Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path Questions in English

(B) Let $x$ be the distance of each particle from the vertex (meeting point) of the rails. The distance between the two particles is $d = 2x \sin \theta$.
If the particles are displaced by a small amount $\Delta x$ from their equilibrium position $x_0$,the change in the length of the spring is $\Delta d = 2 \Delta x \sin \theta$.
The potential energy of the spring is $U = \frac{1}{2} k (\Delta d)^2 = \frac{1}{2} k (2 \Delta x \sin \theta)^2 = 2 k \sin^2 \theta (\Delta x)^2$.
The kinetic energy of the two particles is $K = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 = m (\Delta \dot{x})^2$.
Comparing this with the standard form of energy for a simple harmonic oscillator,$E = \frac{1}{2} k_{eff} x^2 + \frac{1}{2} m_{eff} \dot{x}^2$,we identify $k_{eff} = 4 k \sin^2 \theta$ and $m_{eff} = 2m$.
The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k_{eff}}{m_{eff}}} = \sqrt{\frac{4 k \sin^2 \theta}{2m}} = \sqrt{\frac{2k}{m}} \sin \theta$.

(C) Let $L$ be the length of the stick and $m$ be its mass. The weight $mg$ acts at the center of mass,which is at a distance $L/2$ from the pivot.
Taking torque about the pivot point:
$\tau_{\text{pivot}} = 0$
$mg \left( \frac{L}{2} \sin \theta \right) = T \cdot L \sin \phi$
where $\phi$ is the angle between the stick and the string.
$T = \frac{mg \sin \theta}{2 \sin \phi}$
To minimize $T$,we need to maximize $\sin \phi$. The maximum value of $\sin \phi$ is $1$,which occurs when $\phi = 90^\circ$.
In scenario $B$,the string is horizontal and the stick makes an angle $\theta$ with the wall,so the angle between the stick and the string is $\phi = 90^\circ - \theta$. This is not necessarily $90^\circ$.
In scenario $C$,the string is perpendicular to the stick,so $\phi = 90^\circ$. This gives the minimum tension $T = \frac{mg \sin \theta}{2}$.

(A) Let the coordinates of $A$ be $(0, y)$ and $B$ be $(x, 0)$. The length of the rod is $l = \sqrt{x^2 + y^2}$.
Since $x = l \cos \theta$ and $y = l \sin \theta$,we have $v_0 = \frac{dx}{dt} = -l \sin \theta \frac{d\theta}{dt}$.
Thus,the angular velocity $\omega = |\frac{d\theta}{dt}| = \frac{v_0}{l \sin \theta}$.
For end $A$,$v_A = |\frac{dy}{dt}| = |l \cos \theta \frac{d\theta}{dt}| = l \cos \theta (\frac{v_0}{l \sin \theta}) = v_0 \cot \theta$.
At $\theta = 37^o$,$\cot 37^o = \frac{4}{3}$,so $v_A = \frac{4}{3} v_0$.
Also,$\omega = \frac{v_0}{l \sin 37^o} = \frac{v_0}{l (3/5)} = \frac{5}{3} \frac{v_0}{l}$.
Since $v_A$ and $\omega$ depend on $\theta$,they are not constant. Thus,option $A$ is correct.

(D) $1$. Analyze the rotation of gear $M$: Gear $M$ is rotating in a clockwise direction when viewed from the top.
$2$. Analyze the interaction between $M$ and $I$: As gear $M$ rotates clockwise,it drives gear $I$. Looking at the contact point,gear $I$ will rotate in a counter-clockwise direction when viewed from the right side.
$3$. Analyze the interaction between $I$ and $H$: Gear $I$ is now rotating counter-clockwise. As it meshes with gear $H$ at the bottom,it will drive gear $H$ in a clockwise direction when viewed from the top.
$4$. Conclusion: Gear $I$ rotates counter-clockwise and gear $H$ rotates clockwise. Comparing this with the given options,none of the specific directional combinations match correctly.

(B) The dancer moves at a constant speed $v$ along the path.
In the straight-line segments $PQ$ and $RS$,the direction of motion does not change,so the velocity vector remains constant. Therefore,the acceleration $a = \frac{dv}{dt} = 0$ in these segments.
In the semicircular segments $QR$ and $SP$,the dancer is moving along a curved path with a constant speed $v$. This results in a centripetal acceleration with a magnitude $a = \frac{v^2}{r}$,where $r$ is the radius of the semicircle. Since $v$ and $r$ are constant,the magnitude of the acceleration is constant and non-zero during these segments.
Thus,the acceleration is zero during $PQ$ and $RS$,and constant non-zero during $QR$ and $SP$. This corresponds to the graph in option $B$.

(C) Let the top of the wedge be the reference level for potential energy $(U = 0)$.
In the initial state (Figure-$A$),the chain is divided into two halves,each of length $L/2$ and mass $m/2$,hanging on either side. The center of mass of each half is at a distance $L/4$ from the top along the incline.
The initial potential energy is $U_i = 2 \times [-(m/2)g(L/4) \sin \theta] = -\frac{mgL}{4} \sin \theta$.
In the final state (Figure-$B$),the entire chain of mass $m$ is on the left side. The center of mass of the chain is at a distance $L/2$ from the top along the incline.
The final potential energy is $U_f = -mg(L/2) \sin \theta = -\frac{mgL}{2} \sin \theta$.
By the law of conservation of mechanical energy,the change in potential energy equals the change in kinetic energy: $K_f - K_i = U_i - U_f$.
Since the chain starts from rest,$K_i = 0$.
$K_f = U_i - U_f = -\frac{mgL}{4} \sin \theta - (-\frac{mgL}{2} \sin \theta) = \frac{mgL}{4} \sin \theta$.

(A) For the rod $(AB)$ to be in equilibrium,the net torque about any point must be zero. The forces acting on the rod are:
$1$. The tension $(T)$ in the rope $(AC)$ acting at point $(A)$.
$2$. The weight $(Mg)$ of the rod acting at its center of mass.
$3$. The normal force $(N)$ from the ice at point $(B)$.
Since the ice is frictionless,the normal force $(N)$ must be vertical.
For the rod to be in equilibrium,the lines of action of all three forces must intersect at a single point or be parallel. Since the weight $(Mg)$ acts vertically downwards and the normal force $(N)$ acts vertically upwards,these two forces are parallel. For the system to be in equilibrium,the tension $(T)$ must also be parallel to these forces. Therefore,the rope $(AC)$ must be vertical. This configuration is shown in option $(A)$.

(B) Let $m$ be the mass of the particle and $V$ be its speed. The particle moves in a horizontal circle of radius $r = \sqrt{R^2 - d^2}$.
The forces acting on the particle are the normal reaction $N$ (directed towards the centre of the sphere) and the weight $mg$ (acting downwards).
Resolving the normal reaction $N$ into horizontal and vertical components:
Horizontal component: $N \sin \theta = \frac{mV^2}{r}$ (providing the centripetal force)
Vertical component: $N \cos \theta = mg$ (balancing the weight)
Dividing the two equations: $\tan \theta = \frac{V^2}{rg}$.
From the geometry of the bowl,$\tan \theta = \frac{r}{d}$.
Equating the two expressions for $\tan \theta$: $\frac{V^2}{rg} = \frac{r}{d} \Rightarrow V^2 = \frac{r^2 g}{d}$.
Substituting $r^2 = R^2 - d^2$,we get $V = \sqrt{\frac{g(R^2 - d^2)}{d}}$.

(A) Let $T$ be the tension in the wire. Since the ring is smooth,the tension is the same throughout the wire.
Resolving the forces acting on the ring $C$ along the vertical ($y$-axis) and horizontal ($x$-axis) directions:
Along the vertical direction,the sum of the vertical components of tension balances the weight of the ring:
$T \cos 30^{\circ} + T \cos 60^{\circ} = mg$
$T \left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) = mg$
$T \left( \frac{\sqrt{3} + 1}{2} \right) = mg$ $...(i)$
Along the horizontal direction,the sum of the horizontal components of tension provides the necessary centripetal force:
$T \sin 30^{\circ} + T \sin 60^{\circ} = \frac{mv^2}{r}$
$T \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) = \frac{mv^2}{r}$
$T \left( \frac{\sqrt{3} + 1}{2} \right) = \frac{mv^2}{r}$ $...(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{T \left( \frac{\sqrt{3} + 1}{2} \right)}{T \left( \frac{\sqrt{3} + 1}{2} \right)} = \frac{mv^2/r}{mg}$
$1 = \frac{v^2}{rg}$
$v^2 = rg$
$v = \sqrt{rg}$

(D) Given the speed $v = \alpha \sqrt{x}$.
Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \alpha \sqrt{x}$,which implies $\int x^{-1/2} dx = \int \alpha dt$.
Integrating both sides,we get $2\sqrt{x} = \alpha t$,so $x = \frac{\alpha^2 t^2}{4}$.
The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{d}{dt}(\alpha \sqrt{x}) = \alpha \cdot \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt} = \alpha \cdot \frac{1}{2\sqrt{x}} \cdot (\alpha \sqrt{x}) = \frac{\alpha^2}{2}$.
Since $\alpha$ is a constant,$a_t$ is constant.
The centripetal acceleration is $a_c = \frac{v^2}{r} = \frac{(\alpha \sqrt{x})^2}{r} = \frac{\alpha^2 x}{r}$.
Substituting $x = \frac{\alpha^2 t^2}{4}$,we get $a_c = \frac{\alpha^2}{r} \cdot \frac{\alpha^2 t^2}{4} = \frac{\alpha^4}{4r} t^2$.
Thus,$a_t$ is constant and $a_c$ is proportional to $t^2$ (a parabolic curve starting from the origin).
Comparing this with the given options,the graph where $a_t$ is a horizontal line and $a_c$ is a parabola starting from the origin is the correct one.

(D) Let $\theta$ be the angle that each string makes with the vertical axis. Since the triangle $ABC$ is equilateral (all sides are $l$),$\theta = 60^\circ$.
Resolving forces on mass $m$:
Vertical direction: $(T_1 - T_2) \cos \theta = mg \implies T_1 - T_2 = \frac{mg}{\cos 60^\circ} = 2mg$.
Horizontal direction: $(T_1 + T_2) \sin \theta = m \omega^2 r$,where $r = l \sin 60^\circ = l \frac{\sqrt{3}}{2}$.
$(T_1 + T_2) \frac{\sqrt{3}}{2} = m \omega^2 l \frac{\sqrt{3}}{2} \implies T_1 + T_2 = m \omega^2 l$.
Solving these equations: $T_1 = \frac{m \omega^2 l}{2} + mg$ and $T_2 = \frac{m \omega^2 l}{2} - mg$.
For string $BC$ to remain taut,$T_2 \geq 0 \implies \frac{m \omega^2 l}{2} \geq mg \implies \omega \geq \sqrt{2g/l}$.
Thus,option $D$ is correct.

(B) When the mass on the right is given a horizontal velocity $u$, it starts moving in a circular arc (pendulum motion) because it is constrained by the string connected to the pulley.
As it moves in a circular arc, it gains a centripetal acceleration directed towards the center of the pulley.
For the mass on the right, the equation of motion along the radial direction is $T - mg \cos \theta = \frac{mu^2}{R}$, where $T$ is the tension, $m$ is the mass, $g$ is acceleration due to gravity, $\theta$ is the angle with the vertical, and $R$ is the length of the string.
This implies $T = mg \cos \theta + \frac{mu^2}{R}$.
Since the tension $T$ in the string is greater than the weight $mg$ of the mass on the left (because $T > mg \cos \theta$ and the mass on the right is moving), the mass on the left will be pulled upwards.
Therefore, the mass on the left will rise, making it farther from the ground, while the mass on the right will be nearer to the ground.

(B) Consider a small element of the liquid of mass $dm$ at a distance $r$ from the axis of rotation. The length of this element is $dr$. Since the tube has length $L$ and mass $M$,the linear mass density is $\lambda = M/L$.
Thus,$dm = \lambda dr = (M/L) dr$.
The centripetal force $dF$ required for this element to rotate is $dF = (dm) \omega^2 r = (M/L) \omega^2 r dr$.
To find the total force exerted by the liquid at the axis of rotation,we integrate from $r = 0$ to $r = L$:
$F = \int_0^L \frac{M}{L} \omega^2 r dr = \frac{M \omega^2}{L} \left[ \frac{r^2}{2} \right]_0^L = \frac{M \omega^2}{L} \cdot \frac{L^2}{2} = \frac{1}{2} M \omega^2 L$.
This force represents the total centripetal force required to keep the liquid in circular motion,which is effectively the force exerted by the liquid at the pivot point.

(D) When the car is at the top of the convex bridge,it moves in a circular path of radius $r$. The forces acting on the car are the gravitational force $mg$ acting downwards and the normal force $N$ exerted by the bridge acting upwards.
The net centripetal force required for circular motion is directed towards the center of the circle (downwards in this case). Thus,the equation of motion is:
$mg - N = \frac{mv^2}{r}$
Rearranging the equation to solve for the normal force $N$:
$N = mg - \frac{mv^2}{r}$

(A) The radius of the circular path is $r = l \sin \theta$.
The horizontal component of the tension provides the necessary centripetal force: $T \sin \theta = m \omega^2 r$.
Substituting $r = l \sin \theta$ into the equation,we get: $T \sin \theta = m \omega^2 (l \sin \theta)$.
Dividing both sides by $\sin \theta$ (assuming $\theta \neq 0$),we obtain: $T = m \omega^2 l$.
The vertical component of the tension balances the weight: $T \cos \theta = mg$.

(B) Since the particles are moving in circular orbits,the centripetal force is provided by the given force $F(r)$.
$\frac{mv^2}{r} = |F(r)| = \frac{16}{r} + r^3$
Multiplying both sides by $\frac{r}{2}$,we get the kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}(16 + r^4)$.
For the first particle at $r = 1$:
$K_1 = \frac{1}{2}(16 + 1^4) = \frac{17}{2} = 8.5$.
For the second particle at $r = 4$:
$K_2 = \frac{1}{2}(16 + 4^4) = \frac{1}{2}(16 + 256) = \frac{272}{2} = 136$.
The ratio of kinetic energies is:
$\frac{K_1}{K_2} = \frac{17/2}{272/2} = \frac{17}{272} = \frac{1}{16} = 0.0625$.
Thus,the ratio is approximately $6 \times 10^{-2}$.

(D) Given: $\theta = 45^\circ$,$r = 0.4\,m$,$g = 10\,m/s^2$.
For a conical pendulum,the forces acting on the bob are tension $T$ and weight $mg$.
The horizontal component of tension provides the necessary centripetal force:
$T \sin \theta = \frac{mv^2}{r} \quad \dots(i)$
The vertical component of tension balances the weight:
$T \cos \theta = mg \quad \dots(ii)$
Dividing equation $(i)$ by $(ii)$:
$\tan \theta = \frac{v^2}{rg}$
$v^2 = rg \tan \theta$
Substituting the given values:
$v^2 = 0.4 \times 10 \times \tan(45^\circ)$
$v^2 = 4 \times 1 = 4$
$v = \sqrt{4} = 2\,m/s$.

(A) Given the equations of motion for the particle:
$x = a \sin \omega t \Rightarrow \sin \omega t = \frac{x}{a}$
$y = a \sin 2 \omega t$
Using the trigonometric identity $\sin 2 \theta = 2 \sin \theta \cos \theta$,we can write:
$y = a (2 \sin \omega t \cos \omega t)$
Since $\cos \omega t = \sqrt{1 - \sin^2 \omega t} = \sqrt{1 - (\frac{x}{a})^2} = \frac{\sqrt{a^2 - x^2}}{a}$,we substitute this into the equation for $y$:
$y = 2a (\frac{x}{a}) (\frac{\sqrt{a^2 - x^2}}{a})$
$y = \frac{2x}{a} \sqrt{a^2 - x^2}$
Squaring both sides gives $y^2 = \frac{4x^2}{a^2} (a^2 - x^2)$,which represents a figure-eight curve (Lissajous figure) symmetric about the $x$ and $y$ axes. This corresponds to the shape shown in option $A$.

(C) At point $Y$,the normal reaction $N = 0$. The radial component of the gravitational force provides the necessary centripetal force:
$mg \sin \phi = \frac{mv^2}{r} \implies v^2 = rg \sin \phi$ $...(i)$
Using the principle of conservation of mechanical energy between point $X$ and point $Y$:
$mg(r \cos \theta) = mg(r \sin \phi) + \frac{1}{2} mv^2$
$g r \cos \theta = g r \sin \phi + \frac{1}{2} (rg \sin \phi)$
$g r \cos \theta = \frac{3}{2} rg \sin \phi$
$\cos \theta = \frac{3}{2} \sin \phi$
$\sin \phi = \frac{2}{3} \cos \theta$

(D) Let $m$ be the mass of the bead and $N$ be the normal reaction force exerted by the wire on the bead.
The radius of the circular path of the bead is $R_{path} = r/2$.
The forces acting on the bead are the gravitational force $mg$ (downwards) and the normal reaction $N$ (perpendicular to the wire).
Resolving the forces into horizontal and vertical components:
$N \sin \theta = m R_{path} \omega^2 = m (r/2) \omega^2$ ... $(i)$
$N \cos \theta = mg$ ... (ii)
Dividing $(i)$ by (ii),we get:
$\tan \theta = \frac{(r/2) \omega^2}{g} = \frac{r \omega^2}{2g}$
From the geometry of the circle,the distance from the center $O$ to the vertical axis is $r/2$. The angle $\theta$ that the radius vector $OP$ makes with the vertical is given by $\sin \theta = \frac{r/2}{r} = 1/2$,so $\theta = 30^\circ$.
Thus,$\tan 30^\circ = \frac{1}{\sqrt{3}}$.
Equating the two expressions for $\tan \theta$:
$\frac{1}{\sqrt{3}} = \frac{r \omega^2}{2g}$
$\omega^2 = \frac{2g}{r\sqrt{3}}$

(C) For a particle moving in a circular or curved path with constant speed $v$,the centripetal acceleration is given by $a_c = \frac{v^2}{r}$,where $r$ is the radius of curvature.
In case $(a)$,the path is a uniform helix,meaning the radius of curvature $r$ is constant. Since $v$ and $r$ are constant,the magnitude of acceleration $a_c$ is constant.
In case $(b)$,the path is a spiral where the radius $r$ increases as the particle moves outward. Since $v$ is constant and $r$ increases,the magnitude of acceleration $a_c = \frac{v^2}{r}$ decreases as the particle moves along the spiral path.
Therefore,the magnitude of acceleration is constant in $(a)$ and decreasing in $(b)$.

(A) Given:
Length of the string,$r = 0.1\,m$
Maximum tension,$T = 100\,N$
Mass of the body,$m = 100\,g = 0.1\,kg$
For a body rotating in a horizontal circle,the tension in the string provides the necessary centripetal force.
The formula for centripetal force is $T = m\omega^2r$.
Substituting the given values:
$100 = 0.1 \times \omega^2 \times 0.1$
$100 = 0.01 \times \omega^2$
$\omega^2 = \frac{100}{0.01}$
$\omega^2 = 10000$
$\omega = \sqrt{10000}$
$\omega = 100\,rad/s$.

(B) Let $\theta$ be the angle that the radius vector to the car makes with the vertical at any point on the overbridge.
The forces acting on the car are its weight $mg$ (downwards) and the normal reaction $N$ (upwards).
The net centripetal force required for circular motion is provided by the component of weight towards the center and the normal reaction.
The equation of motion is: $mg \cos \theta - N = \frac{mv^2}{R}$.
Rearranging for the normal reaction: $N = mg \cos \theta - \frac{mv^2}{R}$.
As the car descends from point $B$ (the highest point) to point $C$,the angle $\theta$ increases from $0^\circ$ to $90^\circ$.
As $\theta$ increases,$\cos \theta$ decreases.
Since $N = mg \cos \theta - \frac{mv^2}{R}$,as $\cos \theta$ decreases,the normal reaction $N$ also decreases.

(A) At the highest point of a circular overbridge,the forces acting on the car are the gravitational force $(mg)$ acting downwards and the normal reaction force $(N)$ acting upwards.
The net centripetal force required for circular motion is provided by the difference between these forces:
$mg - N = \frac{mv^2}{r}$
To find the maximum speed at which the car can cross without losing contact,we set the normal reaction force $N$ to zero (the point where the car is just about to leave the surface).
$mg - 0 = \frac{mv^2}{r}$
$g = \frac{v^2}{r}$
$v^2 = rg$
$v = \sqrt{rg}$
Given $r = 20\,m$ and $g = 9.8\,m/s^2$:
$v = \sqrt{20 \times 9.8}$
$v = \sqrt{196}$
$v = 14\,m/s$

(B) The motorcycle moves in a circular path of radius $R$ on the overbridge. The forces acting on the motorcycle are the normal force $N$ (upwards) and the weight $mg$ (downwards).
At any angle $\theta$ with the vertical,the component of weight acting towards the center is $mg \cos \theta$.
The net centripetal force is given by $mg \cos \theta - N = \frac{mv^2}{R}$.
Rearranging for the normal force,we get $N = mg \cos \theta - \frac{mv^2}{R}$.
As the motorcycle ascends the overbridge,the angle $\theta$ with the vertical increases from $0^\circ$ to $90^\circ$.
As $\theta$ increases,$\cos \theta$ decreases.
Since $N = mg \cos \theta - \frac{mv^2}{R}$,as $\cos \theta$ decreases,the normal force $N$ decreases.

(A) The forces acting on the block are the normal force $N$ and the gravitational force $mg$. The block moves in a horizontal circle of radius $x = R \sin \theta$.
For the vertical direction,the forces are balanced:
$N \cos \theta = mg$ ......$(i)$
For the horizontal direction,the normal force provides the necessary centripetal force:
$N \sin \theta = m x \omega^2 = m (R \sin \theta) \omega^2$ ......(ii)
Dividing equation (ii) by equation $(i)$:
$\frac{N \sin \theta}{N \cos \theta} = \frac{m R \sin \theta \omega^2}{mg}$
$\tan \theta = \frac{R \sin \theta \omega^2}{g}$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we have:
$\frac{\sin \theta}{\cos \theta} = \frac{R \sin \theta \omega^2}{g}$
$\frac{1}{\cos \theta} = \frac{R \omega^2}{g}$
$\cos \theta = \frac{g}{R \omega^2}$
$\theta = \cos^{-1}\left(\frac{g}{R \omega^2}\right)$

(D) $1$. Centripetal force is a real force acting on a body moving in a circular path,directed towards the center of the circle. It is required to change the direction of velocity.
$2$. Centrifugal force is a pseudo force (fictitious force) that appears to act on a body in a non-inertial (rotating) frame of reference. It is directed away from the center.
$3$. Since these forces act on different frames of reference,they cannot cancel each other out. Therefore,the $Assertion$ is incorrect.
$4$. Centrifugal force is not a reaction to centripetal force in the context of $Newton's$ third law. $Newton's$ third law applies to forces of the same nature acting on different bodies. Centripetal force is a real force,while centrifugal force is a pseudo force. Therefore,the $Reason$ is also incorrect.
$5$. Thus,both $Assertion$ and $Reason$ are incorrect.

(B) Let the stretch in the spring be $x$. The total length of the spring becomes $r = \ell + x$.
The centripetal force required for the circular motion of the particle is provided by the spring force.
$F_{c} = F_{s}$
$m \omega^{2} r = k x$
$m \omega^{2} (\ell + x) = k x$
$m \omega^{2} \ell + m \omega^{2} x = k x$
$m \omega^{2} \ell = k x - m \omega^{2} x$
$m \omega^{2} \ell = x (k - m \omega^{2})$
$x = \frac{m \ell \omega^{2}}{k - m \omega^{2}}$

(A) The correct answer is $(i) \; T$.
When a particle connected to a string revolves in a circular path on a smooth horizontal table,the centripetal force required for circular motion is provided solely by the tension $T$ in the string.
Since the table is smooth,there is no friction,and the only horizontal force acting on the particle directed towards the centre is the tension $T$.
Therefore,the net force on the particle is $F_{\text{net}} = T = \frac{m v^{2}}{l}$.

(B) Speed of the aircraft,$v = 720 \; km/h = 720 \times \frac{5}{18} = 200 \; m/s$.
Acceleration due to gravity,$g = 10 \; m/s^2$.
Angle of banking,$\theta = 15^{\circ}$.
For the radius $r$ of the horizontal loop,the relation is given by:
$\tan \theta = \frac{v^2}{rg}$.
Rearranging for $r$:
$r = \frac{v^2}{g \tan \theta}$.
Substituting the values:
$r = \frac{200 \times 200}{10 \times \tan 15^{\circ}} = \frac{40000}{10 \times 0.2679} = \frac{4000}{0.2679} \approx 14930 \; m$.
Converting to $km$:
$r \approx 14.93 \; km$.
Rounding to the nearest provided option,the radius is $14.92 \; km$.

(D) Let the radius vector joining the bead $P$ with the centre $O$ make an angle $\theta$ with the vertical downward direction.
Let $m$ be the mass of the bead.
The forces acting on the bead are:
$1$. Weight $mg$ acting vertically downwards.
$2$. Normal reaction $N$ from the wire,acting perpendicular to the tangent at $P$.
Resolving the forces:
The vertical component of the normal reaction $N \cos \theta$ balances the weight $mg$,so $N \cos \theta = mg$.
The horizontal component of the normal reaction $N \sin \theta$ provides the necessary centripetal force $m \omega^2 r$,where $r = R \sin \theta$ is the radius of the circular path of the bead.
Thus,$N \sin \theta = m \omega^2 (R \sin \theta)$.
Since $\sin \theta \neq 0$ for $\theta \neq 0$,we get $N = m R \omega^2$.
Substituting $N$ in the vertical equation: $(m R \omega^2) \cos \theta = mg$.
Therefore,$\cos \theta = \frac{g}{R \omega^2}$.
Since $\cos \theta \leq 1$,the bead remains at the lowermost point $(\theta = 0)$ if $\frac{g}{R \omega^2} \geq 1$,which implies $\omega \leq \sqrt{\frac{g}{R}}$.
For $\omega = \sqrt{\frac{2g}{R}}$,we have $\omega^2 = \frac{2g}{R}$.
Substituting this into the expression for $\cos \theta$:
$\cos \theta = \frac{g}{R (2g/R)} = \frac{1}{2}$.
Thus,$\theta = \cos^{-1}(0.5) = 60^{\circ}$.

(N/A) Consider a stone tied to a string being whirled in a horizontal circular path at a constant speed.
$1$. The momentum of the stone is given by $\vec{p} = m\vec{v}$. Since the speed is constant, the magnitude of momentum remains constant, but its direction is always tangential to the circular path and changes continuously.
$2$. The change in momentum $\Delta \vec{p}$ over a small time interval $\Delta t$ is given by $\Delta \vec{p} = \vec{p}_{final} - \vec{p}_{initial}$. For a circular motion, this change in momentum vector points towards the center of the circle.
$3$. According to Newton's second law, $\vec{F} = \frac{d\vec{p}}{dt}$. The force (tension in the string) is directed towards the center, which is the same direction as the change in momentum.
$4$. Since the momentum vector is tangential and the change in momentum vector is radial (towards the center), they are clearly not in the same direction.

(D) The normal force $N$ provides the necessary centripetal force for circular motion.
$N = F_c = m \omega^2 R$
Since $\omega = \frac{2\pi}{T}$,we have $N = m \left( \frac{2\pi}{T} \right)^2 R = m \frac{4\pi^2}{T^2} R$.
Given values: $m = 200\, g = 0.2\, kg$,$R = 20\, cm = 0.2\, m$,and $T = 40\, s$.
Substituting these values into the formula:
$N = 0.2 \times \frac{4 \times (3.14159)^2}{(40)^2} \times 0.2$
$N = 0.2 \times \frac{4 \times 9.8696}{1600} \times 0.2$
$N = 0.04 \times \frac{39.4784}{1600}$
$N = 0.04 \times 0.024674 = 9.8696 \times 10^{-4}\, N$.
Rounding to the nearest option,we get $N \approx 9.859 \times 10^{-4}\, N$.

(A) The centripetal force required for circular motion is provided by the central force $F$. Given $F \propto \frac{1}{R^{3}}$,we can write $F = \frac{K}{R^{3}}$,where $K$ is a constant.
For a particle of mass $m$ moving with angular velocity $\omega$ in a circle of radius $R$,the centripetal force is $F = m \omega^{2} R$.
Equating the two expressions: $\frac{K}{R^{3}} = m \omega^{2} R$.
Rearranging for $\omega^{2}$: $\omega^{2} = \frac{K}{m R^{4}}$.
Since the time period $T = \frac{2 \pi}{\omega}$,we have $\omega = \frac{2 \pi}{T}$.
Substituting this into the equation: $(\frac{2 \pi}{T})^{2} = \frac{K}{m R^{4}}$.
This simplifies to $\frac{4 \pi^{2}}{T^{2}} = \frac{K}{m R^{4}}$,which implies $T^{2} \propto R^{4}$.
Taking the square root of both sides,we get $T \propto R^{2}$.

(A) This is a case of a conical pendulum.
Let $\theta$ be the angle the string makes with the vertical.
From the geometry of the setup,we have $\sin \theta = \frac{r}{L}$.
Given $r = \frac{L}{\sqrt{2}}$,we get $\sin \theta = \frac{L/\sqrt{2}}{L} = \frac{1}{\sqrt{2}}$.
This implies $\theta = 45^{\circ}$.
The forces acting on the particle are the tension $T$ in the string and the gravitational force $mg$.
Resolving the tension into horizontal and vertical components:
Horizontal component: $T \sin \theta = \frac{mv^2}{r}$ (providing the necessary centripetal force).
Vertical component: $T \cos \theta = mg$ (balancing the weight).
Dividing the two equations: $\frac{T \sin \theta}{T \cos \theta} = \frac{mv^2/r}{mg} \Rightarrow \tan \theta = \frac{v^2}{rg}$.
Since $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$.
Therefore,$1 = \frac{v^2}{rg} \Rightarrow v^2 = rg \Rightarrow v = \sqrt{rg}$.

(C) Let the elongation of the spring be $\Delta x$. The total length of the spring becomes $l = l_{0} + \Delta x$.
The centripetal force required for the circular motion of the mass $m$ is provided by the spring force $F_{s} = k \Delta x$.
Equating the spring force to the centripetal force: $k \Delta x = m \omega^{2} (l_{0} + \Delta x)$.
Expanding the equation: $k \Delta x = m \omega^{2} l_{0} + m \omega^{2} \Delta x$.
Rearranging the terms to solve for $\Delta x$: $k \Delta x - m \omega^{2} \Delta x = m \omega^{2} l_{0}$.
$\Delta x (k - m \omega^{2}) = m \omega^{2} l_{0}$.
Therefore,the elongation is $\Delta x = \frac{m \omega^{2} l_{0}}{k - m \omega^{2}}$.

(B) The apparent weight of the vehicle is equal to the normal reaction $N$ exerted by the track on the vehicle.
At point $A$,the track is horizontal,so the normal reaction is $N_A = mg$.
At point $B$,the vehicle is at the bottom of a concave curve. The net centripetal force is provided by the difference between the normal reaction and gravity: $N_B - mg = \frac{mv^2}{r}$. Thus,$N_B = mg + \frac{mv^2}{r}$.
At point $C$,the vehicle is at the top of a convex curve. The net centripetal force is provided by the difference between gravity and the normal reaction: $mg - N_C = \frac{mv^2}{r}$. Thus,$N_C = mg - \frac{mv^2}{r}$.
Comparing the values,$N_B > N_A > N_C$. Therefore,the apparent weight is maximum at $B$.

(C) The banking of rails is given by the formula $\tan \theta = \frac{h}{d} = \frac{v^2}{rg}$,where $h$ is the elevation,$d$ is the distance between rails,$v$ is the velocity,$r$ is the radius of curvature,and $g$ is the acceleration due to gravity.
Given: $v = 20 \, m/s$,$r = 40,000 \, m$,$d = 1.5 \, m$,and $g = 10 \, m/s^2$.
Substituting the values: $\frac{h}{1.5} = \frac{(20)^2}{40,000 \times 10}$.
$\frac{h}{1.5} = \frac{400}{400,000} = \frac{1}{1000}$.
$h = \frac{1.5}{1000} \, m = 0.0015 \, m$.
Converting to $mm$: $h = 0.0015 \times 1000 \, mm = 1.5 \, mm$.

(C) Let the semi-vertical angle of the cone be $\theta$. From the geometry of the cone,we have $\tan \theta = \frac{r}{h}$.
The forces acting on the particle are its weight $mg$ acting downwards and the normal reaction $N$ from the surface of the cone acting perpendicular to the surface.
Resolving the normal reaction $N$ into horizontal and vertical components:
Vertical component: $N \cos \theta = mg$ --- $(i)$
Horizontal component (providing the centripetal force): $N \sin \theta = \frac{mv^2}{r}$ --- (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{N \sin \theta}{N \cos \theta} = \frac{mv^2/r}{mg}$
$\tan \theta = \frac{v^2}{rg}$
Substituting $\tan \theta = \frac{r}{h}$ into the equation:
$\frac{r}{h} = \frac{v^2}{rg}$
$v^2 = \frac{r^2 g}{h}$
$v = r \sqrt{\frac{g}{h}}$
Wait,re-evaluating the geometry: If $\theta$ is the angle the slant surface makes with the horizontal,then $\tan \theta = \frac{h}{r}$.
Then $\frac{h}{r} = \frac{v^2}{rg} \implies v^2 = gh \implies v = \sqrt{gh}$.

(A) For a conical pendulum,the time period $T$ is given by $T = 2\pi \sqrt{\frac{h}{g}}$,where $h$ is the vertical height of the pendulum bob from the point of suspension.
If the string of length $l$ makes an angle $\alpha$ with the vertical,then $h = l \cos \alpha$.
In this question,the angle $\theta$ is given with the horizontal. Therefore,the angle with the vertical is $\alpha = 90^\circ - \theta$.
Substituting this into the expression for height: $h = l \cos(90^\circ - \theta) = l \sin \theta$.
Thus,the time period is $T = 2\pi \sqrt{\frac{l \sin \theta}{g}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{l \sin \theta}{g}$.
Since $4\pi^2$,$l$,and $g$ are constants,we have $T^2 \propto \sin \theta$.

(A) For an object moving in a circular path with constant speed $v$ and radius $R$,the force $F$ acting on it is the centripetal force.
$F = \frac{m v^2}{R}$
From this,we can express the term $m v^2$ as:
$m v^2 = F R$
The kinetic energy $(KE)$ of the object is given by the formula:
$KE = \frac{1}{2} m v^2$
Substituting the value of $m v^2$ from the force equation into the kinetic energy formula:
$KE = \frac{1}{2} (F R)$
Therefore,the kinetic energy is $\frac{1}{2} F R$.

(B) The centripetal force required for circular motion on a frictionless surface is provided by the horizontal component of the normal reaction when the person leans at an angle $\theta$ with the vertical.
For a body moving in a circle of radius $r$ with speed $v$,the angle of inclination $\theta$ is given by the formula: $\tan \theta = \frac{v^2}{rg}$.
Given values are $v = 10 \ m/s$,$r = 50 \ m$,and $g = 10 \ m/s^2$.
Substituting these values into the formula:
$\tan \theta = \frac{(10)^2}{50 \times 10} = \frac{100}{500} = \frac{1}{5}$.
Therefore,the angle of inclination is $\theta = \tan^{-1}(1/5)$.

(C) The tension in the string provides the necessary centripetal force required to keep the stone moving in a horizontal circular path.
Given mass $m = 0.3\,kg$,radius $R = 1.5\,m$,and velocity $v = 6\,m s^{-1}$.
The formula for centripetal force is $F_c = \frac{mv^2}{R}$.
Substituting the given values:
$T = \frac{0.3 \times (6)^2}{1.5}$
$T = \frac{0.3 \times 36}{1.5}$
$T = \frac{10.8}{1.5} = 7.2\,N$.
Therefore,the tension in the string is $7.2\,N$.

(D) The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T}$.
Given $T = 3.14\,s$ and $\pi \approx 3.14$,we have $\omega = \frac{2 \times 3.14}{3.14} = 2\,rad/s$.
The centrifugal force $F_c$ is given by the formula $F_c = m\omega^2R$.
Substituting the given values: $m = 5\,kg$,$\omega = 2\,rad/s$,and $R = 2\,m$.
$F_c = 5 \times (2)^2 \times 2 = 5 \times 4 \times 2 = 40\,N$.
Therefore,the centrifugal force on the child is $40\,N$.

(A) The formula for centripetal force is given by $F_c = m \omega^2 r$.
Given:
Mass $m = 200\,kg$
Angular velocity $\omega = 0.2\,rad/s$
Radius $r = 70\,m$
Substituting the values into the formula:
$F_c = 200 \times (0.2)^2 \times 70$
$F_c = 200 \times 0.04 \times 70$
$F_c = 8 \times 70 = 560\,N$.
Therefore,the centripetal force acting on the vehicle is $560\,N$.

(B) For banking of rails, the angle of banking $\theta$ is given by $\tan \theta = \frac{v^2}{Rg}$.
Given: speed $v = 12 \,m/s$, radius $R = 400 \,m$, distance between rails $d = 1.5 \,m$, and $g = 10 \,m/s^2$.
Substituting the values: $\tan \theta = \frac{12^2}{400 \times 10} = \frac{144}{4000} = 0.036$.
From the geometry of the banked track, $\tan \theta = \frac{h}{d}$, where $h$ is the height of the outer rail.
Therefore, $h = d \times \tan \theta = 1.5 \,m \times 0.036 = 0.054 \,m$.
Converting to centimeters: $h = 0.054 \times 100 \,cm = 5.4 \,cm$.

(B) The tension $T$ in the string provides the necessary centripetal force for the horizontal circular motion of the ball.
Given mass $m = 0.5 \ kg$,length $\ell = 50 \ cm = 0.5 \ m$,and maximum tension $T_{max} = 400 \ N$.
The formula for centripetal force in a horizontal circle is $T = m \omega^2 \ell$.
Substituting the given values:
$400 = 0.5 \times \omega^2 \times 0.5$
$400 = 0.25 \times \omega^2$
$\omega^2 = \frac{400}{0.25} = 1600$
$\omega = \sqrt{1600} = 40 \ rad/s$.
Thus,the maximum possible angular velocity is $40 \ rad/s$.

(B) Given: Radius $R = 9 \,m$.
The man completes $120$ revolutions in $3$ minutes.
First,calculate the angular velocity $\omega$:
$\omega = \frac{120 \text{ revolutions}}{3 \text{ minutes}} = \frac{120 \times 2\pi \text{ radians}}{3 \times 60 \text{ seconds}} = \frac{240\pi}{180} = \frac{4\pi}{3} \,rad/s$.
The formula for centripetal acceleration is $a_c = \omega^2 R$.
Substituting the values: $a_c = \left(\frac{4\pi}{3}\right)^2 \times 9 = \frac{16\pi^2}{9} \times 9 = 16\pi^2 \,m/s^2$.

(A) For a bob of mass $m$ rotating in a horizontal circle of radius $\ell$ with angular speed $\omega$,the centripetal force is provided by the tension $T$ in the string.
$T = m\ell\omega^2$ --- $(1)$
When the angular speed is increased to $2\omega$ while keeping the radius $\ell$ constant,the new tension $T'$ is given by:
$T' = m\ell(2\omega)^2$
$T' = m\ell(4\omega^2)$
$T' = 4(m\ell\omega^2)$
Substituting equation $(1)$ into this expression,we get:
$T' = 4T$