MCQ Questions for Class 11 Engineering Physics Motion In A Plane Quiz 12

Two balls P and Q are at opposite ends of the diameter of a frictionless horizontal circular groove. P is projected along the groove and at the end of T second, it strikes ball Q. Let difference in their final velocities be proportional to the initial velocity of ball P and coefficient of proportionally is e then second strike occurs at

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$\dfrac{2T}{e}$
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$\dfrac{e}{2T}$
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$2eT$
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$Te$

Explanation

Let u be the initial velocity of ball P, then

$\displaystyle u = \dfrac{2\pi r}{2T}$ and difference in final velocities,

$v^{'} = eu$ (given)
Time for second strike

$\displaystyle = \dfrac{2\pi r}{v^{'}-v} = \displaystyle \dfrac{2\pi r}{eu}$

$\displaystyle = \dfrac{2\pi r}{e\times \displaystyle \dfrac{2\pi r}{2T}} = \dfrac{2T}{e}$

Is the claim of Mr.Kirkpatrick right?

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yes
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No
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cannot say
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may be correct or may be not

Explanation

Consider the jump as a projectile, We know that the range of projectile

$R=\dfrac { { u }^{ 2 }\sin 2\theta }{ g } ,$ i.e. Range is inversely proportional to acceleration due to gravity

${ R }_{ b }=8.09=\dfrac { k }{ 9.8128 } \\ { R }_{ m }=\dfrac { k }{ 9.7999 } \\ \Rightarrow \quad \dfrac { { R }_{ m } }{ { R }_{ b } } =\dfrac { 9.8128 }{ 9.7999 } \\ \Rightarrow { \quad R }_{ m }=\dfrac { 8.09\times 9.8128 }{ 9.7999 } \\ \Rightarrow { \quad R }_{ m }=8.1065$

Find the magnitude of the angular acceleration of the cone.

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$3.3\:rad/s^2$
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$2.6\:rad/s^2$
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$2.3\:rad/s^2$
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$3.6\:rad/s^2$

Explanation

Let the axis of the cone

$(OC)$ rotate in anticlockwise direction with

$\overrightarrow { { w }^{ ' } }$ and thecone about its own axis

$(OO')$ in clockwise direction with

$\overrightarrow { { w }_{ 0 } }$ angular velocity.

Resultant angular velocity of the cone:

$\overrightarrow { w } =\overrightarrow { { w }^{ ' } } +\overrightarrow { { w }_{ 0 } } \longrightarrow (1)$

From the figure:-

${ w }^{ ' }=\cfrac { V }{ R\cot { \alpha } } ,{ w }_{ 0 }=\cfrac { V }{ R } \longrightarrow 2$

Since

$\overrightarrow { { w }^{ ' } } \bot \overrightarrow { { w }_{ 0 } } \\ w=\sqrt { { { w }^{ ' } }^{ 2 }+{ { w }_{ 0 } }^{ 2 } } \\ \quad =\sqrt { { \left( \cfrac { V }{ R\cot { \alpha } } \right) }^{ 2 }+{ \left( \cfrac { V }{ R } \right) }^{ 2 } } \\ \quad =\cfrac { V }{ R\cot { \alpha } } =2.3\quad rad/s$

Angular acceleration:

$\overrightarrow { \beta } =\cfrac { d\overrightarrow { w } }{ dt } =\cfrac { d }{ dt } \left( \overrightarrow { { w }^{ ' } } +\overrightarrow { { w }_{ 0 } } \right)$

$\overrightarrow { { w }_{ 0 } }$ which rotates about

$oo'$ axis with angular velocity

$\overrightarrow { { w }_{ 0 } }$ retains its magnitude. This increment in the time internal

$dt$ is equal to:

$\left| d\overrightarrow { { w }_{ 0 } } \right| =\left( \overrightarrow { { w }^{ ' } } +\overrightarrow { { w }_{ 0 } } \right) dt$

$\therefore \quad \beta =\overrightarrow { { w }^{ ' } } \times \overrightarrow { { w }_{ 0 } }$

Magnitude of

$\beta$

$\beta ={ w }^{ ' }\times { w }_{ 0 }\quad [as\quad \overrightarrow { { w }^{ ' } } \bot \overrightarrow { { w }_{ 0 } } ]\\ \therefore \quad \beta =\cfrac { V }{ R\cot { \alpha } } \cfrac { V }{ R } =\cfrac { { V }^{ 2 } }{ { R }^{ 2 } } \tan { \alpha } \\ \quad \quad \quad \quad \quad \quad \quad \quad =2.3\quad rad/{ s }^{ 2 }$

A projectile is required to hit a target whose coordinates relative horizontal and vertical axes through the point of projection are

$(\alpha,\beta)$ . If the gun velocity is

$\sqrt {2g \alpha}$ , it is impossible to hit the target if

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$\beta >\cfrac { 3 }{ 4 } \alpha$
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$\beta \ge \cfrac { 3 }{ 4 } \alpha$
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$\beta \le \cfrac { 3 }{ 4 } \alpha$
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$\beta < \dfrac{ 3 }{ 4 } \alpha$

Explanation

Equation of trajectory

$y=x\tan { \theta } -\dfrac { g{ x }^{ 2 } }{ 2{ u }^{ 2 } } \sec ^{ 2 }{ \theta }$ and it passes through point

$\left( \alpha ,\beta \right)$

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

$\dfrac { \alpha }{ 4 } \tan ^{ 2 }{ \theta } -\alpha \tan { \theta } +\left( \beta +\dfrac { \alpha }{ 4 } \right) =0$

If the value of discriminant

$\triangle=B^2-4AC <0$ for above quadratic equation then it would be impossible for projectile to hit target

${ \alpha }^{ 2 }-4\left( \dfrac { \alpha }{ 4 } \right) \left( \beta +\dfrac { \alpha }{ 4 } \right) <0\quad \Rightarrow \beta >\dfrac { 3\alpha }{ 4 }$

A vector

$\displaystyle \overline{m}$ of magnitude

$\displaystyle 2\sqrt{101}$ in the direction of internal bisector of the angle between the vector

$\displaystyle \overline{b}=8\hat{i}-6\hat{j}-6\hat{k} \ and \ \overline{c}=4\hat{i}-3\hat{j}+4\hat{k}$ is

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$\displaystyle \frac{16\hat{i}-12\hat{j}+2\hat{k}}{5}$
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$\displaystyle 16\hat{i}-12\hat{j}+2\hat{k}$
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$\displaystyle \frac{-16\hat{i}+12\hat{j}+2\hat{k}}{5}$
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$\displaystyle 16\hat{i}+12\hat{j}+2\hat{k}$

Explanation

$\displaystyle \Rightarrow \overline{b}=8\hat{i}-6\hat{j}-6\hat{k},\overline{c}=-4\hat{i}-3\hat{j}+4\hat{k}$
Now

$\overline{m}$ , is in the direction of internal bisector of angle between

$\overline{b}$ &

$\overline{c}$

$\displaystyle \therefore \overline{m}=\lambda \left ( \hat{b} +\hat{c}\right )=\lambda \left ( \frac{\overline{b}}{\left | \overline{b} \right |}+\frac{\overline{c}}{\left | \overline{c} \right |} \right )$

$\displaystyle \Rightarrow \overline{m}=\lambda \left ( \frac{8\hat{i}-6\hat{j}-6\hat{k}}{10}+\frac{4\hat{i}-3\hat{j}+4\hat{k}}{5} \right )$

$\displaystyle \Rightarrow \overline{m}=\lambda \left ( \frac{4\hat{i}-3\hat{j}+4\hat{k}+4\hat{i}-3\hat{j}+4\hat{k}}{5} \right )$

$\displaystyle \Rightarrow \overline{m}=\lambda \left ( \frac{8\hat{i}-6\hat{j}+\hat{k}}{5} \right )$ ..........(A)

$\displaystyle \therefore \left | \overline{m} \right |=\lambda \sqrt{\frac{101}{25}}$

$\displaystyle \Rightarrow 2\sqrt{101}=\frac{\lambda \sqrt{101}}{5}\Rightarrow \lambda =10$

$\displaystyle \therefore \overline{m}=10\left ( \frac{8\hat{i}-6\hat{j}+\hat{k}}{5} \right )$
Substitute

$\displaystyle \lambda =10$ in equation (A)

$\displaystyle \therefore \overline{m}=16\hat{i}-12\hat{j}+2\hat{k}$

Two particles projected from the same point with same speed

$u$ at angles of projection

$\displaystyle \alpha$ and

$\displaystyle \beta$ strike the horizontal ground at the same point. If

$\displaystyle h_{1}$ and

$\displaystyle h_{2}$ are the maximum heights attained by the projectile,

$R$ is the range for both

$\displaystyle t_{1}$ and

$\displaystyle t_{2}$ are their times of flights, respectively, then

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$\displaystyle \alpha +\beta =\frac{\pi }{2}$
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$\displaystyle R=4\sqrt{h_{1}h_{2}}$
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$\displaystyle \frac{t_{1}}{t_{2}}=\tan \alpha$
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$\displaystyle \tan \alpha =\sqrt{\frac{h_{1}}{h_{2}}}$

Explanation

From above concept

$\alpha+\beta= \dfrac{\pi}{2}$

$R =\dfrac{u^2 \sin 2 \alpha}{g }$

$h_1 = \dfrac{u^2 \sin ^2 \alpha}{2g}$

$h_2 = \dfrac{u^2 \cos ^2 \alpha}{2g}$

multiplying

$h_1$ and

$h_2$

$h_1h_2 = \dfrac{u^2 \sin ^2 \alpha u^2 \cos^2 \alpha}{g^2}$

$h_1h_2 = \dfrac{u^4 \sin ^2 2 \alpha }{4g^2}$

$4h_1h_2 = (\dfrac{R}{2})^2$

$R = 4 \sqrt{h_1h_2}$

$t_1 = \dfrac{2u \sin \alpha}{g}$

$t_2 = \dfrac{2u \cos \alpha}{g}$

$\dfrac{t_1}{t_2} = \tan \alpha$

$\dfrac{h_1}{h_2}= \tan^2 \alpha$

$\tan \alpha = \sqrt{\dfrac{h_1}{h_2}}$

A racing car is travelling along a straight track at a constant velocity of

$40 m/s$ A fixed TV camera is recording the event as shown in figure. In order to keep the car in view, in the position shown, the angular velocity of camera should be

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$3$ rad/s
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$2$ rad/s
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$4$ rad/s
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$1$ rad/s

Explanation

$x=30 \tan \theta$

$\displaystyle \therefore \left( \frac{dX}{dt} \right) =(30 \sec^{2} \theta)\cdot \frac{d \theta}{dt}$
or

$v_{car}=(30 \sec^{2} \theta) \omega$

$\displaystyle \therefore \omega =\frac{v_{car}}{30 \sec^{2} \theta}$

$\displaystyle =\frac{40}{(30) sec^{2} 30 ^{\circ}}=1 rad/s$

A particle is projected at an angle of elevation

$\displaystyle \alpha$ and after t second it appears to have an elevation of

$\displaystyle \beta$ as seen from the point of projection. Find the initial velocity of projection.

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$\displaystyle v= \frac{gt\cos \beta }{2\sin \left ( \alpha -\beta \right )}$
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$\displaystyle v= \frac{2gt\cos \beta }{\sin \left ( \alpha -\beta \right )}$
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$\displaystyle v= \frac{gt\cos \beta }{\sin \left ( \alpha +\beta \right )}$
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$\displaystyle v= \frac{2gt\cos \beta }{\sin \left ( \alpha +\beta \right )}$

Explanation

$x=utcos\alpha ,\quad y=utsin\alpha -\dfrac { 1 }{ 2 } g{ t }^{ 2 }\\ \Rightarrow tan\beta =\dfrac { y }{ x } \Rightarrow tan\beta =\dfrac { utsin\alpha -\dfrac { 1 }{ 2 } g{ t }^{ 2 } }{ utcos\alpha } \\ \Rightarrow tan\alpha -tan\beta =\dfrac { gt }{ 2ucos\alpha } \Rightarrow u=\dfrac { gtcos\beta }{ 2sin(\alpha -\beta ) }$

At a height of

$15\ m$ from ground velocity of a projectile

$\displaystyle \vec{v}= \left ( 10\hat{i}+10\hat{j} \right )$ and (

$g = 10\ ms^{-2}$ )

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particle was projected at an angle of $\displaystyle 45^{\circ}$ with horizontal
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time of flight of projectile is $4\ s$
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horizontal range of projectile is $100\ m$
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maximum height of projectile form ground is $20\ m$

Explanation

Using kinematics IIIrd equation

$v^2 - u^2 = 2gS$

$10^2 - u_y^2 = - 2g\times 15$

$100+300= u_y^2$

$u_y=20 \ m/s$

$u_x = 10\ m/s$

From above values of

$u_x$ and

$u_y$ it is clear that the angle of projection is not

$45^o$

$T=\dfrac{2u_y}{g} = 4 \ s$

$Range = u_x \times T = 10\times 4=40 m$

maximum height of projectile

$h_{max}= \dfrac{u_y^2}{2g} = \dfrac{20^2}{2\times 10} =20\ m$

A projectile of mass m is fired into a liquid at an angle

$\displaystyle \theta _{0}$ with an initial velocity

$\displaystyle v _{0}$ as shown. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity, i.e., F= -kv where k is a positive constant, determine the x and y components of its velocity at any instant. Also find the maximum distance

$\displaystyle x_{max}$ that it travels?

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$\displaystyle v_{x}=v_{0}\cos \theta _{0}e^{-kt/m},v_{y}=\frac{m}{k}\left [ \frac{k}{m}v_{0}\sin\theta _{0}+g^{-\frac{kt}{m}}-g \right ]X_{m}=\frac{mv\cos\theta}{k}$
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$\displaystyle v_{x}=v_{0}\cos \theta _{0}e^{-2kt/m},v_{y}=\frac{m}{k}\left [ \frac{k}{m}v_{0}\sin\theta _{0}+g^{-\frac{kt}{m}}-g \right ]X_{m}=\frac{mv\cos\theta}{2k}$
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$\displaystyle v_{x}=v_{0}\cos \theta _{0}e^{-kt/m},v_{y}=\frac{m}{k}\left [ \frac{k}{m}v_{0}\sin\theta _{0}+g^{-\frac{2kt}{m}}-g \right ]X_{m}=\frac{mv\cos\theta}{k}$
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$\displaystyle v_{x}=v_{0}\cos \theta _{0}e^{-kt/m},v_{y}=\frac{m}{k}\left [ \frac{k}{m}v_{0}\sin\theta _{0}+g^{-\frac{kt}{m}}-g \right ]X_{m}=\frac{mv\cos\theta}{4k}$

Explanation

Given :

$u_x = V_o cos\theta_o$

$u_y = V_o sin\theta_o$

x direction :

$F_x = -k v_x$ where

$F_x = ma_x$

$\therefore$

$a_x = \dfrac{-k}{m} v_x$ where

$a_x = \dfrac{dv_x}{dt}$

$\int^{v_x}_{v_o cos\theta_o} \dfrac{dv_x}{v_x}= \int_{0}^{t} \dfrac{-k}{m} dt$

$\implies$

$ln \dfrac{v_x}{v_o cos\theta_o} = \dfrac{-kt}{m}$

$\implies$

$v_x = v_o cos\theta_o e^{-kt/m}$

Also

$\int^x_0 dx = \int^t_0 (v_o cos\theta_o) e^{-kt/m} dt$ (using

$v_x = dx/dt$ )

$\therefore$

$x =(v_o cos\theta_o) \times \dfrac{e^{-kt/m}}{-k/m} \bigg|_{0}^t$

$\implies x = \dfrac{m v_o cos\theta_o (1 - e^{-kt/m})}{k}$

$\implies$

$x_{max} =\dfrac{mv_o cos\theta_o}{k}$

y direction :

$a_y = - \bigg( \dfrac{k}{m} v_y + g\bigg)$

$\int^{v_y}_{v_o sin\theta_o} \dfrac{dv_y}{\bigg( \dfrac{k}{m} v_y + g \bigg)}= \int_{0}^{t} - dt$

$\implies$

$\dfrac{ln \bigg( \dfrac{k v_y}{m} + g \bigg)}{k/m}\Bigg|_{v_o sin\theta_o}^{v_y} = -t$

$ln \dfrac{\frac{kv_y}{m} + g}{\frac{k}{m}(v_o sin\theta_o) + g} = \dfrac{-kt}{m}$

$\implies$

$v_y = \dfrac{m}{k} \bigg[\bigg( \dfrac{k}{m} v_o sin\theta_o + g \bigg)e^{-kt/m} - g\bigg]$

A particle is moving in a circle of radius

$R$ in such a way that at any instant the normal and tangential component of its acceleration are equal. If its speed at

$t=0$ is

$\displaystyle v_{0}.$ The time taken to complete the first revolution is

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$\displaystyle \frac{R}{v_{0}}$
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$\displaystyle \frac{R}{v_{0}}e^{-2\pi }$
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$\displaystyle \frac{R}{v_{0}}\left ( 1-e^{-2\pi } \right )$
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$\displaystyle \frac{R}{v_{0}}\left ( 1+e^{-2\pi } \right )$

Explanation

Given;

$a_c=a_t$

$Rw^2= \dfrac{dv}{dt}$ where

$v= Rw$

$\implies w^2= \dfrac{dw}{dt}$

Integrating,

$\int^{w}_{w_o} \dfrac{1}{w^2} dw= \int^{t}_{0} dt$

$\dfrac{1}{w}= \dfrac{1}{w_o} - t$

$\dfrac{w}{w_o}= \dfrac{1}{1-w_o t}$ where

$w= \dfrac{d\theta}{dt}$

Integrating again,

$\int^{2\pi}_{0} d\theta= w_o \int^{t}_{0} \dfrac{1}{1-w_o t} dt$

$\implies 2\pi= -ln (1-w_o t)$

$t= \dfrac{1}{w_o}(1-e^{2\pi})$

$as (v_o = R w_o)$

Between

$t=30\;s\;and\;t=40/s$ , the merry-g-round

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rotates clockwise, at a constant rate
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rotates clockwise, and slows down
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rotates counterwise, at a constant rate
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rotates counterclockwise, and slows down

Explanation

The are under the curve between

$t=10s\ and\ t=20s$ is

$\dfrac{1}{2}\times 10\times(-70)=-350^o>-360^o$ .

So the merry-go-round rotates clockwise. As the curve goes towards x-axis, in this time period, the angular velocity goes towards zero that is the merry-go-round slows down.

If the equation for the displacement of a particle moving on a circular path is given by

$(\theta)=2t^3+0.5$ , where

$\theta$ is in radians and

$t$ in seconds, then the angular velocity of the particle after

$2s$ from its start is

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$\;8\;rad/s$
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$\;12\;rad/s$
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$\;24\;rad/s$
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$\;36\;rad/s$

Explanation

Angular displacement of the particle

$\theta(t) =2t^3 + 0.5$ radians

$\therefore$ Angular velocity

$w =\dfrac{d[\theta(t)]}{dt} = 6t^2$ rad/s

Angular velocity at

$t = 2$ s

$w\bigg|_{t =2} =6 \times 2^2 =24$

$rad/s$

In one second, a particle goes from point A to point B moving in a semicircle. Find the magnitude of the average velocity.

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1 m/s
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2 m/s
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0.5 m/s
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None of the above

Explanation

Total displacement of the particle in going from A to B

$S = AB = 2 \times 1 =2$ m

Time taken

$t = 1$ s

$\implies$ Average velocity

$v_{avg} = \dfrac{S}{t} =\dfrac{2}{1} =2$ m/s

Between

$t=0\;and\;t=10\;s$ , about how many revolutions (approx) does the merry-go-round complete?

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$\;1$
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$\;2$
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$\;3$
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$\;4$

Explanation

In between

$t=0$ to 10,

$\omega=70$
As

$\omega=\dfrac{d\theta}{dt}$
so

$\theta=\omega(10-0)=70\times 10=700$
The no of revolution

$=\dfrac{700}{360}\sim 2$

Between

$t=10\;s\;and\;t=20\;s$ , the merry-go-round.

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rotates clockwise, at a constant rate
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rotates clockwise, and slows down
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rotates counterwise, at a constant rate
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rotates counterclockwise, and slows, down

Explanation

The are under the curve between

$t=10s\ and\ t=20s$ is

$\dfrac{1}{2}\times 10\times70=350^o<360^o$ .

So the merry-go-round rotates counterclockwise. As the curve decreases in this time period, the angular velocity decreases that is the merry-go-round slows down.

Single Correct Answer Type
The path of a projectile is given by the equation

$y=ax-bx^2$ , where

$a$ and

$b$ are constants and

$x$ and

$y$ are respectively horizontal and vertical distances of projectile from the point of projection. The maximum height attained by the projectile and the angle of projection are respectively:

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$\displaystyle\frac{2a^2}{b}$ , $\tan^{-1}(a)$
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$\displaystyle\frac{b^2}{2a}$ , $\tan^{-1}(b)$
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$\displaystyle\frac{a^2}{b}$ , $\tan^{-1}(2b)$
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$\displaystyle\frac{a^2}{4b}$ , $\tan^{-1}(a)$

Explanation

The trajectory equation

$y = ax-bx^2$

Differentiating w.r.t

$x$ ,

$\dfrac{dy}{dx} = a - 2bx$

For maximum height,

$\dfrac{dy}{dx} = 0$

$\therefore$

$a - 2bx = 0$

$\implies x = \dfrac{a}{2b}$

Thus maximum height

$H = y\bigg|_{x = \dfrac{a}{2b}}$

$\therefore$

$H = a\times \dfrac{a}{2b} - b\dfrac{a^2}{4b^2} = \dfrac{a^2}{4b}$

Angle of projection is equal to the slope of trajectory at

$x = 0$

$\therefore$

$tan\theta =$ Slope

$= \dfrac{dy}{dx}\bigg|_{x= 0}$

$\therefore$

$tan\theta = a - 2b (0) = a$

$\implies$

$\theta = tan^{-1} (a)$

Consider two children riding on the merry-go-round Child 1 sits near the edge, child 2 sits closer to the centre.
Let

$v_1\;and\;v_2$ denote the linear speed of child 1 and child 2, respectively. Which of the following is true?

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$v_1>v_2$
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$v_1=v_2$
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$v_1\,<\,v_2$
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(4) we cannot determine which is true, without more information

Explanation

Let angular velocity of system be

$\omega$ .

As linear speed is

$R\omega$

so,

$v_1>v_2$

The velocity of a car travelling on a straight road is

$3.5km{ h }^{ -1 }$ at an instant of time. Now travelling with uniform acceleration for

$10s$ the velocity becomes exactly double. If the wheel radius of the car is

$25cm$ , then which of the following is the closest to the number of revolutions that the wheel makes during this

$10s$ ?

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$84$
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$95$
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$126$
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$135$

Explanation

Initial velocity is

$3.5$

$\dfrac { km }{ h }$

Time taken

$= 10\ s$ , Final velocity

$= 7 \dfrac { km }{ h }$

Radius of the wheel (r)

$= 25\ cm$

Using,

$v = u + at$

$\Rightarrow 7 = 3.5 + a \times \dfrac{10}{3600}$

$\Rightarrow a= 3.5 \times 360 km/h$

Angular acceleration

$= \dfrac{a}{r}$

Angular Velocity

$= \dfrac{u}{r}$

From:

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

Net rotation of the wheel will be calculated and

$\dfrac{\theta}{2\pi }$ will give total no. of revolutions 95.

Equation of parabolic trajectory of a projectile can be given by:

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$\dfrac{sin\theta}{cos\theta}x - \dfrac{gx^2}{sin^2\theta+cos^2\theta + cos2\theta}$
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$\dfrac{sin\theta}{cos\theta}x - \dfrac{gx^2}{u^2(sin^2\theta+cos^2\theta + cos2\theta)}$
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$\dfrac{sin\theta}{cos\theta}x - \dfrac{gx^2}{2u^2(sin^2\theta+cos^2\theta + cos2\theta)}$
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$\dfrac{sin\theta}{cos\theta}x - \dfrac{gx^2}{u^3(sin^2\theta+cos^2\theta + cos2\theta)}$

Explanation

Equation of parabolic trajectory:

$y = xtan\theta - \dfrac{gx^2}{2u^2cos^2\theta}$

Using:

$tan\theta = \dfrac{sin\theta}{cos\theta}$ and

$2cos^2\theta = 1+cos2\theta$

$\therefore$

$y = \dfrac{sin\theta}{cos\theta}x - \dfrac{gx^2}{u^2[1+cos2\theta]}$

Now using:

$cos^2\theta+sin^2\theta =1$

$\implies$

$y = \dfrac{sin\theta}{cos\theta}x - \dfrac{gx^2}{u^2[sin^2\theta+cos^\theta+cos2\theta]}$

Two billiard balls are rolling on a flat table. One has velocity components

$V_x = 1 \ m/s, V_y\,=\, \sqrt{3}\, \ m/s$ and the other has component

$V_x \,=\, 2 \ m/s\,$ and

$\, V_y\,=\, 2 \ m/s$ . If both the balls start moving from the same point the angle between their path is:

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$60^{\circ}$
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$45^{\circ}$
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$22.5^{\circ}$
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$15^{\circ}$

Explanation

Components of the relative velocity of the billiard balls are:

$V_{x} = 1ms^{-1}$

$V_{y} = (2 - \sqrt{3})ms^{-1}$

$tan \theta = \dfrac{V_{y}}{V_{x}} = \dfrac{2 - \sqrt{3}}{1}$

$i.e. \theta = tan^{-1} \dfrac{2 - \sqrt{3}}{1} = 15^{\circ}$

$\theta$ can also be found by using the formula of

$\tan(2\theta)$ .

$\tan(2 \theta) = \dfrac{2 \tan(\theta)}{1-\tan^2(\theta)}$

$=\dfrac{2(2-\sqrt 3)}{1-(2-\sqrt 3)^2}$

$=\dfrac{2(2-\sqrt 3)}{4\sqrt 3 -6}$

$=\dfrac{2(2-\sqrt 3)}{2\sqrt 3(2- \sqrt 3)}$

$=\dfrac{1}{\sqrt 3}$

A particle of mass

$10\ g$ moves along a circle of radius

$6.4\ cm$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to

$8 \times {10}^{-4} J$ by the end of the second revolution after the beginning of the motion?

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$0.1\ {m}/{{s}^{2}}$
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$0.15\ {m}/{{s}^{2}}$
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$0.18\ {m}/{{s}^{2}}$
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$0.2 \ {m}/{{s}^{2}}$

Explanation

Solution:

Hint:

· Find the magnitude of velocity by applying the Kinetic energy formula.

· Find the distance for two revolutions.

· Apply third equation of motion to find the acceleration.

Step 1: Find the velocity of the particle from the given kinetic energy.

$\dfrac{1}{2}m{v^2} = 8 \times {10^{ - 4}}$

Here, m is the mass of the particle and v is the velocity.

$\dfrac{1}{2} \times 10 \times {10^{ - 3}} \times {v^2} = 8 \times {10^{ - 4}}$

$v = 0.4\;m/s$

Step 2: Write the total distance covered.

$s = 2(2\pi r)$ ( particle covered two revolutions)

Here, s is the total distance covered.

Step 3: Apply the third equation of motion.

${v^2} -{u^2} = 2as$ (Where v is final velocity, u is initial velocity (given u=0), a is acceleration and s is distance)

$\Rightarrow {(0.4)^2} = 2 \times a \times (4\pi \times 6.4 \times {10^{ - 2}})$

$\therefore a = 0.1\;m/{s^2}$

Hence, option (a) is the correct answer.

A plane flying horizontally at a height of 1500 m with a velocity of

$200ms^{-1}$ passes directly overhead an antiaircraft gun. Then the angle with the horizontal at which the gun should be fired for the shell with a muzzle velocity of 400

$ms^{-1}$ to hit the plane is:

0%
$90^{\circ}$
0%
$60^{\circ}$
0%
$30^{\circ}$
0%
$45^{\circ}$

Explanation

Velocity of plane

$v_p=200m/s$ , velocity of bullet

$v_b=400m/s$ ,

For the shell to hit the plane, horizontal velocity of shell = horizontal velocity of the plane because initial horizontal displacement between plane and shell is zero.

$h=1500m ;\theta=?$

$400cos\theta=200$

$cos\theta=\dfrac{1}{2}$

$\theta =60^0$
Hence option B is correct.

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point P at a horizontal distance

$r$ from the point of projection. If

$t_1$ and

$t_2$ are times taken to reach this point in two possible ways, then the product

$t_1t_2$ is proportional to:

0%
$\displaystyle\frac{1}{r}$
0%
$r$
0%
$r^2$
0%
$\displaystyle\frac{1}{r^2}$

Explanation

The particle will follow two equations:

$r=vcos\theta t$

and

$t=\dfrac{2vsin\theta}{g}$

$\implies \dfrac{r^2}{v^2 t^2}+\dfrac{t^2g^2}{4v^2}=1$

$\implies g^2t^4-4v^2t^2+4r^2=0$

This is a quadratic equation in

$t^2$ with its solutions

$t_1^2, t_2^2$ . For a quadratic equation,

$ax^2+bx+c=0$ , product of roots is given by

$c/a$

$\implies t_1^2t_2^2=\dfrac{4r^2}{g^2}$

$\implies t_1t_2=\dfrac{2r}{g}\propto r$

The x and y co-ordinates of a particle is x = Asin(

$\omega{t}$ ) and y = Asin(

$\omega{t} +\frac{\pi}{2}$ ) . Then the motion of the particle is ( x and y are in the directions shown in the fgure )

0%
Circular anticlockwise
0%
Circular clockwise
0%
Elliptical clockwise
0%
Rectilinear

Explanation

$x=A\sin(\omega t)$

$y=A\sin \left(\omega t+\cfrac{\pi}{2}\right)$

$t=0: x=0, y=A\sin \left(\cfrac{\pi}{2}\right)=A$

$t=\cfrac{T}{2}=\cfrac{2\pi}{2\times \omega}=\cfrac{\pi}{\omega}:$

$x=A\sin (\pi)=0, y=A\sin\left(\pi+\cfrac{\pi}{2}\right)=A\sin\left(\cfrac{3\pi}{2}\right)=-A$

$\therefore$ Motion is circular anti-clockwise

$\vec{A} = 3 \hat{i} - 2 \hat{j} + \hat{k}, \vec{B} = \hat{i} - 3 \hat{j} + 5 \hat{k}\ and\ \vec {C} = 2 \hat{i} + \hat {j} - 4 \hat{k}$ form a right angled triangle then out of the following which one is satisfied?

0%
$\vec{A} = \vec{B} + \vec{C}\ and\ A^2 = B^2 + C^2$
0%
$\vec{A} = \vec{B} + \vec{C}\ and\ B^2 = A^2 + C^2$
0%
$\vec{B} = \vec{A} + \vec{C}\ and\ B^2 = A^2 + C^2$
0%
$\vec{B} = \vec{A} + \vec{C}\ and\ A^2 = B^2 + C^2$

Explanation

$\vec{A}=3\hat{i}-2\hat{j}+\hat{k}$ and

$A^2=9+4+1=14$

$\vec{B}=\hat{i}-3\hat{j}+5\hat{k}$ and

$B^2=1+9+25=35$

$\vec{C}=2\hat{i}+\hat{j}-4\hat{k}$ and

$C^2=4+1+16=21$

Here, we see that,

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

And

$B^2=A^2+C^2$

Hence, answer is option-(B).

A particle is projected in such a way that it follows a curved path with constant acceleration

$\vec{a}$ . For finite interval of motion. Which of the following option(s) may be correct:

$\vec{u}=$ initial velocity,

$\vec{a}=$ acceleration of particle,

$\vec{v}=$ instant velocity for

$t > 0$ .

0%
$|\vec{a}\times \vec{u}|\neq 0$
0%
$|\vec{a}\times \vec{v}|\neq 0$
0%
$|\vec{u}\times \vec{v}|\neq 0$
0%
$\vec{u}\cdot \vec{v}=0$

A stone tied to a string of length

$L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed

$u$ . Determine the magnitude of the change in its velocity as it reaches a position where the string is horizontal is:

0%
$\sqrt { { u }^{ 2 }-2gL }$
0%
$\sqrt { 2gL }$
0%
$\sqrt { { u }^{ 2 }-gL }$
0%
$\sqrt { 2\left( { u }^{ 2 }-2gL \right) }$

Explanation

From energy conservation:

$E_A=E_B$

$\implies \cfrac{1}{2}mu^2=\cfrac{1}{2}mV^2+mgl$

$\implies \cfrac{1}{2}u^2=\cfrac{1}{2}V^2+gl$

$\implies V^2=u^2-2gl$

$\implies V=\sqrt{u^2-2gl}$

Two particles are projected simultaneously from the same point, with the same speed, in the same vertical plane, and at different angles with the horizontal in a uniform gravitational field acting vertically downwards. A frame of the other particle, as observed from this frame, is

$\vec { r }$ . Which of the following statements is correct?

0%
$\vec { r }$ is a constant vector.
0%
$\vec { r }$ changes in magnitude as well as direction with time.
0%
The magnitude of $\vec { r }$ increases linearly with time; its direction does not change.
0%
The magnitude of $\vec { r }$ increases linearly with time; its direction changes.

In the adjacent figure, a uniform disc of mass

$2m$ and radius

$l/2$ is lying at rest on a smooth horizontal surface. A particle

$'A'$ of mass

$m$ is connected to a light string of length

$l$ , whose other end is attached to the circumference of the disc. Initially string is just taut and tangential to the disc, particle

$A$ is at rest. In the same horizontal plane another particle

$B$ of some mass

$m$ moving with velocity

$v_{0}$ perpendicular to string collides elastically with

$A$ . Just after impact which of the following statements will be true

0%
Tension in the string is $\dfrac {2mv_{0}^{2}}{5l}$
0%
Acceleration of the centre of the disc is $\dfrac {v_{0}^{2}}{5l}$
0%
Tension in the string is $\dfrac {mv_{0}^{2}}{5l}$
0%
Acceleration of the centre of the disc is $\dfrac {2v_{0}^{2}}{5l}$

A pulley

$1m$ in diameter rotating at

$600 r.p.m.$ is brought to rest in

$80\ s$ . By constant force of friction on its shaft. How many revolutions does it moves?

0%
$200$
0%
$400$
0%
$600$
0%
$500$

Explanation

Radius,

$r=0.5$

$m$

Initial angular velocity,

$\omega_o=600$

$rpm$

$\implies \omega_o=\cfrac{600}{60}=10$

$rpm$

$\implies \omega_o=10\times 2\pi$

$\implies \omega_o=20\pi$

${rad/s}$

Time,

$t=80$

$s$

Final angular velocity,

$\omega=0$

${rad/s}$

From kinematics:

$\omega-\omega_o=\alpha t$

$\implies \cfrac{-20\pi}{80}=\alpha$

$\implies \alpha=-0.25\pi$

Also,

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

$\implies \theta=20\pi\times 80+\cfrac{1}{2}\times (-0.25\pi)\times 6400$

$\implies \theta=1600\pi-(0.25\times 3200\pi)$

$\implies \theta=1600\pi-800\pi$

$\implies \theta=800\pi$

$\implies \theta=400\times 2\pi$

$\therefore$ The pulley makes

$400$ revolutions.

Two paper screens

$A$ and

$B$ are separated by

$150\ m$ . A bullet pierces

$A$ and

$B$ . The hole in

$B$ is

$15\ cm$ below the hole is

$A$ . If the bullet is travelling horizontally at the time of hitting

$A$ , then the velocity of the bullet at

$A$ is

$(g = 10\ ms^{-2})$ .

0%
$100\sqrt {3}ms^{-1}$
0%
$200\sqrt {3}ms^{-1}$
0%
$300\sqrt {3}ms^{-1}$
0%
$500\sqrt {3}ms^{-1}$

Explanation

The bullet exhibits horizontal projectile motion.

Displacement of bullet in vertical direction

$y = 15 \ cm = 0.15 \ m$

Displacement of bullet in horizontal direction

$x = 150 \ m$

Using equation of trajectory

$y = \dfrac{gx^2}{2u^2}$

$0.15 = \dfrac{10\times (150)^2}{2u^2}$

$\implies \ u = 500\sqrt{3} \ m/s$

Two particles are thrown simultaneously from points

$A$ and

$B$ with velocities

${ u }_{ 1 } = 2\ { ms }^{ -1 }$ and

${ u }_{ 2 } = 14\ { ms }^{ -1 }$ , respectively, as shown in Fig

The direction (angle) with horizontal at which

$B$ will appear to move as seen from

$A$ is

0%
$37^{}$
0%
$53^{}$
0%
$15^{}$
0%
$90^{}$

A particle of mass

$2 kg$ moves with an initial velocity of

$\bar { v } =4\hat { i } +4\hat { j } m{ s }^{ -1 }$ . A constant force of

$\bar { F } =20\hat { j } N$ is applied on the particle. Initially, the particle when its y-coordinate again becomes zero is given by

0%
$1.2 m$
0%
$4.8 m$
0%
$6.0 m$
0%
$3.2 m$

Two particles

$P$ and

$Q$ projected simultaneously away from each other from a point

$A$ as shown in the figure. The velocity of

$P$ relative to

$Q$ in

$m { s }^{ -1 }$ at the instant when the motion of

$P$ is horizontal is

0%
$10\sqrt { 4-\sqrt { 3 } }$
0%
$20\sqrt { 4-\sqrt { 3 } }$
0%
$10\sqrt { 4+\sqrt { 3 } }$
0%
$20\sqrt { 4+\sqrt { 3 } }$

The point from where a ball is projected is taken as the origin of the coordinates axes. The

$x$ and

$y$ components of its displacement are given by

$x = 6t$ and

$y = 8t - 5t^{2}$ . What is the velocity of projection?

0%
$6 \ ms^{-1}$
0%
$8 \ ms^{-1}$
0%
$10 \ ms^{-1}$
0%
$14 \ ms^{-1}$

Explanation

The velocities are given as,

$\dfrac{{dx}}{{dt}} = 6$

${v_x} = 6$

$\dfrac{{dy}}{{dt}} = 6$

${v_y} = 8 - 10t$

The velocity at projection is given as,

$v\left( t \right) = {v_x}\hat i + {v_y}\hat j$

$v\left( t \right) = 6\hat i + \left( {8 - 10t} \right)\hat j$

$v\left( 0 \right) = 6\hat i + 8\hat j$

$\left| {v\left( 0 \right)} \right| = \sqrt {{6^2} + {8^2}}$

$\left| {v\left( 0 \right)} \right| = 10\;{\rm{m/s}}$

The velocity of projection is $10\;{\rm{m/s}}$ .

Which of the following statements is/are correct (figure)?

0%
The sign of the x-component of $\vec{d}_{1}$ is positive and that of $\vec{d}_{2}$ is negative.
0%
The signs of the y-components of $\vec{d}_{1}$ and $\vec{d}_{2}$ are positive and negative, respectively.
0%
The signs of the x- and y-components of $\vec{d}_{1}$ + $\vec{d}_{2}$ are positive.
0%
None of these.

a particle of mass

$2 kg$ moves with an initial velocity of

$(4\hat { i } +2\hat { i } ) m{ s }^{ -1 }$ on the x-y plane. A force

$F=(2\hat { i } +8\hat { j } )N$ acts on the particle. The initial position of the particle is

$(2 m, 3 m)$ . Then for

$y=3 m$ ,

0%
Possible value for x is only $x=2 m$
0%
Possible value of x is not only $x=2 m$ , but there exists some other value of x also
0%
Time taken is $2 s$
0%
All of the above

A uniform disc of mass

$M$ and radius

$R$ is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc and a steady downward pull

$T$ is exerted on the cord. The tangential acceleration of a point on the rim is

0%
$\cfrac { T }{ M }$
0%
$\cfrac { MR }{ T }$
0%
$\cfrac { 2T }{ MR }$
0%
$\cfrac { MR }{ 2T }$

Explanation

$\Rightarrow \alpha=\dfrac{Tosaue}{M.I}$

$=\dfrac{T.R2}{MR^2}$

$=\dfrac{2T}{MR}.$

Hence, the answer is

$\dfrac{2T}{MR}.$

A particle is projected at an angle of elevation

$\alpha$ and after

$t$ second, it appears to have an angle of elevation

$\beta$ as seen from the point of projection. The initial velocity will be

0%
$\dfrac { gt }{ 2\sin { \left( \alpha -\beta \right) } }$
0%
$\dfrac { gt\cos { \beta } }{ 2\sin { \left( \alpha -\beta \right) } }$
0%
$\dfrac { \sin { \left( \alpha -\beta \right) } }{ 2gt }$
0%
$\dfrac { 2\sin { \left( \alpha -\beta \right) } }{ gt\cos { \beta } }$

A disc rotates about its axis with a constant angular acceleration of

$4$ rad

$/s^2$ . Find the radius tangential accelerations of a particle at a distance of

$1$ cm from the axis at the end of the second after the disc starts rotating.

0%
$0.16 m/s$
0%
$1 m/s$
0%
$3 m/s$
0%
$10 m/s$

Explanation

Given :

$\alpha = 4 \ rad/s^2$

$r = 0.01 \ m$

Tangential acceleration

$a_t = r\alpha = 0.01\times 4 = 0.04 \ m/s$

Initial angular velocity

$w_i = 0$

Time

$t = 1 \ s$

Final angular velocity

$w_f = w_i + \alpha t$

$w_f = 0+4\times 1 = 4 \ rad/s$

Radial acceleration

$a_r = rw_f^2 = 0.01\times 4^2 =0 .16 \ m/s$

A disc of mass

$m$ , radius

$R$ is set rolling with angular velocity

$\omega$ , up on a rough inclined plane of inclination

$30$ to the horizontal and is in rolling motion always. The distance it will move up on the inclined plane will be:

0%
$\dfrac{3}{4} \dfrac{R^2 \omega^2}{g}$
0%
$\dfrac{3}{2} \dfrac{R^2 \omega^2}{g}$
0%
$\dfrac{3R^2 \omega^2}{g}$
0%
$\dfrac{9}{8} \dfrac{R^2 \omega^2}{g}$

A wheel is making Revolution about axis with uniform angular acceleration, Starting from rest, it reaches

$100\ rev/sec$ in

$4\ seconds$ . Find the angular rotated during these four seconds.

0%
$200 \pi rad/sec$
0%
$400 \pi rad/sec$
0%
$600 \pi rad/sec$
0%
$800 \pi rad/sec$

Explanation

We know that angular velocity

$(\omega)=2\pi f$

Frequency

$(f)=100rev/sec$

$\implies \omega =200\pi rad/s$

angular acceleration

$(\alpha)=\dfrac{\omega}{t}=\dfrac{200\pi}{4}=50\pi rad/s^2$

angle rotated

$(\theta)=\omega_ot+\dfrac{1}{2}\alpha t^2$

$\omega_o=0$

$\implies \theta=\dfrac{1}{2}50\pi\times 4^2=400\pi rad$

A positive charge moving along the positive x-direction with a speed v traces a non-linear trajectory OA in the x-y plane after passing O as shown in the figure. The combination of electric

$(\vec{E})$ and magnetic field

$(\vec{B})$ fields that would lead to the trajectory OA is (a, b,. c are non-zero positive constants).

0%
$\vec{E}=0; \vec{B}=a\hat{i}+c\hat{k}$
0%
$\vec{E}=a\hat{i}; \vec{B}=b\hat{j}+c\hat{k}$
0%
$\vec{E}=0; \vec{B}=b\hat{j}+c\hat{k}$
0%
$\vec{E}=a\hat{i}; \vec{B}=b\hat{j}-c\hat{k}$

A conducting circular loop of radius r carries a constant current ii. It is placed in a uniform magnetic field

$\vec{B_0}$ such that

$\vec{B_0}$ is perpendicular to the plane of the loop. The magnetic force acting on the loop is?

0%
ir $B_0$
0%
$2\pi$ ir $B_0$
0%
Zero
0%
$\pi$ is $B_0$

You throw a ball which a launch velocity of

$\vec {v}=(3\hat {i}+4\hat {j})\ m/s$ the maximum height attaind by the body is ?

0%
$0.8m$
0%
$0.2 m$
0%
$2.3m$
0%
$5.3m$

Explanation

Given,

$\vec v=(3\hat i+4\hat j)m/s$

$v=\sqrt{3^2+4^2}$

$v=5m/s$

The angle subjected by the velocity vector with the x-axis is

$tan\theta=\dfrac{4}{3}$ and

$sin\theta=\dfrac{4}{5}$

The maximum height attained the body is

$H=\dfrac{v^2sin^2\theta }{2g}$

$H=\dfrac{5\times 5\times (\dfrac{4}{5})^2}{2\times 10}$

$H=0.8m$

The correct option is A.

A particle is moving along a circle such that it completes one revolution in 40 seconds. In 2 minutes 20 seconds, the ratio

$\frac{{\left| {displacment} \right|}}{{dis\tan ce}}$ is

0%
0
0%
1/7
0%
2/4
0%
1/11

Two particles A and B of equal masses have velocities

$\vec{V_a} = 4\hat{i}$ and

$\vec{V_B} = 4\hat{jm}$ . The particles move with accelerations

$\vec{a_A} =(-5\hat{i}+5\hat{j})ms^2$ respectively. while the acceleration of other part is zero.the center of mass of two particles move in a path of

0%
a straight line
0%
a parabola
0%
a circle
0%
an ellipse

Consider the x-axis as representing east, the y-axis as north and z-axis as vertically upwards. Give the vector representing each of the following points.

0%
5 m north east and 2 m up
0%
4 m south east and 3 m up
0%
2 m north west and 4 m up
0%
3 m south east and 5 m up

$R$ and

$H$ are the horizontal range and maximum height attained by a projectile, than its speed of projection is:

0%
$\sqrt{2gR+\dfrac{4R^{2}}{gH}}$
0%
$\sqrt{2gH+\dfrac{R^{2}g}{8H}}$
0%
$\sqrt{2gH+\dfrac{8H}{Rg}}$
0%
$\sqrt{2gH+\dfrac{R^{2}}{H}}$

CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 12 - 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 12 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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