Rotational Kinetic Energy Questions in English

(B) The moment of inertia $I$ of the system about the $X$-axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the $X$-axis.
Given masses and their positions on the $Y$-axis:
$m_1 = 4.00 \, kg$ at $y_1 = 3.00 \, m$,so $r_1 = 3.00 \, m$.
$m_2 = 2.00 \, kg$ at $y_2 = -2.00 \, m$,so $r_2 = 2.00 \, m$.
$m_3 = 3.00 \, kg$ at $y_3 = -4.00 \, m$,so $r_3 = 4.00 \, m$.
Calculating the moment of inertia:
$I = (4.00 \, kg)(3.00 \, m)^2 + (2.00 \, kg)(2.00 \, m)^2 + (3.00 \, kg)(4.00 \, m)^2$
$I = (4.00 \times 9) + (2.00 \times 4) + (3.00 \times 16) = 36 + 8 + 48 = 92 \, kg \cdot m^2$.
The rotational kinetic energy is given by $K.E. = \frac{1}{2} I \omega^2$.
Given $\omega = 2 \, rad/s$:
$K.E. = \frac{1}{2} \times 92 \times (2)^2 = \frac{1}{2} \times 92 \times 4 = 184 \, J$.

(C) Rotational kinetic energy is given by $K E_{R} = \frac{1}{2} I \omega^{2}$.
Since the sphere is rolling without slipping,$\omega = \frac{v}{R}$.
Substituting this,$K E_{R} = \frac{1}{2} I \left(\frac{v}{R}\right)^{2}$.
The moment of inertia of a solid sphere about its center is $I = \frac{2}{5} m R^{2}$.
Substituting $I$ into the equation: $K E_{R} = \frac{1}{2} \left(\frac{2}{5} m R^{2}\right) \frac{v^{2}}{R^{2}} = \frac{1}{5} m v^{2}$.
Translational kinetic energy is $K E_{T} = \frac{1}{2} m v^{2}$.
The ratio is $\frac{K E_{R}}{K E_{T}} = \frac{\frac{1}{5} m v^{2}}{\frac{1}{2} m v^{2}} = \frac{2}{5}$.

(D) The rotational kinetic energy $(K_r)$ of a body is given by the formula $K_r = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
We know that the angular velocity $\omega$ is related to the time period $(T)$ by the relation $\omega = \frac{2\pi}{T}$.
Substituting this into the expression for rotational kinetic energy,we get $K_r = \frac{1}{2} I \left( \frac{2\pi}{T} \right)^2$.
Simplifying this,we get $K_r = \frac{1}{2} I \frac{4\pi^2}{T^2} = \frac{2\pi^2 I}{T^2}$.
Since $I$ and $\pi$ are constants for a given body rotating about a fixed axis,we have $K_r \propto \frac{1}{T^2}$.
Therefore,$K_r \propto (Time\ period)^{-2}$.

(C) The rotational kinetic energy $(K)$ of a body is given by the formula: $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Given that the body completes one revolution in $T = \pi \ s$,the angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2 \ rad/s$.
Substituting the value of $\omega$ into the kinetic energy formula:
$K = \frac{1}{2} I (2)^2 = \frac{1}{2} I (4) = 2I$.
Therefore,$I = \frac{K}{2}$,which means the moment of inertia is half of the rotational kinetic energy.

(C) The rotational kinetic energy $E$ and angular momentum $L$ are related by the formula $E = \frac{L^2}{2I}$,where $I$ is the moment of inertia.
This implies $L = \sqrt{2IE}$,so $L \propto \sqrt{E}$.
Let the initial kinetic energy be $E_1 = E$.
The new kinetic energy $E_2$ is increased by $300\ \%$,so $E_2 = E + 300\% \text{ of } E = E + 3E = 4E$.
The ratio of the new angular momentum $L_2$ to the initial angular momentum $L_1$ is $\frac{L_2}{L_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{\frac{4E}{E}} = 2$.
Thus,$L_2 = 2L_1$.
The percentage increase in angular momentum is given by $\frac{L_2 - L_1}{L_1} \times 100\ \% = \frac{2L_1 - L_1}{L_1} \times 100\ \% = 100\ \%$.

(C) The kinetic energy $(K.E.)$ of a particle is given by $K.E. = \frac{1}{2} M v^2$.
We know that the linear momentum $P = Mv$,so $v = \frac{P}{M}$.
Substituting this into the kinetic energy formula: $K.E. = \frac{1}{2} M \left(\frac{P}{M}\right)^2 = \frac{P^2}{2M}$.
For a particle rotating in a circle of radius $R$,the angular momentum $L$ is given by $L = MvR = PR$.
Therefore,$P = \frac{L}{R}$.
Substituting $P = \frac{L}{R}$ into the kinetic energy expression $K.E. = \frac{P^2}{2M}$,we get:
$K.E. = \frac{(L/R)^2}{2M} = \frac{L^2}{2MR^2}$.

(A) The moment of inertia of a uniform thin rod of mass $m$ and length $l$ about an axis passing through one of its ends is given by $I = \frac{1}{3}ml^2$.
The angular velocity $\omega$ in terms of frequency $f$ is given by $\omega = 2\pi f$.
The rotational kinetic energy $K$ is given by the formula $K = \frac{1}{2}I\omega^2$.
Substituting the values of $I$ and $\omega$ into the formula:
$K = \frac{1}{2} \times (\frac{1}{3}ml^2) \times (2\pi f)^2$
$K = \frac{1}{6}ml^2 \times 4\pi^2 f^2$
$K = \frac{2}{3}\pi^2 f^2 ml^2$.
Thus,the correct option is $A$.

(D) The rotational kinetic energy $K$ of a body rotating about an axis is given by $K = \frac{1}{2}I\omega^2$.
For a ring of mass $M$ and radius $R$ rotating about its diameter,the moment of inertia $I$ is given by $I = \frac{1}{2}MR^2$.
Given: $M = 10 \,kg$,$R = 0.5 \,m$,and $\omega = 20 \,rad/s$.
Substituting the values into the formula:
$I = \frac{1}{2} \times 10 \times (0.5)^2 = 5 \times 0.25 = 1.25 \,kg \cdot m^2$.
Now,calculate the kinetic energy:
$K = \frac{1}{2} \times 1.25 \times (20)^2$
$K = \frac{1}{2} \times 1.25 \times 400$
$K = 1.25 \times 200 = 250 \,J$.

(B) The rotational kinetic energy $E$ is related to angular momentum $L$ and moment of inertia $I$ by the formula $E = \frac{L^2}{2I}$.
Since $I$ remains constant,we have $E \propto L^2$,which implies $\frac{E_2}{E_1} = \left( \frac{L_2}{L_1} \right)^2$.
Given that the angular momentum is increased by $200\ \%$,the new angular momentum $L_2 = L_1 + 200\% \text{ of } L_1 = L_1 + 2L_1 = 3L_1$.
Substituting this into the ratio: $\frac{E_2}{E_1} = \left( \frac{3L_1}{L_1} \right)^2 = 3^2 = 9$.
Therefore,$E_2 = 9E_1$.
The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} \times 100\% = \frac{9E_1 - E_1}{E_1} \times 100\% = 8 \times 100\% = 800\%$.

(C) The rotational kinetic energy $K_R$ of a body is given by the formula $K_R = \frac{1}{2} I \omega^2$.
Given:
Moment of inertia $I = 0.32 \ kg \cdot m^2$
Angular velocity $\omega = 120 \ rad/s$
Substituting the values into the formula:
$K_R = \frac{1}{2} \times 0.32 \times (120)^2$
$K_R = 0.16 \times 14400$
$K_R = 2304 \ J$.

(A) The total kinetic energy $(K_{total})$ is the sum of translational kinetic energy $(K_t)$ and rotational kinetic energy $(K_r)$.
Given that $K_r = 50\% \text{ of } K_{total}$,it implies $K_r = K_t$.
We know $K_r = \frac{1}{2} I \omega^2$ and $K_t = \frac{1}{2} mv^2$.
Equating them: $\frac{1}{2} I \omega^2 = \frac{1}{2} mv^2$.
Since $v = \omega r$,we have $\frac{1}{2} I \omega^2 = \frac{1}{2} m(\omega r)^2$.
Simplifying,$I \omega^2 = m \omega^2 r^2$,which gives $I = mr^2$.
The moment of inertia $I = mr^2$ corresponds to a ring.

(C) The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T}$.
Given $T = 1 \ s$,we have $\omega = 2\pi \ rad/s$.
The rotational kinetic energy is $E_R = \frac{1}{2} I \omega^2$.
Substituting the value of $\omega$: $E_R = \frac{1}{2} I (2\pi)^2 = \frac{1}{2} I (4\pi^2) = 2\pi^2 I$.
Therefore,the moment of inertia $I = \frac{E_R}{2\pi^2}$.

(C) Given: $I_1 = I$ and $I_2 = 2I$.
Rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
Since the rotational kinetic energies are equal,we have $K_1 = K_2$.
Therefore,$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$.
Substituting the values: $\frac{1}{2} (I) \omega_1^2 = \frac{1}{2} (2I) \omega_2^2$.
Simplifying the equation: $\omega_1^2 = 2 \omega_2^2$.
Taking the square root on both sides: $\frac{\omega_1}{\omega_2} = \sqrt{\frac{2}{1}} = \frac{\sqrt{2}}{1}$.
Thus,the ratio of their angular velocities is $\sqrt{2}:1$.

(C) The rotational kinetic energy is given by $K_{rot} = \frac{1}{2} I \omega^2$.
For a thin rod of mass $m$ and length $L$ rotating about one end,the moment of inertia is $I = \frac{mL^2}{3}$.
The angular velocity $\omega$ in terms of revolutions per second $n$ is $\omega = 2 \pi n$.
Substituting these values into the formula:
$K_{rot} = \frac{1}{2} \left( \frac{mL^2}{3} \right) (2 \pi n)^2$
$K_{rot} = \frac{1}{2} \left( \frac{mL^2}{3} \right) (4 \pi^2 n^2)$
$K_{rot} = \frac{2}{3} mL^2 \pi^2 n^2$.

(C) The rotational kinetic energy $K$ is related to angular momentum $L$ and moment of inertia $I$ by the formula: $K = \frac{L^2}{2I}$.
Since the moment of inertia $I$ is constant for a body rotating about a fixed axis, we have $K \propto L^2$.
Let the initial angular momentum be $L_1 = L$ and the final angular momentum be $L_2 = L + 0.10L = 1.10L$.
The ratio of kinetic energies is $\frac{K_2}{K_1} = \left( \frac{L_2}{L_1} \right)^2 = (1.10)^2 = 1.21$.
This means $K_2 = 1.21 K_1$.
The percentage increase in kinetic energy is given by $\frac{K_2 - K_1}{K_1} \times 100\% = (1.21 - 1) \times 100\% = 0.21 \times 100\% = 21\%$.

(D) Let the initial angular momentum be $L_1 = L$.
Since the angular momentum is increased by $200\%$,the new angular momentum is $L_2 = L + 2.00L = 3L$.
The rotational kinetic energy $E$ is related to angular momentum $L$ by the formula $E = \frac{L^2}{2I}$,where $I$ is the moment of inertia.
Assuming $I$ remains constant,the ratio of kinetic energies is $\frac{E_2}{E_1} = \frac{L_2^2}{L_1^2} = \frac{(3L)^2}{L^2} = 9$.
The percentage increase in rotational kinetic energy is given by $\left( \frac{E_2 - E_1}{E_1} \right) \times 100\% = \left( \frac{E_2}{E_1} - 1 \right) \times 100\%$.
Substituting the values,we get $(9 - 1) \times 100\% = 8 \times 100\% = 800\%$.
Thus,the rotational kinetic energy increases by $800\%$.

(A) The rotational kinetic energy $K$ is related to the angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.
Given that the kinetic energies are equal,$K_A = K_B$.
Therefore,$\frac{L_A^2}{2I_A} = \frac{L_B^2}{2I_B}$.
This implies $\frac{L_A^2}{L_B^2} = \frac{I_A}{I_B}$.
Taking the square root on both sides,we get $\frac{L_A}{L_B} = \sqrt{\frac{I_A}{I_B}}$.
Since it is given that $I_B > I_A$,it follows that $\frac{I_A}{I_B} < 1$.
Thus,$\frac{L_A}{L_B} < 1$,which means $L_A < L_B$ or $L_B > L_A$.

(D) The rotational kinetic energy is given by $E = \frac{1}{2} I \omega^2$.
For the solid sphere rotating about its diameter,the moment of inertia is $I_s = \frac{2}{5} m R^2$.
For the solid cylinder rotating about its geometrical axis,the moment of inertia is $I_c = \frac{1}{2} m R^2$.
Given that the angular speed of the cylinder is twice that of the sphere,we have $\omega_c = 2 \omega_s$.
The ratio of their kinetic energies is:
$\frac{E_{sphere}}{E_{cylinder}} = \frac{\frac{1}{2} I_s \omega_s^2}{\frac{1}{2} I_c \omega_c^2} = \frac{I_s \omega_s^2}{I_c (2 \omega_s)^2} = \frac{I_s}{4 I_c}$.
Substituting the values of $I_s$ and $I_c$:
$\frac{E_{sphere}}{E_{cylinder}} = \frac{\frac{2}{5} m R^2}{4 \times \frac{1}{2} m R^2} = \frac{2/5}{2} = \frac{1}{5}$.
Thus,the ratio is $1:5$.

(C) The rotational kinetic energy $(K)$ of a body rotating about a fixed axis is given by the formula: $K = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular speed.
Given:
Kinetic energy $(K)$ = $360 \ J$
Angular speed $(\omega)$ = $30 \ rad/s$
Substituting the values into the formula:
$360 = \frac{1}{2} \times I \times (30)^2$
$360 = \frac{1}{2} \times I \times 900$
$360 = 450 \times I$
$I = \frac{360}{450}$
$I = 0.8 \ kg \ m^2$
Therefore,the moment of inertia of the wheel is $0.8 \ kg \ m^2$.

(C) The rotational kinetic energy $E$ of a rotating body is given by the formula $E = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Rearranging the formula to solve for $L$:
$L^2 = 2EI$
$L = \sqrt{2EI}$
Therefore,the correct option is $C$.

(D) The rotational kinetic energy $(KE)$ is related to the angular momentum $(L)$ and moment of inertia $(I)$ by the formula: $KE = \frac{L^2}{2I}$.
Given that the rotational kinetic energies are equal $(KE_1 = KE_2)$,we have:
$\frac{L_1^2}{2I_1} = \frac{L_2^2}{2I_2}$
Substituting the given values $I_1 = I$ and $I_2 = 2I$:
$\frac{L_1^2}{2I} = \frac{L_2^2}{2(2I)}$
$\frac{L_1^2}{I} = \frac{L_2^2}{2I}$
$\frac{L_1^2}{L_2^2} = \frac{I}{2I} = \frac{1}{2}$
Taking the square root on both sides:
$\frac{L_1}{L_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio of their angular momenta is $1:\sqrt{2}$.

(D) The moment of inertia of a ring about its diameter is given by $I = \frac{1}{2}MR^2$.
Given: $M = 10 \ kg$,$R = 0.5 \ m$,$\omega = 20 \ rad/s$.
Substituting the values: $I = \frac{1}{2} \times 10 \times (0.5)^2 = 5 \times 0.25 = 1.25 \ kg \cdot m^2$.
The rotational kinetic energy $K$ is given by $K = \frac{1}{2}I\omega^2$.
$K = \frac{1}{2} \times 1.25 \times (20)^2$.
$K = \frac{1}{2} \times 1.25 \times 400 = 1.25 \times 200 = 250 \ J$.

(C) The rotational kinetic energy $K$ of a body is related to its angular momentum $L$ and moment of inertia $I$ by the formula: $K = \frac{L^2}{2I}$.
Given that the angular momenta are equal $(L_A = L_B = L)$,the kinetic energy is inversely proportional to the moment of inertia: $K \propto \frac{1}{I}$.
Since $I_A > I_B$,it follows that $\frac{1}{I_A} < \frac{1}{I_B}$.
Therefore,$K_A < K_B$.

(C) The rotational kinetic energy $(K_{rot})$ is given by $\frac{1}{2} I \omega^2$,where $I = \frac{2}{5} mR^2$ for a solid sphere and $\omega = v/R$.
Substituting these,$K_{rot} = \frac{1}{2} (\frac{2}{5} mR^2) (v/R)^2 = \frac{1}{5} mv^2$.
The translational kinetic energy $(K_{trans})$ is $\frac{1}{2} mv^2$.
The ratio of rotational kinetic energy to translational kinetic energy is $\frac{K_{rot}}{K_{trans}} = \frac{\frac{1}{5} mv^2}{\frac{1}{2} mv^2} = \frac{2}{5}$.

(C) The rotational kinetic energy $(K_{rot})$ of a rigid body rotating about a fixed axis is given by the formula:
$K_{rot} = \frac{1}{2} I \omega^2$
where $I$ is the moment of inertia of the object and $\omega$ is its angular velocity.
Comparing this with the given options,the correct expression is $\frac{1}{2} I \omega^2$.

(B) The rotational kinetic energy $K_R$ is given by the formula $K_R = \frac{1}{2} I \omega^2$.
For a solid circular disc,the moment of inertia $I = \frac{1}{2} M R^2$.
The angular velocity $\omega$ in radians per second is $\omega = 2 \pi n$,where $n$ is the frequency in revolutions per second.
Given $M = 72 \ kg$,$R = 0.5 \ m$,and $n = \frac{70}{60} \ rev/s$.
Substituting these values:
$I = \frac{1}{2} \times 72 \times (0.5)^2 = 36 \times 0.25 = 9 \ kg \cdot m^2$.
$\omega = 2 \pi \times \frac{70}{60} = \frac{7 \pi}{3} \ rad/s$.
$K_R = \frac{1}{2} \times 9 \times \left( \frac{7 \pi}{3} \right)^2 = \frac{1}{2} \times 9 \times \frac{49 \pi^2}{9} = \frac{49 \pi^2}{2} \approx \frac{49 \times 9.8696}{2} \approx 241.8 \ J$.
Rounding to the nearest provided option,the value is $240 \ J$.

(A) The rotational kinetic energy $(K_{rot})$ of the rotating body is given by $K_{rot} = \frac{1}{2} I \omega^2$.
Given $I = 3 \ kg \cdot m^2$ and $\omega = 3 \ rad/s$,we have:
$K_{rot} = \frac{1}{2} \times 3 \times (3)^2 = \frac{1}{2} \times 3 \times 9 = 13.5 \ J$.
The translational kinetic energy $(K_{trans})$ of the second body is given by $K_{trans} = \frac{1}{2} m v^2$.
Given $m = 27 \ kg$ and $K_{trans} = K_{rot} = 13.5 \ J$,we have:
$\frac{1}{2} \times 27 \times v^2 = 13.5$.
$v^2 = \frac{13.5 \times 2}{27} = \frac{27}{27} = 1$.
Therefore,$v = 1 \ m/s$.

(D) The rotational kinetic energy $K_R$ of a body is given by the formula $K_R = \frac{1}{2}I\omega^2$.
For a ring of mass $m$ and radius $r$,the moment of inertia $I$ about an axis passing through its center and perpendicular to its plane is $I = mr^2$.
Substituting the value of $I$ into the kinetic energy formula:
$K_R = \frac{1}{2}(mr^2)\omega^2 = \frac{1}{2}mr^2\omega^2$.
Thus,the correct option is $D$.

(D) The rotational kinetic energy $K_R$ is given by the formula $K_R = \frac{1}{2} I \omega^2$.
For a solid sphere,the moment of inertia $I$ about its center is $I = \frac{2}{5} M R^2$.
Given: $M = 1 \, kg$,$R = 3 \, cm = 0.03 \, m$,and $\omega = 50 \, rad/s$.
Substituting the values:
$I = \frac{2}{5} \times 1 \times (0.03)^2 = \frac{2}{5} \times 0.0009 = 0.00036 \, kg \cdot m^2$.
Now,$K_R = \frac{1}{2} \times 0.00036 \times (50)^2$.
$K_R = \frac{1}{2} \times 0.00036 \times 2500$.
$K_R = 0.00018 \times 2500 = 0.45 \, J$.
Since $0.45 = 45/100 = 9/20$,the rotational kinetic energy is $9/20 \, J$.

(B) The rotational kinetic energy $E$ of a body is given by the formula $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Given that the angular velocities are equal $(\omega_1 = \omega_2 = \omega)$,the rotational kinetic energy becomes directly proportional to the moment of inertia: $E \propto I$.
Since it is given that $I_1 > I_2$,it follows that $E_1 > E_2$.

(B) Given that the rotational kinetic energy $K_R$ is $50\%$ of the linear kinetic energy $K_T$.
$K_R = 0.5 K_T$
Using the formulas $K_R = \frac{1}{2} I \omega^2 = \frac{1}{2} (Mk^2) (v/R)^2 = \frac{1}{2} Mv^2 (k^2/R^2)$ and $K_T = \frac{1}{2} Mv^2$,we get:
$\frac{1}{2} Mv^2 (k^2/R^2) = 0.5 \times \frac{1}{2} Mv^2$
$\frac{k^2}{R^2} = 0.5 = \frac{1}{2}$
For a solid cylinder,the moment of inertia is $I = \frac{1}{2} MR^2$,so $k^2 = \frac{1}{2} R^2$,which implies $\frac{k^2}{R^2} = \frac{1}{2}$.
Thus,the object is a solid cylinder.

(A) Given: Mass $M = 6 \ kg$,Radius $R = 40 \ cm = 0.4 \ m$,Frequency $f = 300 \ rpm = \frac{300}{60} \ rps = 5 \ Hz$.
For a rim (hoop),the moment of inertia $I = MR^2 = 6 \times (0.4)^2 = 6 \times 0.16 = 0.96 \ kg \cdot m^2$.
The angular velocity $\omega = 2\pi f = 2\pi \times 5 = 10\pi \ rad/s$.
The rotational kinetic energy $K_R = \frac{1}{2} I \omega^2$.
Substituting the values: $K_R = \frac{1}{2} \times 0.96 \times (10\pi)^2$.
$K_R = 0.48 \times 100\pi^2 = 48\pi^2 \ J$.

(A) The rotational kinetic energy $(K)$ of a rigid body rotating about a fixed axis is given by the formula $K = \frac{1}{2} I \omega^{2}$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a ring of mass $m$ and radius $r$ rotating about an axis passing through its centre and perpendicular to its plane,the moment of inertia is $I = m r^{2}$.
Substituting the value of $I$ into the kinetic energy formula:
$K = \frac{1}{2} (m r^{2}) \omega^{2} = \frac{1}{2} m r^{2} \omega^{2}$.

(A) The number of revolutions per minute is given as $N = \frac{3000}{\pi} \ rpm$.
The frequency $n$ in Hertz is $n = \frac{N}{60} = \frac{3000}{60 \times \pi} = \frac{50}{\pi} \ Hz$.
The angular velocity $\omega$ is given by $\omega = 2 \pi n = 2 \pi \times \frac{50}{\pi} = 100 \ rad/s$.
The moment of inertia $I$ is $400 \ kg \ m^2$.
The rotational kinetic energy $K$ is calculated using the formula $K = \frac{1}{2} I \omega^2$.
Substituting the values: $K = \frac{1}{2} \times 400 \times (100)^2 = 200 \times 10000 = 2 \times 10^6 \ J$.

(D) The rotational kinetic energy $(K.E._{rot})$ is given by the formula: $K.E._{rot} = \frac{1}{2} I \omega^2$.
For a solid sphere,the moment of inertia $(I)$ about an axis passing through its centre is $I = \frac{2}{5} MR^2$.
Given: Mass $(M)$ = $1\,kg$,Radius $(R)$ = $30\,cm = 0.3\,m$,Angular velocity $(\omega)$ = $50\,rad/s$.
Substituting the values into the formula:
$I = \frac{2}{5} \times 1 \times (0.3)^2 = 0.4 \times 0.09 = 0.036\,kg \cdot m^2$.
$K.E._{rot} = \frac{1}{2} \times 0.036 \times (50)^2$.
$K.E._{rot} = 0.5 \times 0.036 \times 2500$.
$K.E._{rot} = 0.018 \times 2500 = 45\,J$.

(B) The rotational kinetic energy $(K.E._{rot})$ is given by the formula: $K.E._{rot} = \frac{1}{2} I \omega^{2}$.
First,calculate the moment of inertia $(I)$ for a solid circular disc: $I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 72 \times (0.5)^{2} = 36 \times 0.25 = 9 \ kg \cdot m^{2}$.
Next,convert the angular velocity $(\omega)$ from $r.p.m.$ to $rad/s$: $\omega = \frac{70 \times 2 \pi}{60} = \frac{7 \pi}{3} \approx 7.33 \ rad/s$.
Now,substitute these values into the kinetic energy formula: $K.E._{rot} = \frac{1}{2} \times 9 \times (7.33)^{2} \approx 0.5 \times 9 \times 53.72 \approx 241.7 \ J$.
Rounding to the nearest provided option,the energy is approximately $240 \ J$.

(D) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5}MR^2$.
Given: $M = 10\,kg$,$R = 3\,cm = 0.03\,m$,and $\omega = 50\,rad/s$.
First,calculate the moment of inertia:
$I = \frac{2}{5} \times 10\,kg \times (0.03\,m)^2 = 4 \times 0.0009\,kg \cdot m^2 = 0.0036\,kg \cdot m^2$.
The rotational kinetic energy is given by $K_R = \frac{1}{2}I\omega^2$.
$K_R = \frac{1}{2} \times 0.0036\,kg \cdot m^2 \times (50\,rad/s)^2$.
$K_R = 0.0018 \times 2500\,J = 4.5\,J$.
Wait,re-evaluating the calculation: $I = 0.4 \times 10 \times 0.0009 = 0.0036$. $K_R = 0.5 \times 0.0036 \times 2500 = 4.5\,J$. Checking options,it seems there might be a unit mismatch or typo in the question's expected magnitude. If $R=30\,cm=0.3\,m$,then $I = 0.4 \times 10 \times 0.09 = 0.36\,kg \cdot m^2$. Then $K_R = 0.5 \times 0.36 \times 2500 = 450\,J$. Thus,the correct option is $D$.

(C) The rotational kinetic energy $E$ is related to angular momentum $L$ by the formula $E = \frac{L^2}{2I}$,where $I$ is the moment of inertia.
Let the initial angular momentum be $L_1 = L$.
Given that the angular momentum is increased by $200\,\%$,the new angular momentum $L_2$ is:
$L_2 = L + 200\% \text{ of } L = L + 2L = 3L$.
Since $E \propto L^2$,the ratio of the new kinetic energy $E_2$ to the initial kinetic energy $E_1$ is:
$\frac{E_2}{E_1} = \left(\frac{L_2}{L_1}\right)^2 = \left(\frac{3L}{L}\right)^2 = 9$.
Therefore,$E_2 = 9E_1$.
The percentage increase in rotational kinetic energy is given by:
$\text{Percentage Increase} = \left(\frac{E_2 - E_1}{E_1}\right) \times 100\% = \left(\frac{9E_1 - E_1}{E_1}\right) \times 100\% = 8 \times 100\% = 800\%$.

(N/A) Given:
Mass of the cylinder,$m = 20 \; kg$
Angular speed,$\omega = 100 \; rad \; s^{-1}$
Radius of the cylinder,$r = 0.25 \; m$
The moment of inertia $(I)$ of a solid cylinder about its axis is given by:
$I = \frac{1}{2} m r^2$
$I = \frac{1}{2} \times 20 \times (0.25)^2 = 10 \times 0.0625 = 0.625 \; kg \; m^2$
$1$. Rotational Kinetic Energy $(K)$:
$K = \frac{1}{2} I \omega^2$
$K = \frac{1}{2} \times 0.625 \times (100)^2$
$K = 0.5 \times 0.625 \times 10000 = 3125 \; J$
$2$. Angular Momentum $(L)$:
$L = I \omega$
$L = 0.625 \times 100 = 62.5 \; kg \; m^2 \; s^{-1}$ (or $J \; s$)

(N/A) The rotational kinetic energy $(K_{rot})$ of a rigid body rotating about a fixed axis is given by the formula:
$K_{rot} = \frac{1}{2} I \omega^2$
Where:
$I$ is the moment of inertia of the body about the axis of rotation.
$\omega$ is the angular velocity of the body.

(C) Let the radius of wheel $Q$ be $R$ and the radius of wheel $P$ be $3R$. Since they are connected by a belt,their tangential speeds at the rim are equal,so $v = \omega_{P} (3R) = \omega_{Q} R$.
This implies $\omega_{P} = \frac{\omega_{Q}}{3}$.
The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^{2}$.
Given that the rotational kinetic energies are equal,we have:
$\frac{1}{2} I_{P} \omega_{P}^{2} = \frac{1}{2} I_{Q} \omega_{Q}^{2}$
$I_{P} \left(\frac{\omega_{Q}}{3}\right)^{2} = I_{Q} \omega_{Q}^{2}$
$I_{P} \left(\frac{1}{9}\right) = I_{Q}$
$\frac{I_{P}}{I_{Q}} = 9$.
Thus,the ratio is $9: 1$,and the value of $x$ is $9$.

(D) The rotational kinetic energy is given by $E = \frac{1}{2} I \omega^2$.
For a thin uniform rod of length $\ell$ rotating about an axis through its center,the moment of inertia is $I = \frac{m \ell^2}{12}$.
The mass $m$ of the rod is given by $m = \text{density} \times \text{volume} = d \times (A \ell) = d A \ell$.
Substituting $m$ into the expression for $I$,we get $I = \frac{(d A \ell) \ell^2}{12} = \frac{d A \ell^3}{12}$.
Now,substitute $I$ into the kinetic energy formula: $E = \frac{1}{2} \left( \frac{d A \ell^3}{12} \right) \omega^2 = \frac{d A \ell^3}{24} \omega^2$.
Given $\ell = 2 \ m$,we have $E = \frac{d A (2)^3}{24} \omega^2 = \frac{8 d A}{24} \omega^2 = \frac{d A}{3} \omega^2$.
Rearranging for $\omega$,we get $\omega^2 = \frac{3 E}{d A}$,which implies $\omega = \sqrt{\frac{3 E}{d A}}$.
Comparing this with the given expression $\omega = \sqrt{\frac{\alpha E}{Ad}}$,we find $\alpha = 3$.

(A) The initial angular velocity is $\omega_i = 60\,rpm = 60 \times \frac{2\pi}{60} = 2\pi\,rad/s$.
The final angular velocity is $\omega_f = 360\,rpm = 360 \times \frac{2\pi}{60} = 12\pi\,rad/s$.
The change in rotational kinetic energy is given by $\Delta K.E. = \frac{1}{2} I (\omega_f^2 - \omega_i^2) = 484\,J$.
Substituting the values: $\frac{1}{2} I ((12\pi)^2 - (2\pi)^2) = 484$.
$\frac{1}{2} I (144\pi^2 - 4\pi^2) = 484$.
$\frac{1}{2} I (140\pi^2) = 484$.
$70\pi^2 I = 484$.
Using $\pi^2 \approx 9.86$,we get $70 \times 9.86 \times I = 484$.
$690.2 I = 484$.
$I = \frac{484}{690.2} \approx 0.701\,kg\cdot m^2$.

(D) The rotational kinetic energy $K$ of a rigid body is related to its angular momentum $L$ and moment of inertia $I$ by the formula: $K = \frac{L^2}{2I}$.
From this, we can express angular momentum as $L = \sqrt{2KI}$.
Let the initial kinetic energy be $K_1 = K$ and the initial angular momentum be $L_1 = L$.
If the kinetic energy is made four times, the new kinetic energy is $K_2 = 4K$.
The new angular momentum $L_2$ will be:
$L_2 = \sqrt{2 K_2 I} = \sqrt{2(4K)I} = \sqrt{4(2KI)} = 2\sqrt{2KI}$.
Since $L = \sqrt{2KI}$, we have $L_2 = 2L$.

(C) The kinetic energy of the rod while passing through the mean position is given by the rotational kinetic energy formula:
$K.E. = \frac{1}{2} I \omega^2$
Here,$I$ is the moment of inertia of the rod about the point of suspension (one end).
The moment of inertia of a uniform rod of mass $m$ and length $l$ about one end is $I = \frac{m l^2}{3}$.
Substituting this value into the kinetic energy formula:
$K.E. = \frac{1}{2} \times \left( \frac{m l^2}{3} \right) \times \omega^2$
$K.E. = \frac{m l^2 \omega^2}{6}$

(A) The kinetic energy $E$ of a particle in rotational motion is given by $E = \frac{1}{2} I \omega^2$.
Angular momentum $L$ is defined as $L = I \omega$,which implies $L^2 = I^2 \omega^2$.
Substituting $\omega^2 = \frac{L^2}{I^2}$ into the kinetic energy equation,we get $E = \frac{1}{2} I \left( \frac{L^2}{I^2} \right) = \frac{L^2}{2I}$.
For a particle of mass '$m$' moving in a circle of radius '$r$',the moment of inertia about the axis passing through the center is $I = mr^2$.
Substituting $I = mr^2$ into the expression for kinetic energy,we get $E = \frac{L^2}{2(mr^2)} = \frac{L^2}{2 mr^2}$.

(C) The rotational kinetic energy $(KE)$ of a body is given by $KE = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Since $I$ remains constant,$KE \propto \omega^2$.
Given that the angular velocity increases by $20 \%$,the new angular velocity $\omega_2$ is:
$\omega_2 = \omega_1 + 0.20 \omega_1 = 1.2 \omega_1$.
The ratio of the new kinetic energy $(KE_2)$ to the initial kinetic energy $(KE_1)$ is:
$\frac{KE_2}{KE_1} = \frac{\omega_2^2}{\omega_1^2} = \frac{(1.2 \omega_1)^2}{\omega_1^2} = (1.2)^2 = 1.44$.
Therefore,$KE_2 = 1.44 KE_1$.
The percentage increase in kinetic energy is:
$\text{Percentage Increase} = \left( \frac{KE_2 - KE_1}{KE_1} \right) \times 100 = (1.44 - 1) \times 100 = 0.44 \times 100 = 44 \%$.

(D) The rotational kinetic energy $K$ is given by the formula $K = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Since the kinetic energies are equal,we have $K_P = K_Q$.
Therefore,$\frac{L_P^2}{2I_P} = \frac{L_Q^2}{2I_Q}$.
Rearranging the terms,we get $\frac{L_Q^2}{L_P^2} = \frac{I_Q}{I_P}$.
Taking the square root on both sides,$\frac{L_Q}{L_P} = \sqrt{\frac{I_Q}{I_P}}$.
Given that $I_Q > I_P$,it follows that $\frac{I_Q}{I_P} > 1$.
Thus,$\sqrt{\frac{I_Q}{I_P}} > 1$,which implies $\frac{L_Q}{L_P} > 1$.
Therefore,$L_Q > L_P$.

(C) The rotational kinetic energy $K$ of a body is given by the formula $K = \frac{L^{2}}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Since the angular momenta of both bodies are equal $(L_{A} = L_{B} = L)$,the kinetic energy is inversely proportional to the moment of inertia,i.e.,$K \propto \frac{1}{I}$.
Given that $I_{A} > I_{B}$,it follows that $\frac{1}{I_{A}} < \frac{1}{I_{B}}$.
Therefore,$K_{A} < K_{B}$.