Free Objective Test 01 Practice Test - 11th and 12th 

Angular velocity of particle P about point A,
${\omega }_A$ = $\frac{v}{r_{AB}}$ = $\frac{v}{2r}$
Angular velocity of particle P about point C,
${\omega }_C$ = $\frac{v}{r_{BC}}$ = $\frac{v}{r}$
Ratio $\frac{{\omega }_A}{{\omega }_C}$ = $\frac{\frac{v}{2r}}{\frac{v}{r}}$ = $\frac{1}{2}$

In 15 second's hand rotate through ${90}^{\circ }$
 Change in velocity |$\overline{△}v$| = 2v sin ($\frac{\theta }{2}$)
=2(r$\omega$) sin ($\frac{{90}^{\circ }}{2}$) = 2 $\times$ 1 $\times$ $\frac{2\pi }{T}$ $\times$ $\frac{1}{\sqrt{2}}$
= $\frac{4\pi }{60\sqrt{2}}$ = $\frac{4\pi }{60\sqrt{2}}$ = $\frac{\pi \sqrt{2}cm}{30sec}$ [As T = 60 sec]

R = mg cos $\theta$ - $\frac{m{v}^2}{r}$

when $\theta$  decreases cos $\theta$ increases i.e.,R increases.

$\frac{m{v}^2}{r}$ = $\frac{k}{r^2}$ $\Rightarrow$ m $v^2$ = $\frac{k}{r}$ $\therefore$ K.E = $\frac{1}{2}m{v}^2$ = $\frac{k}{2r}$
P.E = $\int$F dr = $\int$ $\frac{k}{r^2}$dr = - $\frac{k}{r}$ 
$\therefore$ Total energy = K.E + P.E = $\frac{k}{2r}$ -  $\frac{k}{r}$  =  - $\frac{k}{2r}$ 

The maximum velocity for a baked road with friction,
$v^2$ = gr($\frac{\mu +tan\theta }{1-\mu tan\theta })$
$\Rightarrow$ $v^2$ = 9.8 $\times$ 1000 $\times$ ($\frac{0.5+1}{1-0.5\times 1}$) $\Rightarrow$ v = 172 m/s 

Centripental force = breaking force
$\Rightarrow$ m ${\omega }^2$ r  = Breaking stress $\times$ cross sectional area
$\Rightarrow$ m ${\omega }^2$ r = p $\times$ A $\Rightarrow$ $\omega$ = $\sqrt{\frac{p\times A}{mr}}$ = $\sqrt{\frac{4.8\times {10}^7\times {10}^{-6}}{10\times 0.3}}$
$\therefore$ $\omega$ = 4 rad/sec

Let $\omega$ is the angular speed of revoultion 
$T_3$ =m ${\omega }^2$ 3l
$T_2$  - $T_3$  = m${\omega }^2$ 2l $\Rightarrow$ $T_2$ = m${\omega }^{2}5l$
$T_1$ - $T_2$ = m${\omega }^{2}l$  $\Rightarrow$ $T_1$ = m${\omega }^{2}6l$
$T_3$:$T_2$:$T_1$ = 3 : 5 : 6

The particle is moving in circular path
From the figure, mg = Rsin $\theta$ ..(i)
$\frac{m{v}^2}{r}$ = R cos $\theta$ ..(ii)
From equation (i) and (ii) we get
tan $\theta$ = $\frac{rg}{v^2}$ but tan $\theta$ = $\frac{r}{h}$
$\therefore$ h = $\frac{v^2}{g}$ = $\frac{{\left(0.5\right)}^2}{10}$ = 0.025 m = 2.5 cm

Using relation $\theta$ = ${\omega }_0$t + $\frac{1}{2}$ $\alpha$ $t^2$
${\theta }_1$ = $\frac{1}{2}$ ($\alpha$) $2^2$ = 2 $\alpha$ ..(i) (As ${\omega }_0=0,t=2sec$)
Now using same question for t = 4 sec , ${\omega }_0$ = 0
${\theta }_1$ + ${\theta }_2$ = $\frac{1}{2}$ ${\left(4\right)}^2$ = 8 $\alpha$ ..(ii)
From (i) and (ii), ${\theta }_1$ = 2 $\alpha$ and ${\theta }_2$ = 6$\alpha$ $\therefore$ $\frac{{\theta }_1}{{\theta }_2}$ = 3

When body is released from the position p (inclined at angle $\theta$ from vertical) then velocity at mean position 
v = $\sqrt{2gl(1-cos\theta )}$
$\therefore$ Tension at the lowest point = mg + $\frac{m{v}^2}{l}$
= mg + $\frac{m}{l}$[2gl(1 - cos 60)] = mg + mg = 2mg

By using equation ${\omega }^2$ = ${{\omega }_0}^2$ - 2 $\alpha \theta$
${\frac{{\omega }_0}{2}}^2$ = ${{\omega }_0}^2$ - 2$\alpha \left(2\pi n\right)$ $\Rightarrow$ $\alpha$ = $\frac{3}{4}$ $\frac{{{\omega }_0}^2}{4\pi \times 36}$ , (n = 36) ..(ii)
Now let fan completes total $n^{{}^{\prime }}$ revolution from the starting to come to rest
0 = ${{\omega }_0}^2$ - $2\alpha \left(2\pi n\right)$ $\Rightarrow$ $n^{{}^{\prime }}$ = $\frac{{{\omega }_0}^2}{4\alpha \pi }$
substituting the value of $\alpha$ from equation (i)
$n^{{}^{\prime }}$ = $\frac{{{\omega }_0}^2}{4\pi }$ $\frac{4\times 4\pi \times 36}{3{{\omega }_0}^2}$ = 48 revolution
Number of rotation = 48 - 36 = 12

As we know for hemisphere the particle will leave the sphere at height h = $\frac{2r}{3}$
h = $\frac{2}{3}\times$ 21= 14 m
but from the bottom
H = h +r = 14 + 21 = 35 metre

In this problem it is assuemed that particle although moving in a vertical loop but its speed remain constant.
Tenstion at lowest point $T_{max}$ = $\frac{m{v}^2}{r}$ + mg
Tension at highest point $T_{min}$ = $\frac{m{v}^2}{r}$ - mg
$\frac{T_{max}}{T_{min}}$ = $\frac{\frac{m{v}^2}{r}+mg}{\frac{m{v}^2}{r}-mg}$ = $\frac{5}{3}$
by solving we get , v=$\sqrt{4gr}$ = $\sqrt{4\times 9.8\times 2.5}$ = $\sqrt{98}$ m/s

Vertical component of velocity of ball at point P
$v_v$ = 0 + gt = 10 $\times$ 0.4 = 4 m/s
Horizontal component of velocity = inital velocity
$\Rightarrow$ $v_H$ = 4 m/s
So the speed with which it hits the ground
v = $\sqrt{({v_H}^2+{v_y}^2)}$ = 4$\sqrt{2}$ m/s
and tan $\theta$ = $\frac{v_V}{v_H}$ =$\frac{4}{4}$ = 1 $\Rightarrow$ $\theta$ = ${45}^{\circ }$
It means the ball hits the ground at an angle of ${45}^{\circ }$ to the horizontal.
Height of the table h = $\frac{1}{2}$ g $t^2$ = $\frac{1}{2}$ $\times$ 10 $\times$ ${\left(0.4\right)}^2$ = 0.8m
Horizontal distance travelled by the ball from the edge of table h = ut = 4 $\times$ 0.4 = 1.6m

Horizontal component of velocity $v_x$ = 500 m/s
and vertical components of velocity while striking the ground.
$v_y$ = 0 + 10 $\times$ 10 = 100 m/s
$\therefore$ Angle with which it strikes the ground.
$\theta$ = ${tan}^{-1}$($\frac{v_y}{v_x}$) = ${tan}^{-1}$($\frac{100}{500}$)