Angular Variables and Basic of Uniform Circular Motion Questions in English

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of motion changes continuously.
At any point on the circular path,the instantaneous velocity vector is always directed along the tangent to the circle at that specific point.
This is because the velocity is defined as the rate of change of displacement,and for a circular path,the displacement vector at any instant is tangential to the path.

(A) In uniform circular motion,an object moves along a circular path with a constant speed.
Since speed is the magnitude of velocity,the magnitude of velocity remains constant.
However,the direction of velocity is always tangent to the circular path at any given point.
As the object moves,the direction of the tangent changes continuously.
Therefore,the velocity vector,which includes both magnitude and direction,is not constant.
Thus,only statement $(i)$ is correct.

(N/A) The $SI$ unit of angular velocity $(\omega)$ is $\text{radian per second}$ $(\text{rad } s^{-1})$.
The $SI$ unit of angular acceleration $(\alpha)$ is $\text{radian per second squared}$ $(\text{rad } s^{-2})$.

(A) The statement is correct.
Angular position $\theta$ is defined as the angle subtended by a position vector at the origin,which is a scalar quantity.
Angular displacement $\Delta\theta$ is defined as the change in angular position. For infinitesimal rotations,angular displacement behaves as a vector quantity because it follows the commutative law of vector addition.
Its direction is determined by the right-hand thumb rule,perpendicular to the plane of rotation.

(B) $rpm$ stands for $revolution$ $per$ $minute$.
It is a unit of angular velocity.
$1$ $rpm$ = $1$ revolution / $1$ minute.
Since $1$ revolution = $2\pi$ radians and $1$ minute = $60$ seconds,
$1$ $rpm$ = $\frac{2\pi \text{ rad}}{60 \text{ s}} = \frac{\pi}{30} \text{ rad/s}$.

(A) The angular speed $\omega$ is given by $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the hour hand,the time period $T_h = 12 \text{ hours} = 12 \times 3600 \text{ s}$.
So,$\omega_h = \frac{2\pi}{12 \times 3600} \text{ rad/s}$.
For the minute hand,the time period $T_m = 1 \text{ hour} = 3600 \text{ s}$.
So,$\omega_m = \frac{2\pi}{3600} \text{ rad/s}$.
The ratio of the angular speed of the hour hand to the minute hand is:
$\frac{\omega_h}{\omega_m} = \frac{2\pi / (12 \times 3600)}{2\pi / 3600} = \frac{3600}{12 \times 3600} = \frac{1}{12}$.
Thus,the ratio is $1: 12$.

(A) Yes,this is possible. In uniform circular motion,the magnitude of the linear velocity (speed) remains constant,but the direction of the linear velocity vector changes continuously at every point along the circular path. Since velocity is a vector quantity,a change in direction implies a change in velocity. On the other hand,the angular velocity $\omega$ is defined as $\omega = \frac{d\theta}{dt}$. For a particle moving with constant speed $v$ in a circle of radius $r$,the angular velocity is $\omega = \frac{v}{r}$. Since both $v$ and $r$ are constant,the angular velocity $\omega$ remains constant in both magnitude and direction (along the axis of rotation).

(A) In uniform circular motion,the acceleration has two components: radial (centripetal) and tangential.
$1$. The radial component is given by $a_{r} = r \omega^{2}$. Since the radius $r$ and angular velocity $\omega$ are constant,the magnitude of the radial acceleration is constant. However,its direction is always towards the center,which changes continuously.
$2$. The tangential component $a_{t}$ is given by $a_{t} = r \alpha$. Since the speed is constant,the angular acceleration $\alpha = 0$,so $a_{t} = 0$. $A$ zero vector is constant in both magnitude and direction.
$3$. If the question refers to the non-zero components,the radial component's magnitude is constant,but its direction changes. The tangential component is zero (constant).

(C) In uniform circular motion,the acceleration is the centripetal acceleration,which is always directed towards the center of the circle.
The magnitude of the centripetal acceleration is $a = \frac{v^{2}}{R}$.
For a point $P$ at an angle $\theta$ with the positive $x$-axis,the position vector is directed from the origin to the point. The centripetal acceleration vector $\vec{a}$ is directed from the point $P$ towards the origin.
The unit vector directed from the point $P(R, \theta)$ towards the origin is $-(\cos \theta \hat{i} + \sin \theta \hat{j})$.
Therefore,the acceleration vector is $\vec{a} = -a(\cos \theta \hat{i} + \sin \theta \hat{j}) = -\frac{v^{2}}{R} \cos \theta \hat{i} - \frac{v^{2}}{R} \sin \theta \hat{j}$.

(A) Given equations are $x = 4 \sin \left(\frac{\pi}{2} - \omega t\right)$ and $y = 4 \sin (\omega t)$.
Using the trigonometric identity $\sin \left(\frac{\pi}{2} - \theta\right) = \cos \theta$,we can rewrite the equation for $x$ as $x = 4 \cos (\omega t)$.
Now we have $x = 4 \cos (\omega t)$ and $y = 4 \sin (\omega t)$.
To find the path,square and add both equations:
$x^2 + y^2 = (4 \cos \omega t)^2 + (4 \sin \omega t)^2$
$x^2 + y^2 = 16 \cos^2 \omega t + 16 \sin^2 \omega t$
$x^2 + y^2 = 16 (\cos^2 \omega t + \sin^2 \omega t)$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get $x^2 + y^2 = 4^2$.
This is the equation of a circle with radius $4 \text{ m}$ centered at the origin. Therefore,the path of the particle is circular.

(D) The time period of the ball is $T = 1.5 \, s$.
In $t = 8.3 \, s$,the number of revolutions completed is $n = \frac{8.3}{1.5} = 5.533$ revolutions.
After $5$ complete revolutions $(t = 7.5 \, s)$,the ball returns to its initial position at $t = 0 \, s$.
The remaining time is $\Delta t = 8.3 - 7.5 = 0.8 \, s$.
The angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{1.5} = \frac{4\pi}{3} \, rad/s$.
The angle covered in $\Delta t = 0.8 \, s$ is $\theta = \omega \Delta t = \left(\frac{4\pi}{3}\right) \times 0.8 = \frac{3.2\pi}{3} \approx 1.067\pi \, rad$.
This angle is slightly greater than $\pi \, rad$ $(180^\circ)$.
Since the angle is approximately $192^\circ$,the ball is very close to the point diametrically opposite to the starting position.
The displacement for a diameter is $2R = 2 \times 1 = 2 \, m$.
Thus,the magnitude of displacement is closest to $2 \, m$.

(A) For a particle moving in a circular path with constant speed,the motion is known as uniform circular motion.
In uniform circular motion,the acceleration is purely centripetal,meaning it is directed towards the centre of the circular path.
The position vector of the particle with respect to the centre of the circle is directed radially outward from the centre to the particle's position.
Since the acceleration vector points towards the centre and the position vector points away from the centre,these two vectors are anti-parallel.
Therefore,the angle between the direction of acceleration and the position vector is $180^{\circ}$ or $\pi$ radians.

(A) In uniform circular motion,the acceleration of the particle is the centripetal acceleration,which is directed towards the center of the circular path at every instant.
Let the particle move in a circle of radius $r$ with constant speed $v$ in the $xy$-plane. The acceleration vector at any angle $\theta$ is given by $\vec{a} = -\omega^2 \vec{r} = -\omega^2 (r \cos \theta \hat{i} + r \sin \theta \hat{j})$.
The average acceleration over one complete cycle (from $\theta = 0$ to $\theta = 2\pi$) is $\vec{a}_{avg} = \frac{1}{2\pi} \int_{0}^{2\pi} \vec{a} d\theta$.
$\vec{a}_{avg} = -\frac{\omega^2 r}{2\pi} [\int_{0}^{2\pi} \cos \theta d\theta \hat{i} + \int_{0}^{2\pi} \sin \theta d\theta \hat{j}] = -\frac{\omega^2 r}{2\pi} [0 \hat{i} + 0 \hat{j}] = \vec{0}$.
Thus,the average acceleration vector is a null vector.

(D) Given, time period $T = 40 \, s$.
Total time $t = 2 \, \text{min} \, 20 \, s = 2 \times 60 + 20 = 140 \, s$.
Number of revolutions $n = \frac{t}{T} = \frac{140}{40} = 3.5 = 3 \frac{1}{2}$ revolutions.
Distance travelled in $3.5$ revolutions $= 3.5 \times (2 \pi R) = 7 \pi R$.
After $3.5$ revolutions, the particle is at the diametrically opposite point of the starting position. Thus, the displacement is equal to the diameter of the circle, which is $2R$.
The ratio $\frac{|\text{displacement}|}{\text{distance}} = \frac{2R}{7 \pi R} = \frac{2}{7 \pi}$.
Substituting $\pi = \frac{22}{7}$, we get $\frac{2}{7 \times (22/7)} = \frac{2}{22} = \frac{1}{11}$.

(C) The second hand of a clock completes one full rotation in one minute.
Let the length of the second hand be $R$.
In one minute,the tip of the second hand moves along a circular path and returns to its initial position.
Since the initial and final positions are the same,the displacement,which is the shortest distance between the initial and final positions,is $0$.
The distance covered is the circumference of the circle,which is $2 \pi R$.

(A) In uniform circular motion,the speed of the object remains constant.
Since kinetic energy is given by $K = \frac{1}{2}mv^2$,and both mass $m$ and speed $v$ are constant,the kinetic energy remains constant.
Velocity is a vector quantity; although the speed is constant,the direction of motion changes continuously,so velocity is not constant.
Acceleration in circular motion is centripetal acceleration,which is directed towards the center. Its magnitude is $a = \frac{v^2}{r}$. While the magnitude is constant,the direction changes continuously,so the acceleration vector is not constant.
Displacement is a vector quantity that depends on the position,which changes continuously as the object moves along the circular path.

(A) The frequency of the body is given as $f = 140 \text{ rev/s}$.
The relationship between angular speed $\omega$ and frequency $f$ is given by the formula:
$\omega = 2 \pi f$
Substituting the given value of $f$:
$\omega = 2 \times \pi \times 140$
$\omega = 280 \pi \text{ rad/s}$
Using $\pi \approx 3.14159$:
$\omega = 280 \times 3.14159 \approx 879.64 \text{ rad/s}$
Rounding to the nearest integer,we get:
$\omega \approx 880 \text{ rad/s}$.
Therefore,the correct option is $A$.

(A) The cyclist completes $7$ rounds in $1$ minute $(60 \, s)$.
The circumference of the circular track is $C = 2 \pi r = 2 \times 3.14159 \times 5 \, m \approx 31.416 \, m$.
The total distance covered in $7$ rounds is $D = 7 \times 31.416 \, m = 219.912 \, m$.
The constant speed $v$ is given by $v = \text{Distance} / \text{Time} = 219.912 \, m / 60 \, s \approx 3.665 \, m/s$.
The centripetal acceleration $a_c$ is given by $a_c = v^2 / r$.
Substituting the values: $a_c = (3.665)^2 / 5 \approx 13.432 / 5 \approx 2.686 \, m/s^2$.
Rounding to one decimal place,we get $2.7 \, m/s^2$.
Thus,Option $A$ is correct.

(D) The velocity of the particle at any point is tangent to the circular path.
Let the velocity at point $A$ be $\vec{v}_A$ and at point $B$ be $\vec{v}_B$.
The magnitude of both velocities is $v$,so $|\vec{v}_A| = |\vec{v}_B| = v$.
The angle between the radii $OA$ and $OB$ is $\theta = 60^{\circ}$.
The angle between the velocity vectors $\vec{v}_A$ and $\vec{v}_B$ is also $60^{\circ}$ because the velocity is always perpendicular to the radius.
The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_B - \vec{v}_A|$.
Using the formula for the magnitude of the difference of two vectors of equal magnitude $v$ with an angle $\theta$ between them:
$|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos \theta} = \sqrt{2v^2(1 - \cos \theta)} = \sqrt{2v^2(2 \sin^2(\theta/2))} = 2v \sin(\theta/2)$.
Substituting $\theta = 60^{\circ}$:
$|\Delta \vec{v}| = 2v \sin(60^{\circ}/2) = 2v \sin(30^{\circ}) = 2v \times (1/2) = v$.
Therefore,the magnitude of the change in velocity is $v$.

(A) The angular speed $\omega$ is given by the formula $\omega = \frac{2\pi}{T}$.
Here,$T$ is the time period of rotation of the Earth about its own axis,which is $24 \text{ hours}$.
Converting the time period into seconds: $T = 24 \times 60 \times 60 \text{ s} = 86400 \text{ s}$.
Substituting the value of $T$ in the formula: $\omega = \frac{2\pi}{86400} \text{ rad/s}$.
Simplifying the expression: $\omega = \frac{\pi}{43200} \text{ rad/s}$.
Therefore,the correct option is $A$.

(B) The velocity of a particle moving in a circular path with constant speed $v$ is a vector quantity. Let the velocity at point $A$ be $\vec{v}_A$ and at point $B$ be $\vec{v}_B$. The magnitude of both is $v$.
The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_B - \vec{v}_A| = 2v \sin(\theta/2)$,where $\theta = 60^{\circ}$.
$|\Delta \vec{v}| = 2v \sin(60^{\circ}/2) = 2v \sin(30^{\circ}) = 2v \times (1/2) = v$.
The magnitude of the velocity is constant $(v)$ throughout the motion. Therefore,the change in the magnitude of the velocity is $v - v = 0$.
Thus,the magnitude of the change in velocity is $v$ and the change in the magnitude of velocity is $0$.

(D) In uniform circular motion,an object moves along a circular path with a constant speed.
Although the speed is constant,the direction of motion changes at every point on the path.
Since velocity is a vector quantity (having both magnitude and direction),the change in direction implies that the velocity is not constant.
Similarly,momentum $(p = mv)$ changes because velocity changes.
Centripetal acceleration $(a_c = v^2/r)$ is directed towards the center of the circle,and its direction changes continuously as the object moves.
Therefore,only the speed remains constant throughout the motion.

(B) The object moves in a circular path with radius $r = 2\,m$ and constant speed $v = 4\,m/s$.
The magnitude of centripetal acceleration is $a_c = \frac{v^2}{r} = \frac{4^2}{2} = \frac{16}{2} = 8\,m/s^2$.
At $x = +2\,m$,the velocity is $-4 \hat{j}\,m/s$,which implies the object is moving in the clockwise direction.
At $x = -2\,m$,the object will be at the diametrically opposite point.
Since it is moving clockwise,at $x = -2\,m$,the velocity vector will be directed upwards,so $v = 4 \hat{j}\,m/s$.
The centripetal acceleration is always directed towards the centre $(0,0)$. At $x = -2\,m$,the centre is to the right,so the acceleration is directed along the positive $x$-axis,$a = 8 \hat{i}\,m/s^2$.

(A) Let the constant speed of the particle be $v$ and the radius of the circular path be $R$.
When the particle turns by an angle of $90^{\circ}$ (or $\pi/2$ radians),the distance traveled along the arc is $s = R \theta = R(\pi/2) = \pi R / 2$.
The time taken for this motion is $t = s / v = (\pi R / 2) / v = \pi R / (2v)$.
The displacement is the straight-line distance between the initial and final positions,which is the chord length $AB = \sqrt{R^2 + R^2} = R\sqrt{2}$.
The average velocity is defined as the total displacement divided by the total time taken:
$\langle v \rangle = \frac{\text{Displacement}}{\text{Time}} = \frac{R\sqrt{2}}{\pi R / (2v)} = \frac{R\sqrt{2} \cdot 2v}{\pi R} = \frac{2\sqrt{2}v}{\pi}$.
The ratio of instantaneous velocity $v$ to average velocity $\langle v \rangle$ is:
$\frac{v}{\langle v \rangle} = \frac{v}{2\sqrt{2}v / \pi} = \frac{\pi}{2\sqrt{2}}$.
Comparing this with the given ratio $\pi : x\sqrt{2}$,we have $\frac{\pi}{2\sqrt{2}} = \frac{\pi}{x\sqrt{2}}$.
Therefore,$x = 2$.

(D) The magnitude of the average velocity is given by the formula: $|\vec{v}_{avg}| = \frac{|\vec{r}_B - \vec{r}_A|}{\Delta t}$.
The displacement $|\vec{r}_B - \vec{r}_A|$ is the chord length $AB$. In a circle of radius $R$,the chord length for an angle $\theta = 120^{\circ}$ is $2R \sin(\theta/2) = 2R \sin(60^{\circ}) = 2R(\sqrt{3}/2) = R\sqrt{3}$.
The time taken $\Delta t$ is the distance along the arc divided by speed $v$. The arc length is $s = R\theta = R(2\pi/3)$.
Thus,$\Delta t = s/v = (2\pi R / 3) / \pi = 2R/3$.
Therefore,$|\vec{v}_{avg}| = \frac{R\sqrt{3}}{2R/3} = \frac{3\sqrt{3}}{2} = 1.5\sqrt{3} \, m/s$.

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of motion changes at every point along the circular path.
Since velocity is a vector quantity (having both magnitude and direction),a change in direction implies a change in velocity. Therefore,the velocity is varying.
Furthermore,the centripetal acceleration,given by $a_c = v^2/r$,is directed towards the center of the circle. As the particle moves,the direction of this acceleration vector changes continuously to remain pointing towards the center.
Thus,both velocity and acceleration are varying.

(A) Given the position vector $\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$.
To find the force,we first calculate the acceleration $\vec{a} = \frac{d^2\vec{r}}{dt^2}$.
Velocity $\vec{v} = \frac{d\vec{r}}{dt} = a\omega(-\hat{i} \sin \omega t + \hat{j} \cos \omega t)$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2(\hat{i} \cos \omega t + \hat{j} \sin \omega t) = -\omega^2\vec{r}$.
Since $\vec{F} = m\vec{a}$,we have $\vec{F} = -m\omega^2\vec{r}$.
This indicates that the force $\vec{F}$ is directed opposite to the position vector $\vec{r}$.

(D) In uniform circular motion,the speed of the body remains constant,but the direction of velocity changes continuously at every instant. Since velocity is a vector quantity,a change in direction implies that the velocity is varying.
Similarly,the centripetal acceleration is always directed towards the center of the circular path. As the body moves,the direction of the center relative to the body changes,meaning the direction of acceleration also changes continuously. Thus,both velocity and acceleration are varying.
The magnitude of centripetal acceleration is given by $a = v^2/r$,where $v$ is the speed and $r$ is the radius. Both statements are correct,and the reason explains why the acceleration is varying (because its direction changes even if its magnitude $v^2/r$ remains constant).

(C) In uniform circular motion,the velocity vector $\vec{v}$ is always perpendicular to the angular velocity vector $\vec{\omega}$.
Therefore,their dot product must be zero: $\vec{v} \cdot \vec{\omega} = 0$.
Given $\vec{v} = (6 \hat{i} - 2 \hat{j}) \ m/s$ and $\vec{\omega} = (a \hat{i} + b \hat{j}) \ rad/s$.
Substituting these into the dot product equation:
$(6 \hat{i} - 2 \hat{j}) \cdot (a \hat{i} + b \hat{j}) = 0$
$6a - 2b = 0$
$6a = 2b$
$\frac{b}{a} = \frac{6}{2} = 3$.

(C) The radius of the circular path is $r = \frac{\pi}{2} \,m$.
The particle makes $x$ revolutions in time $t$.
The total distance covered in $x$ revolutions is $d = x \times (2\pi r)$.
Substituting the value of $r$: $d = x \times (2\pi \times \frac{\pi}{2}) = x \times \pi^2 = \pi^2 x \,m$.
The tangential velocity $v$ is defined as the total distance covered divided by the time taken:
$v = \frac{d}{t} = \frac{\pi^2 x}{t} \,m/s$.
Therefore, the correct option is $C$.

(B) In uniform circular motion,the particle moves with a constant angular speed $\omega$.
$1$. The velocity vector $\vec{v}$ is always tangent to the circular path at any point. Thus,statement $A$ is true.
$2$. Since the speed is constant,there is no tangential acceleration. The only acceleration present is the centripetal acceleration $\vec{a}_c$,which is directed towards the centre of the circle. Thus,statement $D$ is true.
$3$. Because the acceleration vector $\vec{a}_c$ points towards the centre and the velocity vector $\vec{v}$ is tangent to the circle (perpendicular to the radius),the velocity and acceleration vectors are always perpendicular to each other. Thus,statement $C$ is true.
$4$. Since the acceleration vector points towards the centre,it is $NOT$ tangent to the circle. Therefore,statement $B$ is false.

(B) In uniform circular motion,a particle moves along a circular path with a constant speed.
Since the speed is constant,the magnitude of the linear velocity remains unchanged.
The direction of the linear velocity at any point is always tangent to the circular path at that point.
Therefore,the linear velocity is always tangential to the circular path,and its magnitude remains constant.

(D) The angular speed $\omega$ is defined as the angle covered per unit time. Since both objects complete one revolution ($2\pi$ radians) in the same time $t$,their angular speeds are equal: $\omega_1 = \omega_2 = \frac{2\pi}{t}$.
Linear speed $v$ is related to angular speed $\omega$ and radius $r$ by the formula $v = r\omega$.
For the first object: $v_1 = r_1\omega_1$.
For the second object: $v_2 = r_2\omega_2$.
The ratio of the linear speed of $m_2$ to that of $m_1$ is $\frac{v_2}{v_1} = \frac{r_2\omega_2}{r_1\omega_1}$.
Since $\omega_1 = \omega_2$,the ratio simplifies to $\frac{v_2}{v_1} = \frac{r_2}{r_1}$.

(D) In $U.C.M.$ (Uniform Circular Motion),the speed $V$ of the particle remains constant.
Since the radius $R$ is also constant,the magnitude of angular velocity $\omega = V/R$ remains constant.
Angular acceleration $\alpha$ is defined as the rate of change of angular velocity,i.e.,$\alpha = d\omega/dt$.
Since $\omega$ is constant,its derivative with respect to time is zero.
Therefore,the angular acceleration of the particle is $0$.

(B) In uniform circular motion,the speed of the body remains constant,but the direction of motion changes at every point along the path.
Since velocity $\vec{v}$ is a vector quantity,its direction changes continuously,so velocity is not constant.
Since momentum $\vec{p} = m\vec{v}$ depends on the velocity vector,it also changes continuously.
Acceleration in uniform circular motion is centripetal acceleration,given by $a_c = \frac{v^2}{r}$. Although its magnitude is constant,its direction is always towards the center of the circle,which changes continuously. Thus,acceleration is not constant.
Kinetic energy is given by $K = \frac{1}{2}mv^2$. Since mass $m$ and speed $v = |\vec{v}|$ are constant,the kinetic energy remains constant throughout the motion.
Therefore,the correct option is kinetic energy.

(D) The angular velocity $\omega$ is given by the formula $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the hour hand of a clock,the time period $T_1 = 12 \text{ hours}$.
For the Earth,the time period $T_2 = 24 \text{ hours}$.
Therefore,the angular velocity of the hour hand is $\omega_1 = \frac{2\pi}{12}$ and the angular velocity of the Earth is $\omega_2 = \frac{2\pi}{24}$.
The ratio is $\frac{\omega_1}{\omega_2} = \frac{2\pi / 12}{2\pi / 24} = \frac{24}{12} = 2$.
Thus,the ratio $\omega_1 : \omega_2$ is $2 : 1$.

(B) Given: Diameter $d = 50 \ cm = 0.5 \ m$.
Radius $r = d/2 = 0.25 \ m = 25 \times 10^{-2} \ m$.
Frequency $f = 2 \ Hz$.
Angular velocity $\omega = 2 \pi f = 2 \pi \times 2 = 4 \pi \ rad/s$.
The centripetal acceleration $a$ in $U.C.M.$ is given by $a = r \omega^2$.
Substituting the values: $a = (25 \times 10^{-2}) \times (4 \pi)^2$.
$a = 0.25 \times 16 \pi^2$.
$a = 4 \pi^2 \ m/s^2$.

(D) The centripetal acceleration '$a$' of a particle moving in a circular path of radius '$r$' with speed '$v$' is given by the formula: $a = \frac{v^2}{r}$.
From this relation,it is clear that the centripetal acceleration is directly proportional to the square of the speed: $a \propto v^2$.
Let the initial speed be '$v_1 = v$' and the initial acceleration be '$a_1 = a$'.
Let the final speed be '$v_2 = 2v$' and the final acceleration be '$a_2$'.
Taking the ratio of the final acceleration to the initial acceleration:
$\frac{a_2}{a_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{2v}{v}\right)^2 = (2)^2 = 4$.
Therefore,the ratio of the acceleration after and before the change is $4:1$.

(B) The minute hand of a clock completes one full revolution $(360^{\circ})$ in $60$ minutes.
Time taken for one revolution $T = 60 \text{ minutes} = 60 \times 60 \text{ seconds} = 3600 \text{ seconds}$.
Angular speed $\omega$ is defined as the angle covered per unit time: $\omega = \frac{\Delta \theta}{\Delta t}$.
Substituting the values: $\omega = \frac{360^{\circ}}{3600 \text{ s}} = \frac{1}{10} \text{ deg/s} = 0.1 \text{ deg/s}$.

(D) In uniform circular motion,the speed of the body remains constant over time.
Tangential acceleration $(a_t)$ is defined as the rate of change of the magnitude of velocity (speed) with respect to time.
Mathematically,$a_t = \frac{dv}{dt}$.
Since the speed '$v$' is uniform (constant),its derivative with respect to time is zero.
Therefore,$a_t = 0$.

(B) $1$. The linear velocity $v$ of a particle moving in a circular path is always directed along the tangent to the circle at that point.
$2$. The angular velocity $\omega$ is a vector quantity whose direction is determined by the right-hand thumb rule. For a particle revolving in an anticlockwise sense in the $xy$-plane,the angular velocity vector $\omega$ points along the axis of rotation,which is perpendicular to the plane of the circle (i.e.,along the $z$-axis).
$3$. Since the linear velocity $v$ lies in the plane of the circle and the angular velocity $\omega$ is perpendicular to the plane of the circle,the angle between them is always $90^{\circ}$.

(B) The centripetal acceleration $a_c$ for a body in uniform circular motion is given by the formula $a_c = R \omega^2$.
Here,$R$ is the radius of the circular path and $\omega$ is the angular velocity.
The relationship between angular velocity $\omega$ and frequency $n$ is $\omega = 2 \pi n$.
Substituting this value into the acceleration formula:
$a_c = R (2 \pi n)^2$
$a_c = R (4 \pi^2 n^2)$
$a_c = 4 \pi^2 n^2 R$.
Therefore,the correct option is $B$.

(B) In Uniform Circular Motion,the velocity vector $v$ is always tangent to the circular path.
Let the velocity at time $t$ be $v_1$ and at time $t + \delta t$ be $v_2$. The change in velocity is $\delta v = v_2 - v_1$.
Since the speed is constant in $U$.$C$.$M$.,$|v_1| = |v_2| = v$.
The vector $\delta v$ forms the base of an isosceles triangle with sides $v_1$ and $v_2$ having an angle $\theta$ between them.
The angle $\phi$ between the change in velocity $\delta v$ and the initial velocity $v_1$ is given by $\phi = \frac{180^{\circ} - \theta}{2} = 90^{\circ} - \frac{\theta}{2}$.
As the time interval $\delta t \rightarrow 0$,the angle $\theta$ between the velocity vectors also approaches $0^{\circ}$.
Substituting $\theta \approx 0^{\circ}$ into the expression,we get $\phi = 90^{\circ} - 0^{\circ} = 90^{\circ}$.
Thus,the change in velocity is perpendicular to the instantaneous velocity vector.

(A) The linear speed $V$ of an object in circular motion is given by $V = r\omega$,where $r$ is the radius and $\omega$ is the angular speed.
Since both cars complete one revolution in the same time $t$,their angular speeds are equal,i.e.,$\omega_{1} = \omega_{2} = \frac{2\pi}{t}$.
Therefore,the ratio of their linear speeds is:
$\frac{V_{1}}{V_{2}} = \frac{r_{1}\omega_{1}}{r_{2}\omega_{2}}$
Since $\omega_{1} = \omega_{2}$,the ratio simplifies to:
$\frac{V_{1}}{V_{2}} = \frac{r_{1}}{r_{2}}$
Thus,the ratio of the linear speed of $m_{1}$ to the linear speed of $m_{2}$ is $r_{1}: r_{2}$.

(C) The relationship between linear velocity $(v)$ and angular velocity $(\omega)$ for a body moving in a circular path of radius $(r)$ is given by the formula:
$v = r \omega$
To find the angular velocity $(\omega)$,we rearrange the formula:
$\omega = \frac{v}{r}$
Therefore,the angular velocity is $v / r$.

(C) The hour hand of a clock completes one full rotation $(360^{\circ})$ in $12$ hours.
First,convert the time into seconds: $12 \text{ hours} = 12 \times 60 \text{ minutes} = 12 \times 60 \times 60 \text{ seconds} = 43200 \text{ seconds}$.
The angular speed $\omega$ is given by the formula $\omega = \frac{\theta}{t}$.
Substituting the values: $\omega = \frac{360^{\circ}}{43200 \text{ s}}$.
Simplifying the fraction: $\omega = \frac{360}{43200} = \frac{36}{4320} = \frac{1}{120} \text{ degrees per second}$.

(A) The kinetic energy $(E)$ of a particle of mass $m$ moving with speed $v$ is given by $E = \frac{1}{2} m v^2$.
For uniform circular motion,the centripetal acceleration $(a)$ is given by $a = \frac{v^2}{r}$,which implies $v^2 = a r$.
Substituting $v^2 = a r$ into the kinetic energy equation:
$E = \frac{1}{2} m (a r)$
Rearranging the equation to solve for acceleration $(a)$:
$2 E = m a r$
$a = \frac{2 E}{m r}$

(D) For a particle performing uniform circular motion, the centripetal acceleration is given by $a = \omega^2 r$.
Given, diameter $d = 1 \ m$, so the radius $r = d/2 = 0.5 \ m$.
The frequency $f = 4 \ Hz$.
The angular velocity $\omega = 2 \pi f = 2 \pi (4) = 8 \pi \ rad/s$.
Substituting these values into the acceleration formula:
$a = (8 \pi)^2 \times 0.5$
$a = 64 \pi^2 \times 0.5$
$a = 32 \pi^2 \ m/s^2$.