Answer: Option D. -> 250 J
:
D
Rotational kinetic energy $\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\omega }^2=\frac{1}{2}(\frac{1}{2}\times 10\times (0.5{)}^2)(20{)}^2=250J$

Answer: Option A. -> vL
:
A
Initial angular momentum of the system about point O
= Linear momentum $\times$ Perpendicular distance of linear momentum from the axis of rotation $=Mv\left(\frac{L}{2}\right)...\left(i\right)$
Final angular momentum of the system about point $O=I_1\omega +I_2\omega =(I_1+I_2)\omega$
$=[M{\left(\frac{L}{2}\right)}^2+M{\left(\frac{L}{2}\right)}^2]\omega$ .....(ii)
Applying the law of conservation of angular momentum
$\Rightarrow Mv\left(\frac{L}{2}\right)=2M{\left(\frac{L}{2}\right)}^2\omega$ $\Rightarrow \omega =\frac{v}{L}$

Answer: Option B. -> (1+μ2k)ω20R8πμk(1+μk)g
:
B
As the centre of mass of the cylinder does not accelerate, hence $\sum F=0$
$\sum F_x=0,N_2-{\mu }_k{N}_1=0$ ...(1)
$\sum F_x=0,N_1-{\mu }_k{N}_2-mg=0$ ...(2)
Solving these equations: $N_1=\frac{mg}{1+{\mu }_k^2},N_2=\frac{{\mu }_{k}mg}{1+{\mu }_k^2}$

The torque on the cylinder about the axis of rotation
The moment of inertia about axis of rotation $1_{cm}=\frac{1}{2}m{R}^2$
The torque equation $T=1\alpha$
Using equation ${\omega }^2={\omega }_0^2+2\alpha \theta$, Calculate the angular displacement $\theta$,
Revolution accomplished,

Answer: Option B. -> 800%
:
B
As $E=\frac{L^2}{2I}\Rightarrow \frac{E_2}{E_1}={\left(\frac{L_2}{L_1}\right)}^2={\left(\frac{3L_1}{L_1}\right)}^2$ $[As{L}_2=L_1+200\%.L_1=3L_1]$
$\Rightarrow E_2=9E_1=E_1+800\%of{E}_1$

Answer: Option B. -> mv302√2g(−^k)
:
B
Let us take the origin at $P$, $X$-axis along the horizontal and $Y$-axis along the vertically upward direction as shown in figure. For 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,
$v_y$ = $v_{0}sin{45}^{\circ }-gt$ = $\frac{v_0}{\sqrt{2}}-v_0$ = $\frac{(1-\sqrt{2})}{\sqrt{2}}{v}_0$
and $y$ = $\left(v_{0}sin{45}^{\circ }\right)t-\frac{1}{2}g{t}^2$
= $\frac{v_0^2}{\sqrt{2}g}$ - $\frac{v_0^2}{2g}$ = $\frac{v_0^2}{2g}(\sqrt{2}-1)$
The angular momentum of the particle at time $t$ about the origin is
$L$ = $\overrightarrow{r}x\overrightarrow{p}$ = $m\overrightarrow{r}x\overrightarrow{v}$
= $m(\overrightarrow{i}x+\overrightarrow{j}y)x(\overrightarrow{i}{v}_x+\overrightarrow{j}{v}_y)$
= $m(\overrightarrow{k}x{v}_y-\overrightarrow{k}y{v}_x)$
= $m\overrightarrow{k}\left[\left(\frac{v_0^2}{\sqrt{2}g}\right)\frac{v_0}{\sqrt{2}}\right(1-\sqrt{2})-\frac{v_0^2}{2g}(\sqrt{2}-1\left)\frac{v_0}{\sqrt{2}}\right]$
= $-\overrightarrow{k}\frac{m{v}_0^3}{2\sqrt{2}g}$.
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 B. -> Angular acceleration of the body has to be zero
:
B
Angular acceleration of the body may not necessarily be zero about any axis about which net external torque is non-zero.

Answer: Option B. -> 513
:
B
$I=\frac{5}{8}m{r}^2$
$\frac{R.K.E}{T.E}=\frac{1}{1+\frac{R^2}{K^2}}=\frac{1}{1+\frac{8}{5}}=\frac{5}{13}$

Answer: Option C. -> 30 J
:
C
${\omega }_1=20$ rad\sec, ${\omega }_2=0,$ t = 4 sec. So angular retardation $\alpha =\frac{{\omega }_1-{\omega }_2}{t}=\frac{20}{4}=5rad{\sec }^2$
Now angular speed after 2 sec ${\omega }_2={\omega }_1-\alpha t=20-5\times 2=10rad\sec$
Work done by torque in 2 sec = loss in kinetic energy $=\frac{1}{2}I({\omega }_1^2-{\omega }_2^2)=\frac{1}{2}\left(0.20\right)\left(\right(20{)}^2-\left(10{)}^2\right)$
$=\frac{1}{2}\times 0.2\times 300=30J.$

Answer: Option C. -> 2304 J
:
C
Kinetic energy,K.E. $=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(0.32\right)(120{)}^2=2304J.$

Answer: Option D. -> 1 m/s
:
D
Rotational kinetic energy of the body $=\frac{1}{2}I{\omega }^2$ and translatory kinetic energy $=\frac{1}{2}m{v}^2$
According to problem $=\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2\Rightarrow \frac{1}{2}\times 3\times (2{)}^2=\frac{1}{2}\times 12\times v^2\Rightarrow v=1m/s.$