Answer: Option B. -> √43gh
:
B
Velocity at the bottom (v)$=\sqrt{\frac{2gh}{1+\frac{K^2}{R^2}}}=\sqrt{\frac{2gh}{1+\frac{1}{2}}}=\sqrt{\frac{4}{3}gh}.$

Answer: Option A. -> 5mT2M
:
A
$I_1{\omega }_1=I_2{\omega }_2$
$\frac{2}{5}M{R}^2\times \left(\frac{2\pi }{T_1}\right)=(\frac{2}{5}M{R}^2+m{R}^2)\frac{2\pi }{T_2}$
$\frac{T_2}{T_1}=1+\frac{5m}{2M}$
$\frac{T_2-T_1}{T_1}=\frac{5m}{2M}$
$T_2-T_1=\frac{5mT}{2M}$ $[i.e;T_1=T]$

Answer: Option C. -> g4
:
C
$a=\frac{g\sin \theta }{1+\frac{k^2}{R^2}}=\frac{g\sin {30}^{\circ }}{1+1}=\frac{g}{4}$ $[As\frac{k^2}{R^2}=1and\theta ={30}^{\circ }]$

Answer: Option B. -> √34gl sinθ
:
B
Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it reaches the bottom is v. As the cylinder rolls without slipping, its angular speed about its axis is $\omega =\frac{v}{r}$. The kinetic energy at the bottom will be= $\frac{3}{4}m{v}^2$
Note: Try reducing this result on your own for practice.
Thus, $\frac{3}{4}m{v}^2=mglsin\theta$
or, v=$\sqrt{\frac{4}{3}glsin\theta }$

Answer: Option B. -> mv302√2g^(−j)
:
B
Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t,

$v_x=v_{0}cos{45}^{\circ }=\frac{v_0}{\sqrt{2}}$
and $x=v_{x}t=\frac{v_0}{\sqrt{2}}.\frac{v_0}{g}=\frac{v_0^2}{\sqrt{2}g}$
For vertical motion,
$v_{\gamma }=v_{0}sin{45}^{\circ }-gt=\frac{v_0}{\sqrt{2}}-v_0=\frac{1-\sqrt{2}}{\sqrt{2}}{v}_0$
and $y=\left(v_{0}sin{45}^{\circ }\right)t-\frac{1}{2}g{t}^2$
$y=\frac{v_0^2}{\sqrt{2}g}-\frac{v_0^2}{2g}=\frac{v_0^2}{2g}(\sqrt{2}-1)$
The angular momentum of the particle at time t about the origin is
$L=\vec{r}\times \vec{p}=m\vec{r}\times \vec{v}$
=$m(\overrightarrow{ix}+\overrightarrow{jy})\times (\overrightarrow{i{v}_x}+\overrightarrow{j{v}_y})$
=$m(\vec{k}x{v}_y-\overrightarrow{k_y}{v}_x)$
$=m\vec{k}\left[\left(\frac{v_0^2}{\sqrt{2}g}\right)\frac{v_0}{\sqrt{2}}\right(1-\sqrt{2})-\frac{v_0^2}{2g}(\sqrt{2}-1\left)\frac{v_0}{\sqrt{2}}\right]$
$=-\vec{k}\frac{m{v}_0^3}{2\sqrt{2}g}$
Thus, the angular momentum of the particle is $\frac{m{v}_0^3}{2\sqrt{2}g}$ in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.

Answer: Option C. -> 0.9√2(dm)g
:
C
Let I = length of the rod, and m = mass of the rod.
Applying energy principle

$\left(\frac{1}{2}\right)l{\omega }^2-O=mg\left(\frac{1}{2}\right)(cos{37}^{\circ }-cos{60}^{\circ })$
$\Rightarrow \frac{1}{2}\times \frac{m{l}^2}{3}{\omega }^2=mg\times \frac{1}{2}(\frac{4}{5}-\frac{1}{2})$
$\Rightarrow {\omega }^2=\frac{9g}{10l}=0.9\left(\frac{g}{l}\right)$
Again $\left(\frac{\frac{m}{2}}{3}\right)\propto =mg\left(\frac{1}{2}\right)sin{37}^{\circ }=\frac{mgl}{2}\times \frac{3}{5}$
$\therefore \alpha =0.9\left(\frac{g}{l}\right)=$ angular acceleration.
So, to find out the force on the particle at the tip of the rod
$F_1-$centrifugal force=(dm) $\omega 2l=0.9\left(dm\right)g$
$F_t-$tangential force=(dm) $\omega l=0.9\left(dm\right)g$
So, total force $F=\sqrt{(F_1^2+F_t^2)}=0.9\sqrt{2}\left(dm\right)g$

Answer: Option D. -> √14:√15
:
D
Time of descent $t=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}(1+\frac{k^2}{R^2})}\therefore Ratio=\sqrt{\frac{{(1+\frac{k^2}{R^2})}_{sphere}}{{(1+\frac{k^2}{R^2})}_{disc}}}=\sqrt{\frac{1+\frac{2}{5}}{1+\frac{1}{2}}}=\sqrt{\frac{7}{5}\times \frac{2}{3}}=\sqrt{\frac{14}{15}}$

Answer: Option A. -> 3v4a
:
A
Conserving Angular momentum before and after collision
$mv\times \frac{a}{2}=I\omega$
Now, $I=\frac{m(\sqrt{2}a{)}^2}{12}+m{\left(\frac{a}{\sqrt{2}}\right)}^2$
$=m{a}^2(\frac{1}{6}+\frac{1}{2})$
$=\frac{2}{3}m{a}^2.$
$\therefore \omega =\frac{mva}{2}\times \frac{3}{2m{a}^2}=\frac{3v}{4a}.$

Answer: Option D. -> 2mgtR(M+2m) 
:
D
We know the tangential acceleration $a=\frac{g}{1+\frac{I}{m{R}^2}}=\frac{g}{1+\frac{1/2M{R}^2}{m{R}^2}}=\frac{2mg}{2m+M}$ $[AsI=\frac{1}{2}M{R}^{2}forcylinder]$
After time t, linear velocity of mass $m,v=u+at=0+\frac{2mgt}{2m+M}$
So angular velocity of the cylinder $\omega =\frac{v}{R}=\frac{2mgt}{R(M+2m)}.$

Answer: Option C. -> 3v2L
:
C
Initial angular momentum of the system = Angular momentum of bullet before collision $=Mv\left(\frac{L}{2}\right)$
.....(i) let the rod rotates with angular velocity $\omega$.
Final angular momentum of the system $=\left(\frac{M{L}^2}{12}\right)\omega +M{\left(\frac{L}{2}\right)}^2\omega$ ....(ii)
By equation (i) and (ii) $Mv\frac{L}{2}=(\frac{M{L}^2}{12}+\frac{M{L}^2}{4})\omega or\omega =\frac{3v}{2L}$ ​​​​​​​