JEE Circular Motion Questions

Mass of Removed Part ($$m$$):

$$The\ mass\ of\ a\ uniform\ disc\ is\ proportional\ to\ its\ area\ (\text{Area}=\pi R^2):$$

$$m = \frac{\pi (R/2)^2}{\pi R^2} \times M = \frac{M}{4}$$

Moment of Inertia of the Removed Part ($$I_{rem}$$):

The center of the removed part is at a distance $$d = R/2$$ from the center of the original disc. Using the Parallel Axis Theorem:

$$I_{rem} = I_{cm} + md^2$$

$$I_{rem} = \frac{1}{2} m \left(\frac{R}{2}\right)^2 + m \left(\frac{R}{2}\right)^2 = \frac{3}{2} m \left(\frac{R}{2}\right)^2$$

$$I_{rem} = \frac{3}{2} \left(\frac{M}{4}\right) \frac{R^2}{4} = \frac{3}{32} MR^2$$

Moment of Inertia of the Remaining Part ($$I_{final}$$):

$$I_{final} = I_{original} - I_{rem}$$

$$I_{final} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2$$

$$I_{final} = \frac{16}{32} MR^2 - \frac{3}{32} MR^2 = \frac{13}{32} MR^2$$

The torque $$\vec{\tau}$$ is given by $$\vec{\tau} = \vec{r} \times \vec{F}$$, where $$\vec{r} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$$.

Computing the cross product using the determinant formula:

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$$

$$\hat{i}$$ component: $$(1)(2) - (1)(1) = 2 - 1 = 1$$

$$\hat{j}$$ component: $$-[(1)(2) - (1)(2)] = -(2 - 2) = 0$$

$$\hat{k}$$ component: $$(1)(1) - (1)(2) = 1 - 2 = -1$$

$$\vec{\tau} = \hat{i} - \hat{k}$$

The correct answer is Option A: $$\hat{i} - \hat{k}$$.

$$T^2 \propto R^3 \implies T \propto R^{3/2}$$ (Kepler’s Third Law)

Since the increase in radius is small ($$R$$ to $$1.03R$$, a $$3\%$$ change), we can use the approximation method for small percentage changes.

The percentage change in the radius is:

$$\frac{\Delta R}{R} \times 100 = \frac{1.03R - R}{R} \times 100 = 3\%$$

$$\text{Percentage change in } T \approx \frac{3}{2} \times (\text{Percentage change in } R)$$

$$\text{Percentage change in } T \approx \frac{3}{2} \times 3\% = 4.5\%$$

$$\vec{F} = m\vec{a}$$

$$\vec{a} = \frac{d^2\vec{r}}{dt^2}$$

$$\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$$

$$\vec{v} = \frac{d\vec{r}}{dt} = a(-\omega \hat{i} \sin \omega t + \omega \hat{j} \cos \omega t)$$

$$\vec{v} = a\omega(-\hat{i} \sin \omega t + \hat{j} \cos \omega t)$$

$$\vec{a} = \frac{d\vec{v}}{dt} = a\omega(-\omega \hat{i} \cos \omega t - \omega \hat{j} \sin \omega t)$$

$$\vec{a} = -\omega^2 \cdot a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$$

$$\vec{a} = -\omega^2 \vec{r}$$

$$\vec{F} = m\vec{a} = -m\omega^2 \vec{r}$$

Since mass ($$m$$) and squared angular frequency ($$\omega^2$$) are strictly positive scalar quantities, the negative sign indicates that the force vector $$\vec{F}$$ points directly in the direction opposite to the position vector $$\vec{r}$$.

We begin with a satellite orbiting Earth that is 9 times closer than the Moon and seek its time period.

We know from Kepler’s third law that $$T^2 \propto r^3$$.

Since the satellite is 9 times closer: $$r_s = \frac{r_m}{9}$$.

Substituting into the proportionality gives $$\frac{T_s^2}{T_m^2} = \frac{r_s^3}{r_m^3} = \left(\frac{1}{9}\right)^3 = \frac{1}{729}$$.

Taking the square root yields $$T_s = T_m \times \frac{1}{\sqrt{729}} = 27 \times \frac{1}{27} = 1$$ day.

The correct answer is Option 2: 1 day.

Given the angular position:

$$\theta(t) = 5t^2 - 8t$$

Differentiating with respect to time to find the angular velocity:

$$\omega = \frac{d\theta}{dt} = 10t - 8$$

$$At\ t=2\text{ s},\ \omega=10(2)-8=12\text{ rad/s}$$

$$\alpha = \frac{d\omega}{dt} = 10 \text{ rad/s}^2$$

$$\tau = I\alpha$$

$$I=\frac{1}{2}MR^2$$ for a circular disk rotating about an axis perpendicular to its plane through the center

$$\tau = \left( \frac{1}{2}MR^2 \right) \times 10 = 5MR^2$$ 

Power Delivered ($$P$$):

$$P = \tau \cdot \omega$$

$$P = (5MR^2) \times 12$$
$$P = 60MR^2$$

Given frequency of rotation 

$$f=\frac{3}{\pi\ }$$

Now calculate angular velocity : -

$$\omega\ =2\pi\ f$$

$$\omega\ =2\pi\ \times\ \frac{3}{\pi\ }$$

$$\omega\ =2\times\ 3\ =\ 6$$

We know that centrifugal force will be equal to the horizontal component of the T(tension of the wire)

$$\frac{mv^2}{r}=T\sin\theta\ $$

$$v=\omega\ r$$

so 

$$m\omega^2r=T\sin\theta\ $$

Here $$r=L\sin\theta\ $$

$$m\omega^2L\sin\theta\ =T\sin\theta\ $$

$$m\omega^2L\ =T$$

substitute values

$$M\left(6\right)^2L\ =T$$

$$T=36ML\ $$

so answer is 36

We need to find the magnitude of torque in the z-direction when a force $$\vec{F} = \hat{i} - \hat{j} + \hat{k}$$ acts on a particle at position $$\vec{r} = \hat{i} + \hat{j} + \hat{k}$$ (coordinates (1, 1, 1) m).

Recall that torque is given by $$\vec{\tau} = \vec{r} \times \vec{F}$$.

Substituting the given vectors into the determinant,
$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$
$$= \hat{i}(1 \cdot 1 - 1 \cdot (-1)) - \hat{j}(1 \cdot 1 - 1 \cdot 1) + \hat{k}(1 \cdot (-1) - 1 \cdot 1)$$
$$= \hat{i}(1 + 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$$
$$= 2\hat{i} - 0\hat{j} - 2\hat{k}$$

The z-component of torque is $$\tau_z = -2$$, so the magnitude of torque in the z-direction is $$|\tau_z| = 2$$.

The answer is 2.

We need to find the magnitude of angular momentum of particle A with respect to the position of particle B at t = 1s.

$$\vec{r_A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k})$$ m

$$\vec{r_B} = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k})$$ m

$$\alpha_1 = 1, \alpha_2 = 3n, \alpha_3 = 2, \beta_1 = 2, \beta_2 = -1, \beta_3 = 4p$$

$$\vec{V_A} = \frac{d\vec{r_A}}{dt} = (2\alpha_1 t \hat{i} + \alpha_2 \hat{j} + \alpha_3 \hat{k})$$

At t = 1: $$\vec{V_A} = (2 \hat{i} + 3n \hat{j} + 2 \hat{k})$$ m/s

$$\vec{V_B} = \frac{d\vec{r_B}}{dt} = (\beta_1 \hat{i} + 2\beta_2 t \hat{j} + \beta_3 \hat{k})$$

At t = 1: $$\vec{V_B} = (2 \hat{i} - 2 \hat{j} + 4p \hat{k})$$ m/s

$$4 + 9n^2 + 4 = 4 + 4 + 16p^2$$

$$9n^2 + 4 = 4 + 16p^2$$

$$9n^2 = 16p^2$$ ... (i)

$$\vec{V_A} \cdot \vec{V_B} = 0$$

$$2(2) + 3n(-2) + 2(4p) = 0$$

$$4 - 6n + 8p = 0$$

$$6n - 8p = 4$$

$$3n - 4p = 2$$ ... (ii)

From (i): $$3n = 4p$$ (taking positive root, but let's check both)

If $$3n = 4p$$, substituting in (ii): $$4p - 4p = 2 \Rightarrow 0 = 2$$ (contradiction)

If $$3n = -4p$$, substituting in (ii): $$-4p - 4p = 2 \Rightarrow -8p = 2 \Rightarrow p = -1/4$$

Then $$3n = -4(-1/4) = 1 \Rightarrow n = 1/3$$

$$\vec{r_A}(1) = (1 \hat{i} + 3(1/3) \hat{j} + 2 \hat{k}) = (1 \hat{i} + 1 \hat{j} + 2 \hat{k})$$ m

$$\vec{r_B}(1) = (2 \hat{i} + (-1) \hat{j} + 4(-1/4) \hat{k}) = (2 \hat{i} - 1 \hat{j} - 1 \hat{k})$$ m

$$\vec{V_A}(1) = (2 \hat{i} + 1 \hat{j} + 2 \hat{k})$$ m/s

$$\vec{r} = \vec{r_A} - \vec{r_B} = (1-2)\hat{i} + (1+1)\hat{j} + (2+1)\hat{k} = (-1\hat{i} + 2\hat{j} + 3\hat{k})$$ m

$$\vec{L} = m(\vec{r} \times \vec{V_A}) = 1 \times \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & 1 & 2 \end{vmatrix}$$

$$= \hat{i}(2 \times 2 - 3 \times 1) - \hat{j}((-1)(2) - 3(2)) + \hat{k}((-1)(1) - 2(2))$$

$$= \hat{i}(4-3) - \hat{j}(-2-6) + \hat{k}(-1-4)$$

$$= \hat{i}(1) - \hat{j}(-8) + \hat{k}(-5)$$

$$= (1\hat{i} + 8\hat{j} - 5\hat{k})$$

$$|\vec{L}| = \sqrt{1 + 64 + 25} = \sqrt{90}$$

Since $$|\vec{L}| = \sqrt{L}$$, we get $$L = 90$$.

The answer is 90.

The angular momentum of a planet in circular orbit is $$L = mv r$$, where $$v$$ is the orbital velocity.

For a circular orbit we have $$\frac{GMm}{R^2} = \frac{mv^2}{R}$$, so $$v = \sqrt{\frac{GM}{R}}$$.

Substituting this into the expression for angular momentum gives $$L = mR\sqrt{\frac{GM}{R}} = m\sqrt{GMR}$$.

We then consider the ratio of the angular momenta of bodies B and A:

$$ \frac{L_B}{L_A} = \frac{m_B\sqrt{GMR_B}}{m_A\sqrt{GMR_A}} = \frac{m_B}{m_A}\sqrt{\frac{R_B}{R_A}} $$

Using the given relations $$m_B = 4\sqrt{2} \cdot m_A$$ and $$R_B = 2R_A$$, we find

$$ \frac{L_B}{L_A} = 4\sqrt{2} \times \sqrt{2} = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8 $$

The answer is 8.

A satellite of mass $$\frac{M}{2}$$ revolves around the Earth in a circular orbit at a height of $$\frac{R}{3}$$ from the Earth's surface. The angular momentum is given as $$M\sqrt{\frac{GMR}{x}}$$. We need to find $$x$$.

First we find the orbital radius, which is the distance from the center of the Earth to the satellite:

$$ r = R + \frac{R}{3} = \frac{3R + R}{3} = \frac{4R}{3} $$

Next we find the orbital velocity by equating the gravitational force to the centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} $$

Here $$M$$ is Earth's mass, $$m = \frac{M}{2}$$ is the satellite mass, and $$r$$ is the orbital radius. Solving for $$v$$ gives

$$ v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}} $$

Then the angular momentum of the satellite is $$L = mvr$$, where $$m = \frac{M}{2}$$, so

$$ L = \frac{M}{2} \times \sqrt{\frac{3GM}{4R}} \times \frac{4R}{3} $$

We simplify this expression step by step. First, multiply the constant factors:

$$ L = \frac{M}{2} \times \frac{4R}{3} \times \sqrt{\frac{3GM}{4R}} $$

$$ L = \frac{2MR}{3} \times \sqrt{\frac{3GM}{4R}} $$

Since

$$ \sqrt{\frac{3GM}{4R}} = \frac{\sqrt{3GM}}{2\sqrt{R}}, $$

substituting back gives

$$ L = \frac{2MR}{3} \times \frac{\sqrt{3GM}}{2\sqrt{R}} = \frac{MR}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} = \frac{M\sqrt{R} \times \sqrt{R}}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} $$

which simplifies to

$$ L = \frac{M\sqrt{R}}{3} \times \sqrt{3GM} = \frac{M}{3}\sqrt{3GMR}. $$

We now rewrite this result in the required form:

$$ L = \frac{M}{3}\sqrt{3GMR} = M \times \frac{\sqrt{3GMR}}{3} = M \times \sqrt{\frac{3GMR}{9}} = M\sqrt{\frac{GMR}{3}} $$

Here we used

$$\frac{\sqrt{3GMR}}{3} = \frac{\sqrt{3GMR}}{\sqrt{9}} = \sqrt{\frac{3GMR}{9}} = \sqrt{\frac{GMR}{3}}.$$

Comparing $$L = M\sqrt{\frac{GMR}{3}}$$ with $$L = M\sqrt{\frac{GMR}{x}}$$ shows

$$ x = 3 $$

Thus the value of $$x$$ is $$\boxed{3}$$.

We need to find $$\alpha$$ where the angular velocity is $$\sqrt{\frac{F}{\alpha M}}$$ for a tube of length 1 m filled with liquid of mass 2M, rotated about one end.

Set up the problem

A tube of length $$L = 1$$ m is filled with liquid of mass $$2M$$. It rotates about one end with angular velocity $$\omega$$. We need the force at the other end (the far end).

Find the centripetal force

Linear mass density: $$\lambda = \frac{2M}{L} = 2M$$ (since $$L = 1$$).

Consider an element at distance $$r$$ from the axis: $$dm = \lambda \, dr = 2M \, dr$$.

The centrifugal force on this element: $$dF = dm \cdot \omega^2 r = 2M\omega^2 r \, dr$$.

The force at the far end ($$r = L = 1$$) is due to the liquid beyond some point — but since it's the end, the force is exerted by the liquid on the closed end.

Actually, the force at the far end is the integral of centripetal acceleration of all the liquid:

$$F = \int_0^L \lambda \omega^2 r \, dr = 2M\omega^2 \int_0^1 r \, dr = 2M\omega^2 \times \frac{1}{2} = M\omega^2$$

$$\omega^2 = \frac{F}{M} \Rightarrow \omega = \sqrt{\frac{F}{M}}$$

So $$\alpha = 1$$.

Conclusion

$$\alpha = 1$$.

Therefore, the answer is 1.

Two iron solid discs have radii $$R_1$$ and $$R_2$$ with $$R_2 = 2R_1$$. We first determine the ratio $$I_1/I_2$$.

Next, we recall that the moment of inertia of a solid disc is given by
$$I = \frac{1}{2}MR^2$$

Since both discs are made of iron (same density $$\rho$$) and have negligible thickness $$t$$, their mass can be expressed as
$$M = \rho \cdot \pi R^2 \cdot t$$ so $$M \propto R^2$$.

Substituting into the inertia formula gives
$$I = \frac{1}{2}MR^2 = \frac{1}{2}(\rho \pi R^2 t)R^2 = \frac{\rho \pi t}{2} R^4$$ and hence $$I \propto R^4$$.

It follows that
$$\frac{I_1}{I_2} = \frac{R_1^4}{R_2^4} = \frac{R_1^4}{(2R_1)^4} = \frac{R_1^4}{16R_1^4} = \frac{1}{16}\,. $$

Therefore, $$I_1/I_2 = 1/x$$ where $$x = \boxed{16}$$.

To compare the moments of inertia of a disc, ring, and sphere (all of mass $$M$$ and radius $$R$$) rotating about a diameter, we first recall the standard formulas for each.

For a ring about its diameter we have $$I_{\text{ring}} = \frac{1}{2}MR^2$$, for a disc about its diameter $$I_{\text{disc}} = \frac{1}{4}MR^2$$, and for a sphere about its diameter $$I_{\text{sphere}} = \frac{2}{5}MR^2$$.

Next, we check the condition given in the problem by computing the ratio

$$ \frac{I_{\text{disc}}}{I_{\text{ring}}} = \frac{MR^2/4}{MR^2/2} = \frac{1}{2} $$.

Since the problem states that $$I_{\text{disc}} = 2.5 \times I_{\text{ring}}$$, this standard result does not hold unless the disc and ring have different radii. We are told, however, that the sphere has the same radius as the disc.

Let us denote the ring’s radius by $$r$$ and the disc’s (and sphere’s) radius by $$R$$. Imposing $$I_{\text{disc}} = 2.5 \times I_{\text{ring}}$$ gives

$$ \frac{1}{4}MR^2 = 2.5 \times \frac{1}{2}Mr^2 = \frac{5}{4}Mr^2 $$.

From this we find

$$ R^2 = 5r^2 $$.

Since the sphere has the same radius as the disc, its moment of inertia is

$$ I_{\text{sphere}} = \frac{2}{5}MR^2 = \frac{2}{5}M(5r^2) = 2Mr^2 $$.

Consequently, the ratio of the sphere’s moment of inertia to the ring’s is

$$ n = \frac{I_{\text{sphere}}}{I_{\text{ring}}} = \frac{2Mr^2}{\frac{1}{2}Mr^2} = 4 $$.

Hence the answer is $$n = 4$$.

We are given the momentum vector at time t as $$\vec{p}(t)=\hat{i}\cos(kt)-\hat{j}\sin(kt)$$, and we wish to find the angle between the force $$\vec{F}$$ and the momentum $$\vec{p}$$.

Since Newton’s second law states that the force is the time derivative of the momentum, differentiating $$\vec{p}(t)$$ with respect to t yields $$\vec{F}=\frac{d\vec{p}}{dt}=-k\hat{i}\sin(kt)-k\hat{j}\cos(kt)$$.

Next we compute the dot product of $$\vec{F}$$ and $$\vec{p}$$, which gives

$$\vec{F}\cdot\vec{p}=(-k\sin kt)(\cos kt)+(-k\cos kt)(-\sin kt)=-k\sin kt\cos kt+k\cos kt\sin kt=0$$.

Because the dot product is zero, the vectors are perpendicular and hence the angle between them is $$\frac{\pi}{2}$$.

Therefore, the correct answer is Option (3): $$\frac{\pi}{2}$$.

For an electron orbiting a nucleus with charge $$Ze$$:

The potential energy of the electron is given by $$U = -\frac{kZe^2}{r}$$.

Using the centripetal force condition $$\frac{mv^2}{r} = \frac{kZe^2}{r^2}$$, we find the kinetic energy as

$$ KE = \frac{1}{2}mv^2 = \frac{kZe^2}{2r} = -\frac{U}{2} $$

Hence the total energy is

$$ E = KE + U = -\frac{U}{2} + U = \frac{U}{2} $$

It follows that

$$ 2E = U $$

The correct answer is Option (2): \boxed{2E = U}.

given:
radius of circle = 2 km
start point = P (top of circle)
end point = S (left side of circle)

step 1: what is displacement?

displacement = shortest straight-line distance between start and end
NOT the curved path along the circle

so we just need straight line PS

step 2: visualize positions

center = O

P is directly above O → distance 2
S is directly left of O → distance 2

so OP = 2 km (vertical)
OS = 2 km (horizontal)

and OP ⟂ OS (90° angle)

step 3: triangle OPS

you now have a right triangle:

OP = 2
OS = 2
PS = hypotenuse

apply Pythagoras:

PS² = OP² + OS²
PS² = 2² + 2²
PS² = 4 + 4 = 8

PS = √8 km

step 4: final understanding

even though cyclist moved along a curved path, displacement ignores path
it only depends on start and end points

so straight-line distance = √8 km

We need to find $$\alpha$$ where the time for the first revolution is $$\frac{1}{\alpha}(1 - e^{-2\pi})$$ s.

Considering radius $$R = 0.5$$ m and initial speed $$v_0 = 4$$ m/s, and using $$a_t = a_n$$, we have $$\frac{dv}{dt} = \frac{v^2}{R}\,. $$

Separating variables gives $$\frac{dv}{v^2} = \frac{dt}{R}\,,$$ and integrating yields $$-\frac{1}{v} = \frac{t}{R} + C\,. $$ At $$t = 0$$, $$C = -\frac{1}{v_0} = -\frac{1}{4}$$, so $$v = \frac{4}{1 - 8t}\,. $$

The distance for one revolution is $$s = 2\pi R = \pi$$ and also $$s = \int_0^T v\,dt = \int_0^T \frac{4}{1-8t}\,dt = -\frac{1}{2}\ln(1-8T)\,. $$ Setting this equal to $$\pi$$ gives $$-\frac{1}{2}\ln(1-8T) = \pi\,, $$ so $$1-8T = e^{-2\pi}$$ and $$T = \frac{1 - e^{-2\pi}}{8} = \frac{1}{8}(1 - e^{-2\pi})\,. $$

Comparing this result with the given expression $$\frac{1}{\alpha}(1-e^{-2\pi})$$ shows that $$\alpha = 8\,. $$

Option: 8

We need to find the angular momentum of a 5 kg body moving with speed $$3\sqrt{2}$$ m/s along the line $$y = x + 4$$ about the origin.

The formula for angular momentum is $$L = mvd$$, where $$d$$ is the perpendicular distance from the point (origin) to the line of motion.

Rewriting the line $$y = x + 4$$ in standard form gives $$x - y + 4 = 0$$. Hence the perpendicular distance from $$(0, 0)$$ to this line is $$d = \frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$ m.

Substituting into the angular momentum formula yields $$L = mvd = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 5 \times 3 \times 2 \times (\sqrt{2} \times \sqrt{2}) = 5 \times 6 \times 2 = 60$$ kg m$$^2$$/s.

The answer is 60 kg m$$^2$$ s$$^{-1}$$.

For a body rolling down an inclined plane without slipping, the linear acceleration ($$a$$) of the COM is given by:

$$a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$$

For a solid cylinder, the moment of inertia about its central axis is $$I = \frac{1}{2}MR^2$$.

$$a = \frac{g \sin 60^\circ}{1 + \frac{1}{2}} = \frac{g \sin 60^\circ}{3/2} = \frac{2}{3} g \sin 60^\circ$$

$$a = \frac{2}{3} \times 10 \times \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{3} = \frac{10}{\sqrt{3}} \text{ m s}^{-2}$$

$$x = 10$$

The given central force is $$F(r)=-kr$$, so the corresponding potential energy is $$V(r)=\dfrac{1}{2}kr^{2}$$.

For a stable circular orbit of radius $$r$$ and speed $$v$$, the required centripetal force is provided entirely by the central force:

$$\frac{mv^{2}}{r}=kr \qquad -(1)$$

From Bohr’s quantisation rule, the magnitude of the angular momentum is

$$L = n\hbar \qquad -(2)$$

But for circular motion, angular momentum is also

$$L = mvr \qquad -(3)$$

We now combine equations $$(1)$$ and $$(3)$$ to obtain relations among $$r$$, $$v$$, and $$L$$.

Step 1: Eliminate $$v$$ between $$(1)$$ and $$(3)$$
From $$(3)$$, $$v=\dfrac{L}{mr}$$. Substitute this in $$(1)$$:

$$\frac{m}{r}\left(\frac{L}{mr}\right)^{2}=kr \;\;\Longrightarrow\;\; \frac{L^{2}}{m r^{3}}=kr$$

$$\Longrightarrow\; L^{2}=km\,r^{4} \qquad -(4)$$

Step 2: Express $$r^{2}$$ in terms of $$L$$
From $$(4)$$,

$$r^{4}=\frac{L^{2}}{km}\;\;\Longrightarrow\;\; r^{2}=\frac{L}{\sqrt{km}} \qquad -(5)$$

Using $$(2)$$ ( $$L=n\hbar$$ ) in $$(5)$$ gives

$$r^{2}=n\hbar\sqrt{\frac{1}{mk}}$$

This is exactly Option A.

Step 3: Obtain $$v^{2}$$
Return to equation $$(1)$$: $$v^{2}=\frac{k}{m}r^{2}$$. Insert the value of $$r^{2}$$ from $$(5)$$:

$$v^{2}=\frac{k}{m}\left(n\hbar\sqrt{\frac{1}{mk}}\right) =n\hbar\sqrt{\frac{k}{m^{3}}}$$

This is Option B.

Step 4: Evaluate $$\dfrac{L}{mr^{2}}$$
Using $$(2)$$ and $$(5)$$,

$$\frac{L}{mr^{2}}=\frac{n\hbar}{m\left(n\hbar/\sqrt{km}\right)} =\frac{\sqrt{km}}{m} =\sqrt{\frac{k}{m}}$$

This matches Option C.

Step 5: Compute the total mechanical energy $$E$$
Kinetic energy: $$T=\dfrac{1}{2}mv^{2}$$. Using $$(1)$$, $$v^{2}=\dfrac{k}{m}r^{2}$$ gives

$$T=\frac{1}{2}m\left(\frac{k}{m}r^{2}\right)=\frac{1}{2}kr^{2}$$

Potential energy: $$V=\dfrac{1}{2}kr^{2}$$.

Total energy: $$E=T+V=kr^{2}$$.

Substituting $$r^{2}=n\hbar/\sqrt{km}$$:

$$E=k\left(\frac{n\hbar}{\sqrt{km}}\right)=n\hbar\sqrt{\frac{k}{m}}$$

The factor is $$1$$, not $$\tfrac{1}{2}$$, so Option D is incorrect.

Therefore the correct statements are:
Option A, Option B, and Option C.

A uniform rod AB of mass 2 kg and length 30 cm (0.3 m) is initially at rest on a smooth horizontal surface. When an impulse of 0.2 N s is applied horizontally at end B, the rod acquires both linear and angular motion, and we seek the time required for it to rotate by 90° (π/2 radians), expressed in the form π/x seconds.

The impulse-momentum theorem gives the change in linear momentum of the rod’s center of mass according to $$J = M \cdot v_{\text{cm}}$$, where $$J = 0.2$$ N s and $$M = 2$$ kg. Substituting the values, $$0.2 = 2 \cdot v_{\text{cm}}$$, so $$v_{\text{cm}} = \frac{0.2}{2} = 0.1\ \text{m/s}$$. Thus the center of mass moves at 0.1 m/s in the direction of the impulse.

The same impulse also imparts angular momentum about the center of mass. The perpendicular distance from the center of mass to B is $$L/2 = 0.15$$ m, so the angular impulse is $$J \cdot \frac{L}{2} = 0.2 \cdot 0.15 = 0.03\ \text{N s m}$$. The moment of inertia of a uniform rod about its center of mass is $$I = \frac{M L^2}{12} = \frac{2 \cdot (0.3)^2}{12} = \frac{2 \cdot 0.09}{12} = \frac{0.18}{12} = 0.015\ \text{kg m}^2\,. $$ Equating angular impulse to the change in angular momentum via $$J \cdot \frac{L}{2} = I \cdot \omega$$ gives $$0.03 = 0.015 \cdot \omega$$ and hence $$\omega = \frac{0.03}{0.015} = 2\ \text{rad/s}\,.$$

Since no external torques act after the impulse, the rotation proceeds at constant angular velocity $$\omega = 2\ \text{rad/s}$$. The angle rotated satisfies $$\theta = \omega \cdot t$$, so $$\frac{\pi}{2} = 2 \cdot t$$ and thus $$t = \frac{\pi}{4}\ \text{seconds}\,, $$ giving $$x = 4$$.

The translation of the center of mass does not affect the rotation angle, as the angular motion is independent; hence the time taken for the rod to rotate by 90° is $$\pi/4$$ seconds.

Four particles each of mass 1 kg at corners of a square with side 2 m. The axis passes through one vertex, perpendicular to the plane.

Taking the vertex through which the axis passes as origin, the distances of the four particles from the axis are:

- Particle at the vertex: $$r_1 = 0$$

- Two adjacent particles: $$r_2 = r_3 = 2$$ m (side length)

- Diagonally opposite particle: $$r_4 = 2\sqrt{2}$$ m (diagonal)

$$I = m(r_1^2 + r_2^2 + r_3^2 + r_4^2) = 1(0 + 4 + 4 + 8) = 16 \text{ kg m}^2$$

The answer is $$\boxed{16}$$ kg m$$^2$$.

Second hand: in 30 min = 30 revolutions. Distance = 30×2π×0.75 = 45π m.

Minute hand: in 30 min = 0.5 revolution. Distance = 0.5×2π×0.60 = 0.6π m.

x = 45π - 0.6π = 44.4π = 44.4×3.14 = 139.4 m.

The correct answer is Option (3): 139.4.

120 rev in 3 min = 40 rev/min = 2/3 rev/s. ω=2π·2/3=4π/3 rad/s.

a=ω²r=(4π/3)²·9=16π²/9·9=16π² m/s².

The answer is Option (4): 16π² m/s².

In uniform circular motion, speed $$v$$ is given by dividing the circumference by the period:
$$v = \frac{2\pi R}{T}$$ $$-(1)$$

The maximum height $$H$$ reached by a projectile with initial speed $$u$$ at angle $$\theta$$ is given by the formula:
Formula: $$H = \frac{u^2 \sin^2 \theta}{2g}$$.

Setting $$u = v$$, we have:
$$H = \frac{v^2 \sin^2 \theta}{2g}$$ $$-(2)$$.

Given that this height equals $$4R$$, substitute (1) into (2):
$$4R = \frac{\Bigl(\frac{2\pi R}{T}\Bigr)^2 \sin^2 \theta}{2g}$$.

Simplify the square:
$$\Bigl(\frac{2\pi R}{T}\Bigr)^2 = \frac{4\pi^2 R^2}{T^2}$$. Thus
$$4R = \frac{4\pi^2 R^2 \sin^2 \theta}{2g T^2} = \frac{2\pi^2 R^2 \sin^2 \theta}{g T^2}$$.

Divide both sides by $$R$$ and rearrange:
$$4 = \frac{2\pi^2 R \sin^2 \theta}{g T^2} \quad\Longrightarrow\quad \sin^2 \theta = \frac{4gT^2}{2\pi^2 R} = \frac{2g T^2}{\pi^2 R}$$ $$-(3)$$.

Taking the positive square root (angle is acute):
$$\sin \theta = \sqrt{\frac{2g T^2}{\pi^2 R}}\quad\Longrightarrow\quad \theta = \sin^{-1}\!\sqrt{\frac{2g T^2}{\pi^2 R}}\!.$$

Therefore, the required angle of projection is
$$\theta = \sin^{-1}\Bigl(\frac{2gT^2}{\pi^2R}\Bigr)^{1/2},$$ which corresponds to Option A.

The mass of the stone is $$m = 900\text{ g} = 0.9\text{ kg}$$ and the radius of the vertical circle is $$r = 1\text{ m}$$.

First convert the speed of revolution from revolutions per minute (rpm) to radians per second (rad s−1).
A body that makes $$N$$ revolutions per minute has an angular speed
$$\omega = 2\pi N \text{ rad per minute}$$.
Given $$N = 10$$ rpm:

$$\omega = 2\pi \times 10 = 20\pi \text{ rad min}^{-1}$$.
To express this in seconds, divide by $$60$$:
$$\omega = \frac{20\pi}{60} = \frac{\pi}{3} \text{ rad s}^{-1}$$.

The linear speed $$v$$ of the stone is related to angular speed by $$v = \omega r$$.
Hence
$$v = \frac{\pi}{3} \times 1 = \frac{\pi}{3} \text{ m s}^{-1}$$.

The square of the speed is therefore
$$v^{2} = \left(\frac{\pi}{3}\right)^{2} = \frac{\pi^{2}}{9}$$.
Given $$\pi^{2} = 9.8$$, we get
$$v^{2} = \frac{9.8}{9} = 1.0889 \text{ m}^{2}\text{ s}^{-2} \approx 1.09 \text{ m}^{2}\text{ s}^{-2}$$.

At the lowest point of the vertical circle, the tension $$T$$ in the string must provide the centripetal force towards the centre in addition to balancing the weight acting downward. Therefore

$$T - mg = \frac{mv^{2}}{r}$$.
Rearranging gives
$$T = mg + \frac{mv^{2}}{r}$$.

Substituting the values:
$$mg = 0.9 \times 9.8 = 8.82 \text{ N}$$,
$$\frac{mv^{2}}{r} = 0.9 \times 1.09 = 0.98 \text{ N}$$.

Hence
$$T = 8.82 + 0.98 = 9.80 \text{ N}$$.

Thus, the tension in the string at the lowest point is approximately $$9.8 \text{ N}$$.

Option B is correct.

For banking of tracks, the formula for the height of the outer rail is:

$$\tan\theta = \frac{v^2}{Rg}$$

For small angles, $$\tan\theta \approx \sin\theta \approx \frac{h}{l}$$, where $$h$$ is the height difference and $$l$$ is the distance between rails.

$$\frac{h}{l} = \frac{v^2}{Rg}$$

$$h = \frac{v^2 l}{Rg} = \frac{(12)^2 \times 1.5}{400 \times 10} = \frac{144 \times 1.5}{4000} = \frac{216}{4000} = 0.054 \text{ m} = 5.4 \text{ cm}$$

The answer is $$5.4$$ cm, which corresponds to Option (2).

We need to find the ratio of velocities of two particles with the same mass, given that their radii of curvature are in the ratio 3:4 and the centripetal force is constant.

The centripetal force on a particle of mass $$m$$ moving with speed $$v$$ along a circular path of radius $$r$$ is:

$$F = \frac{mv^2}{r}$$

Since both particles have the same mass and the same centripetal force:

$$\frac{mv_1^2}{r_1} = \frac{mv_2^2}{r_2}$$

The mass cancels:

$$\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2}$$

$$\frac{v_1^2}{v_2^2} = \frac{r_1}{r_2}$$

Taking the square root:

$$\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$

Therefore, $$v_1 : v_2 = \sqrt{3} : 2$$.

The correct answer is Option (1): $$\sqrt{3} : 2$$.

At the extreme position, the velocity is zero. The acceleration is purely tangential (gravity component): $$a_{ext} = g\sin\theta$$.

At the lowest position, the acceleration is centripetal: $$a_{low} = \frac{v^2}{L}$$.

Using energy conservation from extreme to lowest position:

$$\frac{1}{2}mv^2 = mgL(1 - \cos\theta)$$

$$v^2 = 2gL(1 - \cos\theta)$$

So $$a_{low} = \frac{2gL(1 - \cos\theta)}{L} = 2g(1 - \cos\theta)$$.

Setting $$a_{ext} = a_{low}$$:

$$g\sin\theta = 2g(1 - \cos\theta)$$

$$\sin\theta = 2(1 - \cos\theta)$$

Using half-angle: $$2\sin(\theta/2)\cos(\theta/2) = 2 \cdot 2\sin^2(\theta/2)$$

$$\cos(\theta/2) = 2\sin(\theta/2)$$

$$\tan(\theta/2) = \frac{1}{2}$$

$$\theta = 2\tan^{-1}\left(\frac{1}{2}\right)$$

The answer corresponds to Option (2).

Start from force resolution on a banked road at maximum speed. At this condition, friction acts down the plane (towards the center).

Forces:

• Normal reaction N (perpendicular to surface)
• Friction f = μN (down the slope)
• Weight mg (vertical)

Resolve along horizontal (towards center) and vertical.

Horizontal:
$$N\sinθ+μN\cosθ=mv^2/r$$

Vertical:
$$N\cosθ−μN\sinθ=mg$$

Now divide the horizontal equation by the vertical equation:

$$(N\sinθ+μN\cosθ)/(N\cosθ−μN\sinθ)=(mv^2/r)/(mg)$$

Cancel N and m:

$$(\sinθ+μ\cosθ)/(\cosθ−μ\sinθ)=v^2/(rg)$$

Now divide numerator and denominator by cosθ:

$$(\tanθ+μ)/(1−μ\tanθ)=v^2/(rg)$$

So,

$$\frac{v^2}{r}=g\cdot\frac{\tan\theta+\mu}{1-\mu\tan\theta}$$

Now substitute the given values

$$v^2=300\times\ 10\times\ \frac{\frac{1}{\sqrt{\ 3}}+0.2}{1-\frac{1}{\sqrt{\ 3}}\times\ 0.2}$$

$$v^2=300\times\ 10\times\ \frac{1.3464}{1.532}$$

$$v^2=3000\times0.8788$$

$$v^2=2636.4$$

$$v\approx\ 51.4\ \frac{m}{s}$$

Condition at the Topmost Point ($$B$$)

For the bob to just complete the circle, the tension in the string at point B must be zero. The centripetal force at this point is provided entirely by the gravitational force:

$$mg = \frac{mv_B^2}{L}$$

$$v_B^2 = gL$$

Kinetic Energy at Point $$B$$

The kinetic energy at the highest point is: 

$$(K.E.)_B=\frac{1}{2}mv_B^2=\frac{1}{2}mgL$$

Conservation of Mechanical Energy

$$(K.E.)_A = (K.E.)_B + (P.E.)_B$$

$$(K.E.)_A = \frac{1}{2}mgL + mg(2L)$$

$$(K.E.)_A = \frac{5}{2}mgL$$

The ratio is $$5 : 1$$.

A ball of mass 0.5 kg on a string of length 50 cm rotates in a horizontal circle. Maximum tension = 400 N. Find maximum angular velocity.

For circular motion, the string tension provides the centripetal force needed to keep the ball moving in a circle:

$$ T = m\omega^2 r $$

where $$T$$ is tension, $$m$$ is mass, $$\omega$$ is angular velocity, and $$r$$ is radius of the circular path.

Rearranging for $$\omega$$ gives:

$$ \omega^2 = \frac{T}{mr} $$

Substituting the values $$T = 400\text{ N}$$, $$m = 0.5\text{ kg}$$, and $$r = 50\text{ cm} = 0.5\text{ m}$$ yields:

$$ \omega^2 = \frac{400}{0.5 \times 0.5} = \frac{400}{0.25} = 1600 $$

$$ \omega = \sqrt{1600} = 40\text{ rad/s} $$

The correct answer is Option B: 40 rad/s.

A coin is placed on a rotating disc at a distance $$r$$ from the center. The coefficient of friction between the coin and the disc is $$\mu$$. We need to find the maximum angular velocity so the coin does not slip.

For the coin to move in a circle on the disc, it needs a centripetal force directed toward the center. The only horizontal force acting on the coin is the static friction between the coin and the disc.

The centripetal force required for circular motion is $$F_{centripetal} = m\omega^2 r$$, where $$m$$ is the mass of the coin and $$\omega$$ is the angular velocity. The maximum static friction force available is $$f_{max} = \mu N = \mu mg$$, where $$N = mg$$ is the normal force (since the disc is horizontal).

For the coin not to slip, the required centripetal force must not exceed the maximum friction, which gives $$m\omega^2 r \leq \mu mg$$. At the maximum angular velocity, the two forces are equal: $$m\omega_{max}^2 r = \mu mg$$.

The mass $$m$$ cancels from both sides, yielding $$\omega_{max}^2 = \dfrac{\mu g}{r}$$ and hence $$\omega_{max} = \sqrt{\dfrac{\mu g}{r}}$$.

The correct answer is Option (3): $$\sqrt{\dfrac{\mu g}{r}}$$.

A planet revolves around a star in a circular orbit of radius $$R$$ with period $$T$$, and the gravitational force is proportional to $$R^{-3/2}$$.

For a circular orbit, the gravitational force must provide the centripetal force, so we have $$F = \frac{mv^2}{R}$$.

Since the force varies as $$F \propto R^{-3/2}$$, it can be expressed as $$F = \frac{k}{R^{3/2}}$$ for some constant $$k$$.

Equating these expressions gives $$\frac{k}{R^{3/2}} = \frac{mv^2}{R}$$, which simplifies to $$v^2 = \frac{k}{m\sqrt{R}}$$ and hence $$v = \sqrt{\frac{k}{m}}\,R^{-1/4}$$.

The period $$T$$ is the orbital circumference divided by the speed, so $$T = \frac{2\pi R}{v} = 2\pi R \sqrt{\frac{m}{k}}\,R^{1/4} = 2\pi\sqrt{\frac{m}{k}}\,R^{5/4}$$.

It follows that $$T \propto R^{5/4}$$, and squaring this relationship yields $$T^2 \propto R^{5/2}$$.

The correct answer is $$T^2 \propto R^{5/2}$$ (Option A).

$$T_p=6$$h, $$M_p=M_E/4$$. For geostationary orbit: $$T²\propto r³/M$$.

$$r_p³=\frac{M_p}{M_E}\cdot\frac{T_p²}{T_E²}\cdot r_E³=\frac{1}{4}\cdot\frac{36}{576}\cdot r_E³=\frac{1}{4}\cdot\frac{1}{16}\cdot r_E³=\frac{r_E³}{64}$$.

$$r_p=r_E/4=4.2\times10^4/4=1.05\times10^4$$ km.

The answer is Option (2): $$1.05\times10^4$$ km.

Derive the correct formula for the height of a satellite above Earth's surface.

For a satellite in circular orbit at distance $$r$$ from Earth's center, the gravitational force provides centripetal force:

$$ \frac{GMm}{r^2} = \frac{mv^2}{r} = m\omega^2 r $$

where $$\omega = \frac{2\pi}{T}$$.

$$ \frac{GM}{r^2} = \omega^2 r = \frac{4\pi^2}{T^2}r $$

$$ GM = \frac{4\pi^2 r^3}{T^2} $$

Since $$g = \frac{GM}{R^2}$$, we have $$GM = gR^2$$. Substituting:

$$ gR^2 = \frac{4\pi^2 r^3}{T^2} $$

$$ r^3 = \frac{gR^2T^2}{4\pi^2} = \frac{T^2R^2g}{4\pi^2} $$

$$ r = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} $$

Since $$r = R + h$$:

$$ h = r - R = \left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R $$

The correct answer is Option B: $$\left(\frac{T^2R^2g}{4\pi^2}\right)^{1/3} - R$$.

The angular speed ($$\omega$$) of an orbiting body is defined by its time period ($$T$$): $$\omega = \frac{2\pi}{T}$$

The Moon completes its orbit around the Earth in $$T_m \approx 27.3$$ days, whereas the Earth takes $$T_e \approx 365.25$$ days to orbit the Sun. Because the Moon has a much shorter time period ($$T_m < T_e$$), its angular speed is significantly higher: $$\omega_{\text{moon}} > \omega_{\text{earth}}$$

The Assertion is true, and the Reason provides the factual basis (shorter time period) that mathematically explains why the Moon's angular speed is greater.

Correct Option: (B)

Two satellites A and B with orbital radii 4R and R. Speed of A = 3v. Find speed of B.

For a circular orbit, the orbital speed formula is $$v = \sqrt{\frac{GM}{r}}$$, so $$v \propto \frac{1}{\sqrt{r}}$$.

Using this relation, $$\frac{v_B}{v_A}=\sqrt{\frac{r_A}{r_B}}=\sqrt{\frac{4R}{R}}=2$$ and hence $$v_B=2v_A=2\times3v=6v$$.

The correct answer is Option (2): 6v.

For a planet of mass $$m$$ moving in a circular orbit of radius $$r$$ around the Sun (mass $$M_\odot$$), the necessary centripetal force is provided by gravitation:

$$\frac{G M_\odot m}{r^{2}} = \frac{m v^{2}}{r}$$

Solving for the orbital speed,

$$v = \sqrt{\frac{G M_\odot}{r}} \quad -(1)$$

Angular momentum about the Sun is

$$L = m v r$$

Substituting $$v$$ from $$(1)$$,

$$L = m r \sqrt{\frac{G M_\odot}{r}} = m \sqrt{G M_\odot\, r} \quad -(2)$$

Applying $$(2)$$ to planets $$A$$ and $$B$$:

$$L_A = m_1 \sqrt{G M_\odot\, r_1}$$
$$L_B = m_2 \sqrt{G M_\odot\, r_2}$$

Given $$L_B = 3L_A$$,

$$m_2 \sqrt{G M_\odot\, r_2} = 3 m_1 \sqrt{G M_\odot\, r_1}$$

Canceling the common factor $$\sqrt{G M_\odot}$$,

$$m_2 \sqrt{r_2} = 3 m_1 \sqrt{r_1} \quad -(3)$$

From $$(3)$$,

$$\frac{\sqrt{r_2}}{\sqrt{r_1}} = \frac{3 m_1}{m_2}$$
Squaring both sides: $$\frac{r_2}{r_1} = \frac{9 m_1^{2}}{m_2^{2}} \quad -(4)$$

Kepler’s third law for circular orbits gives the time period:

$$T = 2\pi \sqrt{\frac{r^{3}}{G M_\odot}} \propto r^{3/2} \quad -(5)$$

Therefore,

$$\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} \quad -(6)$$

Insert $$\frac{r_1}{r_2}$$ from $$(4)$$ into $$(6)$$:

$$\frac{T_A}{T_B} = \left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2}$$

Compute the exponent:

$$\left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2} = \left(\frac{m_2}{m_1}\right)^{3} \left(\frac{1}{9}\right)^{3/2}$$

Since $$9 = 3^{2}$$, we have $$9^{3/2} = 3^{3} = 27$$. Thus

$$\frac{T_A}{T_B} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$$

Hence, the correct ratio is $$\frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$$, which corresponds to Option B.

The projection of a particle moving in a circle on the x-axis gives simple harmonic motion. We have the general expression for the x-projection as:

$$x(t) = r\cos(\omega t + \phi_0)$$

where $$\phi_0$$ is the initial phase angle (the angle at $$t = 0$$).

Now, from the figure, at $$t = 0$$, the particle P is at an angle of $$\frac{\pi}{6}$$ (30°) above the positive x-axis. So the initial phase is $$\phi_0 = \frac{\pi}{6}$$, and we get:

$$x(t) = r\cos\left(\omega t + \frac{\pi}{6}\right)$$

Hence, the correct answer is Option 2.

We have a particle moving with constant speed $$v$$ in a circular path of radius $$r$$. When it turns by $$90°$$, we need to find the ratio of instantaneous velocity to average velocity.

The instantaneous speed at any point is $$v$$. For a $$90°$$ turn, the displacement is the chord subtending $$90°$$ at the centre. If the particle moves from one point to another $$90°$$ apart, the displacement is

$$|\vec{d}| = \sqrt{r^2 + r^2} = r\sqrt{2}$$

Now, the time taken for a quarter circle is

$$t = \frac{\text{arc length}}{v} = \frac{\frac{2\pi r}{4}}{v} = \frac{\pi r}{2v}$$

So the magnitude of average velocity becomes

$$v_{avg} = \frac{|\vec{d}|}{t} = \frac{r\sqrt{2}}{\frac{\pi r}{2v}} = \frac{2v\sqrt{2}}{\pi}$$

The ratio of instantaneous velocity to average velocity is then

$$\frac{v}{v_{avg}} = \frac{v}{\frac{2v\sqrt{2}}{\pi}} = \frac{\pi}{2\sqrt{2}}$$

This gives the ratio as $$\pi : 2\sqrt{2}$$. Comparing with $$\pi : x\sqrt{2}$$, we get $$x = 2$$. So, the answer is $$2$$.

The object is in uniform circular motion with a radius of $$r=+2$$ meters (since it happens to be at $$x=+2$$ at an instant when the center of the motion is origin). At $$x=+2$$ meters, the velocity $$\vec{v}$$ is $$-4\hat{j}$$ m/s. This indicates that the object is moving clockwise.

Thus, at $$x=-2$$ meters, the velocity vector will be equal in magnitude but opposite in direction, and will therefore be $$\vec{v'} = 4\hat{j}$$. 

Furthermore, the magnitude of acceleration at $$x=-2$$ will be given by $$a= \dfrac{v^2}{r} = \dfrac{4^2}{2} = \dfrac{16}{2} = 8 m/s^2$$

And the direction at $$x=-2$$ will be towards the origin (centripetal force), i.e. along $$+\hat{i}$$, thus, we have, $$\vec{a}=8\hat{i}$$.

Option B is the correct answer.

Find the maximum speed of a car on a horizontal curved road with radius 50 m and friction coefficient 0.34.

The friction provides the centripetal force:

$$\mu mg = \frac{mv^2}{R}$$

$$v_{max} = \sqrt{\mu R g}$$

$$v_{max} = \sqrt{0.34 \times 50 \times 10}$$

$$v_{max} = \sqrt{170}$$

$$\sqrt{170} \approx \sqrt{169} = 13$$ m/s (approximately)

The correct answer is Option C: $$13$$ m s$$^{-1}$$.

Given: mass $$m = 5\;\text{kg}$$, time period $$T = 3.14\;\text{s}$$, radius $$r = 2\;\text{m}$$.

The angular velocity is:

$$ \omega = \frac{2\pi}{T} = \frac{2\pi}{3.14} = 2\;\text{rad/s} $$

The centrifugal force (centripetal force in magnitude) on the child is:

$$ F = m\omega^2 r = 5 \times (2)^2 \times 2 = 5 \times 4 \times 2 = 40\;\text{N} $$

The centrifugal force on the child is 40 N.

For a coin on a rotating table, the coin just slips when the centripetal force equals the maximum static friction force:

$$f_s = m\omega^2 r$$

At the point of slipping:

$$\mu m g = m\omega^2 r$$

$$\mu g = \omega^2 r$$

We are given that The coin just slips at $$r_1 = 1$$ cm when angular velocity is $$\omega$$.

So: $$\mu g = \omega^2 \times 1$$ ... (i)

When the angular velocity is halved to $$\frac{\omega}{2}$$, let the coin just slip at distance $$r_2$$:

$$\mu g = \left(\frac{\omega}{2}\right)^2 \times r_2 = \frac{\omega^2}{4} \times r_2$$ ... (ii)

From equations (i) and (ii):

$$\omega^2 \times 1 = \frac{\omega^2}{4} \times r_2$$

$$r_2 = 4 \text{ cm}$$

The coin will just slip when placed at a distance of 4 cm from the centre.

A stone of mass 1 kg tied to a string of length 1 m is rotated in a horizontal plane. The breaking tension is 400 N. We need the maximum linear velocity.

We start by noting that

For circular motion in a horizontal plane, the tension in the string provides the centripetal force needed to keep the stone moving in a circle:

$$ T = \frac{mv^2}{r} $$

where $$m$$ is the mass, $$v$$ is the linear velocity, and $$r$$ is the radius of the circle (equal to the string length).

Next,

The string breaks when $$T > 400$$ N. For the maximum velocity without breaking:

$$ T_{\max} = \frac{mv_{\max}^2}{r} $$

$$ v_{\max} = \sqrt{\frac{T_{\max} \times r}{m}} = \sqrt{\frac{400 \times 1}{1}} = \sqrt{400} = 20\;\text{m/s} $$

The maximum linear velocity is 20 m s$$^{-1}$$.

The correct answer is Option 1: 20 m s$$^{-1}$$.

Given: mass $$m = 200 \text{ g} = 0.2 \text{ kg}$$, spring constant $$k = 12.5 \text{ N/m}$$, angular speed $$\omega = 5 \text{ rad/s}$$.

Let the natural length of the spring be $$L$$ and the extension be $$x$$. The body moves in a circular path of radius $$r = L + x$$.

The spring force provides the centripetal force:

$$kx = m\omega^2 r = m\omega^2(L + x)$$

$$kx = m\omega^2 L + m\omega^2 x$$

$$kx - m\omega^2 x = m\omega^2 L$$

$$x(k - m\omega^2) = m\omega^2 L$$

$$\frac{x}{L} = \frac{m\omega^2}{k - m\omega^2}$$

Substituting values:

$$m\omega^2 = 0.2 \times 25 = 5 \text{ N/m}$$

$$k - m\omega^2 = 12.5 - 5 = 7.5 \text{ N/m}$$

$$\frac{x}{L} = \frac{5}{7.5} = \frac{2}{3}$$

The ratio of extension to natural length is $$2 : 3$$.

The correct answer is Option 3.

Centripetal force is given by $$F = m\omega^2 r$$.

Given: $$m = 200$$ kg, $$r = 70$$ m, $$\omega = 0.2$$ rad/s.

$$F = 200 \times (0.2)^2 \times 70 = 200 \times 0.04 \times 70 = 560 \text{ N}$$

The centripetal force is $$\mathbf{560}$$ N.

We have a mass $$m = 100\;\text{g} = 0.1\;\text{kg}$$ attached to a spring of constant $$k = 7.5\;\text{N/m}$$ and natural length $$L_0 = 20\;\text{cm} = 0.2\;\text{m}$$, rotating with angular velocity $$\omega = 5\;\text{rad/s}$$.

Let the extension of the spring be $$x$$. The total radius of the circular path is $$r = L_0 + x$$. The spring force provides the centripetal force (since the restoring force of the spring is what keeps the mass in circular motion), so

$$kx = m\omega^2(L_0 + x)$$

Rearranging to isolate $$x$$:

$$kx - m\omega^2 x = m\omega^2 L_0$$

$$x(k - m\omega^2) = m\omega^2 L_0$$

Now substituting the values:

$$x(7.5 - 0.1 \times 25) = 0.1 \times 25 \times 0.2$$

$$x(7.5 - 2.5) = 0.5$$

$$x \times 5 = 0.5$$

$$x = 0.1\;\text{m}$$

So the tension in the spring is $$T = kx = 7.5 \times 0.1 = 0.75\;\text{N}$$. Hence, the correct answer is 0.75 N.

We need to find the magnitude of the displacement of a person who moves from point $$A$$ to point $$B$$ along a circular path, covering a distance of $$60\text{ m}$$.

1. Relate Distance to the Radius ($$R$$) of the Circular Path

From the problem , let's assume the angle subtended by the arc $$AB$$ at the center of the circular path is $$\theta = 135^\circ$$.

The arc length (distance traveled) is given by the formula:

$$\text{Distance} = R\theta \quad (\text{where } \theta \text{ is in radians})$$

Convert the angle from degrees to radians:

$$\theta = 135^\circ \times \frac{\pi}{180^\circ} = \frac{3\pi}{4}\text{ radians}$$

Substitute the given distance ($$60\text{ m}$$) to find the radius ($$R$$):

$$60 = R \times \frac{3\pi}{4}$$

$$R = \frac{60 \times 4}{3\pi} = \frac{80}{\pi}\text{ m}$$

2. Calculate the Magnitude of Displacement

The straight-line distance (displacement, $$d$$) between two points on a circle of radius $$R$$ separated by an angle $$\theta$$ can be calculated using the law of cosines or the standard shortcut formula:

$$d = 2R \sin\left(\frac{\theta}{2}\right)$$

Alternatively, using the given value of $$\cos 135^\circ = -0.7$$ directly via the vector subtraction formula:

$$d = \sqrt{R^2 + R^2 - 2R^2 \cos\theta} = \sqrt{2R^2(1 - \cos 135^\circ)}$$

Substitute $$\cos 135^\circ = -0.7$$ into the equation:

$$d = \sqrt{2R^2(1 - (-0.7))} = \sqrt{2R^2(1 + 0.7)} = \sqrt{2R^2(1.7)} = \sqrt{3.4R^2}$$

$$d = R\sqrt{3.4}$$

3. Substitute $$R$$ to Find the Final Value

Now, substitute $$R = \frac{80}{\pi}$$ and approximate $$\pi \approx 3.14$$ along with $$\sqrt{3.4} \approx 1.844$$:

$$d = \left(\frac{80}{3.14}\right) \times 1.844$$

$$d \approx 25.48 \times 1.844 \approx 47\text{ m}$$

Conclusion

The magnitude of the displacement is approximately 47 m, which corresponds to Option B.

We are given the equations of motion: $$x = 4\sin\left(\frac{\pi}{2} - \omega t\right)$$ and $$y = 4\sin(\omega t)$$. Using the identity $$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$$, this simplifies to $$x = 4\cos(\omega t)$$, so we have $$x = 4\cos(\omega t)$$ and $$y = 4\sin(\omega t)$$.

Squaring and adding both equations gives:

$$x^2 + y^2 = 16\cos^2(\omega t) + 16\sin^2(\omega t)$$

$$x^2 + y^2 = 16(\cos^2(\omega t) + \sin^2(\omega t))$$

$$x^2 + y^2 = 16$$

This is the equation of a circle with radius 4 m centered at the origin. Therefore, the path of the particle is circular. The correct answer is Option A.

For the first curved road the radius is $$R_1 = 75$$ m and the maximum speed is $$v_1 = 30$$ m/s, while for the second curved road the radius is $$R_2 = 48$$ m and the maximum speed $$v_2$$ is to be determined.

On a level curved road without skidding the centripetal force must equal the frictional force, so $$\frac{mv^2}{R} = \mu mg$$, which gives $$v_{\max} = \sqrt{\mu g R}$$.

From the first case we have $$v_1 = \sqrt{\mu g R_1}$$, and substituting $$30 = \sqrt{\mu \times g \times 75}$$ allows the determination of the coefficient of friction µ, but since µ and g remain the same for both curves, one can write the speeds ratio as $$\frac{v_2}{v_1} = \sqrt{\frac{R_2}{R_1}}$$.

Substituting $$v_1 = 30$$ m/s, $$R_1 = 75$$ m and $$R_2 = 48$$ m gives $$v_2 = v_1 \sqrt{\frac{R_2}{R_1}} = 30 \times \sqrt{\frac{48}{75}}$$, which simplifies to $$v_2 = 30 \times \sqrt{\frac{16}{25}} = 30 \times \frac{4}{5}$$, yielding $$v_2 = 24 \text{ m/s}$$.

Hence, the maximum allowed speed is 24 m/s.

We have a metal wire of length $$L = 0.5$$ m, cross-sectional area $$A = 10^{-4}$$ m$$^2$$, and breaking stress $$\sigma = 5 \times 10^8$$ N/m$$^2$$. A block of mass $$m = 10$$ kg is attached at one end and is rotating in a horizontal circle.

The maximum tension the wire can withstand before breaking is $$T_{\max} = \sigma \times A = 5 \times 10^8 \times 10^{-4} = 5 \times 10^4$$ N.

For circular motion, the tension provides the centripetal force: $$T = \frac{mv^2}{L}$$. At the maximum velocity, $$T_{\max} = \frac{mv_{\max}^2}{L}$$.

Solving for $$v_{\max}$$: $$v_{\max}^2 = \frac{T_{\max} \cdot L}{m} = \frac{5 \times 10^4 \times 0.5}{10} = \frac{2.5 \times 10^4}{10} = 2500$$

Hence $$v_{\max} = \sqrt{2500} = 50$$ m/s.

Hence, the correct answer is 50.

We have a rigid body of mass $$m = 2 \text{ kg}$$ attached to a string of cross-sectional area $$A = 4 \text{ mm}^2 = 4 \times 10^{-6} \text{ m}^2$$ and length $$0.5 \text{ m}$$. The body rotates in a vertical circle of radius $$r = 0.5 \text{ m}$$ and has a speed of $$v = 5 \text{ m/s}$$ at the bottom. We need to find the strain in the string at the bottom.

At the bottom of the circular path, the net upward force provides the centripetal acceleration. Applying Newton's second law along the radial direction (towards the centre, which is upward at the bottom), we get $$T - mg = \frac{mv^2}{r}$$.

So the tension is $$T = m\left(g + \frac{v^2}{r}\right) = 2\left(10 + \frac{25}{0.5}\right) = 2(10 + 50) = 2 \times 60 = 120 \text{ N}$$.

Now, from the definition of Young's modulus, $$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\text{Strain}}$$, so $$\text{Strain} = \frac{T}{AY}$$.

Substituting the values, $$\text{Strain} = \frac{120}{4 \times 10^{-6} \times 10^{11}} = \frac{120}{4 \times 10^{5}} = \frac{120}{400000} = 3 \times 10^{-4} = 30 \times 10^{-5}$$.

Hence, the strain produced is $$\textbf{30} \times 10^{-5}$$.

The radii of the circular orbits of the two satellites are given by $$R_1 = 3200$$ km for satellite $$S_1$$ and $$R_2 = 800$$ km for satellite $$S_2$$.

Since the orbital speed of a satellite in a circular orbit of radius $$R$$ is given by $$v = \sqrt{\frac{GM}{R}}$$, where $$M$$ is the mass of the planet, we can derive the ratio of their speeds.

Substituting the given radii into the speed formula yields $$\frac{v_1}{v_2} = \sqrt{\frac{R_2}{R_1}} = \sqrt{\frac{800}{3200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\,.$$

Comparing this result with the form $$\frac{1}{x}$$, we have $$\frac{v_1}{v_2} = \frac{1}{2}$$, which implies $$x = 2$$. Therefore, the value of $$x$$ is 2.

We are given: mass of stone $$m = 100$$ g $$= 0.1$$ kg, length of string (radius) $$r = 2$$ m, maximum tension $$T_{max} = 80$$ N.

Apply the condition for circular motion: for horizontal circular motion, the tension provides the centripetal force:

$$ T = m\omega^2 r $$

Find the maximum angular velocity: $$ \omega_{max} = \sqrt{\frac{T_{max}}{mr}} = \sqrt{\frac{80}{0.1 \times 2}} = \sqrt{\frac{80}{0.2}} = \sqrt{400} = 20 \text{ rad/s} $$

Convert angular velocity to revolutions per minute: the relationship between angular velocity and frequency in rev/min is:

$$ \omega = \frac{2\pi n}{60} $$

where $$n$$ is in rev/min.

$$ n = \frac{60\omega}{2\pi} = \frac{60 \times 20}{2\pi} = \frac{1200}{2\pi} = \frac{600}{\pi} \text{ rev/min} $$

Find $$K$$: given that the maximum speed is $$\frac{K}{\pi}$$ rev/min:

$$ \frac{K}{\pi} = \frac{600}{\pi} $$

$$ K = 600 $$

Therefore, the correct answer is Option C.

A stone of mass $$m is whirled in a vertical circle of radius r with uniform speed v$$. At any point in the circle, the net radial force provides the centripetal acceleration. Let $$\theta$$ be the angle measured from the top of the circle.

At the highest point of the circle, both weight $$mg and tension T$$ act toward the center (downward is toward the center): $$T_{\text{top}} + mg = \frac{mv^2}{r} leading to T_{\text{top}} = \frac{mv^2}{r} - mg$$.

At the lowest point of the circle, tension acts upward (toward the center) and weight acts downward (away from the center): $$T_{\text{bottom}} - mg = \frac{mv^2}{r} giving T_{\text{bottom}} = \frac{mv^2}{r} + mg$$.

At a horizontal position, the weight has no radial component, so $$T_{\text{horizontal}} = \frac{mv^2}{r}$$.

Comparing these tensions, we have $$T_{\text{top}} = \frac{mv^2}{r} - mg\quad(\text{smallest}),\quad T_{\text{horizontal}} = \frac{mv^2}{r}\quad(\text{intermediate}),\quad T_{\text{bottom}} = \frac{mv^2}{r} + mg\quad(\text{largest}). Since the speed is uniform, the tension is minimum at the highest position of the circular path, where gravity assists in providing the centripetal force.

The correct answer is Option C.

$$

For a particle in uniform circular motion of radius $$R$$, the position at angle $$\theta$$ from the positive x-axis is:

$$\vec{r} = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}$$

In uniform circular motion, the acceleration is centripetal (directed toward the center), i.e., opposite to the position vector from the center:

$$\vec{a} = -\frac{v^2}{R}\hat{r}$$

where $$\hat{r}$$ is the unit vector along the radial direction (outward from center).

The unit radial vector at angle $$\theta$$ is:

$$\hat{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}$$

Therefore:

$$\vec{a} = -\frac{v^2}{R}(\cos\theta\,\hat{i} + \sin\theta\,\hat{j})$$

$$\vec{a} = -\frac{v^2}{R}\cos\theta\,\hat{i} - \frac{v^2}{R}\sin\theta\,\hat{j}$$

Hence, the correct answer is Option C.

Here we know that Normal reaction is equal to the centrifugal force

Centrifugal Force , $$F_c=\frac{mv^2}{r}$$

so $$ N=F_c=\frac{mv^2}{r}$$

Therefore N is a function of $$v^2$$

so the curve is parabolic, the same as $$kx^2$$

So Option A is correct

A massless spring of spring constant $$k$$ and natural length $$l_0$$ has one end fixed. A mass $$m$$ at the other end rotates on a frictionless table with angular velocity $$\omega$$.

Set up the equation of motion.

Let the elongation of the spring be $$x$$. The total length of the spring is $$l_0 + x$$.

The mass moves in a circle of radius $$r = l_0 + x$$.

The centripetal force required is provided by the spring force:

$$kx = m\omega^2(l_0 + x)$$

Solve for $$x$$.

$$kx = m\omega^2 l_0 + m\omega^2 x$$

$$kx - m\omega^2 x = m\omega^2 l_0$$

$$x(k - m\omega^2) = m\omega^2 l_0$$

$$x = \frac{m\omega^2 l_0}{k - m\omega^2}$$

The elongation of the spring is $$\dfrac{m\omega^2 l_0}{k - m\omega^2}$$.

The correct answer is Option C.

We are given two planets A and B with equal mass, revolving in circular orbits with periods $$T_A = 2T_B$$. According to Kepler’s third law, $$T^2 \propto r^3$$, or more precisely $$T^2 = \frac{4\pi^2}{GM}r^3$$, so for the two planets we have $$\frac{T_A^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$.

Substituting $$T_A = 2T_B$$ into this relation gives $$\frac{(2T_B)^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$, which simplifies to $$\frac{4T_B^2}{T_B^2} = \frac{r_A^3}{r_B^3}$$, or $$4 = \frac{r_A^3}{r_B^3}$$. Therefore, $$r_A^3 = 4r_B^3$$. The correct answer is Option C.

We need to find the ratio of total mechanical energy of satellites A and B. The total mechanical energy of a satellite of mass $$m$$ in a circular orbit of radius $$r$$ around Earth (mass $$M$$) is given by $$E = -\frac{GMm}{2r}$$.

For satellite A, which has mass $$4m_0$$ and orbits at radius $$3r$$, the energy becomes $$E_A = -\frac{GM(4m_0)}{2(3r)} = -\frac{4GMm_0}{6r} = -\frac{2GMm_0}{3r}$$. Similarly, for satellite B with mass $$3m_0$$ at radius $$4r$$, one finds $$E_B = -\frac{GM(3m_0)}{2(4r)} = -\frac{3GMm_0}{8r}$$.

Taking the ratio yields $$\frac{E_A}{E_B} = \frac{-\frac{2GMm_0}{3r}}{-\frac{3GMm_0}{8r}} = \frac{2}{3} \times \frac{8}{3} = \frac{16}{9}$$, so the correct answer is Option B: $$16:9$$.

We use Kepler’s Third Law, which states that the square of the orbital period is proportional to the cube of the semi-major axis (orbital radius): $$T^2 \propto R^3$$.

For the original distance $$R$$ with period $$T_1 = 1$$ year: $$T_1^2 \propto R^3$$. For the new distance $$3R$$ with period $$T_2$$: $$T_2^2 \propto (3R)^3$$.

Taking the ratio gives $$\frac{T_2^2}{T_1^2} = \frac{(3R)^3}{R^3} = 27$$, so $$T_2^2 = 27 \times T_1^2 = 27$$ and hence $$T_2 = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \text{ years}$$.

The duration of the year when the distance becomes $$3R$$ is $$3\sqrt{3}$$ years. The correct answer is Option B.

The time period of a satellite is 7 hours. The radius is increased to 3 times its previous value. Find the new time period.

$$T^2 \propto r^3$$, so:

$$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left(\frac{r_2}{r_1}\right)^3 = 3^3 = 27$$

$$T_2 = T_1\sqrt{27} = 7 \times 3\sqrt{3} = 21\sqrt{3} \approx 21 \times 1.732 \approx 36.37 \text{ hours}$$

The approximate new time period is 36 hours.

Hence, the correct answer is Option A: 36 hours.

Analysis of the Assertion:

The assertion states that for four points $$A,\ B, \ C, \ D$$ on a semi-circular arc with center $$O$$ such that $$|\vec{AB}|=|\vec{BC}|=|\vec{CD}|$$:

$$\vec{AB}+\vec{AC}+\vec{AD} = 4\vec{AO} + \vec{OB} + \vec{OC}$$

We can rewrite the vectors on the left side in terms of the origin (center) $$O$$:

1. $$\vec{AB} = \vec{OB} - \vec{OA}$$
2. $$\vec{AC} = \vec{OC} - \vec{OA}$$
3. $$\vec{AD} = \vec{OD} - \vec{OA}$$

Summing these gives,

$$\vec{AB}+\vec{AC}+\vec{AD} = (\vec{OB}+\vec{OC}+\vec{OD}) - 3\vec{OA}$$

We find that, since $$\vec{OA}$$ and $$\vec{OD}$$ are opposite radii originating from $$O$$, $$\vec{OD} = -\vec{OA} = \vec{AO}$$, since $$\vec{OA} = -\vec{AO}$$ (same length but opposite direction), we therefore get,

$$\vec{AB}+\vec{AC}+\vec{AD} = (\vec{OB}+\vec{OC}+\vec{AO}) + 3\vec{AO}$$

or,

$$\vec{AB}+\vec{AC}+\vec{AD} = 4\vec{AO} + \vec{OB} + \vec{OC}$$

Assertion A is true.

Analysis of the Reason:

By the polygon law of vector addition, $$\vec{AB}+\vec{BC}+\vec{CD} = \vec{AD}$$, and we also see that $$\vec{AD} = 2\vec{AO}$$, since $$\vec{AD}$$ is the diameter, and $$\vec{AO}$$ is the radius.

Thus, the reason R is correct.

But the polygon law of vector addition has not been useful in examining the assertion A while we derived equality in assertion A. Thus, although both the assertion A and the reason R are correct, R is not the correct explanation of A.

We are given a block of mass $$m = 200 \text{ g} = 0.2 \text{ kg}$$ moving with uniform speed in a horizontal circular groove of radius $$r = 20 \text{ cm} = 0.2 \text{ m}$$, completing one round in $$T = 40 \text{ s}$$.

Since the block moves in a horizontal circular path, the normal force from the side walls provides the centripetal force. The angular velocity is $$\omega = \frac{2\pi}{T} = \frac{2\pi}{40} = \frac{\pi}{20} \text{ rad/s}$$.

The centripetal force (which equals the normal force) is $$N = m\omega^2 r = 0.2 \times \left(\frac{\pi}{20}\right)^2 \times 0.2$$.

Computing step by step: $$\omega^2 = \frac{\pi^2}{400}$$, so $$N = 0.2 \times \frac{\pi^2}{400} \times 0.2 = \frac{0.04 \pi^2}{400} = \frac{\pi^2}{10000}$$.

Using $$\pi^2 \approx 9.8596$$, we get $$N = \frac{9.8596}{10000} = 9.859 \times 10^{-4} \text{ N}$$.

We need to determine the magnitude of the downward negative lift $$F_L$$ acting on a racing car moving along a flat circular track.

1. Identify the Forces Acting on the Car

From the problem statement on the page, the car is moving in a horizontal circle of radius $$R$$ with a speed $$v$$. Let's analyze the forces acting in both vertical and horizontal directions:

• Vertical Direction:
  The forces keeping the car vertically balanced are:
  • Downward gravitational force: $$mg$$
  • Downward negative lift force (aerodynamic downforce): $$F_L$$
  • Upward normal reaction force from the track: $$N$$
  Since there is no vertical acceleration, the net vertical force is zero:

  $$N = mg + F_L \quad \text{--- (Equation 1)}$$

• Horizontal Direction:
  The centripetal force required to keep the car moving in a circle of radius $$R$$ at speed $$v$$ is provided entirely by the static friction ($$f_s$$) between the tires and the track:

  $$f_s = \frac{m v^2}{R}$$

2. Relate Friction to the Normal Force

To safely navigate the turn at speed $$v$$, the required centripetal friction must be provided by the maximum limiting static friction available ($$f_s = \mu_s N$$):

$$\mu_s N = \frac{m v^2}{R}$$

Solving for the normal force $$N$$ gives:

$$N = \frac{m v^2}{\mu_s R} \quad \text{--- (Equation 2)}$$

3. Solve for the Negative Lift ($$F_L$$)

Now, equate Equation 1 and Equation 2 to eliminate $$N$$:

$$mg + F_L = \frac{m v^2}{\mu_s R}$$

Isolate the negative lift force $$F_L$$:

$$F_L = \frac{m v^2}{\mu_s R} - mg$$

Factoring out the mass $$m$$ from both terms yields the final expression:

$$F_L = m \left( \frac{v^2}{\mu_s R} - g \right)$$

Conclusion

The magnitude of the negative lift acting downwards on the car is $$m \left( \frac{v^2}{\mu_s R} - g \right)$$, which corresponds to Option B.

At the equator, bodies start floating when the centripetal acceleration equals the gravitational acceleration, i.e., $$\omega^2 R = g$$. This gives $$\omega = \sqrt{\frac{g}{R}}$$.

The duration of the day is $$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g}}$$.

Substituting the given values: $$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}} = 6.28 \times \sqrt{6.4 \times 10^5} = 6.28 \times 800 = 5024$$ seconds.

Converting to minutes: $$T = \frac{5024}{60} \approx 84$$ minutes.

Required velocity of the bob immediately after collision:

$$v_b = \sqrt{5gL} = \sqrt{5 \times 10 \times 0.5} = 5\text{ m s}^{-1}$$

Using conservation of linear momentum: $$m v = m v' + M v_b$$

$$0.01v = 0.01(-100) + 1(5)$$

$$0.01v = -1 + 5 \implies 0.01v = 4 \implies v = 400\text{ m s}^{-1}$$

For a bob of mass $$m$$ tied to a string of length $$L = 1$$ m describing a vertical circle, the tension at any point depends on the speed and the component of gravity along the string.

At the lowest point, the tension is maximum: $$T_{max} = \frac{mv_{bot}^2}{L} + mg$$.

At the highest point, the tension is minimum: $$T_{min} = \frac{mv_{top}^2}{L} - mg$$.

Using energy conservation between the top and bottom of the circle (height difference $$= 2L$$):

$$\frac{1}{2}mv_{bot}^2 = \frac{1}{2}mv_{top}^2 + mg(2L)$$

$$v_{bot}^2 = v_{top}^2 + 4gL$$

Given $$\frac{T_{max}}{T_{min}} = 5$$:

$$\frac{\frac{mv_{bot}^2}{L} + mg}{\frac{mv_{top}^2}{L} - mg} = 5$$

$$\frac{v_{bot}^2 + gL}{v_{top}^2 - gL} = 5$$

Substituting $$v_{bot}^2 = v_{top}^2 + 4gL$$:

$$\frac{v_{top}^2 + 4gL + gL}{v_{top}^2 - gL} = 5$$

$$v_{top}^2 + 5gL = 5v_{top}^2 - 5gL$$

$$10gL = 4v_{top}^2$$

$$v_{top}^2 = \frac{10 \times 10 \times 1}{4} = 25$$

$$v_{top} = 5$$ m s$$^{-1}$$

The velocity of the bob at the highest position is $$5$$ m s$$^{-1}$$.

The two satellites move in uniform circular motion round the same planet and both turn anticlockwise. Let us name the nearer satellite as 1 and the farther satellite as 2. We are given

$$T_1 = 1 \text{ h}, \qquad T_2 = 8 \text{ h}, \qquad r_1 = 2 \times 10^{3}\ \text{km}$$

First we write their angular speeds about the planet. For uniform circular motion the relation is

$$\omega = \frac{2\pi}{T}.$$

So we have

$$\omega_1 = \frac{2\pi}{T_1} = \frac{2\pi}{1} = 2\pi \ \text{rad h}^{-1},$$

$$\omega_2 = \frac{2\pi}{T_2} = \frac{2\pi}{8} = \frac{\pi}{4} \ \text{rad h}^{-1}.$$

The radius of the outer orbit, $$r_2,$$ is not given directly, but the satellites circle the same planet, so Kepler’s third law applies:

$$\frac{T^2}{r^3} = \text{constant for the same planet}.$$

Hence

$$\frac{T_2^2}{r_2^3} = \frac{T_1^2}{r_1^3}\; \Longrightarrow\; r_2 = r_1\biggl(\frac{T_2}{T_1}\biggr)^{2/3}.$$

Substituting $$T_2 = 8 \text{ h},\; T_1 = 1 \text{ h}$$ and $$r_1 = 2 \times 10^{3} \text{ km}$$ gives

$$r_2 = 2 \times 10^{3}\ \text{km} \;\bigl(8\bigr)^{2/3}.$$

Because $$8 = 2^3,$$ we get $$8^{2/3} = (2^3)^{2/3} = 2^{2} = 4,$$ therefore

$$r_2 = 2 \times 10^{3}\ \text{km} \times 4 = 8 \times 10^{3}\ \text{km}.$$

To find how fast the farther satellite seems to move across the sky of an observer on the nearer satellite, we look at the instantaneous rate of change of the angle that the line joining the two satellites makes. Let us place the planet at the origin. At the instant of closest approach (both satellites on the same straight line through the planet) put that line along the $$x$$-axis. A little time $$t$$ later their angular displacements about the planet are $$\theta_1 = \omega_1 t$$ and $$\theta_2 = \omega_2 t.$$ The position vectors are

$$\vec r_1 = r_1(\cos\theta_1\,\hat i + \sin\theta_1\,\hat j),$$

$$\vec r_2 = r_2(\cos\theta_2\,\hat i + \sin\theta_2\,\hat j).$$

The vector from satellite 1 to satellite 2 is

$$\vec R = \vec r_2 - \vec r_1.$$

For small $$t$$ we expand $$\cos\theta \approx 1-\theta^2/2$$ and $$\sin\theta \approx \theta.$$ Then to first order in $$t$$

$$\vec R \approx \bigl(r_2 - r_1\bigr)\hat i + \bigl(r_2\omega_2 - r_1\omega_1\bigr)t\,\hat j.$$

The $$x$$-component is essentially constant, while the $$y$$-component grows linearly. The apparent angle $$\phi$$ measured from the initial line is therefore

$$\tan\phi \approx \frac{(r_2\omega_2 - r_1\omega_1)t}{r_2 - r_1}\; \Longrightarrow\; \phi \approx \frac{(r_2\omega_2 - r_1\omega_1)t}{r_2 - r_1}.$$

Hence the instantaneous angular speed of satellite 2 as seen from satellite 1 is

$$\frac{d\phi}{dt} = \frac{r_2\omega_2 - r_1\omega_1}{\,r_2 - r_1\,}.$$

Substituting the numerical values:

$$r_2\omega_2 = (8 \times 10^{3}\ \text{km})\left(\frac{\pi}{4}\ \text{rad h}^{-1}\right) = 2000\pi\ \text{km h}^{-1},$$

$$r_1\omega_1 = (2 \times 10^{3}\ \text{km})(2\pi\ \text{rad h}^{-1}) = 4000\pi\ \text{km h}^{-1},$$

$$r_2 - r_1 = 8 \times 10^{3}\ \text{km} - 2 \times 10^{3}\ \text{km} = 6 \times 10^{3}\ \text{km}.$$

Therefore

$$\left|\frac{d\phi}{dt}\right| = \left|\frac{2000\pi - 4000\pi}{6 \times 10^{3}}\right| = \left|\frac{-2000\pi}{6000}\right| = \frac{\pi}{3}\ \text{rad h}^{-1}.$$

This matches the form $$\dfrac{\pi}{x}\ \text{rad h}^{-1},$$ so we identify $$x = 3.$$

Hence, the correct answer is Option 3.

We need to determine the correct relationship between the radius $$r$$ and the orbit velocity $$v$$ of a particle moving in a stable circular orbit under a given central potential field.

1. Relate Potential Energy to Force

From the problem, the central potential field is given by:

$$U(r) = \frac{C}{r} = C \cdot r^{-1}$$

Where $$C$$ is a positive constant.

The conservative central force $$\vec{F}$$ acting on the particle is related to the gradient of this potential energy by $$F = -\frac{dU}{dr}$$:

$$F = -\frac{d}{dr}\left(C \cdot r^{-1}\right) = -C \cdot (-1 \cdot r^{-2}) = \frac{C}{r^2}$$

This tells us that the particle experiences an attractive central force directed towards the origin whose magnitude is inversely proportional to the square of the radius.

2. Apply the Circular Orbit Condition

For a particle of mass $$m$$ maintaining a stable circular orbit of radius $$r$$ with a constant tangential velocity $$v$$, this conservative central force provides the necessary centripetal force:

$$F_{\text{centripetal}} = \frac{m v^2}{r}$$

Equating the two forces gives:

$$\frac{C}{r^2} = \frac{m v^2}{r}$$

3. Determine the Radius-Velocity ($$r-v$$) Relationship

Simplify the equation by multiplying both sides by $$r$$:

$$\frac{C}{r} = m v^2$$

Isolate the radius $$r$$:

$$r = \frac{C}{m v^2}$$

Since both $$C$$ and $$m$$ are positive constants, we can express this relationship as a proportionality:

$$r \propto \frac{1}{v^2}$$

4. Identify the Correct Graph

Let's analyze how the radius behaves as a function of velocity based on $$r = \frac{\text{constant}}{v^2}$$:

• As the velocity $$v$$ approaches zero ($$v \to 0$$), the orbit radius shoots up towards infinity ($$r \to \infty$$).
• As the velocity $$v$$ becomes very large ($$v \to \infty$$), the orbit radius decays asymptotically towards zero ($$r \to 0$$).

This inverse-square relationship yields a decreasing, non-linear curve that stays within the first quadrant, asymptotically approaching both the vertical $$r$$-axis and horizontal $$v$$-axis. This behavior matches the curve shown in the first graph option.

Conclusion

The correct radius-velocity graph of the particle's motion is represented by Option A.

We need to evaluate the validity of the two statements regarding a cyclist negotiating a circular curve under different road conditions.

1. Analyze Statement I (Unbanked Road)

For a cyclist to safely cross a flat, unbanked circular curve without slipping, the maximum safe speed ($$v_{\text{max}}$$) is determined by the limiting static friction:

$$v_{\text{max}} = \sqrt{\mu_s g r}$$

Given parameters for Statement I:

• Radius ($$r$$) = $$2\text{ m}$$
• Static friction coefficient ($$\mu_s$$) = $$0.2$$
• Acceleration due to gravity ($$g$$) = $$9.8\text{ m s}^{-2}$$

Calculate $$v_{\text{max}}$$:

$$v_{\text{max}} = \sqrt{0.2 \times 9.8 \times 2} = \sqrt{3.92} \approx 1.98\text{ m s}^{-1}$$

Convert this maximum safe speed to kilometers per hour ($$\text{km h}^{-1}$$):

$$v_{\text{max}} = 1.98 \times \frac{18}{5} = 7.128\text{ km h}^{-1}$$

Since the cyclist’s actual speed ($$7\text{ km h}^{-1}$$) is less than the maximum safe threshold ($$7.128\text{ km h}^{-1}$$), the cyclist will successfully pass the curve without slipping. Thus, Statement I is true.

2. Analyze Statement II (Banked Road)

When a road is banked at an angle $$\theta$$ and the contribution of friction is neglected to find the ideal safe speed ($$v_0$$) without relying on tire grip:

$$v_0 = \sqrt{g r \tan\theta}$$

Given parameters for Statement II:

• Banking angle ($$\theta$$) = $$45^\circ$$
• Radius ($$r$$) = $$2\text{ m}$$

Calculate $$v_0$$:

$$v_0 = \sqrt{9.8 \times 2 \times \tan 45^\circ} = \sqrt{19.6 \times 1} \approx 4.427\text{ m s}^{-1}$$

Convert this optimal speed to kilometers per hour ($$\text{km h}^{-1}$$):

$$v_0 = 4.427 \times \frac{18}{5} \approx 15.94\text{ km h}^{-1}$$

If we factor in the friction ($$\mu_s = 0.2$$) to find the absolute maximum possible speed before sliding up the bank ($$v_{\text{max, banked}}$$):

$$v_{\text{max, banked}} = \sqrt{g r \left( \frac{\tan\theta + \mu_s}{1 - \mu_s \tan\theta} \right)}$$

$$v_{\text{max, banked}} = \sqrt{9.8 \times 2 \times \left( \frac{1 + 0.2}{1 - 0.2 \times 1} \right)} = \sqrt{19.6 \times \frac{1.2}{0.8}} = \sqrt{19.6 \times 1.5} = \sqrt{29.4} \approx 5.422\text{ m s}^{-1}$$

Converting this absolute limit to kilometers per hour ($$\text{km h}^{-1}$$):

$$v_{\text{max, banked}} = 5.422 \times \frac{18}{5} \approx 19.52\text{ km h}^{-1}$$

Since the stated speed of $$18.5\text{ km h}^{-1}$$ is strictly below this limit ($$19.52\text{ km h}^{-1}$$), the cyclist can securely navigate the curve without slipping. Thus, Statement II is true.

Conclusion

Both Statement I and Statement II are true, which matches Option D .

Balancing forces vertically:  $$N \cos \theta - mg - f \sin \theta = 0$$

Using limiting friction:  $$f = \mu N$$, 

$$N \cos 30^\circ - \mu N \sin 30^\circ = mg$$

$$\implies N(\cos 30^\circ - \mu \sin 30^\circ) = mg$$

$$\implies N = \frac{mg}{\cos 30^\circ - \mu \sin 30^\circ}$$

$$\implies N = \frac{800 \times 10}{0.87 - 0.2(0.5)} = \frac{8000}{0.87 - 0.1} = \frac{8000}{0.77}$$

$$\implies N \approx 10389\ \text{N} \approx 10.2 \times 10^3\ \text{N}$$

We first recall the relation between the arc-length $$s$$, the radius $$r$$ of the circle and the angle $$\theta$$ subtended by the arc at the centre: the formula is $$s = r\,\theta.$$

Here the given arc-length is $$4.4\ \text{ly}$$, and the angle is stated as $$4$$ arc-seconds. We must put every quantity in compatible units.

One light-year is given as $$1\ \text{ly}=9.46 \times 10^{15}\ \text{m},$$ so

$$s = 4.4\ \text{ly}=4.4 \times 9.46 \times 10^{15}\ \text{m} = 41.624 \times 10^{15}\ \text{m} = 4.1624 \times 10^{16}\ \text{m}.$$

Next, we convert the very small angle of $$4$$ arc-seconds to radians. We know

$$1^\circ = \frac{\pi}{180}\ \text{rad},$$

$$1'\ (\text{one arc-minute}) = \frac{1^\circ}{60},$$

$$1''\ (\text{one arc-second}) = \frac{1'}{60}= \frac{1^\circ}{3600}.$$

Therefore

$$4'' = 4 \times \frac{1^\circ}{3600} = \frac{4}{3600}\ ^\circ = \frac{1}{900}\ ^\circ,$$

and in radians

$$\theta = \frac{1}{900}\ ^\circ \times \frac{\pi}{180} = \frac{\pi}{900 \times 180} = \frac{\pi}{162000}\ \text{rad}.$$

Evaluating

$$\theta = \frac{3.1415926}{162000}\ \text{rad} \approx 1.939 \times 10^{-5}\ \text{rad}.$$

Now we can find the radius $$r$$ from $$r = \dfrac{s}{\theta}$$:

$$r = \frac{4.1624 \times 10^{16}}{1.939 \times 10^{-5}}\ \text{m} = 4.1624 \times 10^{16} \times \frac{1}{1.939 \times 10^{-5}}\ \text{m}.$$

The reciprocal $$\dfrac{1}{1.939 \times 10^{-5}} \approx 5.158 \times 10^{4},$$ so

$$r \approx 4.1624 \times 10^{16} \times 5.158 \times 10^{4}\ \text{m} \approx 2.15 \times 10^{21}\ \text{m}.$$

With the radius known, the full circumference of the circle is given by $$C = 2\pi r$$:

$$C = 2\pi (2.15 \times 10^{21})\ \text{m} = 6.283 \times 2.15 \times 10^{21}\ \text{m} \approx 1.35 \times 10^{22}\ \text{m}.$$

The body is required to complete four revolutions, so the total path-length traversed is

$$D = 4C = 4 \times 1.35 \times 10^{22}\ \text{m} \approx 5.40 \times 10^{22}\ \text{m}.$$

The speed of the body is stated as $$8\ \text{AU s}^{-1}.$$ With $$1\ \text{AU} = 1.5 \times 10^{11}\ \text{m},$$ the speed in SI units is

$$v = 8 \times 1.5 \times 10^{11}\ \text{m s}^{-1} = 1.2 \times 10^{12}\ \text{m s}^{-1}.$$

The time required is obtained from the basic relation $$t = \dfrac{D}{v}.$$ Substituting the values just found,

$$t = \frac{5.40 \times 10^{22}}{1.2 \times 10^{12}}\ \text{s} = 4.50 \times 10^{10}\ \text{s}.$$

Expressed in scientific notation this is $$4.5 \times 10^{10}\ \text{s},$$ matching Option B.

Hence, the correct answer is Option B.

Let us denote by $$\theta$$ the constant angle that the string makes with the vertical while the particle is executing uniform circular motion in a horizontal plane. Because the string has length $$L$$ and the horizontal projection of that length is the radius $$r$$, simple right-triangle geometry gives

$$\sin\theta \;=\;\frac{\text{opposite side}}{\text{hypotenuse}}\;=\;\frac{r}{L}.$$

We are told that $$r=\dfrac{L}{\sqrt2}$$, so

$$\sin\theta \;=\;\frac{L/\sqrt2}{L}\;=\;\frac1{\sqrt2}.$$

The value $$\sin\theta=\dfrac1{\sqrt2}$$ corresponds to

$$\theta=45^\circ,\qquad\text{and hence}\qquad \cos\theta=\frac1{\sqrt2},\qquad\tan\theta=1.$$

Now we analyse the forces on the particle. Two forces act: its weight $$mg$$ vertically downward and the tension $$T$$ along the string at an angle $$\theta$$ from the vertical. We resolve the tension into horizontal and vertical components.

Vertical equilibrium: the particle neither rises nor falls, so the vertical components balance. Thus

$$T\cos\theta \;=\;mg\quad\text{(1)}.$$

Horizontal motion: the particle executes uniform circular motion of radius $$r$$, so the required centripetal force is provided by the horizontal component of the tension. Using the standard centripetal‐force formula $$F_c=\dfrac{mv^2}{r}$$, we write

$$T\sin\theta \;=\;\frac{mv^2}{r}\quad\text{(2)}.$$

From equation (1) we solve for the tension:

$$T=\frac{mg}{\cos\theta}.$$

We substitute this value of $$T$$ into equation (2):

$$\frac{mg}{\cos\theta}\;\sin\theta \;=\;\frac{mv^2}{r}.$$

Cancel the common factor $$m$$ and rearrange step by step:

$$g\,\frac{\sin\theta}{\cos\theta}\;=\;\frac{v^2}{r},$$

$$g\,\tan\theta \;=\;\frac{v^2}{r},$$

$$v^2 \;=\;r\,g\,\tan\theta.$$

We already found $$\tan\theta=1$$, so

$$v^2=r\,g\,(1)=r\,g.$$

Taking the positive square root gives the speed:

$$v=\sqrt{r\,g}.$$

This matches Option A. Hence, the correct answer is Option A.

The angular velocity of a satellite is related to its period of revolution by $$\omega = \frac{2\pi}{T}$$.

For satellite $$S_1$$, the period is $$T_1 = 1$$ hr, so $$\omega_1 = \frac{2\pi}{1}$$.

For satellite $$S_2$$, the period is $$T_2 = 8$$ hr, so $$\omega_2 = \frac{2\pi}{8}$$.

Now the ratio of angular velocities is $$\frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} = \frac{8}{1}$$.

So $$\omega_1 : \omega_2 = 8 : 1$$.

Hence, the correct answer is Option A.

For a geostationary satellite orbiting at a height of $$11R$$ above the surface, the orbital radius is $$r_1 = R + 11R = 12R$$ and its time period is $$T_1 = 24$$ hours.

For the second satellite at a height of $$2R$$, the orbital radius is $$r_2 = R + 2R = 3R$$.

By Kepler's third law, $$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$$, which gives $$T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{12R}\right)^{3/2} = 24 \left(\frac{1}{4}\right)^{3/2}$$.

Now $$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{4\sqrt{4}} = \frac{1}{8}$$, so $$T_2 = \frac{24}{8} = 3$$ hours.

Each particle experiences gravitational forces from the other three.

Net Gravitational Force ($$F_{net}$$)

A single particle is pulled by two adjacent particles at a distance $$d = \sqrt{2}R$$ and one opposite particle at a distance $$D = 2R$$. The resultant force acting towards the center is the sum of the components of these forces:

$$F_{net} = 2 \left( \frac{Gm^2}{(\sqrt{2}R)^2} \right) \cos 45^\circ + \frac{Gm^2}{(2R)^2}$$

$$F_{net} = \frac{Gm^2}{R^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Gm^2}{R^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$$

This mutual attraction provides the centripetal force required for the circular motion of each particle:

$$\frac{mv^2}{R} = \frac{Gm^2}{R^2} \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

$$v^2 = G \left( \frac{1 + 2\sqrt{2}}{4} \right)$$

$$v = \frac{\sqrt{G(1 + 2\sqrt{2})}}{2}$$

We have a planet of mass $$m$$ revolving around the Sun under the action of the gravitational force. Because this force is always directed towards the Sun (the centre of force), it is a central force. For motion under a central force, the angular momentum about the centre remains conserved. We therefore begin with the statement of the law of conservation of angular momentum:

$$L \;=\; mvr \;=\; \text{constant}.$$

Here $$v$$ is the linear speed of the planet at any instant and $$r$$ is the corresponding Sun-planet distance. Since the mass $$m$$ of the planet does not change, the product $$vr$$ must stay the same at every point along the orbit.

The orbit is elliptical. Let

$$x_1 = \text{minimum distance (perihelion)},$$

$$x_2 = \text{maximum distance (aphelion)}.$$

At aphelion the distance is greatest, so the speed is least; this minimum speed is given to be $$v_0$$. At perihelion the distance is least, so the speed is greatest; let us call this unknown maximum speed $$v_{\max}$$. Using the conserved quantity $$mvr$$ at the two extreme positions:

$$m\,v_{\max}\,x_1 \;=\; m\,v_0\,x_2.$$

Now we cancel the common factor $$m$$ from both sides:

$$v_{\max}\,x_1 \;=\; v_0\,x_2.$$

To isolate $$v_{\max}$$ we divide both sides by $$x_1$$:

$$v_{\max} \;=\; \dfrac{v_0\,x_2}{x_1}.$$

This is the required expression for the maximum speed of the planet. Comparing with the given options, we notice that this matches option D.

Hence, the correct answer is Option D.

Solution & Explanation

1. Establish the Time Period Formula

The time period ($$T$$) of a satellite revolving in a circular orbit around the Earth depends strictly on its orbital radius ($$r = R + h$$) and the mass of the Earth ($$M$$). It is completely independent of the satellite's own mass:

$$T = \frac{2\pi r^{3/2}}{\sqrt{GM}} = 2\pi \sqrt{\frac{r^3}{GM}}$$

Let's pre-calculate the constant value $$2\pi \frac{1}{\sqrt{GM}}$$ using SI units:

• $$G = 6.67 \times 10^{-11} \,\, \text{N m}^2/\text{kg}^2$$
• $$M = 6 \times 10^{24} \,\, \text{kg}$$
• $$\sqrt{GM} = \sqrt{(6.67 \times 10^{-11}) \times (6 \times 10^{24})} = \sqrt{4.002 \times 10^{14}} \approx 2 \times 10^7 \,\, \text{m}^{3/2}\text{s}^{-1}$$

2. Calculate the Time Period for Satellite A ($$T_A$$)

The total orbital radius for satellite $$A$$ is:

$$r_A = R + h_A = 6400 \,\, \text{km} + 600 \,\, \text{km} = 7000 \,\, \text{km} = 7 \times 10^6 \,\, \text{m}$$

Substitute this value into the time period equation:

$$T_A = 2\pi \sqrt{\frac{(7 \times 10^6)^3}{2 \times 10^7 \times 2 \times 10^7}} = \frac{2\pi}{2 \times 10^7} \times (7 \times 10^6)^{1.5}$$

$$T_A \approx 3.14 \times 10^{-7} \times 1.852 \times 10^{10} \approx 5815 \,\, \text{s}$$

3. Calculate the Time Period for Satellite B ($$T_B$$)

The total orbital radius for satellite $$B$$ is:

$$r_B = R + h_B = 6400 \,\, \text{km} + 1600 \,\, \text{km} = 8000 \,\, \text{km} = 8 \times 10^6 \,\, \text{m}$$

Substitute this value into the time period equation:

$$T_B = \frac{2\pi}{2 \times 10^7} \times (8 \times 10^6)^{1.5}$$

$$T_B \approx 3.14 \times 10^{-7} \times 2.263 \times 10^{10} \approx 7106 \,\, \text{s}$$

4. Determine the Difference ($$T_B - T_A$$)

Subtract the two time values to find the final interval:

$$\Delta T = T_B - T_A = 7106 \,\, \text{s} - 5815 \,\, \text{s} = 1291 \,\, \text{s} \approx 1.33 \times 10^3 \,\, \text{s}$$

Correct Option: C ($$1.33 \times 10^3 \,\, \text{s}$$)

Let us consider the two identical particles, each of mass $$m = 1\text{ kg}$$. Because the particles are identical, their centre of mass lies exactly at the midpoint of the line joining them. We are told that each particle moves in a circle of radius $$R$$ around this midpoint, so the distance between the two particles is twice this radius:

$$d = 2R.$$

First, we write the universal law of gravitation:

Gravitational force between two point masses is $$F = \dfrac{G\,m_1 m_2}{d^2}.$$

Substituting $$m_1 = m_2 = 1\text{ kg}$$ and $$d = 2R$$, we get

$$F = \dfrac{G \,(1)(1)}{(2R)^2} = \dfrac{G}{4R^2}.$$

This gravitational force is the only force acting on each particle, and it points along the line joining the particles, which also passes through the centre of the circular path. Therefore, this same force provides the necessary centripetal force for uniform circular motion.

The formula for centripetal force is

$$F_{\text{centripetal}} = m\,\omega^2\,r,$$

where $$\omega$$ is the angular speed and $$r$$ is the radius of the circular path. For each particle, $$m = 1\text{ kg}$$ and $$r = R$$, so

$$F_{\text{centripetal}} = (1)\,\omega^2\,R = \omega^2 R.$$

Because the gravitational force supplies the required centripetal force, we equate the two:

$$\dfrac{G}{4R^2} = \omega^2 R.$$

We now solve this equation for $$\omega$$ step by step.

First, isolate $$\omega^2$$ by dividing both sides by $$R$$:

$$\omega^2 = \dfrac{G}{4R^2} \cdot \dfrac{1}{R} = \dfrac{G}{4R^3}.$$

Now take the positive square root to obtain $$\omega$$ (angular speed is taken as positive):

$$\omega = \sqrt{\dfrac{G}{4R^3}} = \dfrac{1}{2}\sqrt{\dfrac{G}{R^3}}.$$

This matches Option B.

Hence, the correct answer is Option B.

By Kepler's third law, $$T \propto R^{3/2}$$. If the first satellite has orbital radius $$R$$ with period $$T$$, and the second has radius $$1.02R$$, then $$T' = T(1.02)^{3/2}$$.

The percentage difference is $$\frac{T' - T}{T} \times 100 = \left[(1.02)^{3/2} - 1\right] \times 100$$.

Using the binomial approximation $$(1+x)^n \approx 1 + nx$$ for small $$x$$: $$(1.02)^{3/2} \approx 1 + \frac{3}{2}(0.02) = 1 + 0.03 = 1.03$$.

Therefore the percentage difference is approximately $$3.0\%$$.

$$\omega = \frac{\theta}{\Delta t} = \frac{\pi / 6}{0.1} = \frac{5\pi}{3}\text{ rad s}^{-1}$$

When projection $$P$$ touches extreme point $$M$$, displacement equals the radius: $$x = R = 0.36\text{ m}$$

$$a = \omega^2 R \implies a = \left(\frac{5\pi}{3}\right)^2 \times 0.36 = \frac{25\pi^2}{9} \times 0.36 = 25\pi^2 \times 0.04 = \pi^2\text{ N kg}^{-1}$$

$$a \approx 9.87\text{ N kg}^{-1}$$

We start by noting that the tip of the second’s hand performs uniform circular motion with radius equal to the length of the hand. Thus, for all instants its speed is constant in magnitude and the only acceleration it possesses is the centripetal acceleration directed toward the centre of the clock face.

The length (radius) of the second’s hand is given as $$r = 0.1\;\text{m}.$$

First we need the angular speed. A second’s hand completes one full revolution in $$T = 60\;\text{s}.$$ The standard formula connecting time-period and angular speed is

$$\omega = \frac{2\pi}{T}.$$

Substituting $$T = 60\;\text{s},$$ we obtain

$$\omega = \frac{2\pi}{60} = \frac{\pi}{30}\;\text{rad s}^{-1}.$$

To find the linear speed $$v$$ of the tip we use the relation

$$v = \omega r.$$

Putting $$\omega = \dfrac{\pi}{30}\;\text{rad s}^{-1}$$ and $$r = 0.1\;\text{m},$$ we get

$$v = \left(\frac{\pi}{30}\right)\!(0.1) = \frac{0.1\pi}{30} = \frac{\pi}{300} \;\text{m s}^{-1}.$$

Taking $$\pi \approx 3.14,$$ this yields

$$v \approx \frac{3.14}{300} = 0.01047\;\text{m s}^{-1}.$$

For uniform circular motion the instantaneous (centripetal) acceleration magnitude is

$$a_c = \frac{v^{2}}{r}.$$

Substituting $$v = 0.01047\;\text{m s}^{-1}$$ and $$r = 0.1\;\text{m},$$ we have

$$a_c = \frac{(0.01047)^{2}}{0.1} = \frac{0.0001097}{0.1} = 0.001097\;\text{m s}^{-2}.$$

This value is approximately $$1.1 \times 10^{-3}\;\text{m s}^{-2}.$$ Since the direction of the centripetal acceleration keeps changing but its magnitude stays the same, the average magnitude over any interval is of the same order. Hence, the required order of the average acceleration is $$10^{-3}\;\text{m s}^{-2}.$$

Hence, the correct answer is Option A.

Let the spring of natural length $$l$$ elongate by an amount $$x$$ when the system starts rotating. Thus the actual radial distance of the mass from the fixed support becomes $$r = l + x$$.

First we write the two physical relations that govern the situation.

1. Hooke’s law for a spring tells us that the magnitude of the tension (restoring force) in the spring equals the product of its force constant and its extension: $$T = kx$$.

2. For uniform circular motion the required centripetal force is provided entirely by this tension. The standard formula for centripetal force is $$F_{\text{c}} = m\omega^{2}r$$, where $$r$$ is the radius of the circular path.

Because the tension supplies the centripetal force, we equate the two expressions:

$$kx = m\omega^{2}r.$$

Now we substitute $$r = l + x$$ into this equality:

$$kx = m\omega^{2}(l + x).$$

Next, we expand the right-hand side to separate the terms:

$$kx = m\omega^{2}l + m\omega^{2}x.$$

To collect all the terms containing $$x$$ on the left side, we subtract $$m\omega^{2}x$$ from both sides:

$$kx - m\omega^{2}x = m\omega^{2}l.$$

We factor out $$x$$ from the left-hand side:

$$x\,(k - m\omega^{2}) = m\omega^{2}l.$$

Finally, to isolate $$x$$ we divide both sides by $$(k - m\omega^{2})$$, yielding

$$x = \frac{m\omega^{2}l}{k - m\omega^{2}}.$$

This fraction represents the stretch produced in the spring when the system rotates with angular speed $$\omega$$ in the absence of gravity.

Hence, the correct answer is Option B.

Let the natural (unstretched) length of the spring be $$l$$. Because the free end is fixed at the centre of the disc, the mass of the spring-mass system always lies on a radius of the disc. When the disc starts rotating with a constant angular velocity $$\omega$$, the mass experiences an outward centrifugal force.

In the rotating (non-inertial) frame the only two radial forces on the mass are (1) the restoring force of the spring, directed towards the centre and equal in magnitude to $$k\Delta l$$, where $$\Delta l$$ is the extension, and (2) the centrifugal force, directed away from the centre and equal in magnitude to $$m\omega^{2}r$$, where $$r$$ is the distance of the mass from the centre after extension.

At the new static equilibrium these two forces must balance, so we write the condition $$k\Delta l = m\omega^{2}r.$$

The new length of the spring is clearly $$r = l + \Delta l.$$

Substituting this expression for $$r$$ into the force-balance equation we obtain $$k\Delta l = m\omega^{2}(l + \Delta l).$$

We now wish to express every quantity in terms of the natural length $$l$$. Define the relative (fractional) change in length as $$x = \frac{\Delta l}{l}.$$

Then $$\Delta l = xl,$$ and substituting this into the previous equation gives $$k(xl) = m\omega^{2}\bigl(l + xl\bigr).$$

We can cancel the common factor $$l$$ on both sides:

$$kx = m\omega^{2}(1 + x).$$

To isolate $$x$$, first divide both sides by $$k$$:

$$x = \frac{m\omega^{2}}{k}(1 + x).$$

Now expand the right-hand side and bring all terms involving $$x$$ to the left:

$$x - \frac{m\omega^{2}}{k}x = \frac{m\omega^{2}}{k}.$$

Factor out $$x$$ on the left:

$$x\left(1 - \frac{m\omega^{2}}{k}\right) = \frac{m\omega^{2}}{k}.$$

Finally, divide by the factor in parentheses:

$$x = \frac{\dfrac{m\omega^{2}}{k}}{1 - \dfrac{m\omega^{2}}{k}}.$$

The problem statement tells us that $$k \gg m\omega^{2}$$, which means $$\frac{m\omega^{2}}{k} \ll 1.$$ For such a small quantity we can safely neglect its square and higher powers, so we approximate $$\frac{1}{1 - \dfrac{m\omega^{2}}{k}} \approx 1.$$

Under this first-order approximation the relative change reduces to

$$x \approx \frac{m\omega^{2}}{k}.$$

Because $$x$$ was defined as $$\dfrac{\Delta l}{l}$$, we have found

$$\frac{\Delta l}{l} \approx \frac{m\omega^{2}}{k}.$$

This expression exactly matches Option C.

Hence, the correct answer is Option 3.

Equation of the parabolic wire profile: $$y = 4Cx^2$$

Slope of the tangent at any point on the parabola: $$\tan\theta = \frac{dy}{dx} = 8Cx$$

At point $$P(a,b)$$, the horizontal distance from the axis of rotation is $$x = a$$: $$\tan\theta = 8Ca$$

Balancing forces in the rotating frame of the wire: $$N\sin\theta = m\omega^2 a$$

$$N\cos\theta = mg$$

Dividing the horizontal equilibrium equation by the vertical one:

$$\tan\theta = \frac{m\omega^2 a}{mg} = \frac{\omega^2 a}{g}$$

$$8Ca = \frac{\omega^2 a}{g} \implies \omega^2 = 8gC$$

$$\omega = \sqrt{8gC} = 2\sqrt{2gC}$$

We have the position vector of the particle as $$\vec r(t)=\cos\omega t\,\hat i+\sin\omega t\,\hat j,$$ where $$\omega$$ is a constant and $$t$$ is time.

By definition, the velocity vector is the time derivative of the position vector, that is $$\vec v(t)=\dfrac{d\vec r}{dt}.$$

Now, differentiating each component separately and using the standard derivatives $$\dfrac{d}{dt}\bigl(\cos\omega t\bigr)=-\omega\sin\omega t$$ and $$\dfrac{d}{dt}\bigl(\sin\omega t\bigr)=\ \ \omega\cos\omega t,$$ we obtain

$$\vec v(t)=\dfrac{d}{dt}\Bigl(\cos\omega t\,\hat i+\sin\omega t\,\hat j\Bigr)=\bigl(-\omega\sin\omega t\bigr)\hat i+\bigl(\ \ \omega\cos\omega t\bigr)\hat j.$$

Next, we verify whether the velocity is perpendicular to the position vector by taking their dot product. Using the formula $$\vec A\cdot\vec B=A_xB_x+A_yB_y$$ for two‐dimensional vectors, we write

$$\vec r(t)\cdot\vec v(t)=\bigl(\cos\omega t\bigr)\bigl(-\omega\sin\omega t\bigr)+\bigl(\sin\omega t\bigr)\bigl(\ \ \omega\cos\omega t\bigr).$$

Simplifying each term, we have

$$\vec r(t)\cdot\vec v(t)=-\omega\cos\omega t\sin\omega t+\omega\sin\omega t\cos\omega t=0.$$

Because the dot product is zero, $$\vec v(t)\perp\vec r(t).$$ So the velocity is indeed perpendicular to the position vector at every instant.

Now, by definition, the acceleration vector is the time derivative of the velocity vector, that is $$\vec a(t)=\dfrac{d\vec v}{dt}.$$

Differentiating $$\vec v(t)=\bigl(-\omega\sin\omega t\bigr)\hat i+\bigl(\omega\cos\omega t\bigr)\hat j$$ term by term and again using the same trigonometric derivatives, we get

$$\vec a(t)=\dfrac{d}{dt}\Bigl(-\omega\sin\omega t\,\hat i+\omega\cos\omega t\,\hat j\Bigr)=\bigl(-\omega^2\cos\omega t\bigr)\hat i+\bigl(-\omega^2\sin\omega t\bigr)\hat j.$$

Notice that every component of $$\vec a(t)$$ is $$-\omega^2$$ times the corresponding component of $$\vec r(t).$$ Writing this compactly, we have

$$\vec a(t)=-\omega^2\Bigl(\cos\omega t\,\hat i+\sin\omega t\,\hat j\Bigr)=-\omega^2\vec r(t).$$

The negative sign means that $$\vec a(t)$$ is opposite in direction to $$\vec r(t).$$ Since $$\vec r(t)$$ points outward from the origin to the particle, $$-\vec r(t)$$ points inward toward the origin. Hence the acceleration is directed toward the origin, i.e., toward the centre of the circular path.

We have thus shown that the velocity is perpendicular to the position vector and the acceleration is directed toward the origin. Among the given options, this description corresponds to Option D.

Hence, the correct answer is Option D.

We start with the satellite in a very low, almost circular orbit whose radius is practically the earth’s radius itself. Hence the initial orbital radius is $$r_1 = R_e$$.

For a circular orbit, the standard formula for orbital speed is

$$v_c = \sqrt{\frac{GM}{r}},$$

where $$G$$ is the universal gravitational constant and $$M$$ is the mass of the earth. Substituting $$r = R_e$$ gives the initial speed

$$v_1 = \sqrt{\frac{GM}{R_e}}.$$

Immediately after the rockets are fired, the speed is increased tangentially to

$$v_2 = \sqrt{\frac{3}{2}}\;v_1 = \sqrt{\frac{3}{2}}\;\sqrt{\frac{GM}{R_e}} = \sqrt{\frac{3GM}{2R_e}}.$$

The moment after the impulse, the satellite is still at the same position $$r = R_e$$ but now with the higher speed $$v_2$$. Its subsequent path is no longer circular; it becomes an ellipse with the present point as the perigee (closest point). We shall find the apogee distance $$R$$ using conservation laws.

1. Specific mechanical energy (total energy per unit mass)
The formula is

$$\varepsilon = \frac{v^2}{2} - \frac{GM}{r}.$$

At perigee just after firing, we substitute $$v = v_2$$ and $$r = R_e$$:

$$\varepsilon = \frac{1}{2}\left(\frac{3GM}{2R_e}\right) - \frac{GM}{R_e} = \frac{3GM}{4R_e} - \frac{GM}{R_e} = \frac{3GM}{4R_e} - \frac{4GM}{4R_e} = -\frac{GM}{4R_e}.$$

For an elliptical orbit, the same energy is also expressed as

$$\varepsilon = -\frac{GM}{2a},$$

where $$a$$ is the semi-major axis. Equating the two expressions, we get

$$-\frac{GM}{2a} = -\frac{GM}{4R_e} \;\;\Longrightarrow\;\; 2a = 4R_e \;\;\Longrightarrow\;\; a = 2R_e.$$

2. Relation between perigee, apogee, and semi-major axis
For any ellipse,

$$r_p + r_a = 2a,$$

where $$r_p$$ is the perigee distance and $$r_a$$ is the apogee distance. Here

$$r_p = R_e,\qquad r_a = R.$$

Substituting these values and $$a = 2R_e$$, we obtain

$$R_e + R = 2a = 4R_e \;\;\Longrightarrow\;\; R = 4R_e - R_e = 3R_e.$$

So the farthest distance from the earth’s centre reached by the satellite is three times the earth’s radius.

Hence, the correct answer is Option C.

Let the mass of the object be $$m$$. Whenever a spring balance is used, it actually registers the normal reaction on the object, which we call the apparent weight.

First we consider the object kept on the equator at the earth’s surface. At the equator the object is moving in a horizontal circle of radius $$R$$ with the earth, so it experiences a centrifugal acceleration directed radially outward. The magnitude of this acceleration is $$\omega^{2}R$$, where $$\omega$$ is the earth’s angular velocity.

Hence the effective or apparent acceleration acting downward on the object at the equator is the vector difference of the true gravitational acceleration $$g$$ and the centrifugal acceleration $$\omega^{2}R$$. Therefore the apparent weight at the equator is

$$W_{e}=m\left(g-\omega^{2}R\right).$$

Now we take the same object to a point directly above the pole at a small height $$h$$ (with $$h\ll R$$). Exactly at the pole the distance of the object from the earth’s axis is zero, so it has no circular motion about the axis and hence no centrifugal effect. The only change comes from the fact that we are now at a larger distance $$R+h$$ from the earth’s centre.

The law of gravitation gives the gravitational acceleration at distance $$r$$ as $$g(r)=\dfrac{GM}{r^{2}}$$, where $$G$$ is the gravitational constant and $$M$$ is the earth’s mass. Putting $$r=R+h$$ we get

$$g_{p}=g\left(\dfrac{R}{R+h}\right)^{2}.$$

Because $$h\ll R$$ we expand by the binomial theorem. For any small $$x$$, $$(1+x)^{-2}\approx1-2x$$. Taking $$x=\dfrac{h}{R}$$ we have

$$\left(1+\dfrac{h}{R}\right)^{-2}\approx1-\dfrac{2h}{R}.$$

So

$$g_{p}\approx g\left(1-\dfrac{2h}{R}\right).$$

Thus the apparent weight at the height $$h$$ above the pole is simply

$$W_{p}=m\,g_{p}=m\,g\left(1-\dfrac{2h}{R}\right).$$

The problem states that the two spring-balance readings are the same, i.e.

$$W_{e}=W_{p}.$$

Substituting the expressions we have obtained:

$$m\left(g-\omega^{2}R\right)=m\,g\left(1-\dfrac{2h}{R}\right).$$

We cancel the common factor $$m$$ on both sides, giving

$$g-\omega^{2}R=g\left(1-\dfrac{2h}{R}\right).$$

Expanding the right-hand side:

$$g-\omega^{2}R=g-\dfrac{2gh}{R}.$$

Now we subtract $$g$$ from both sides to isolate the terms that contain $$h$$:

$$-\omega^{2}R=-\dfrac{2gh}{R}.$$

Multiplying by $$-1$$ removes the minus signs:

$$\omega^{2}R=\dfrac{2gh}{R}.$$

Finally we solve for $$h$$ by multiplying both sides by $$\dfrac{R}{2g}$$:

$$h=\dfrac{\omega^{2}R^{2}}{2g}.$$

This matches Option A.

Hence, the correct answer is Option A.

Let the mass of the planet be $$M$$. At the instant just before the collision body $$A$$ of mass $$m$$ is moving in a circular path of radius $$R$$ with speed $$v$$. For a circular orbit we first recall the condition obtained by equating the required centripetal force to the gravitational attraction

$$\frac{m v^{2}}{R}= \frac{G M m}{R^{2}}\; ,\qquad\Rightarrow\qquad v^{2}= \frac{G M}{R}$$

The gravitational potential energy of $$A$$ in that orbit is

$$U_{A}= -\,\frac{G M m}{R}= -m v^{2}$$

and its kinetic energy is

$$K_{A}= \frac12 m v^{2}$$

so the total mechanical energy of $$A$$ is

$$E_{A}=K_{A}+U_{A}=\frac12 m v^{2}-m v^{2}= -\frac12 m v^{2}$$

Now a second body $$B$$ of mass $$\dfrac{m}{2}$$ approaches and collides head-on (along the same tangential direction) with $$A$$ with speed $$\dfrac{v}{2}$$. The collision is completely inelastic, hence the two bodies stick together. During the very short collision the only significant forces are their mutual internal forces; the external gravitational force acts radially and therefore exerts no impulse in the tangential direction. Consequently the linear (tangential) momentum is conserved.

Initial tangential momentum:

$$p_{\text{initial}} \;=\; m\,v \;+\;\frac{m}{2}\,\frac{v}{2}\;=\;m v+\frac14 m v \;=\;\frac54\,m v$$

Total mass after collision:

$$m_{\text{tot}} = m+\frac{m}{2}= \frac32\,m$$

If $$V$$ is the common speed immediately after sticking together, conservation of momentum gives

$$\frac54\,m v = \frac32\,m\,V \quad\Rightarrow\quad V = \frac{\frac54}{\frac32}\,v = \frac{5}{4}\times\frac{2}{3}\,v = \frac56\,v$$

Thus the speed of the combined body has fallen to $$\dfrac56\,v$$.

Its kinetic energy right after the collision is

$$K' = \frac12\Bigl(\frac32\,m\Bigr) V^{2} =\frac34\,m\,\Bigl(\frac56\,v\Bigr)^{2} =\frac34\,m\,\frac{25}{36}\,v^{2} =\frac{25}{48}\,m v^{2}$$

The collision occurs at the same distance $$R$$ from the planet, so the gravitational potential energy of the combined body is

$$U' = -\,\frac{G M}{R}\Bigl(\frac32\,m\Bigr) =-\frac32\,m\,v^{2}$$ (because $$v^{2}= \dfrac{G M}{R}$$)

Hence the total mechanical energy after collision is

$$E' = K' + U' = \frac{25}{48}\,m v^{2} \;-\;\frac32\,m v^{2} = \frac{25}{48}\,m v^{2}-\frac{72}{48}\,m v^{2} = -\frac{47}{48}\,m v^{2}$$

We immediately notice that $$E'$$ is negative, so the system is still gravitationally bound; therefore it cannot escape (escape requires $$E\ge 0$$).

Next, compare the new speed $$V=\dfrac56\,v$$ with the speed required for a circular orbit at radius $$R$$. The circular-orbit speed does not depend on the mass of the moving body, and we have already found it to be $$v$$. Because

$$V=\frac56\,v \;\lt \; v$$

the combined body has less speed than needed for a circular path, so it cannot continue in the same circular orbit. However, its angular momentum about the planet is not zero:

$$L' = m_{\text{tot}} V R = \Bigl(\frac32\,m\Bigr)\Bigl(\frac56\,v\Bigr)R = \frac54\,m v R \ne 0$$

Because angular momentum is non-zero the body cannot fall straight towards the planet; instead it must follow another bound orbit. For a gravitational potential, bound non-circular orbits are elliptical. Hence the new path is an ellipse with the point of collision lying on that ellipse (specifically at the apogee, since the speed there is lower than the circular value).

Therefore the combined body starts moving in an elliptical orbit around the planet.

Hence, the correct answer is Option 4.

At the north pole the spring balance reads the true gravitational weight, so we have

$$W_{\text{pole}} = m\,g$$

The reading is given as $$W_{\text{pole}} = 196\ \text{N}$$ and the acceleration due to gravity at the pole is $$g = 10\ \text{m s}^{-2}$$. Substituting these values we can find the mass of the box:

$$m = \frac{W_{\text{pole}}}{g} = \frac{196}{10} = 19.6\ \text{kg}$$

When the box is taken to the equator it revolves with the Earth, so a centrifugal force acts outward. A spring balance measures the normal reaction, which equals the effective weight. The effective acceleration $$g'$$ at the equator is therefore reduced by the centrifugal term. The relation is

$$g' = g - \omega^{2}R$$

Here $$\omega$$ is the angular speed of the Earth and $$R$$ is the radius of the Earth.

First, we calculate $$\omega$$. The Earth completes one rotation in 24 hours:

$$\omega = \frac{2\pi}{T}, \qquad T = 24 \times 3600\ \text{s}$$

$$\omega = \frac{2\pi}{24 \times 3600} = 7.27 \times 10^{-5}\ \text{rad s}^{-1}$$

Next, we compute $$\omega^{2}R$$. The radius is given as $$R = 6400\ \text{km} = 6.4 \times 10^{6}\ \text{m}$$.

$$\omega^{2} = (7.27 \times 10^{-5})^{2} = 5.29 \times 10^{-9}\ \text{s}^{-2}$$

Multiplying by the radius,

$$\omega^{2}R = 5.29 \times 10^{-9}\ \text{s}^{-2} \times 6.4 \times 10^{6}\ \text{m} = 33.856 \times 10^{-3}\ \text{m s}^{-2} = 0.033856\ \text{m s}^{-2}$$

Now we obtain the effective acceleration at the equator:

$$g' = 10 - 0.033856 = 9.966144\ \text{m s}^{-2}$$

The spring balance reading at the equator is

$$W_{\text{equator}} = m\,g' = 19.6\ \text{kg} \times 9.966144\ \text{m s}^{-2}$$

Performing the multiplication,

$$W_{\text{equator}} = 195.329\ \text{N} \approx 195.32\ \text{N}$$

Among the given choices, this value matches Option D.

Hence, the correct answer is Option D.

For a satellite moving under the central gravitational force of a planet, there is no external torque about the planet. Therefore, the angular momentum of the satellite about the planet remains conserved throughout the orbit. The statement of conservation of angular momentum is:

$$m\,r\,v = \text{constant}$$

Here $$m$$ is the mass of the satellite, $$r$$ is the instantaneous distance from the planet, and $$v$$ is the corresponding linear speed. We compare two special points of the elliptical orbit:

• Closest point (periapsis): distance $$r_c$$ and speed $$v_c$$.
• Farthest point (apoapsis): distance $$r_f$$ and speed $$v_f$$.

Applying the conservation law between these two points, we have

$$m\,r_c\,v_c \;=\; m\,r_f\,v_f.$$

Because the satellite’s mass $$m$$ is common to both terms, it cancels:

$$r_c\,v_c \;=\; r_f\,v_f.$$

The problem states that when the satellite is farthest from the planet, its speed is six times smaller than when it is closest. Expressing this in symbols,

$$v_f \;=\;\dfrac{v_c}{6}.$$

Substituting this value of $$v_f$$ into the conserved‐angular‐momentum equation gives

$$r_c\,v_c \;=\; r_f\left(\dfrac{v_c}{6}\right).$$

Now the speed $$v_c$$ appears on both sides, so it divides out:

$$r_c \;=\;\dfrac{r_f}{6}.$$

Rearranging, the ratio of distances becomes

$$\dfrac{r_c}{r_f} \;=\;\dfrac{1}{6}.$$

Thus, the distance of the satellite from the planet at the closest point is one‐sixth of its distance at the farthest point. Writing this as a ratio closest : farthest, we get

$$r_c : r_f \;=\; 1 : 6.$$

Hence, the correct answer is Option A.

We begin by recalling that the speed of a satellite in a circular orbit of radius $$r$$ around a planet of mass $$M$$ is obtained by equating the gravitational attraction to the centripetal force.

Gravitational force: $$F_g=\dfrac{G M m}{r^{2}}.$$

Centripetal force needed for circular motion: $$F_c=\dfrac{m v_{\text{orbit}}^{2}}{r}.$$

Setting $$F_g=F_c$$ gives

$$\dfrac{G M m}{r^{2}}=\dfrac{m v_{\text{orbit}}^{2}}{r}.$$

The mass $$m$$ of the body cancels out, and one factor of $$r$$ in the denominator cancels as well, so we have

$$\dfrac{G M}{r}=v_{\text{orbit}}^{2}.$$

Taking the square root on both sides, we obtain the orbital speed,

$$v_{\text{orbit}}=\sqrt{\dfrac{G M}{r}}.$$

In this problem the orbit is said to be “low,” meaning its radius is practically the planet’s radius itself, $$r\approx R.$$ Substituting $$r=R$$ we get

$$v_{\text{orbit}}=\sqrt{\dfrac{G M}{R}}.$$

Next, we state the formula for escape velocity. A body will just escape the planet’s gravitational field when its kinetic energy at the surface equals the magnitude of its gravitational potential energy (taking zero potential energy at infinity):

$$\dfrac{1}{2}m v_{\text{escape}}^{2}=\dfrac{G M m}{R}.$$

Again $$m$$ cancels, and solving for $$v_{\text{escape}}$$ we obtain

$$v_{\text{escape}}=\sqrt{\dfrac{2 G M}{R}}.$$

We are asked for the ratio $$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}.$$ Substituting the two expressions derived above:

$$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}=\dfrac{\sqrt{\dfrac{G M}{R}}}{\sqrt{\dfrac{2 G M}{R}}}.$$

The common factor $$\sqrt{\dfrac{G M}{R}}$$ in numerator and denominator simplifies, leaving

$$\dfrac{v_{\text{orbit}}}{v_{\text{escape}}}=\dfrac{1}{\sqrt{2}}.$$

Hence, the correct answer is Option A.

Let us consider an element of the liquid situated at a distance $$r$$ from the axis of the cylindrical vessel which is rotating with a uniform angular speed $$\omega$$ rad s$$^{-1}$$. Because of the rotation, the element experiences a centrifugal acceleration $$\omega^{2}r$$ directed horizontally outward, while the usual gravitational acceleration $$g$$ acts vertically downward.

On the free surface of the liquid the pressure is the same everywhere (equal to the atmospheric pressure). For any two points on that free surface the total change in hydrostatic pressure must therefore be zero. We can write this balance in differential form. If the surface rises by a vertical amount $$dz$$ when we move radially outward by an infinitesimal distance $$dr$$, the pressure change due to gravity is $$\rho g\,dz$$ downward, and the pressure change due to the centrifugal field is $$\rho\,\omega^{2}r\,dr$$ outward. Equating the two (and cancelling the common factor $$\rho$$) we obtain

$$g\,dz \;=\; \omega^{2}r\,dr.$$

We now integrate this differential relation from the axis ($$r=0$$) to any general point on the surface at radius $$r$$. The left-hand side integrates over the corresponding vertical rise $$z$$ (measured from the level at the axis), while the right-hand side integrates over the radial distance:

$$\displaystyle\int_{0}^{z}\! g\,dz \;=\; \int_{0}^{r}\! \omega^{2}r\,dr.$$

Carrying out the integrations term by term, we get

$$g\,z \;=\; \omega^{2}\,\frac{r^{2}}{2}.$$

Solving for the height $$z$$ of the liquid surface above the centre level at a distance $$r$$ gives the well-known parabolic profile

$$z \;=\; \frac{\omega^{2}r^{2}}{2g}.$$

Now we are interested in the difference in height between the surface at the extreme side of the vessel ($$r = R$$) and the surface at the axis ($$r = 0$$). At the centre, $$r = 0$$ so $$z = 0$$. At the wall, $$r = R = 5\;{\rm cm}$$. Substituting this value into the expression for $$z$$, we obtain

$$h \;=\; z(R) - z(0) \;=\; \frac{\omega^{2}R^{2}}{2g} \;=\; \frac{\omega^{2}(5\;{\rm cm})^{2}}{2g}.$$

Evaluating the square of the radius,

$$(5\;{\rm cm})^{2} = 25\;{\rm cm}^{2},$$

and substituting back, we have

$$h \;=\; \frac{25\,\omega^{2}}{2g}\;{\rm cm}.$$

This expression matches Option C in the given list.

Hence, the correct answer is Option C.

$$M(R) = \int_{0}^{R} \rho(r) \cdot 4\pi r^2 \, dr$$

$$M(R) = \int_{0}^{R} \frac{K}{r} \cdot 4\pi r^2 \, dr = 4\pi K \int_{0}^{R} r \, dr$$

$$M(R) = 4\pi K \left[ \frac{r^2}{2} \right]_{0}^{R} = 2\pi K R^2 \implies M(R) \propto R^2$$

$$\frac{m v^2}{R} = \frac{G M(R) m}{R^2}$$ $$\implies$$ $$v^2 = \frac{G M(R)}{R}$$

$$v^2 \propto \frac{R^2}{R} \implies v^2 \propto R \implies v \propto R^{1/2}$$

$$T = \frac{2\pi R}{v}$$ $$\implies$$ $$T^2 = \frac{4\pi^2 R^2}{v^2}$$

$$T^2 \propto \frac{R^2}{R} \implies T^2 \propto R$$

We are told that the instantaneous position of the particle is described, as a function of time $$t,$$ by the three co-ordinate equations

$$x = a \cos \omega t,$$

$$y = a \sin \omega t,$$

and

$$z = a \, \omega t.$$

To obtain the speed we must first find the velocity vector. By definition, the velocity components are the time derivatives of the corresponding position components. In symbols, if $$x(t),\,y(t),\,z(t)$$ are known, then

$$v_x = \dfrac{dx}{dt}, \qquad v_y = \dfrac{dy}{dt}, \qquad v_z = \dfrac{dz}{dt}.$$

Carrying out these differentiations one by one, we have

$$v_x = \dfrac{d}{dt}\!\bigl(a \cos \omega t\bigr).$$

Because the derivative of $$\cos (\omega t)$$ with respect to $$t$$ is $$-\omega \sin (\omega t),$$ we obtain

$$v_x = a\;(-\omega \sin \omega t) = -\,a \omega \sin \omega t.$$

Next, for the $$y$$ component,

$$v_y = \dfrac{d}{dt}\!\bigl(a \sin \omega t\bigr) = a \,\omega \cos \omega t,$$

since the derivative of $$\sin (\omega t)$$ is $$\omega \cos (\omega t).$$

Finally, for the $$z$$ component,

$$v_z = \dfrac{d}{dt}\!\bigl(a \omega t\bigr) = a \omega,$$

because the derivative of a linear function $$\omega t$$ is simply $$\omega.$$

Thus the full velocity vector is

$$\vec v = \bigl(-a\omega \sin \omega t\,,\; a\omega \cos \omega t\,,\; a\omega\bigr).$$

The speed $$v$$ is the magnitude of this vector. The magnitude (or Euclidean norm) of a vector $$\!(v_x,v_y,v_z)$$ is given by the formula

$$v = \sqrt{\,v_x^2 + v_y^2 + v_z^2\,}.$$

Now we substitute the three components just calculated:

$$v = \sqrt{\bigl(-a\omega \sin \omega t\bigr)^2 + \bigl(a\omega \cos \omega t\bigr)^2 + \bigl(a\omega\bigr)^2}.$$

Expanding each squared term inside the radical gives

$$v = \sqrt{a^2\omega^2 \sin^2 \omega t + a^2\omega^2 \cos^2 \omega t + a^2\omega^2}.$$

We notice that the first two terms share the common factor $$a^2\omega^2,$$ so we group them:

$$v = \sqrt{a^2\omega^2 \bigl(\sin^2 \omega t + \cos^2 \omega t\bigr) + a^2\omega^2}.$$

From trigonometry we know the fundamental identity $$\sin^2 \theta + \cos^2 \theta = 1.$$ Applying this identity (with $$\theta = \omega t$$) converts the grouped bracket to $$1$$, giving

$$v = \sqrt{a^2\omega^2 \cdot 1 + a^2\omega^2}.$$

So we simply have two identical terms inside the square root:

$$v = \sqrt{a^2\omega^2 + a^2\omega^2} = \sqrt{2\,a^2\omega^2}.$$

Finally, we take the square root of the product. Because $$a$$ and $$\omega$$ are positive constants (length and angular frequency respectively), their squares inside the root come out as absolute values, which are just themselves:

$$v = \sqrt{2}\; a \omega.$$

This expression is a constant, showing that the particle moves with uniform speed of magnitude $$\sqrt{2}\,a\omega.$$

Hence, the correct answer is Option A.

We start by recalling that in uniform circular motion the speed is constant, but the velocity vector is continuously changing its direction. At any point on the circle, the velocity is tangent to the path and perpendicular to the radius drawn to that point. Let the constant speed be denoted by $$v = 10\ \text{m s}^{-1}$$.

Suppose the particle is at point $$P_{1}$$ on the circle. Its velocity vector at this point is $$\vec v_{1}$$, tangent to the circle. After the particle has moved through a central angle of $$\theta = 60^{\circ}$$ (which is $$\theta = \dfrac{\pi}{3}\ \text{radians}$$) it reaches point $$P_{2}$$. The speed is still $$v$$, and the velocity vector there is $$\vec v_{2}$$, also tangent to the circle.

The two velocity vectors $$\vec v_{1}$$ and $$\vec v_{2}$$ have the same magnitude $$v$$ but they are directed along tangents that make an angle $$\theta$$ with each other, because each tangent is perpendicular to its corresponding radius and the radii themselves are separated by the same central angle $$\theta$$.

We want the magnitude of the change in velocity:

$$|\Delta \vec v| = |\vec v_{2} - \vec v_{1}|$$.

For two vectors of equal magnitude $$v$$ with an included angle $$\theta$$ between them, we can use the law of cosines (vector form) to find the magnitude of their difference. The formula is stated as:

$$|\vec a - \vec b| = \sqrt{a^{2} + b^{2} - 2ab\cos\theta}$$.

Here $$a = b = v$$ and $$\theta = 60^{\circ}$$, so

$$|\Delta \vec v| = \sqrt{v^{2} + v^{2} - 2v\,v\cos\theta}$$.

Substituting $$v = 10\ \text{m s}^{-1}$$ and $$\cos 60^{\circ} = \dfrac{1}{2}$$, we get

$$|\Delta \vec v| = \sqrt{10^{2} + 10^{2} - 2 \cdot 10 \cdot 10 \cdot \dfrac{1}{2}}$$ $$= \sqrt{100 + 100 - 100}$$ $$= \sqrt{100}$$ $$= 10\ \text{m s}^{-1}.$$

Many students also memorize a simpler equivalent relation for uniform circular motion: $$|\Delta \vec v| = 2v\sin\left(\dfrac{\theta}{2}\right)$$. Stating and using that formula gives the same result. We have $$\theta/2 = 30^{\circ}$$ and $$\sin 30^{\circ} = \dfrac{1}{2}$$, so $$|\Delta \vec v| = 2 \times 10 \times \dfrac{1}{2} = 10\ \text{m s}^{-1}$$.

Either way, the magnitude of the change in velocity is $$10\ \text{m s}^{-1}$$. Hence, the correct answer is Option D.

For the angular displacement of both particles in time $$t = \frac{\pi}{2\omega}$$:

$$\theta = \omega t = \omega \left(\frac{\pi}{2\omega}\right) = \frac{\pi}{2}\text{ rad} = 90^\circ$$

For particle A (initially at $$(R_1, 0)$$ moving in the counter-clockwise direction along $$+\hat{j}$$):

$$\text{After rotating by } 90^\circ\text{, its position is on the positive y-axis, and its velocity points along } -\hat{i}:$$ 

$$\vec{V}_A = -\omega R_1 \hat{i}$$

For particle B (initially at $$(R_2, 0)$$ moving in the clockwise direction along $$-\hat{j}$$):

$$\text{After rotating by } 90^\circ\text{, its position is on the negative y-axis, and its velocity points along } -\hat{i}:$$

$$\vec{V}_B = -\omega R_2 \hat{i}$$

For relative velocity $$\vec{V}_A - \vec{V}_B$$:

$$\vec{V}_A - \vec{V}_B = (-\omega R_1 \hat{i}) - (-\omega R_2 \hat{i}) = \omega(R_2 - R_1)\hat{i}$$

We begin with the Lorentz force law, which gives the magnetic force on a moving charge:

$$\vec F = q(\vec v \times \vec B).$$

Because the magnetic force is always perpendicular to the instantaneous velocity, the particle moves in a circular path of constant speed. The magnitude of the centripetal force equals the magnetic force, so

$$\frac{mv^{2}}{r}=qvB \quad\Longrightarrow\quad r=\frac{mv}{qB}.$$

The problem states that the strip of field extends from $$y=0$$ to $$y=d$$ with

$$d=\frac{mv}{2qB}=\frac{r}{2}.$$

Thus the particle travels only the lower half of the circle’s radius in the $$+y$$-direction while it is inside the field.

To follow the motion geometrically, let us place the centre of the circle at the point $$C(0,r)$$. The particle enters the field at the point $$P_{0}(0,0)$$ with velocity $$\vec v=v\hat i$$. Relative to the centre, the initial radius vector is $$\vec{CP}_{0}= -r\hat j$$, i.e. straight down. We measure an angular coordinate $$\theta$$ from this downward radius, increasing anticlockwise (toward $$+y$$).

Relative to the centre, the position vector is therefore

$$\vec r(\theta)=r\bigl(\sin\theta\,\hat i-\cos\theta\,\hat j\bigr).$$

Differentiating with respect to time gives the velocity direction:

$$\vec v(\theta)=\frac{d\vec r}{d\theta}\frac{d\theta}{dt} =r\dot\theta\bigl(\cos\theta\,\hat i+\sin\theta\,\hat j\bigr) =v\bigl(\cos\theta\,\hat i+\sin\theta\,\hat j\bigr),$$

because $$r\dot\theta=v$$ (the speed is unchanged).

The particle leaves the field when its $$y$$-coordinate equals $$d=r/2$$. The absolute $$y$$-coordinate is

$$y = r - r\cos\theta = r(1-\cos\theta).$$

Setting this equal to $$r/2$$ gives

$$r(1-\cos\theta)=\frac{r}{2}\;\;\Longrightarrow\;\;1-\cos\theta=\frac12 \;\;\Longrightarrow\;\;\cos\theta=\frac12 \;\;\Longrightarrow\;\;\theta=\frac{\pi}{3}\;(60^\circ).$$

Substituting $$\theta=\pi/3$$ into the velocity direction obtains

$$\vec v_{\text{exit}} =v\!\left(\frac12\,\hat i+\frac{\sqrt3}{2}\,\hat j\right).$$

Now we calculate the acceleration at the instant of exit. Using $$\vec a=\dfrac{q}{m}(\vec v\times\vec B)$$ with $$\vec B=B\hat k$$,

$$\vec v_{\text{exit}}\times\vec B =v\!\left(\frac12\,\hat i+\frac{\sqrt3}{2}\,\hat j\right)\!\times B\hat k =vB\!\left[ \frac12(\hat i\times\hat k)+\frac{\sqrt3}{2}(\hat j\times\hat k) \right].$$

Using the right-hand rule $$\hat i\times\hat k=-\hat j$$ and $$\hat j\times\hat k=\hat i$$, we get

$$\vec v_{\text{exit}}\times\vec B =vB\!\left[ \frac12(-\hat j)+\frac{\sqrt3}{2}\hat i \right] =vB\!\left( \frac{\sqrt3}{2}\,\hat i-\frac12\,\hat j \right).$$

Therefore

$$\boxed{\; \displaystyle \vec a_{\text{exit}} =\frac{qvB}{m}\left( \frac{\sqrt3}{2}\,\hat i-\frac12\,\hat j \right) \;} .$$

This vector is not identical to any of the three vectors listed in Options A, B and C, so none of those options is correct.

Hence, the correct answer is Option 4.

Let the mass of the earth be denoted by $$M_E$$ and the universal gravitational constant by $$G$$. For a satellite of mass $$M$$ moving in a circular orbit of radius $$R$$ the velocity is obtained from the condition for uniform circular motion

$$\displaystyle \frac{G\,M_E\,M}{R^2}= \frac{M v^2}{R}\;\;\Longrightarrow\;\;v=\sqrt{\frac{G\,M_E}{R}}\;.$$

The meteorite has the same mass $$M$$ and, just before impact, is given to have the same speed $$v$$. At the instant of collision the two velocity directions are orthogonal: the satellite velocity is tangential while the meteorite velocity is radially inward. Hence the angle between $$\vec v_{\!s}$$ (satellite) and $$\vec v_{\!m}$$ (meteorite) is $$90^\circ\;.$$

The collision is perfectly inelastic, so the two bodies stick together to form a single body of mass $$2M$$. Because external forces on the system are purely central (gravity through the earth’s centre), the angular momentum about the earth’s centre is conserved during the collision.

Initially, only the satellite possesses angular momentum because the meteorite is moving radially. Using $$\vec L=M\;\vec r\times\vec v$$ with $$r=R$$ we have

$$L_i = M\,R\,v\;.$$

After the collision the linear momenta add vectorially:

$$\vec p_f = M\vec v_{\!s}+M\vec v_{\!m} = M(\vec v_{\!s}+\vec v_{\!m}).$$

Since the two velocities are at right angles and equal in magnitude, their vector sum has magnitude

$$|\vec v_{\!s}+\vec v_{\!m}| = v\sqrt{2}\;.$$

The velocity of the fused body (mass $$2M$$) is therefore

$$\vec v'=\frac{\vec p_f}{2M}= \frac{\vec v_{\!s}+\vec v_{\!m}}{2},\qquad |\vec v'|=\frac{v\sqrt{2}}{2}= \frac{v}{\sqrt2}\;.$$

Because $$\vec v'$$ bisects the right angle between the original velocities, it makes $$45^\circ$$ with the tangent. Its tangential and radial components are each

$$v_t=\frac{v}{2},\qquad v_r=\frac{v}{2}\;.$$

The conserved angular momentum after collision equals

$$L_f = 2M\,R\,v_t = 2M\,R\left(\frac{v}{2}\right)=M\,R\,v=L_i\;,$$

so angular momentum conservation is satisfied.

The specific (per-unit-mass) angular momentum of the combined body is

$$h=\frac{L_f}{2M}=R\,v_t = R\left(\frac{v}{2}\right)=\frac{R\,v}{2}\;.$$

Next we evaluate the specific mechanical energy $$\varepsilon=\dfrac{v'^2}{2}-\dfrac{G\,M_E}{R}\;.$$ Using $$v'^2=\dfrac{v^2}{2}$$ and $$v^2=\dfrac{G\,M_E}{R}$$ we obtain

$$\varepsilon=\frac{1}{2}\left(\frac{v^2}{2}\right)-\frac{v^2}{1}= \frac{v^2}{4}-v^2=-\frac{3\,v^2}{4}\;.$$

The energy is negative, indicating a bound orbit.

To decide whether the bound orbit is circular or elliptical we employ the standard formula for the eccentricity $$e$$ of a Keplerian orbit expressed in terms of specific energy $$\varepsilon$$ and specific angular momentum $$h$$ :

$$e=\sqrt{\,1+\frac{2\varepsilon h^{2}}{(G\,M_E)^{2}}\,}\;.$$

Because $$v^2=\dfrac{G\,M_E}{R}$$ we write $$G\,M_E = v^2 R\;.$$ Now compute the term inside the square root:

$$\frac{2\varepsilon h^{2}}{(G\,M_E)^{2}} =\frac{2\left(-\dfrac{3v^{2}}{4}\right)\left(\dfrac{R^{2}v^{2}}{4}\right)}{(v^{2}R)^{2}} =\frac{-3R^{2}v^{4}/8}{R^{2}v^{4}} =-\frac{3}{8}\;.$$

Therefore

$$e^{2}=1-\frac{3}{8}=\frac{5}{8}\quad\Longrightarrow\quad e=\sqrt{\frac{5}{8}}<1\;.$$

An eccentricity less than unity but greater than zero corresponds to an ellipse that is not a circle. Thus the fused body of mass $$2M$$ moves in an elliptical orbit whose apogee is at the collision point (radius $$R$$) and whose perigee lies closer to the earth.

Hence, the correct answer is Option A.

$$h \ll R \implies r = R + h \approx R$$

Orbital speed of the satellite: $$v_o = \sqrt{\frac{GM}{r}} \approx \sqrt{\frac{GM}{R}}$$

Escape speed from that orbit: $$v_e = \sqrt{\frac{2GM}{r}} \approx \sqrt{\frac{2GM}{R}}$$

Acceleration due to gravity at the surface: $$g = \frac{GM}{R^2}$$

$$v_o = \sqrt{gR}$$

$$v_e = \sqrt{2gR}$$

$$\Delta v = v_e - v_o = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2} - 1)$$

We place the four identical particles at the corners of a square ABCD of side $$a$$. Each particle has mass $$M$$ and all of them are made to revolve together, keeping the square rigid, about the centre $$O$$ of the square. Because they keep their relative positions unchanged, every particle describes a circle whose centre is $$O$$ and whose radius is the distance from a corner to the centre.

The diagonal of the square is $$a\sqrt 2$$, so half of this diagonal gives the radius of the circular path,

$$R=\dfrac{a\sqrt 2}{2}=\dfrac{a}{\sqrt 2}.$$

We now pick one particle, say the one at corner A. It experiences gravitational attraction from the three remaining particles situated at B, C and D. According to Newton’s law of gravitation, the magnitude of the gravitational force exerted by a particle of mass $$M$$ on another particle of mass $$M$$ separated by a distance $$r$$ is

$$F=\dfrac{G M^2}{r^2},$$

where $$G$$ is the universal gravitational constant.

1. Attraction by the two adjacent particles B and D    • Distance AB = AD = $$a$$    • Magnitude of each force

$$F_1=\dfrac{G M^2}{a^2}.$$

These two forces act along AB and AD, which are perpendicular to each other. The direction OA (from A towards the centre O) bisects the right angle at A, therefore it makes an angle of $$45^{\circ}$$ with each side. Resolving each adjacent force towards the centre, we obtain the radial component

$$F_{1r}=F_1\cos45^{\circ}=F_1\dfrac{1}{\sqrt 2}.$$

Because there are two identical adjacent forces, the total radial pull supplied by them is

$$F_{adj}=2F_{1r}=2F_1\dfrac{1}{\sqrt 2}=\sqrt 2\,F_1.$$

(The tangential components cancel each other because they are equal in magnitude and opposite in direction.)

2. Attraction by the particle at the opposite corner C    • Distance AC (full diagonal) $$=a\sqrt 2$$    • Magnitude of the force

$$F_2=\dfrac{G M^2}{(a\sqrt 2)^2}=\dfrac{G M^2}{2a^2}=\dfrac{F_1}{2}.$$

This force acts exactly along the diagonal AC, i.e. along AO, hence it is already directed towards the centre and needs no resolution.

Therefore the net gravitational force on the particle at A directed towards the centre is

$$F_{\text{net}}=F_{adj}+F_2=\sqrt 2\,F_1+\dfrac{F_1}{2}=\left(\sqrt 2+\dfrac12\right)F_1.$$

Substituting $$F_1=\dfrac{G M^2}{a^2}$$ gives

$$F_{\text{net}}=\left(\sqrt 2+\dfrac12\right)\dfrac{G M^2}{a^2}.$$

For uniform circular motion, the required centripetal force is provided exactly by this net inward force. The centripetal force needed for a particle of mass $$M$$ moving with speed $$v$$ in a circle of radius $$R$$ is

$$F_c=M\dfrac{v^2}{R}.$$

Equating the two forces,

$$M\dfrac{v^2}{R}=\left(\sqrt 2+\dfrac12\right)\dfrac{G M^2}{a^2}.$$

We cancel one factor of $$M$$ from both sides and substitute $$R=\dfrac{a}{\sqrt 2}$$:

$$\dfrac{v^2}{a/\sqrt 2}=\left(\sqrt 2+\dfrac12\right)\dfrac{G M}{a^2}.$$

Multiplying both sides by $$\dfrac{a}{\sqrt 2}$$ gives

$$v^2=\dfrac{1}{\sqrt 2}\left(\sqrt 2+\dfrac12\right)\dfrac{G M}{a}.$$

Simplify the numerical factor:

$$\dfrac{1}{\sqrt 2}\left(\sqrt 2+\dfrac12\right)=\dfrac{\sqrt 2}{\sqrt 2}+\dfrac{1}{2\sqrt 2}=1+\dfrac{1}{2\sqrt 2}.$$

The term $$\dfrac{1}{2\sqrt 2}\approx0.3536$$, so the bracket equals $$1.3536$$. Hence

$$v^2\approx1.3536\,\dfrac{G M}{a}.$$

Taking the square root,

$$v\approx1.16\,\sqrt{\dfrac{G M}{a}}.$$

This numerical coefficient (1.16 rounded to two decimal places) matches option D.

Hence, the correct answer is Option D.

First, recall the formula for the speed of a satellite moving in a stable circular orbit of radius $$r$$ around the Earth (mass $$M$$, gravitational constant $$G$$). For circular motion, the required centripetal force is supplied entirely by gravity:

$$\frac{mv^{2}}{r}= \frac{GMm}{r^{2}}.$$

On cancelling the mass $$m$$ of the satellite (or of the object before ejection) and one factor of $$r$$, we get:

$$v^{2}= \frac{GM}{r}, \qquad\text{so}\qquad v = \sqrt{\frac{GM}{r}}.$$

Next, recall the escape-speed formula. The minimum speed required at a distance $$r$$ from the centre of the Earth for an object to reach infinity with zero final speed is

$$v_{e} = \sqrt{\frac{2GM}{r}}.$$

Notice that we can relate this escape speed to the satellite’s orbital speed by substituting $$\displaystyle v^{2}=\frac{GM}{r}$$ just obtained:

$$v_{e} = \sqrt{2}\,\sqrt{\frac{GM}{r}} = \sqrt{2}\,v.$$

The problem says the object is “ejected … such that it just escapes the gravitational pull of the Earth.” Therefore, right after ejection its speed relative to the Earth must be exactly the escape speed $$v_{e}=\sqrt{2}\,v$$.

The kinetic energy of the object of mass $$m$$ at that instant is, by definition,

$$K = \frac{1}{2} m v_{e}^{2}.$$

Substituting $$v_{e}^{2}=2v^{2}$$ obtained above, we get

$$K = \frac{1}{2} m \bigl(2v^{2}\bigr) = m v^{2}.$$

So the kinetic energy equals $$mv^{2}$$.

Hence, the correct answer is Option B.

We start by collecting all the numerical data given in the question.

Mass of the planet: $$M = 8 \times 10^{22}\,\text{kg}$$

Radius of the planet: $$R = 2 \times 10^{6}\,\text{m}$$

Height of the circular orbit above the surface: $$h = 20\,\text{km} = 20 \times 10^{3}\,\text{m} = 2.0 \times 10^{4}\,\text{m}$$

Universal gravitational constant: $$G = 6.67 \times 10^{-11}\,\text{N\,m}^2/\text{kg}^2$$

The distance of the spaceship from the centre of the planet is the orbital radius

$$r = R + h = 2 \times 10^{6}\,\text{m} + 2 \times 10^{4}\,\text{m} = 2.02 \times 10^{6}\,\text{m}.$$

For a body in a circular orbit, the gravitational force supplies the required centripetal force. We write this fundamental equality explicitly:

$$\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}.$$

Cancelling the common factor $$m$$ and one power of $$r$$, we obtain the orbital speed $$v$$:

$$v^{2} = \frac{G M}{r} \quad\Longrightarrow\quad v = \sqrt{\frac{G M}{r}}.$$

The time period $$T$$ of one revolution is the distance travelled in one orbit divided by the speed. The path length for one revolution is the circumference $$2\pi r$$, so

$$T = \frac{2\pi r}{v}.$$

Substituting the expression for $$v$$ that we just derived, we have

$$T = \frac{2\pi r}{\sqrt{\dfrac{G M}{r}}} = 2\pi r \sqrt{\frac{r}{G M}} = 2\pi \frac{r^{3/2}}{\sqrt{G M}}.$$

Now we evaluate each factor step by step.

First, compute $$r^{3}$$:

$$r = 2.02 \times 10^{6}\,\text{m},$$ $$r^{3} = (2.02 \times 10^{6})^{3} = 2.02^{3} \times 10^{18} = 8.244 \times 10^{18}\,\text{m}^{3}.$$

Taking the square root gives $$r^{3/2}$$:

$$r^{3/2} = \sqrt{r^{3}} = \sqrt{8.244 \times 10^{18}} = \sqrt{8.244}\,\times\,10^{9} \approx 2.87 \times 10^{9}\,\text{m}^{3/2}.$$

Next, calculate $$G M$$:

$$G M = 6.67 \times 10^{-11}\,\text{N\,m}^{2}\!/\text{kg}^{2}\; \times 8 \times 10^{22}\,\text{kg} = 53.36 \times 10^{11}\,\text{m}^{3}\!/\text{s}^{2} = 5.336 \times 10^{12}\,\text{m}^{3}\!/\text{s}^{2}.$$

We now need $$\sqrt{G M}$$:

$$\sqrt{G M} = \sqrt{5.336 \times 10^{12}} = \sqrt{5.336}\,\times\,10^{6} \approx 2.309 \times 10^{6}\,\text{m}^{3/2}\!/\text{s}.$$

With these two intermediate results, we substitute back into the formula for the period:

$$T = 2\pi\,\frac{r^{3/2}}{\sqrt{G M}} = 2\pi\,\frac{2.87 \times 10^{9}}{2.309 \times 10^{6}} = 2\pi\,\times\,1.243 \times 10^{3} = 6.2832 \times 1.243 \times 10^{3} \approx 7.81 \times 10^{3}\,\text{s}.$$

Thus one revolution takes approximately

$$T \approx 7.81 \times 10^{3}\,\text{s} = 7810\,\text{s}.$$

We convert this period into hours for convenience:

$$1\,\text{hour} = 3600\,\text{s},\quad T = \frac{7810}{3600}\,\text{h} \approx 2.17\,\text{h}.$$

The spaceship completes one revolution in about 2.17 hours. The total time we are interested in is 24 hours, so the number of complete revolutions $$N$$ is

$$N = \frac{24\,\text{h}}{2.17\,\text{h}} \approx 11.06.$$

Since only whole (complete) revolutions count, the spaceship makes 11 complete orbits in 24 hours.

Hence, the correct answer is Option D.

We consider a planet of mass $$m$$ describing a circular orbit of radius $$r$$ around the Sun. For such a motion the angular velocity about the Sun is denoted by $$\omega$$ (this is the rate of change of the polar angle $$\theta$$).

First we recall the definition of angular momentum about the centre of the Sun for a particle of mass $$m$$:

$$\displaystyle L = m\,r^{2}\,\omega.$$

Next we determine the areal velocity, i.e. the rate at which area is swept out by the radius vector. In a short time interval $$dt$$, the radius vector turns through a small angle $$d\theta$$, and the infinitesimal sector (of a circle) swept out has area

$$\displaystyle dA = \tfrac12\,r^{2}\,d\theta.$$

Dividing both sides by $$dt$$ gives the areal velocity:

$$\displaystyle \frac{dA}{dt} = \tfrac12\,r^{2}\,\frac{d\theta}{dt}.$$

But $$\frac{d\theta}{dt} = \omega$$, so

$$\displaystyle \frac{dA}{dt} = \tfrac12\,r^{2}\,\omega.$$

Now we substitute for $$r^{2}\,\omega$$ from the angular momentum expression $$L = m\,r^{2}\,\omega$$. Solving that equation for $$r^{2}\,\omega$$, we have

$$\displaystyle r^{2}\,\omega = \frac{L}{m}.$$

Placing this into the formula for the areal velocity:

$$\displaystyle \frac{dA}{dt} = \tfrac12\left(\frac{L}{m}\right).$$

Simplifying, we obtain

$$\displaystyle \frac{dA}{dt} = \frac{L}{2m}.$$

Hence, the correct answer is Option C.

Let the mass of the earth be denoted by $$M_E$$ and the universal gravitational constant by $$G$$. For a satellite moving in a circular orbit, the necessary centripetal force comes entirely from the gravitational attraction of the earth. Hence we write the force-centripetal balance equation:

$$\frac{G\,M_E\,m_s}{r^2} = \frac{m_s\,v^2}{r}.$$

Here $$m_s$$ is the mass of the satellite, $$r$$ is the orbital radius and $$v$$ is the orbital speed. Cancelling the common factor $$m_s$$ and one power of $$r$$ we get

$$v^2 = \frac{G\,M_E}{r}.$$

The kinetic energy of a body of mass $$m_s$$ moving with speed $$v$$ is given by the standard formula $$K = \tfrac12 m_s v^2$$. Substituting the expression for $$v^2$$ obtained above, we have

$$K = \tfrac12 m_s \left(\frac{G\,M_E}{r}\right) = \frac{G\,M_E\,m_s}{2\,r}.$$

This shows that for satellites around the same planet, the kinetic energy is directly proportional to the satellite mass and inversely proportional to the orbital radius:

$$K \propto \frac{m_s}{r}.$$

Now we apply this result to the two satellites mentioned in the question.

Satellite A has mass $$m_A = m$$ and orbital radius $$r_A = R$$. Therefore,

$$K_A = \frac{G\,M_E\,m}{2\,R}.$$

Satellite B has mass $$m_B = 2m$$ and orbital radius $$r_B = 2R$$. Therefore,

$$K_B = \frac{G\,M_E\,(2m)}{2\,(2R)} = \frac{2\,G\,M_E\,m}{4\,R} = \frac{G\,M_E\,m}{2\,R}.$$

We observe that the factors $$G$$, $$M_E$$ and $$2R$$ neatly cancel to give the same numerator and denominator as in the expression for $$K_A$$. Hence

$$K_A = K_B.$$

Taking the ratio,

$$\frac{K_A}{K_B} = \frac{K_B}{K_B} = 1.$$

Hence, the correct answer is Option C.

We are told that both particles are of the same mass $$m$$ and are moving in circular orbits under a central force

$$F(r)= -\frac{16}{r}-r^{3}.$$

The negative sign merely shows that the force is attractive, so for the purpose of balancing the centripetal requirement we will use its magnitude

$$|F(r)|=\frac{16}{r}+r^{3}.$$

For a body of mass $$m$$ in uniform circular motion of radius $$r$$ and speed $$v$$, the centripetal force is given by the standard relation

$$F_{\text{centripetal}}=\frac{m v^{2}}{r}.$$

Because the circular motion is maintained by the given force, we equate the magnitudes:

$$\frac{m v^{2}}{r}= \frac{16}{r}+r^{3}.$$

Multiplying both sides by $$r$$ to clear the denominator, we get

$$m v^{2}= 16 + r^{4}.$$

Now we solve for $$v^{2}$$:

$$v^{2}= \frac{16 + r^{4}}{m}.$$

The kinetic energy of a particle is expressed by the familiar formula

$$K=\frac{1}{2} m v^{2}.$$

Substituting the value of $$v^{2}$$ obtained above, we obtain

$$K=\frac{1}{2} m\left(\frac{16 + r^{4}}{m}\right)=\frac{1}{2}(16 + r^{4}).$$

Notice that the mass $$m$$ cancels out, so the kinetic energy depends only on the radius:

$$K(r)=8 + \frac{r^{4}}{2}.$$

For the first particle, the radius is $$r_{1}=1$$. Substituting $$r=1$$ in the expression for $$K(r)$$, we have

$$K_{1}=8 + \frac{1^{4}}{2}=8 + \frac{1}{2}=8.5.$$

For the second particle, the radius is $$r_{2}=4$$. Substituting $$r=4$$, we get

$$K_{2}=8 + \frac{4^{4}}{2}=8 + \frac{256}{2}=8 + 128=136.$$

We need the ratio of the kinetic energies of the first particle to the second particle:

$$\frac{K_{1}}{K_{2}}=\frac{8.5}{136}.$$

Dividing numerator and denominator by 136, or using a calculator, gives

$$\frac{K_{1}}{K_{2}}\approx 0.0625=6.25 \times 10^{-2}.$$

Among the options provided, $$6 \times 10^{-2}$$ is closest to this numerical value.

Hence, the correct answer is Option C.

A particle of mass $$ M $$ is moving in a circle of fixed radius $$ R $$ with centripetal acceleration given by $$ n^2 R t^2 $$, where $$ n $$ is a constant. We need to find the power delivered to the particle by the force acting on it.

First, recall that centripetal acceleration $$ a_c $$ is related to the tangential velocity $$ v $$ by the formula:

$$ a_c = \frac{v^2}{R} $$

Given that $$ a_c = n^2 R t^2 $$, we substitute:

$$ \frac{v^2}{R} = n^2 R t^2 $$

Solving for $$ v^2 $$, multiply both sides by $$ R $$:

$$ v^2 = n^2 R t^2 \times R $$

$$ v^2 = n^2 R^2 t^2 $$

Taking the square root of both sides (and considering speed as positive, so we take the positive root):

$$ v = n R t $$

Now, centripetal acceleration changes with time, which means the speed is changing. Therefore, there must be a tangential acceleration $$ a_t $$ responsible for changing the speed. Tangential acceleration is the derivative of velocity with respect to time:

$$ a_t = \frac{dv}{dt} $$

Substitute $$ v = n R t $$:

$$ a_t = \frac{d}{dt} (n R t) $$

Since $$ n $$ and $$ R $$ are constants:

$$ a_t = n R \frac{d}{dt}(t) = n R \times 1 = n R $$

The tangential force $$ F_t $$ is given by Newton's second law:

$$ F_t = M a_t = M \times n R = M n R $$

The centripetal force $$ F_c $$ is perpendicular to the velocity and does no work, so it delivers no power. Power is delivered only by the tangential force component, which is parallel to the velocity. Power $$ P $$ is the dot product of force and velocity:

$$ P = \vec{F} \cdot \vec{v} = F_t \times v \quad (\text{since they are parallel}) $$

Substitute $$ F_t = M n R $$ and $$ v = n R t $$:

$$ P = (M n R) \times (n R t) $$

$$ P = M n R \times n R t $$

$$ P = M n^2 R^2 t $$

Comparing with the options:

A. $$ \frac{1}{2} M n^2 R^2 t^2 $$

B. $$ M n^2 R^2 t $$

C. $$ M n R^2 t^2 $$

D. $$ M n R^2 t $$

Option B matches our result. Hence, the correct answer is Option B.

The formula for centripetal acceleration is $$a = \frac{v^2}{r}$$

Since $$v^2$$ is constant, the relationship becomes $$a \propto \frac{1}{r}$$

This represents a rectangular hyperbola.

Option (D) represents this correctly.