ will be
$$=\dfrac{\vec A.\vec B}{|\vec B|}=\dfrac{(2\hat i+3\hat j-\hat k).(-\hat i+3\hat j+4\hat k)}{\sqrt{(-1)^2+(3)^2+(4)^2}}$$
$$=\dfrac{-2+9-4}{\sqrt{26}}$$
$$=\dfrac{3}{\sqrt{26}}$$
The correct option is B.
A merry-go-round, made of a ring-like platform of radius R and mass M. is revolving with angular speed $$\omega$$? A person of mass M is standing on it. At one instant., the person jumps off the
round, radially away from the center of the round (as seen from the round). The speed of the round afterwards is:
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$$2\omega$$
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$$\omega$$
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$$\dfrac{\omega}{2}$$
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0
Explanation
Here angular momentum is converted
$$L_i = L_f$$
$$(I_{ring} + MR^2) \omega = (L_{man} + L_{ring}) \omega'$$ $$L_{man} = L_{ring}$$
[A man is assumed to at circumstances of ring and is final angular speed of ring after men jump off]
$$(MR^2 +MR^2) \omega = O + MR^2 \omega'$$
$$2 \omega = \omega'$$
Hance (A) option is correct
If $$v_1\cos \theta_1 = v_2\cos \theta_2$$, then choose the incorrect statement.
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One particle will remain exactly below or above the other particle
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The trajectory of one with respect to other will be a vertical straight line
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Both will have the same range
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None of these
A body thrown vertically up with initial velocity $$52 \ m/s$$ from the ground passes twice a point at $$h$$ height above at an interval of $$10 \ s$$. The height $$h$$ is $$(g = 10 \ m/s^2)$$:
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$$22 \ m $$
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$$10.2 \ m$$
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$$11.2 \ m$$
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$$15 \ m$$
For a particle in a uniform circular motion (speed is $$v$$ and radius of revolution is $$R$$)
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$$\vert \dfrac{d \vec{v}}{dt} \vert = \dfrac{d}{dt} (\vert \vec{v} \vert)$$
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$$\dfrac{dv}{dt} = v^2/R$$
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Average acceleration in half revolution is $$\dfrac{4 v^2}{\pi R}$$
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Ratio of average acceleration and average velocity is independent of the angular displacement.
Find the resultant of three vectors $$\vec{OA}, \vec{OB}$$ and $$\vec{OC}$$ shown in the following figure. The radius of the circle is R.
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$$2R$$
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$$R(1 + \sqrt 2)$$
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$$R \sqrt 2$$
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$$R(\sqrt 2 -1)$$
Explanation
$$\textbf{Step 1: Resolving Vector}$$
$$\textbf{[Ref. Fig. 2]}$$
Resolve vector $$\overrightarrow{OB}$$ in $$x$$ and $$y$$ direction
$$\textbf{Step 2: Resultant of vectors in x and y direction[Ref. Fig. 3]}$$
Magnitude of vector $$\vec{OA}, \vec{OB}$$ and $$\vec{OC}$$ are same and equal to $$R$$
In $$x$$ direction $$R_x =\vec{OA} +\vec{OB}\cos 45^o= R + \dfrac{R}{\sqrt{2}}$$
In $$y$$ direction $$R_y =\vec{OC}+ \vec{OB}\sin 45^o= R + \dfrac{R}{\sqrt{2}}$$
$$\textbf{Step 3: Find resultant}$$
Resultant $$= \sqrt{R_x^2 + R_y^2 + 2R_xR_y\cos 90^o}$$
$$= \sqrt{\left[R \left(1 + \dfrac{1}{\sqrt{2}}\right)\right]^2 + \left[R \left(1 + \dfrac{1}{\sqrt{2}} \right)\right]^2+0}$$
$$= R \left(1 + \dfrac{1}{\sqrt{2}} \right) \sqrt{2}$$
Resultant $$= R (\sqrt{2} + 1)$$
Hence , Option(B) is correct
If a vector $$\overrightarrow{OP} = 3 \hat{i} + 3 \hat{j}$$ is turned clockwise by an angle of $$15^\circ$$, then the y-component of rotated vector would be:
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$$2 \sqrt 9$$
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$$\dfrac{3}{2} \sqrt 2$$
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$$\dfrac{2}{\sqrt 9}$$
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$$\dfrac{\sqrt 9}{2}$$
The unit vector parallel to the resultant of the vectors $$\vec{A} = \hat{i} + 2 \hat{j} - \hat{k}$$ and
$$\vec{B} = 2 \hat{i} + 4 \hat{j} - \hat{k}$$ is:
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$$\dfrac{1}{49} ( 7 \hat{i} + 6 \hat{j} - 2 \hat{k})$$
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$$\dfrac{1}{7} ( 3 \hat{i} + 6 \hat{j} - 2 \hat{k})$$
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$$\dfrac{1}{49} ( 3 \hat{i} + 6 \hat{j} - 2 \hat{k})$$
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$$\dfrac{1}{7} ( 7 \hat{i} + 6 \hat{j} - 2 \hat{k})$$
A ball is projected from a point A with a speed $${v_o} = 30m/s$$ at an angle of $${30^ \circ }$$ with the horizontal. Consider a target at point B at height $$10\ m$$ in the same vertical plane. The possible time instants at which ball may hit the target are ($$g = 10\ m/{s^2}$$)
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$$1s,2s$$
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$$\frac{1}{2}s,\frac{5}{2}s$$
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$$\frac{3}{2}s,\frac{5}{2}s$$
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$$3s,4s$$
A bird flies at a angle $$60^{0}$$ to the horizontal. its horizontal component of velocity is$$ 10 m s^{-1}$$, find the vertical component of velocity
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$$10 \sqrt 3$$
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$$10 \sqrt 4$$
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$$5$$
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$$26$$
Explanation
Lets consider, $$u=$$ initial velocity of the particle
Given,
$$\theta=60^0$$
Horizontal component of velocity, $$u_x=ucos\theta=10m/s$$
$$ucos60^0=10$$
$$u=\dfrac{10}{cos60^0}=\dfrac{10}{0.5}$$
$$u=20m/s$$
Vertical component of velocity, $$u_y=usin\theta$$
$$u_y=20\times sin60^0=20\times \dfrac{\sqrt{3}}{2}$$
$$u_y=10\sqrt{3}m/s$$
The correct option is A.
A person in an aeroplane which is coming down at acceleration $$a$$ releases a coin. After release, the acceleration of coin with respect to observer on ground and in aeroplane both will be respectively,
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$$g$$ and $$(g-a)$$
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$$(g-a),g$$
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$$(g+a),g$$
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$$g,(g+a)$$
A particle is moving on a straight line according to the given acceleration time graph. If particle starts with velocity $$3 ms^{-2}$$, then velocity at t = 6 second is
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$$23 ms^{-1}$$
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$$13 ms^{-1}$$
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$$20 ms^{-1}$$
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$$10 ms^{-1}$$
A ball is thrown with speed 20m/s at an angle $$30^0$$ with vertical. Its height above the point of projection when it is at horizontal distance of 5 m from the thrower is (g = 10 $$m/s^2$$)
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7.41 m
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8.41 m
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6.14 m
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5.14 m
a projectile is thrown with initial velocity $$v_0$$ and angle $$30^o$$ with the horizontal. if it remains in the air for $$1 s$$. what was its initial velocity?
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$$19.6 m/s$$
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$$9.8 m/s$$
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$$4.9 m/s$$
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$$1 m/s$$
Explanation
Given,
$$T=1s$$
$$\theta=30^0$$
$$u=v_0$$ Initial velocity
Take, $$g=9.8m/s^2$$
Time of flight in projectile motion,
$$T=\dfrac{2usin\theta}{g}$$
$$1=\dfrac{2v_0\ sin30^0}{9.8}$$
$$v_0=\dfrac{9.8}{2\times sin30^0}$$
$$v_0=9.8\ m/s$$
A particle is moving on x-y plane so that its position vector with time $$\vec { r } =10\sqrt { 3t } \hat { i } +\left( 10t-{ t }^{ 2 } \right) \hat { j } $$. If initial velocity of particle is inclined at angle $$\theta $$ from vertical then $$\theta $$, is
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$$30$$
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$$40$$
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$$50$$
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$$\tan ^{ -1 }{ \left( \cfrac { 3 }{ 4 } \right) } $$
Find the angular velocity of a body rotating with an acceleration of $$2 rev/s^2$$ as it complete the $$5th$$ revolution after the start.
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$$3\sqrt 5 \,rev/s$$
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$$2\sqrt 5 \,rev/s$$
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$$2\sqrt 6 \,rev/s$$
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$$2\sqrt 7 \,rev/s$$
The stone of mass 1 kg is whirled in a horizontal circle at the end of 1.3 m long string. If the string makes an angle of $$30^0$$ with the vertical then necessary centripetal force is given by
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5.659 N
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0.5659 N
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4.353 N
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0.4353 N
A body rotates about a fixed axis with an angular acceleration of $$1 rad/s^2$$. Through what angle does it rotate during the time in which its angular velocity increases from $$5\,\,rad/s$$ to
$$15\,\,rad/s$$
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$$10\,\,rad/s$$
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$$500\,\,rad/s$$
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$$50\,\,rad/s$$
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$$100\,\,rad/s$$
A ball (solid sphere) of mass $$m$$ is rolling on a smooth horizontal surface as shown in figure. At an instant the magnitude of the velocity of the centre of mass is $$v_0$$ and its angular velocity is $$\omega \dfrac{v_0}{2R}$$, where $$R$$ is the radius of the ball. The total kinetic energy of the rolling ball at this instant is:-
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$$\dfrac{7}{5} mR^2 \omega^{2}_{0}$$
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$$\dfrac{11}{5} mR^2 \omega^{2}_{0}$$
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$$\dfrac{11}{20} mv^{2}_{0}$$
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$$\dfrac{7}{20} mv^{2}_{0}$$
Explanation
$$KE = \dfrac{1}{2}m{v_0}^2 + \dfrac{1}{2}\left( {\dfrac{{2m{R^2}}}{5}} \right){w^2}$$
$$ = \dfrac{1}{2}m{v_0}^2 + \dfrac{{m{R^2}}}{5} \cdot \dfrac{{{v_0}^2}}{{4{R^2}}}$$
$$ = \dfrac{1}{2}m{v_0}^2 + \dfrac{{m{v_0}^2}}{{20}}$$
$$ = \dfrac{{11m{v_0}^2}}{{20}}$$
Hence,
option $$(C)$$ is correct answer.
A wheel is subject to uniform angular acceleration about its axis. Initially its angular velocity is zero .In first two seconds it rotates through an angle
$$\Theta _{1}$$
. In the next 2 second, it rotates through an additional angle $$\Theta _{2}$$. The ratio of$$\Theta _{1}$$/$$\Theta _{2}$$ is:
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2
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5
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3
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1
A particle of mass m is projected with velocity $$v_0$$ making an angle of $$45^0$$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is:
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zero
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$$mv^3/(4\sqrt{2}g)$$
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$$mv^3/(\sqrt{2}g)$$
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$$m\sqrt{2gh^3}$$
Figure shows two identical particles $$1$$ and $$2,$$ each of mass $$m,$$ moving in opposite directions with same speed $$\vec { v }$$ along parallel lines. At a particular instant, $$\vec { r } _ { 1 }$$ and $$\vec { r } _ { 2 }$$ are their respective position vectors drawn from point $$A$$ which is in the plane of the parallel lines. Choose the correct option(s).
[$$\hat{e}$$ represents a unit vector coming out of the page]
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Angular momentum of $$particle 1$$ about $$A$$ is $$\operatorname { mvd } _ { 1 } ( - \hat { e } )$$
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Angular momentum of $$particle 2$$ about $$A$$ is $$\operatorname { mvd } _ { 2 } ( \hat { e } )$$
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Total angular momentum of the system about $$A$$ is $$mv \left( \mathrm { d } _ { 2 } - \mathrm { d } _ { 1 } \right) \hat e$$
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Total angular momentum of the system about $$A$$ is $$mv \left( \mathrm { d } _ { 1 } + \mathrm { d } _ { 2 } \right)$$ ($$ \hat {-e}$$)
A circular hoop of mass $$m$$ and radius $$R$$ rests flat on a horizontal smooth surface. A bullet of same mass $$m$$ and moving with velocity $$v$$ strikes the hoop tangentially and gets embedded in it at the lowest position of hoop. Neglect the thickness of hoop is comparison to radius of hoop. Answer the following questions:
Mark the correct option
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Kinetic energy of system $$(hoop + bullet)$$ is conserved
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Momentum of system is conserved
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Momentum of system is not conserved
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Angular momentum is not conserved due to torque of gravitational force
The position vector of a point $$C$$ with respect to $$B$$ is $$\hat { i } + \hat { j }$$ and that of B with respect to A is $$\hat { i } - \hat { j }$$. The position vector of $$C$$ with respect to $$A$$ is
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$$\hat { 2i }$$
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$$-\hat { 2i }$$
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$$\hat { 2j }$$
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$$-\hat { 2j }$$
All straight wires are very long. Both $$AB$$ and $$CD$$ are arcs of the same circle, both subtending right angles at the centre $$O$$. Then the magnetic field at $$O$$ is-
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$$\frac { \mu _ { 0 } \mathbf { i } } { 4 \pi R }$$
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$$\frac { \mu _ { 0 } \mathbf { i } } { 4 \pi R } \sqrt { 2 }$$
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$$\frac { \mu _ { 0 } \mathbf { i } } { 2 \pi R }$$
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$$\frac { \mu _ { 0 } i } { 2 \pi R } ( \pi + 1 )$$
Explanation
$$\begin{array}{l} B=\dfrac { { { \mu _{ 0 } }i } }{ { 4\pi R } } +\dfrac { { { \mu _{ 0 } }i } }{ { 4\pi R } } +\dfrac { { { \mu _{ 0 } }i\left( { \dfrac { \pi }{ 2 } } \right) } }{ { 4\pi R } } -\dfrac { { { \mu _{ 0 } }i\left( { \dfrac { \pi }{ 2 } } \right) } }{ { 4\pi R } } \\ =\dfrac { { { \mu _{ 0 } }i } }{ { 2\pi R } } \end{array}$$
Hence,
option $$(C)$$ is correct answer.
A projectile is thrown with some initial velocity u at an angle q to the horizontal. Its speed when it is at the highest point is$$\sqrt{\cfrac{2}{5}}$$ times the speed $$V$$ when it is at height half of the maximum height. Then ratio $$\cfrac{v}{u}=\sqrt{\cfrac{n}{8}}$$ Find the value of $$n.$$
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5
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3
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2
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1
A particle moving along a circular path in the XY plane (see figure). When it crosses X-axis it has an acceleration along the path of $$1.5m/{s}^{2}$$ and is moving with a speed of $$10m/s$$ in the negative direction. The net acceleration of the particle is
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$$(50\hat{i}-1.5\hat{j})m/{s}^{2}$$
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$$(-50\hat{i}-1.5\hat{j})m/{s}^{2}$$
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$$(10\hat{i}-1.5\hat{j})m/{s}^{2}$$
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$$(1.5\hat{i}-50\hat{j})m/{s}^{2}$$
A stone projected at angle $$'\theta'$$ with horizontal from the roof of a tall building falls on the ground after three second. Two second after the projection it was again at the level of projection. Then the height of the building is$$(g=10 m/s^{2})$$
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$$5m$$
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$$25m$$
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$$20m$$
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$$15m$$
The position vector of two points A and B are 6a+2b and a-3b. If a point C divides AB in the ratio 3 : 2, then the position vector of C is
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3a-b
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3a+b
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a+b
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a-b
A point moves in a straight line so that its displacement $$x$$ m at time t sec is given by $$ x^2= 1+t^2$$. Its acceleration in
$$ m/s^2 $$
at a time t sec is:
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$$1/x^3$$
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$$-t/x^3$$
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$$\cfrac {1} {x} - \cfrac {t^2} {x^3}$$
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$$\cfrac {1} {x} - \cfrac {1} {x^2}$$
The graph below shows the velocity with respect to time of an object moving in a straight line. The positive direction is to the right and the negative direction is to the left. Which of the following statements best describes the motion of this object ?
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The object starts at a location to the left of the origin and travels at a constant speed toward the right.
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The object starts at a location to the left of the origin at a slow speed and speeds up as it moves to the right.
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The object slows down as it moves to the left, stops, and starts moving to the right.
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The object slows down as it moves to the right, stops, and continues moving to the right.
The path of a projectile is a parabola.
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True
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False
A toy cannon ball is launched form a cannon on top of a platform. The equation $$h\left(t\right)=-5t^{2}+10t+22$$ gives the height $$h$$, in meters, of the ball $$t$$ seconds after it is launched. What equation can be used to tell whether the ball reaches a height of $$35\ m$$?
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$$-5t^{2}+20t+4=0$$
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$$-5t^{2}+10t+22=35$$
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$$-5t^{2}+20t+4+12=0$$
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$$-5t^{2}+20t+4=x+12$$
The linear speed $$V$$ of a particle moving in a circle of radius $$R$$ varies with time t as $$V=V_0-Kt$$, where $$K$$ is a positive constant and $$V_0$$ is initial velocity. The time at which magnitude of tangential acceleration and angular velocity becomes equal is :
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$$\cfrac {(V_0+KR)} K$$
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$$\cfrac {(V_0-KR)} K$$
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$$\cfrac {(V_0-KR)} K^2$$
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$$\cfrac {(V_0-KR)} {K^2}$$
A tiny ball of mass $$2kg$$ is made to move along the positive x-axis under a force described by the potential energy $$U$$ shown below. It is projected towards positive x-axis from the origin with a speed $${v}_{0}$$. What is the minimum value of $${v}_{0}$$ for which the ball will escape infinitely far away from the origin.
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$$8m/sec$$
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$$4m/sec$$
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$$2m/sec$$
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It is not possible
A particle is moving on a circular path of radius 5 m with uniform speed 10 $$ms^-1$$ from point A to B. The magnitude of average velocity of particle in moving from A to B is
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$$\dfrac{2}{\pi } ms^{-1} $$
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$$\dfrac{10}{\pi } ms^{-1} $$
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$$\dfrac{20}{\pi } ms^{-1} $$
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Zero
Explanation
Ref . Image.
i) = change in velocity = 5+5 = 10 m/s.
Time required to distance covered = $$ \frac{\pi r}{5} = \frac{5\pi }{5} = \pi $$
(in semicircle)
Average acceleration = $$\frac{change in velocity}{time}$$
$$ = \frac{10}{\pi } m/s^{2}$$
If the position vectors of the vertices $$A,B$$ and $$C$$ of a $$\Delta ABC$$ are respectively $$4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k}$$ and $$2\hat{i}+5\hat{j}+7\hat{k}$$, then the position vector of the point, where the bisector of $$\angle A$$ meets $$BC$$ is:
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$$\frac{1}{2}(4\hat{i}+8\hat{j}+11\hat{k})$$
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$$\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$$
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$$\frac{1}{4}(8\hat{i}+14\hat{j}+19\hat{k})$$
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$$\frac{1}{3}(6\hat{i}+8\hat{j}+15\hat{k})$$
If the position vectors of $$A, B, C, D$$ are $$3\hat{i} + 2\hat{j} + \hat{k}, 4\hat{i} + 5\hat{j} + 5\hat{k}, 4\hat{i} + 2\hat{j} - 2\hat{k}, 6\hat{i} + 5\hat{j} - \hat{k}$$ respectively then the position vector of the point of intersection of $$\bar{AB}$$ and $$\bar{CD}$$ is
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$$2\hat{i} + \hat{j} - 3\hat{k}$$
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$$2\hat{i} - \hat{j} + 3\hat{k}$$
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$$2\hat{i} + \hat{j} + 3\hat{k}$$
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$$2\hat{i} - \hat{j} - 3\hat{k}$$
Explanation
$$ \begin{aligned} \overrightarrow{A B} &=\vec{b}-\vec{a} \\ &=(4 i+5 \hat{\jmath}+5 \hat{k})-(3 \hat{i}+2 \hat{\jmath}+\hat{k}) \\ &=\hat{\imath}+3 \hat{\jmath}+4 \hat{k} , and \end{aligned} $$
$$\begin{aligned} \overrightarrow{C D} &=\vec{d}-\vec{c} \\ &=(6 \hat{\imath}+5 \hat{\jmath}-\hat{k})-(4 \hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \\ &=\overrightarrow{2} \hat{\imath}+3 \hat{\jmath}+\hat{k} \\ \text { Let } \overrightarrow{A B} & \text { and } \overrightarrow{C D} \text { intersect at } \vec{P} . \\ \text { Line } A B=(3 i+2 j+\hat{k})+\lambda(i+3 \hat{j}+4 \hat{k}) & \text { and } \end{aligned}$$
Line $$C D=(4 i+2 \hat{\imath}-2 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$$
W.r.t. line $$A_{0}$$,
$$P=(3+\lambda, 2+3 \lambda, 1+4 \lambda)$$
and $$w \cdot r \cdot t \cdot$$ line $$C D$$,
$$P=(4+2 \mu, 2+3 \mu,-2+\mu)$$
Comparing both points,
$$\begin{aligned} \Rightarrow & 3+\lambda^{\prime}=4+2 \mu, \\ & 2+3 \lambda=2+3 \mu \\ & 1+4 \lambda=-2+\mu \end{aligned}$$
Solving these equs.,
$$\therefore \vec{p}=2 \hat{\imath}-\hat{\jmath}-3 \hat{k}$$
$$\therefore$$ option $$D$$ is correct.
Two particles P and Q are moving with velocities of $$\left( \overset { \wedge }{ i } \quad +\quad \overset { \wedge }{ j } \right) $$ respectively. At time t = 0, P is at origin and Q is at a point with position vector $$\left( 2\overset { \wedge }{ i } \quad +\quad \overset { \wedge }{ j } \right) $$. Then the shortest distance between P & Q is:-
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$$\frac { 2\sqrt { 5 } }{ 5 } $$
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$$\frac { 4\sqrt { 5 } }{ 5 } $$
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$$\sqrt { 5 } $$
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$$\frac { 3\sqrt { 5 } }{ 5 } $$
The vector along the incident ray on a mirror is $$ -2 \hat { i } + 3 \hat { j } + 4 \hat { k } $$. considering the x-axis to be along the normal, find the unit vector along the reflected ray.
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$$ \dfrac { -2 \hat { i } + 3 \hat { j } + 4 \hat { k }}{ \sqrt {29}} $$
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$$ \dfrac { -2 \hat { i } - 3 \hat { j } + 4 \hat { k }}{ \sqrt {29}} $$
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$$ \dfrac { -2 \hat { i } - 3 \hat { j } - 4 \hat { k }}{ \sqrt {29}} $$
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None of these
Two particles having position vectors $$\vec {r}_{1} = (3\hat {i} + 5\hat {j})$$ metres and $$\vec {r}_{2} = (-5\hat {i} -3\hat {j})$$ metres are moving with velocities $$\vec {v}_{1} = (4\hat {i} + 3\hat {j})m/s$$ and $$\vec {v}_{2} = (\alpha \hat {i} + 7\hat {j})m/s$$. If they collide after $$2$$ seconds, the value of $$'\alpha'$$ is
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$$2$$
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$$4$$
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$$6$$
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$$8$$
For a moving particle, the relation between time and position is given by t = ax^2 + Bx.where A and B are constants. then the velocity of the particle as a function of position
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$$\dfrac { 2A }{ xB }$$
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$$\dfrac { 1 }{ (2Ax + B) }$$
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$$\dfrac { 1 }{ (2Ax - B) }$$
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$$\dfrac { 1 }{ (Ax + B) }$$
A projectile can have the same range $$R$$ for two angles of projection. If $${t}_{1}$$ and $${t}_{2}$$ are the times of flight in the two cases, then
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$${t}_{1}{t}_{2}\propto {R}^{2}$$
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$${t}_{1}{t}_{2}\propto \cfrac{1}{ {R}^{2}}$$
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$${t}_{1}{t}_{2}\propto {R}^{}$$
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$${t}_{1}{t}_{2}\propto \cfrac{1}{ {R}^{}}$$
Two vectors $$\vec {F_{1}}$$ and $$\vec {F_{2}}$$ each of magnitude inclined to each other such that their resultant is to $$\sqrt {3}F$$ then the resultant of $$\vec {F_{1}}$$ and $$-\vec {F_{2}}$$ is
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$$F$$`
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$$\sqrt {2}F$$
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$$\sqrt {3}F$$
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$$2F$$
The position vectors of two particles change with time as $$\vec {r}_{1}=(t^ {2}-3t+4)\hat {i}+(t^ {2}-t)\hat {j}$$ and $$\vec {r}_{2}=(2t-2)\hat {i}+t\hat {j}$$
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Particles will collide at $$=2s$$
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Particles will collide at $$=3s$$
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Particles will collide at $$=0s$$
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Particles will collide at $$=1s$$
In the cube of side $$'a'$$ shown in the figure, the vector from the central point of the face $$ABOD$$ to the central point of the face $$BEFO$$ will be
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$$\dfrac {1}{2} a(\hat {i} - \hat {k})$$
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$$\dfrac {1}{2} a(\hat {j} - \hat {i})$$
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$$\dfrac {1}{2} a(\hat {k} - \hat {i})$$
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$$\dfrac {1}{2} a(\hat {j} - \hat {k})$$
Explanation
$$\vec {r}_{g} = \dfrac {a}{2}\hat {i} + \dfrac {a}{2}\hat {k}$$
$$\vec {r}_{H} = \dfrac {a}{2} \hat {j} + \dfrac {a}{2}\hat {k}$$
$$\vec {r}_{H} - \vec {r}_{g} = \dfrac {a}{2} (\hat {j} - \hat {i})$$.
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range-
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$$\tan\theta=2$$
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$$\tan\theta=4$$
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$$\tan\theta=3$$
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$$\tan\theta=6$$
A bus is moving with a constant acceleration $$a = \dfrac{3g}{4}$$ towards right. In the bus, a ball is tied with a rope and is rotated in a vertical circle at constant speed $$v_0$$ as shown. The tension in the rope will be minimum, when the rope makes an angle $$\theta$$ =?
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$$0^o < \theta < 90^o$$
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$$90^o < \theta < 180^o$$
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$$180^o < \theta < 270^o$$
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$$270^o < \theta < 360^o$$
Two particles $$P$$ and $$Q$$ of masses $$m$$ and $$3m$$ respectively are initially held at rest in free space with separation $$d$$ between them. At $$t=0$$, $$P$$ is projected with speed $${v}_{0}$$ perpendicular to the line joining $$P$$ and $$Q$$ and $$Q$$ is simultaneously released. The minimum value of $${v}_{0}$$ for which $$P$$ will escape the gravitational field of $$Q$$ is
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$$\sqrt{\cfrac{6Gm}{d}}$$
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$$\sqrt{\cfrac{2Gm}{d}}$$
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$$2\sqrt{\cfrac{2Gm}{d}}$$
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$$\sqrt{\cfrac{Gm}{d}}$$
A particle is moving along a circular path with a constant speed of $$10 ms^{1}$$. What is the magnitude of the change is velocity of the particle, when it moves through an angle of $$ 60$$ around the centre of the circle?
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$$0$$
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$$10m/s$$
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$$10\sqrt{3}m/s$$
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$$10\sqrt{2}m/s$$
Explanation
Change in velocity of a particle in circle in given by,
$$|\Delta \bar{v}|=\sqrt{v_1^2+v_2^2+2v_1v_2\cos (\pi -\theta)}$$
Since $$[|\bar{v_1}|=|\bar{v_2}|]=v$$
$$|\Delta \bar{v}|=2v\sin \dfrac{\theta }{2}$$
On putting the values of $$v\, and\, \theta$$
$$=(2\times 10)\times \sin(30^o)$$
$$=10m/s$$
Hence the change in velocity is $$10m/s$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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