Physics M.C.Q (1 Marks)

$T=m / \omega^2$

(image)

$T^{\prime}=m \ell(2 \omega)^2$

$T^{\prime}=4$

$T=m / \omega^2$

(image)

$T^{\prime}=m \ell(2 \omega)^2$

$T^{\prime}=4$

$=\frac{(280)^2\left(\sin 30^{\circ}\right)^2}{2(9.8)}$

$=1000\,m$

$=\frac{(280)^2\left(\sin 30^{\circ}\right)^2}{2(9.8)}$

$=1000\,m$

$u _{ x }= u \cos \theta =10 \cos 30^{\circ}$

$=5 \sqrt{3} ms ^{-1}$

$u _{ x }= u \cos \theta =10 \cos 30^{\circ}$

$=5 \sqrt{3} ms ^{-1}$

$=5$

$=5$

$\mathrm{a}=5$

$\mathrm{t}=4$

$V=u+a t$

$V=0+5 \times 4$

$V=20$

$\mathrm{V}_{\mathrm{x}}=20\, \mathrm{~m} / \mathrm{sec}$

$\mathrm{V}_{\mathrm{y}}=\mathrm{u}+\mathrm{ut}$

$=10 \times 2$

$\mathrm{~V}_{\mathrm{y}}=20\, \mathrm{~m} / \mathrm{sec}$

$\mathrm{V}=20 \sqrt{2}$

and $\mathrm{a}=10\, \mathrm{~m} / \mathrm{sec}^{2}$

$V=\frac{2 \pi R}{T}$

$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$

$4 R=\frac{4 \pi^{2} R^{2} \sin ^{2} \theta}{T^{2} 2 g}$

$\sin ^{2} \theta=\frac{8 R T^{2} g}{4 \pi^{2} R^{2}}$

$\sin \theta=\sqrt{\frac{2 T^{2} g}{\pi^{2} R}}$

$\theta=\sin ^{-1}\left(\frac{2 T^{2} g}{\pi^{2} R}\right)^{1 / 2}$

$\mathrm{a}=5$

$\mathrm{t}=4$

$V=u+a t$

$V=0+5 \times 4$

$V=20$

$\mathrm{V}_{\mathrm{x}}=20\, \mathrm{~m} / \mathrm{sec}$

$\mathrm{V}_{\mathrm{y}}=\mathrm{u}+\mathrm{ut}$

$=10 \times 2$

$\mathrm{~V}_{\mathrm{y}}=20\, \mathrm{~m} / \mathrm{sec}$

$\mathrm{V}=20 \sqrt{2}$

and $\mathrm{a}=10\, \mathrm{~m} / \mathrm{sec}^{2}$

$V=\frac{2 \pi R}{T}$

$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$

$4 R=\frac{4 \pi^{2} R^{2} \sin ^{2} \theta}{T^{2} 2 g}$

$\sin ^{2} \theta=\frac{8 R T^{2} g}{4 \pi^{2} R^{2}}$

$\sin \theta=\sqrt{\frac{2 T^{2} g}{\pi^{2} R}}$

$\theta=\sin ^{-1}\left(\frac{2 T^{2} g}{\pi^{2} R}\right)^{1 / 2}$

$\mathrm{u}=10 \mathrm{m} / \mathrm{s}$

$\sin \theta=\frac{u}{v}=\frac{10}{20}=\frac{1}{2}$

$\Rightarrow \theta=30^{\circ}$ west

$\mathrm{s}_{\mathrm{y}}=-\frac{1}{2} \mathrm{gt}^{2}=-\frac{1}{2} \times 10 \times 4=-20$

Height $=180 \mathrm{m}$

For $y-$projection,

$y=A \cos \omega t$

$\Rightarrow y=3 \cos \left(\frac{\pi t}{2}\right)$

$\Rightarrow \frac{2 \pi}{\omega_{A}}=\frac{2 \pi}{\omega_{B}}$

$\Rightarrow \frac{\omega_{A}}{\omega_{B}}=1: 1$

$\Rightarrow \mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}}=\frac{\mathrm{u}^{2}}{2 \mathrm{gsin} \theta}$

$\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}$

$\Rightarrow \frac{x_{1}}{x_{2}}=\frac{1}{\sqrt{3}}$

$\alpha=\frac{\omega^{2}-\omega_{0}^{2}}{2 \theta}=\frac{\left(\mathrm{V}_{0} / \mathrm{r}\right)^{2}-0}{2(2 \pi \mathrm{n})}$$=\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{nr}^{2}}$

$\mathrm{u}=10 \mathrm{m} / \mathrm{s}$

$\sin \theta=\frac{u}{v}=\frac{10}{20}=\frac{1}{2}$

$\Rightarrow \theta=30^{\circ}$ west

$\mathrm{s}_{\mathrm{y}}=-\frac{1}{2} \mathrm{gt}^{2}=-\frac{1}{2} \times 10 \times 4=-20$

Height $=180 \mathrm{m}$

For $y-$projection,

$y=A \cos \omega t$

$\Rightarrow y=3 \cos \left(\frac{\pi t}{2}\right)$

$\Rightarrow \frac{2 \pi}{\omega_{A}}=\frac{2 \pi}{\omega_{B}}$

$\Rightarrow \frac{\omega_{A}}{\omega_{B}}=1: 1$

$\Rightarrow \mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}}=\frac{\mathrm{u}^{2}}{2 \mathrm{gsin} \theta}$

$\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}$

$\Rightarrow \frac{x_{1}}{x_{2}}=\frac{1}{\sqrt{3}}$

$\alpha=\frac{\omega^{2}-\omega_{0}^{2}}{2 \theta}=\frac{\left(\mathrm{V}_{0} / \mathrm{r}\right)^{2}-0}{2(2 \pi \mathrm{n})}$$=\frac{\mathrm{V}_{0}^{2}}{4 \pi \mathrm{nr}^{2}}$

$T _{2}=\frac{2 u _{2} \sin \theta}{ g }$

$T _{1}= T _{2}$

$\Rightarrow \frac{2}{ g } u _{1}=\frac{2 u _{2} \sin \theta}{ g } \Rightarrow u _{1}= u _{2} \sin \theta$

$\text { Height }=\frac{ u ^{2} \sin ^{2} \theta}{2 g } \Rightarrow \frac{ H _{1}}{ H _{2}}=\frac{u_{1}^{2}}{u_{2}^{2} \sin ^{2} \theta}$

$\Rightarrow \frac{ H _{1}}{ H _{2}}=1$

$T _{2}=\frac{2 u _{2} \sin \theta}{ g }$

$T _{1}= T _{2}$

$\Rightarrow \frac{2}{ g } u _{1}=\frac{2 u _{2} \sin \theta}{ g } \Rightarrow u _{1}= u _{2} \sin \theta$

$\text { Height }=\frac{ u ^{2} \sin ^{2} \theta}{2 g } \Rightarrow \frac{ H _{1}}{ H _{2}}=\frac{u_{1}^{2}}{u_{2}^{2} \sin ^{2} \theta}$

$\Rightarrow \frac{ H _{1}}{ H _{2}}=1$

$ = 2v \times \frac{1}{{\sqrt 2 }} = \sqrt 2 v$

$ = \sqrt 2 \times r\omega = \sqrt 2 \times 1 \times \frac{{2\pi }}{{60}} = \frac{{\sqrt 2 \pi }}{{`30}}\;cm/s$
 

Velocity along $Y-$axis ${v_y} = \frac{{dy}}{{dt}} = 2bt$

Magnitude of velocity of the particle,

$v = \sqrt {v_x^2 + v_y^2} = 2t\sqrt {{a^2} + {b^2}} $

At $t = 1,\;{v_x} = 0$

${v_y} = \frac{{dy}}{{dt}} = \frac{d}{{dt}}({t^2} - 2t) = 2t - 2$.

At $t = 1,\;{v_y} = 0$

Hence $v = \sqrt {v_x^2 + v_y^2} = 0$

at the time of projection i.e. ${v_y} = \frac{{dy}}{{dt}} = 8$and ${v_x} = 6$

The angle of projection is given by $\theta = {\tan ^{ - 1}}\left( {\frac{{{v_y}}}{{{v_x}}}} \right) $

$= {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$

${a_y} = \frac{d}{{dt}}({v_y}) = - 10\;m/{s^2}$

Net acceleration $a = \sqrt {a_x^2 + a_y^2} $=$\sqrt {{0^2} + {{10}^2}} =10m/s^2$

$y = {u_y}t + \frac{1}{2}{a_y}{t^2} = \frac{1}{2} \times 8 \times {(4)^2} = 64\,m$

$d = \sqrt {{x^2} + {y^2}} = \sqrt {{{(48)}^2} + {{(64)}^2}} = 80\,m$

$ = 2 \times \frac{1}{{\sqrt 2 }}\, = \,\sqrt 2 \,km$

Net movement along y-direction $Sy = (6 + 4) sin 45^o $$\hat j$ $ = 10 \times \frac{1}{{\sqrt 2 }} = 5\sqrt 2 \,km$

Net movement from starting point

$|\overrightarrow S |\, = \sqrt {{S_x}^2 + {S_y}^2} $$ = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {5\sqrt 2 } \right)}^2}} $=$\sqrt {52} \,\,km$

Angle which makes with the east direction

$\tan \theta = \frac{{{\rm{Y}} - {\rm{component}}}}{{{\rm{X}} - \,{\rm{component}}}}$ $ = \frac{{5\sqrt 2 }}{{\sqrt 2 }}$ $\therefore \theta = {\tan ^{ - 1}}(5)$

$\sin (t)$ Slope at $t=0$ is $- ve.$ 

Hence, options $(A),$ $(B)$ and $(D)$ are not possible.

$t=\frac{x}{v_{e s c l a \rightarrow r}+v_{\operatorname{man}}} \Rightarrow t=\frac{x}{x+\frac{x}{3}}$

$\Rightarrow t=\frac{3}{4} \min \Rightarrow t=45 \mathrm{sec}$

So, here $a=m v$ [where $m$ is negative] $\Rightarrow \frac{d v}{d t}=m v$

$\Rightarrow \frac{d v}{v}=m d t$

$\Rightarrow v=e^{m t}$

as $m$ is negative. So, $v=e^{-k t}$

Required graph will be Option$'D’$

Now, The average velocity of the region from point $P$ till the line will be nullified by the vertical component of velocity of region $BCD.$

Thus, the average velocity will only be in the right direction. Thus, direction of average velocity will be same as instantaneous velocity of point $\mathrm{C}$.

Which is max. in region $4$ to $6 s$