JEE Laws of Motion Questions

Let the centre of the ring be $$O$$ and let the centre of the disk be $$C$$. Because the disk rolls without slipping on the inner wall of the ring, the distance $$OC$$ remains constant and is equal to the inner radius available for the centre,

$$a = R - r$$

Choose the small angular displacement $$\theta$$ of the centre $$C$$ from the lowest (vertical) position as the generalised coordinate. Arc‐length travelled by the centre is then

$$s = a\theta \qquad (|\theta| \ll 1)$$

1. Constraint of pure rolling
Pure rolling gives a fixed relation between the translational speed of the centre and the angular speed of the disk about its own axis:

$$v_{\text{cm}} = a\dot\theta = r\dot\phi \qquad -(1)$$

where $$\dot\phi$$ is the angular velocity of the disk about its diameter.

2. Kinetic energy
Translational kinetic energy: $$T_{\text{trans}} = \tfrac12 m v_{\text{cm}}^{2} = \tfrac12 m (a\dot\theta)^{2}$$

Moment of inertia of a solid disk about its centre: $$I = \tfrac12 m r^{2}$$

Using the rolling relation $$(1)$$, rotational kinetic energy is

$$T_{\text{rot}} = \tfrac12 I\dot\phi^{2} = \tfrac12\!\left(\tfrac12 m r^{2}\right)\! \left(\tfrac{a\dot\theta}{r}\right)^{2} = \tfrac14 m a^{2}\dot\theta^{2}$$

Thus total kinetic energy is

$$T = T_{\text{trans}} + T_{\text{rot}} = \left(\tfrac12 + \tfrac14\right)m a^{2}\dot\theta^{2} = \tfrac34 m a^{2}\dot\theta^{2} \qquad -(2)$$

3. Potential energy

(i) Gravitational: For a small angular displacement, the rise in the centre’s height is $$h = a(1-\cos\theta) \approx \tfrac12 a\theta^{2}$$ Therefore

$$U_g = mgh = \tfrac12 m g a\theta^{2} \qquad -(3)$$

(ii) Spring: The spring is attached tangentially along the periphery, so its extension/compression equals the arc length $$s = a\theta$$. Hooke’s law gives

$$U_s = \tfrac12 k s^{2} = \tfrac12 k a^{2}\theta^{2} \qquad -(4)$$

Total potential energy: $$U = U_g + U_s$$

$$U = \tfrac12\left(m g a + k a^{2}\right)\theta^{2} \qquad -(5)$$

4. Small-oscillation equation
Using the Lagrangian $$L = T - U$$ with $$T$$ from $$(2)$$ and $$U$$ from $$(5)$$:

$$L = \tfrac34 m a^{2}\dot\theta^{2} - \tfrac12\left(m g a + k a^{2}\right)\theta^{2}$$

Applying the Euler-Lagrange equation $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta}\right) - \frac{\partial L}{\partial\theta}=0$$ gives

$$\frac{d}{dt}\!\left(\tfrac34\cdot 2 m a^{2}\dot\theta\right) + \left(m g a + k a^{2}\right)\theta = 0$$

$$\tfrac32 m a^{2}\ddot\theta + m g a\theta + k a^{2}\theta = 0 \qquad -(6)$$

Divide by $$\tfrac32 m a^{2}$$:

$$\ddot\theta + \left[\frac{2}{3}\left(\frac{g}{a} + \frac{k}{m}\right)\right]\theta = 0$$

This is the standard SHM form $$\ddot\theta + \omega^{2}\theta = 0$$ with

$$\omega^{2} = \frac{2}{3}\left(\frac{g}{a} + \frac{k}{m}\right) = \frac{2}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right) \qquad -(7)$$

5. Time period

$$T = \frac{2\pi}{\omega}$$ with $$\omega$$ given by $$(7)$$.

Hence the correct expression is

$$\boxed{\displaystyle \omega = \sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)}}$$

Option A perfectly matches this result.

A massless spring elongates by $$x_1$$ under 5N and $$x_2$$ under 7N. Find the tension for elongation $$(5x_1 - 2x_2)$$.

By Hooke's law:

$$F = kx$$, where k is the spring constant.

$$5 = kx_1 \Rightarrow x_1 = 5/k$$

$$7 = kx_2 \Rightarrow x_2 = 7/k$$

Find the elongation:

$$5x_1 - 2x_2 = 5 \times \frac{5}{k} - 2 \times \frac{7}{k} = \frac{25 - 14}{k} = \frac{11}{k}$$

Find the tension:

$$F = k \times \frac{11}{k} = 11 \text{ N}$$

The correct answer is Option 3: 11 N.

First step is to find out $$v_f$$ which is the final velocity with which both the blocks along with the spring move

Using conservation of momentum:-

so $$m_1v_1=\left(m_1+m_2\right)v_f$$

$$v_f=\frac{m_1v_1}{m_1+m_2}$$

On substituting the given values

$$v_f=\frac{10\times\ 3}{10+5}$$

$$v_f=\frac{30}{15}=2\ \frac{m}{s}$$

Now lets conserve Energy

Initital K.E = $$\frac{1}{2}m_1v_1^2$$

$$=\frac{1}{2}\times\ 10\times\left(\ 3^2\right)$$

$$=5\times9$$

$$K.E_i=45$$

Final K.E is $$\frac{1}{2}\left(m_1+m_2\right)v_f^2$$

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

$$=15\times\ 2$$

$$K.E_f=30$$

Now difference between these is the energy stored in the spring

$$K.E_i-K.E_f$$$$K.E_i-K.E_f\ =\ \ 45-\ 30\ =15$$

Now 

$$\frac{1}{2}kx^2=15$$

$$\frac{1}{2}\times\ 3000\times\ x^2=15$$

$$\ x^2=\frac{30}{3000}$$

$$\ x^2=\frac{1}{100}=10^{-2}$$

$$\ x=10^{-1}=0.1m$$

Given:

$$T_1=\sqrt{\ 3}T_2$$

Horizontal equilibrium

$$T_1\cos\theta_1=T_2\cos\theta_2$$

Substitute:

$$\sqrt{3}T_2\cos\theta_1=T_2\cos\theta_2$$​

Cancel $$T_2$$​:

$$\sqrt{3}\cos\theta_1=\cos\theta_2$$

From the options

We know:

• $$\cos60^{\circ}=\frac{1}{2}$$
• $$\cos30^{\circ}=\frac{\sqrt{3}}{2}$$

Try:

$$\theta_1=60^{\circ\ }$$

Then:

$$\sqrt{3}\cos60^{\circ}=\sqrt{3}\cdot\frac{1}{2}=\frac{\sqrt{3}}{2}$$​​

So:

$$\cos\ \theta\ _2=\frac{\sqrt{\ 3}}{2}$$

$$\theta\ _2=30^{\circ\ }$$

$$T_1\sin\theta_1+T_2\sin\theta_2=mg$$

Substitute:

$$\sqrt{3}T_2\cdot\frac{\sqrt{3}}{2}+T_2\cdot\frac{1}{2}=mg$$

$$\frac{3}{2}T_2+\frac{1}{2}T_2=mg\Rightarrow2T_2=mg$$

$$T_2 = \dfrac{mg}{2}$$

Initial velocity of the body is given as $$\vec v_{in}=3\hat i+4\hat j$$ ms$$^{-1}$$, so

$$u_x = 3$$ ms$$^{-1},\; u_y = 4$$ ms$$^{-1},\; u_z = 0$$ ms$$^{-1}$$.

The constant force acting on the mass is $$\vec F = 6\hat k$$ N (along +z-axis).

Using Newton’s second law, $$\vec F = m\vec a \Rightarrow \vec a = \frac{\vec F}{m}$$.

The mass of the body is $$m = 2$$ kg, therefore

$$\vec a = \frac{6\hat k}{2} = 3\hat k$$ ms$$^{-2}$$.

The body stays in this field for $$t = \frac{5}{3}$$ s. The kinematics relation $$\vec v = \vec u + \vec a t$$ gives the final velocity:

Along x-axis: $$v_x = u_x + a_x t = 3 + 0 \times t = 3$$ ms$$^{-1}$$ (no force in x).
Along y-axis: $$v_y = u_y + a_y t = 4 + 0 \times t = 4$$ ms$$^{-1}$$ (no force in y).
Along z-axis: $$v_z = u_z + a_z t = 0 + 3 \times \frac{5}{3} = 5$$ ms$$^{-1}$$.

Hence the velocity vector when the body emerges is

$$\vec v_{out}=3\hat i + 4\hat j + 5\hat k$$ ms$$^{-1}$$.

Therefore, the correct option is Option B.

Initially: F_buoy - Mg = Ma, so F_buoy = M(g+a).

After releasing mass m: F_buoy - (M-m)g = (M-m)(3a).

M(g+a) = (M-m)(g+3a), M(g+a) = M(g+3a) - m(g+3a)

m(g+3a) = M(g+3a-g-a) = 2Ma, m = 2Ma/(g+3a).

The correct answer is Option 1.

Here as the system is in equilibrium 

We have to equate horizontal components of both $$T_{1\ }and\ T_2$$ as there is no external Force acting in the horizontal direction

Here horizontal component of $$T_{1\ }is\ T_1\cos60^{\circ\ }$$

Here horizontal component of $$T_{2\ }is\ T_2\cos30^{\circ\ }$$

Equate these 2

$$T_2\cos60^{\circ\ }=\ T_1\cos30^{\circ\ }$$

$$T_2\times\ \frac{\sqrt{\ 3}}{2}=\ T_1\times\ \frac{1}{2}$$

$$T_1\ =\ \sqrt{\ 3}T_2$$

Now coming to the Vertical direction 

There is gravitational force downwards which is Mg and vertical components of $$T_{1\ }and\ T_2$$ add up and are opposing the force due to gravitation

so $$Mg=\ T_1\sin\theta\ \ +\ T_2\sin\ \ \theta\ $$

$$Mg=\ T_1\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$$

Substituting  $$T_1\ =\ \sqrt{\ 3}T_2$$ in that equation

$$Mg=\ \sqrt{\ 3}T_2\times\ \frac{\sqrt{\ 3}}{2}\ +\ T_2\ \times\ \frac{1}{2}$$

$$Mg=\frac{3T_2}{2}+\frac{T_2}{2}$$
$$Mg=2T_2$$

$$T_2=\frac{g}{2}\ \ \left(as\ M=1\right)$$

$$T_2=5N$$

as $$T_1\ =\ \sqrt{\ 3}T_2$$

$$T_1=5\sqrt{\ 3}N$$

The speed of the object varies with its position as $$v = 4\sqrt{x}$$, where $$x$$ is the coordinate (in metres) measured along the +x-axis.

First convert the mass to SI units: given mass $$m = 500\,$$g $$= 0.5\,$$kg.

To find the force we need the acceleration. For one-dimensional motion, acceleration can be expressed in terms of position as
$$a = v\,\frac{dv}{dx}$$

STEP 1 - Differentiate the speed with respect to $$x$$:
$$v = 4x^{1/2} \; \Rightarrow \; \frac{dv}{dx} = 4 \cdot \frac{1}{2}x^{-1/2} = 2x^{-1/2}$$

STEP 2 - Use $$a = v\,\dfrac{dv}{dx}$$:
Substitute $$v = 4x^{1/2}$$ and $$\dfrac{dv}{dx} = 2x^{-1/2}$$:
$$a = \left(4x^{1/2}\right)\left(2x^{-1/2}\right) = 8 \text{ m\,s}^{-2}$$

The acceleration turns out to be a constant $$8 \text{ m\,s}^{-2}$$ (independent of $$x$$).

STEP 3 - Apply Newton’s second law $$F = ma$$:
$$F = 0.5 \times 8 = 4 \text{ N}$$

Hence the force acting on the object is $$4 \text{ N}$$.

Therefore, Option D is correct.

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

Two cars P and Q move on a road in the same direction. Car P has an acceleration that increases linearly with time, and car Q has constant acceleration. They cross each other at $$t = 0$$ for the first time, and we need to find the maximum possible number of crossings.

We begin by setting up the position equations. Let both cars be at the same position at $$t = 0$$ and define the relative position $$\Delta x = x_P - x_Q$$.

For car P the acceleration is $$a_P = \alpha t$$ (linearly increasing, where $$\alpha > 0$$ is a constant). Integrating gives the velocity $$v_P = v_{P0} + \frac{\alpha t^2}{2}$$ and the position $$x_P = x_0 + v_{P0}t + \frac{\alpha t^3}{6}$$.

Similarly, for car Q the constant acceleration is $$a_Q = \beta$$. Integrating gives the velocity $$v_Q = v_{Q0} + \beta t$$ and the position $$x_Q = x_0 + v_{Q0}t + \frac{\beta t^2}{2}$$.

Subtracting these positions, the relative position becomes

$$\Delta x = x_P - x_Q = (v_{P0} - v_{Q0})t + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$$

Letting $$c = v_{P0} - v_{Q0}$$ (the initial relative velocity) simplifies this to

$$\Delta x = ct + \frac{\alpha t^3}{6} - \frac{\beta t^2}{2}$$.

Crossings occur when $$\Delta x = 0$$, so we set

$$t\left(c + \frac{\alpha t^2}{6} - \frac{\beta t}{2}\right) = 0$$

This equation yields the trivial root $$t = 0$$ (the first crossing) and the quadratic

$$\frac{\alpha}{6}t^2 - \frac{\beta}{2}t + c = 0$$.

This quadratic can have at most two real roots. In order for both of these roots to be positive (since $$t > 0$$), three conditions must be satisfied:

- Discriminant $$\left(\frac{\beta}{2}\right)^2 - 4 \cdot \frac{\alpha}{6} \cdot c \geq 0$$, i.e., $$\frac{\beta^2}{4} \geq \frac{2\alpha c}{3}$$

- Sum of roots $$\frac{\beta/2}{\alpha/6} = \frac{3\beta}{\alpha} > 0$$ (true since $$\alpha, \beta > 0$$)

- Product of roots $$\frac{c}{\alpha/6} = \frac{6c}{\alpha} > 0$$, which requires $$c > 0$$

When these conditions are met for appropriate values of $$\alpha$$, $$\beta$$, and $$c$$, the quadratic has two positive real roots, corresponding to two additional crossings beyond the one at $$t = 0$$.

To see that no more crossings are possible, note that $$\Delta x / t$$ is a quadratic function, so after factoring out the root at $$t = 0$$ the remaining quadratic factor can have at most two positive zeros. Therefore, the total number of crossings cannot exceed three.

In conclusion, the maximum possible number of crossings (including the one at $$t = 0$$) is $$\boxed{3}$$.

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): $$\frac{\pi}{2}$$.

Let $$F_1 = F$$ and $$F_2 = 3F$$. The resultant equals the larger force: $$R = 3F$$.

Using the formula: $$R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$$

$$9F^2 = F^2 + 9F^2 + 6F^2\cos\theta$$

$$0 = F^2 + 6F^2\cos\theta$$

$$\cos\theta = -\frac{1}{6}$$

So $$\theta = \cos^{-1}\left(\frac{1}{n}\right)$$ where $$n = -6$$, and $$|n| = 6$$.

The answer is 6.

We have three identical spheres each of mass $$2M$$ at points O(0,0), A(4,0) and B(0,4).

Formula for centre of mass coordinates: $$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}\quad -(1)$$ and $$y_{cm} = \frac{\sum m_i y_i}{\sum m_i}\quad -(2)$$.

Total mass = $$2M + 2M + 2M = 6M$$.

From $$(1)$$:
$$x_{cm} = \frac{2M\cdot0 + 2M\cdot4 + 2M\cdot0}{6M} = \frac{8M}{6M} = \frac{4}{3}$$.

From $$(2)$$:
$$y_{cm} = \frac{2M\cdot0 + 2M\cdot0 + 2M\cdot4}{6M} = \frac{8M}{6M} = \frac{4}{3}$$.

Position vector magnitude of centre of mass:
Formula: $$r = \sqrt{x_{cm}^2 + y_{cm}^2}\quad -(3)$$.

Substituting:
$$r = \sqrt{\Bigl(\frac{4}{3}\Bigr)^2 + \Bigl(\frac{4}{3}\Bigr)^2} = \frac{4}{3}\sqrt{2} = \frac{4\sqrt{2}}{3}$$.

Comparing with $$\frac{4\sqrt{2}}{x}$$ gives $$x = 3$$.

Answer: 3.

All three masses move upward with acceleration a=$$2 \text{ ms}^{-2}$$

Total mass:

$$m_{total}=4+6+10=20kg$$

Now treat the three blocks as a single body.

Forces on the system:

• Upward: $$T_1$$
• Downward: total weight $$=20g$$

Apply Newton’s second law (upward positive):

$$T_1-20g=20\times2$$

Substitute $$g = 10 \text{ m/s}^2$$

$$T_1-200=40$$

$$T_1=240N$$

Final answer: 240 N

Masses$$:m₁=2kg,m₂=4kg$$

For Atwood machine:

Acceleration:
$$a=\frac{(m₂−m₁)g}{(m₁+m₂)}$$
= $$\frac{\left((4−2)\times10\right)}{(2+4)}$$

$$=\frac{20}{6}=\frac{10}{3}$$

Tension:
$$T=m₁(g+a)$$
$$=2(10+10/3)$$
=$$2\times(40/3)$$
$$=\frac{80}{3}N$$

Now strain:

$$strain=stress/Y$$

$$Stress=T/A$$

Area of wire:
$$A=\pi r^2=\pi(4\times10⁻⁵)^2$$
$$=\pi\times16\times10⁻^1⁰$$
$$=16\pi\times10⁻^1⁰$$

So:

$$strain=T/(AY)$$

$$=(80/3)/[(16\pi\times10⁻^1⁰)(2\times10^{11})]$$

Simplify denominator:

$$16\times2=32$$
$$10⁻^1⁰\times10^{11}=10$$

$$Sodeno\min ator=32\pi\times10=320\pi$$

Now:

$$strain=(80/3)/(320\pi)$$
$$=80/(960\pi)$$
$$=1/(12\pi)$$

So given:

$$strain=1/(α\pi)$$

$$⇒α=12$$

Final answer:

$$12$$

In half-deflection method,

Current through galvanometer:
$$I=\frac{E}{(R+G)}$$

Deflection θ ∝ I
⇒ $$θ=kI$$
⇒$$θ=\frac{kE}{(R+G)}$$

So:
$$\frac{1}{θ}=\frac{\left(R+G\right)}{(kE)}$$

⇒$$1/θ=(1/kE)R+(G/kE)$$

This is of the form:
$$y=mx+c$$

So slope $$=1/(kE)$$

From graph:

Take two points:
$$(R=2,1/θ=1)$$

$$(R=6,1/θ=2)$$

Slope:
$$m=\frac{(2−1)}{(6−2)}=1/4$$

So:
$$1/(kE)=1/4$$

Look at the markings:
the vertical values like $$1,1\frac{1}{2},2$$ are not raw values — they are compressed scale values

Because of this, the slope you calculate  is actually 4 times smaller than the true slope

So real slope $$=4\times(1/4)=1$$

Now:

$$1/(kE)=1$$
$$⇒kE=1$$
$$⇒k\times2=1$$
$$⇒k=\frac{1}{2}A/div$$

Now convert to required form:

$$k=0.5A/div=5\times10⁻^1A/div$$

So answer = 5

First lets write the Forces acting in Vertical and Horizontal direction

In the Vertical direction,

Forces are $$F_g=mg\ due\ to\ gravity\ and\ T\sin\theta\ \ \left(\ tension\ of\ the\ rope\right)$$

and in Horizontal , The forces are $$F\left(appplied\ force\right)\ and\ T\cos\theta\ \left(tension\ of\ the\ rope\right)$$

First lets equate the Vertical Forces

$$mg\ =\ T\sin\ \theta\ $$

$$10=\ T\times\ \frac{1}{\sqrt{\ 2}}$$

Now substitute this and equal Horizontal forces

$$F=\ T\cos\theta$$

$$F=\ \sqrt{\ 2}\times\ 10\times\ \cos45^{\circ\ }\ $$

$$F=\ \frac{10}{\sqrt{\ 2}}\times\ \sqrt{\ 2}$$

$$F=\ 10N$$

Given that the block begins to slide 

so $$f_l=Force\ acting\ down\ the\ incline$$

We know that $$f_l=\mu\ _s\times\ N$$ , Where N is the normal force acting on the block

Here the force acting down the incline is $$Mg\sin\theta\ $$

so $$f_l=Mg\sin\theta\ $$

Now 

$$f_l=Mg\sin\theta\ =\mu\ _s\times\ N$$

N is normal force which is, $$N=Mg\cos\theta\ $$

$$f_l=Mg\sin\theta\ =\mu\ _s\times\ Mg\cos\theta\ $$

$$\mu\ _s=\tan\ \theta\ $$

Its given that $$\theta\ =45^{\circ\ }$$

so 

$$\mu\ _s=1$$

We need to find the force with which the branch pulls the chain.

Weight of body = 200 N, mass of chain = 10 kg, $$g = 10$$ m/s$$^2$$.

$$W_{chain} = mg = 10 \times 10 = 100 \text{ N}$$

The branch must support both the chain and the body. The total downward force at the point where the chain meets the branch is:

$$F_{total} = W_{body} + W_{chain} = 200 + 100 = 300 \text{ N}$$

The chain pulls the branch downward with 300 N. By Newton's third law, the branch pulls the chain upward with an equal and opposite force of 300 N.

The correct answer is Option 3: 300 N.

A cricket player catches a ball of mass $$m = 120 \text{ g} = 0.12 \text{ kg}$$ moving with velocity $$v = 25 \text{ m/s}$$. The catching process takes $$\Delta t = 0.1 \text{ s}$$.

The initial momentum of the ball is calculated as $$p_i = mv = 0.12 \times 25 = 3 \text{ kg m/s}$$, and since the ball comes to rest, the final momentum is $$p_f = 0$$. Therefore, the change in momentum is $$\Delta p = |p_f - p_i| = 3 \text{ kg m/s}$$.

The average force exerted during the catch is then $$F = \frac{\Delta p}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$$.

Hence, the answer is Option D: $$30$$.

We compare the descent of an object along a smooth incline with that along a rough incline, both inclined at 45° and covering the same distance s. In the absence of friction, the component of gravitational acceleration along the plane is $$a_s = g\sin 45° = \frac{g}{\sqrt{2}}$$, so that from $$s = \tfrac{1}{2}a_s t_s^2$$ it follows that $$t_s = \sqrt{\frac{2s}{a_s}}\,. $$

When kinetic friction is present, the net acceleration down the plane becomes $$a_r = g\sin 45° - \mu_k g\cos 45° = \frac{g}{\sqrt{2}}\,(1 - \mu_k)\,, $$ and the time to travel the same distance s satisfies $$t_r = \sqrt{\frac{2s}{a_r}}\,. $$

Because the object takes n times as long on the rough incline, we have $$\frac{t_r}{t_s} = n\,. $$ Squaring both sides yields $$\frac{t_r^2}{t_s^2} = n^2\quad\Longrightarrow\quad \frac{a_s}{a_r} = n^2\,. $$ Substituting the expressions for $$a_s$$ and $$a_r$$ gives

$$\frac{g/\sqrt{2}}{(g/\sqrt{2})(1-\mu_k)} = \frac{1}{1-\mu_k} = n^2\,, $$

from which $$1 - \mu_k = \frac{1}{n^2}\quad\Longrightarrow\quad \mu_k = 1 - \frac{1}{n^2}\,. $$

Therefore, the coefficient of kinetic friction is $$\mu_k = 1 - \frac{1}{n^2}\,.$$

Lets start writing equations for both the blocks

For $$m_1$$ : -

$$m_1g-T=m_1a$$

For $$m_2$$: - 

$$T-m_2g=m_2a$$

Given that $$a=\frac{g}{8}$$

On substituting a in both equations and adding them we get , 

$$m_1g-\ m_2g\ =\ m_1\left(\frac{g}{8}\right)+m_2\left(\frac{g}{8}\right)$$

$$\left(m_1-\ m_2\right)g\ =\ \left(m_1+m_2\right)\left(\frac{g}{8}\right)$$

$$\frac{\left(m_1-\ m_2\right)}{m_1+m_2}=\frac{1}{8}$$

On dividing Numerator and denominator with $$m_2$$

$$\frac{\left(\frac{m_1}{m_2}-1\right)}{\left(\frac{m_1}{m_2}+1\right)}=\frac{1}{8}$$

consider $$\frac{m_1}{m_2}$$ as x and solve for x

$$\frac{\left(x-1\right)}{\left(x+1\right)}=\frac{1}{8}$$

$$\left(x-1\right)=\frac{1}{8}\left(x+1\right)$$

$$x-\frac{1}{8}x=\frac{1}{8}+1$$

$$\frac{7}{8}x=\frac{9}{8}$$

$$x=\frac{9}{7}$$

For an Atwood machine, the acceleration is given by:

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$

(assuming $$m_2 > m_1$$)

Given $$a = \frac{g}{8}$$:

$$\frac{m_2 - m_1}{m_1 + m_2} = \frac{1}{8}$$

$$8(m_2 - m_1) = m_1 + m_2$$

$$8m_2 - 8m_1 = m_1 + m_2$$

$$7m_2 = 9m_1$$

$$\frac{m_2}{m_1} = \frac{9}{7}$$

The correct answer is Option 4: $$9:7$$.

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).

A train starts from rest, accelerates uniformly to 80 km/h in time $$t$$, then moves at constant speed for time $$3t$$. We need the average speed.

Calculate the distance during acceleration.

During uniform acceleration from 0 to 80 km/h over time $$t$$, the average speed is $$\frac{0 + 80}{2} = 40$$ km/h.

$$ d_1 = \text{average speed} \times \text{time} = 40t \text{ (in km, with } t \text{ in hours)} $$

Calculate the distance during constant speed.

$$ d_2 = 80 \times 3t = 240t \text{ km} $$

Calculate the average speed.

$$ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d_1 + d_2}{t + 3t} = \frac{40t + 240t}{4t} = \frac{280t}{4t} = 70 \text{ km/h} $$

The correct answer is Option (4): 70.

You have:

• a 2 kg block on a 30° incline
• connected by a string over a pulley
• then a movable pulley holding a 4 kg mass

Key idea: movable pulley changes both force and motion relation

Step 1: understand tension

Let tension in string = T

For the 4 kg block:
It is supported by two segments of the string
So upward force = 2T

Equation:
$$2T−4g=4a₄.$$

Step 2: relation of acceleration

Very important part

If the 2 kg block moves up the incline by x
the movable pulley (and 4 kg mass) moves down by x/2

So:
$$a₂=2a₄$$

Now lets write equation for the 2 Kg block

$$m_2g\sin\ \theta\ -T=m_2a_2$$

 $$2g\times\ \frac{1}{2}-T=2a_2$$

$$g-T=4a_4$$

$$2T−4g=4a₄.$$

now on multiply and adding both so that T gets cancelled

$$-2g=12a_4$$

$$\ a_4=-\frac{g}{6}\left(negative\ as\ we\ considered\ opposite\ direction\right)$$

so now we know 

$$a₂=2a₄$$

so 

$$a₂=2\times\ \frac{g}{6}$$

$$a₂=\ \frac{g}{3}$$

Statement I: The limiting force of static friction depends on the area of contact and independent of materials. This is incorrect. Static friction depends on the coefficient of friction (which depends on materials) and normal force, but is independent of area of contact.

Statement II: The limiting force of kinetic friction is independent of the area of contact and depends on materials. This is correct. Kinetic friction depends on the coefficient of kinetic friction (material-dependent) and normal force, but not on area.

Statement I is incorrect but Statement II is correct. The answer 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): $$\sqrt{3} : 2$$.

Let us write equations for all three blocks A,B and C

For A:-

$$F-T_2=2a$$

$$80-T_2=2a$$

For B:-

$$T_2-T_1=3a$$

and For C:-

$$T_1=5a$$

Now on adding all 3 equations

We get $$80\ =\ 10a$$

$$a = 8 \text{ m s}^{-2}$$

now substitute value of a in equation 1

$$80-T_2=2a$$

$$80-T_2=2\times\ 8$$

$$80-16\ =T_2$$

$$T_2=64N$$

Substitute the value of a in equation 3

$$T_1=5a$$

$$T_1=5\times\ 8$$

$$T_1=40N$$

We need to find the work done against friction when a 100 kg block slides 10 m on a horizontal surface with a coefficient of friction of 0.4.

The kinetic friction force on a horizontal surface is:

$$f = \mu m g$$

where $$\mu$$ is the coefficient of friction, $$m$$ is the mass, and $$g$$ is the acceleration due to gravity.

$$f = 0.4 \times 100 \times 10 = 400 \text{ N}$$

(using $$g = 10$$ m/s$$^2$$)

Work done against friction is:

$$W = f \times d = 400 \times 10 = 4000 \text{ J}$$

where $$d = 10$$ m is the distance.

Note: The work done against friction is positive because the friction force opposes the direction of motion. The energy is converted into heat.

The correct answer is Option (3): 4000 J.

$$m=5kg,θ=30°,μ=0.1=1/10,g=10$$

Normal reaction:
$$N=mg\cosθ=5\times10\times\cos30°=50\times(\sqrt{3}/2)=25\sqrt{3}$$

Friction:
$$f=μN=(1/10)\times25\sqrt{3}=(5\sqrt{3})/2$$

Now case 1: just moving up the plane

Friction acts down the plane

$$F₁=mg\sinθ+f$$
$$=5\times10\times(1/2)+(5\sqrt{3})/2$$
$$=25+(5\sqrt{3})/2$$

Now case 2: just preventing sliding down

Friction acts up the plane

$$F₂=mg\sinθ−f$$
$$=25−(5\sqrt{3})/2$$

Now difference:

$$|F₁|−|F₂|$$
$$=[25+(5\sqrt{3})/2]−[25−(5\sqrt{3})/2]$$

$$\frac{5\sqrt{3}}{2}-\left(-\frac{5\sqrt{3}}{2}\right)=\frac{5\sqrt{3}}{2}+\frac{5\sqrt{3}}{2}$$

$$=5\sqrt{3}$$

Final answer:

$$=5\sqrt{3}$$

Surface is given by
$$y=\frac{x^2}{4}$$

At any point, slope:
$$\frac{dy}{dx}=\frac{x}{2}$$

So,
$$\tanθ=\frac{x}{2}$$

For no slipping:
$$\tanθ\leμ$$

Given $$μ=0.5$$

So:
$$\frac{x}{2}\le0.5$$

$$x\le1$$

Now find maximum height:

$$y=\frac{x^2}{4}$$
$$=\frac{(1)^2}{4}$$
=

$$\frac{1}{4}$$

Final answer:

$$\frac{1}{4}$$

Given:

F₁ = $$5\hat{i} + 8\hat{j} + 7\hat{k}$$
F₂ = $$3\hat{i} - 4\hat{j} - 3\hat{k}$$
mass m = 4 kg

First find net force:

$$F_{net}$$ = F₁ + F₂
= $$(5+3)\hat{i} + (8-4)\hat{j} + (7-3) \hat{k}$$
= $$8\hat{i} + 4\hat{j} + \hat{k}$$

Now acceleration:

a =$$\frac{F_{net}}{m}$$
= $$(8\hat{i} + 4\hat{j} + \hat{k})$$ / 4
= $$2\hat{i} + \hat{j} + \hat{k}$$

So the acceleration is:

$$2\hat{i} + \hat{j} + \hat{k}$$

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

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

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

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

$$v^2=2636.4$$

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

Find the kinetic friction force on a 50 kg box on a horizontal surface with $$\mu_k=0.3$$.

The normal force on the horizontal surface is $$N = mg = 50 \times 9.8 = 490$$ N.

From this, the kinetic friction force is $$f_k = \mu_k N = 0.3 \times 490 = 147$$ N.

The correct answer is Option (2): 147 N.

$$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{g}{\sqrt{2}}$$.

$$\frac{m_2 - m_1}{m_1 + m_2} = \frac{1}{\sqrt{2}}$$.

Let $$r = m_1/m_2$$: $$\frac{1-r}{1+r} = \frac{1}{\sqrt{2}}$$.

$$\sqrt{2}(1-r) = 1+r \Rightarrow \sqrt{2} - \sqrt{2}r = 1 + r \Rightarrow r(1+\sqrt{2}) = \sqrt{2}-1 \Rightarrow r = \frac{\sqrt{2}-1}{\sqrt{2}+1}$$.

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

A wooden block (5 kg) and iron cylinder (25 kg) go down together with acceleration $$a = 0.1$$ m/s$$^2$$.

The total mass is $$m = 5 + 25 = 30$$ kg.

According to Newton's second law, the system accelerates downward at $$0.1$$ m/s$$^2$$. The forces acting on the system are weight (downward) and the normal reaction from the floor (upward).

- Weight (downward): $$mg = 30 \times 9.8 = 294$$ N

- Normal reaction from floor (upward): $$N$$

According to Newton's second law,

$$ mg - N = ma $$

Therefore,

$$ N = mg - ma = m(g - a) = 30(9.8 - 0.1) = 30 \times 9.7 = 291 \text{ N} $$

By Newton's third law, the action force of the system on the floor equals $$N = 291$$ N downward.

The correct answer is Option (2): 291 N.

Lets write equations for both the blocks

First for the 6 Kg block 

$$mg-T=ma$$

$$6g-T=6a$$

Now for the 20 Kg block

now our frictional force is $$f_l=\mu\ _kN$$

$$f_l=\mu\ _kmg\ $$

$$f_l=0.04\times\ 200$$

$$f_l=8N$$

Now equation for the 20 Kg block is

$$T-f_l=ma$$

$$T-f_l=20a$$

$$6g-T=6a$$

Now ad both equations

$$60-f_l=26a$$

$$60-8=26a=52$$

$$a=2\ \frac{m}{s^2}$$

By conservation of momentum (assuming the extra mass is added with zero velocity in the horizontal direction):

$$m_1 v_1 = (m_1 + m_2)v_f$$

$$1000 \times 6 = (1000 + 200) \times v_f$$

$$6000 = 1200 \times v_f$$

$$v_f = 5 \text{ m s}^{-1}$$

The answer is $$5$$ m/s, which corresponds to Option (4).

A spherical body of mass $$m = 100 \text{ g} = 0.1 \text{ kg}$$ is dropped from height $$10 \text{ m}$$ and rebounds to height $$5 \text{ m}$$.

Find the velocity just before hitting the ground: using $$v = \sqrt{2gh}$$:

$$v_1 = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \text{ m/s}$$ (downward)

Find the velocity just after rebounding: $$v_2 = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 7\sqrt{2} \text{ m/s}$$ (upward)

Calculate the impulse: impulse equals the change in momentum. Taking upward as positive:

$$J = m(v_2 - (-v_1)) = m(v_2 + v_1)$$

$$J = 0.1 \times (7\sqrt{2} + 14) = 0.1 \times (9.899 + 14) = 0.1 \times 23.899$$

$$J \approx 2.39 \text{ kg m s}^{-1}$$

The correct answer is Option (4): $$2.39 \text{ kg m s}^{-1}$$.

Given

initially at rest and then Linearly increasing force is applied

$$F=ma$$

as m is constant

F is directly proportional to a

So even $$a=0$$ at first then starts increasing linearly

so option D is correct

Now lets write the equations for both the block

So for the 10 Kg block 

$$Mg\sin\theta\ _1-\mu\ _sMg\cos\theta\ _1-T=Ma$$

$$Mg\sin\ 53^{\circ\ }-0.25Mg\cos53^{\circ\ }-T=Ma$$

$$100\times\ \frac{4}{5}-0.25\times\ 100\times\ \frac{3}{5}-T=20$$

$$80-T=15$$

$$T=45$$

Now for the block of mass m
$$T-mg\sin\theta\ _2-\mu\ _smg\cos\ \theta\ _2=ma$$
$$T-mg\sin37^{\circ\ }-0.25mg\cos37^{\circ\ }=m\left(2\right)$$
$$45-mg\times\ \frac{3}{5}-0.25mg\times\ \frac{4}{5}\ =m\times\ 2$$
$$mg\times\ \frac{3}{5}+0.25mg\times\ \frac{4}{5}\ +m\times\ 2=45$$
$$6m+2m+2m=45$$
$$10m=45$$
$$m=4.5\ Kg$$

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 cricket ball of mass 150 g moving at 20 m/s is caught in 0.1 s. Find the force on the player's hand.

Recall Newton's second law in terms of impulse: $$F = \frac{\Delta p}{\Delta t}$$ where $$\Delta p$$ is the change in momentum and $$\Delta t$$ is the time interval.

The initial momentum is $$p_i = mv = 0.15 \times 20 = 3$$ kg m/s, and the final momentum is $$p_f = 0$$ (ball is brought to rest). Hence, $$|\Delta p| = |p_f - p_i| = 3 \text{ kg m/s}$$.

Substituting into the formula gives $$F = \frac{|\Delta p|}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$$.

By Newton's third law, the force exerted by the ball on the player's hand equals the force exerted by the hand on the ball.

The correct answer is Option (4): 30 N.

F₂ and F₃ are perpendicular. Their resultant = $$\sqrt{8^2 + 6^2} = \sqrt{100} = 10$$ N = F₁.

When F₁ is removed, the net force = resultant of F₂ and F₃ = 10 N.

$$a = F/m = 10/5 = 2$$ m/s²

Given: mass = 10 kg, force F at 30° to horizontal, $$\mu_s = 0.25$$, $$g = 10$$ m/s².

The normal force: $$N = mg - F\sin 30° = 100 - \frac{F}{2}$$

For the block to just start moving, the horizontal component of F equals the maximum static friction:

$$ F\cos 30° = \mu_s N $$

$$ F \times \frac{\sqrt{3}}{2} = 0.25\left(100 - \frac{F}{2}\right) $$

$$ \frac{\sqrt{3}}{2}F = 25 - \frac{F}{8} $$

$$ F\left(\frac{\sqrt{3}}{2} + \frac{1}{8}\right) = 25 $$

$$ F \times \frac{4\sqrt{3} + 1}{8} = 25 $$

$$ F = \frac{200}{4\sqrt{3} + 1} = \frac{200}{7.928} \approx 25.2 \text{ N} $$

Therefore, $$F \approx 25.2$$ N.

A spherical body of mass $$m = 2$$ kg starts from rest and acquires kinetic energy $$KE = 10000$$ J at the end of 5 seconds.

Find the velocity at $$t = 5$$ s.

$$KE = \frac{1}{2}mv^2 \implies 10000 = \frac{1}{2}(2)v^2 \implies v^2 = 10000 \implies v = 100 \text{ m/s}$$

Find the acceleration.

Starting from rest: $$v = at$$

$$100 = a \times 5 \implies a = 20 \text{ m/s}^2$$

Find the force.

$$F = ma = 2 \times 20 = 40 \text{ N}$$

The answer is $$\boxed{40}$$ N.

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$$^{-1}$$.

A car moves in a circular horizontal track and a bob is suspended from the roof by a string. We need to find the angle the string makes with the vertical.

We are given that Speed $$v = 20$$ m/s, radius $$R = 40$$ m, $$g = 10$$ m/s$$^2$$.

To begin,

When the car moves in a circular path, it undergoes centripetal acceleration directed toward the centre of the circle. The bob, being inside the car, must also undergo this centripetal acceleration. The only forces on the bob are its weight ($$mg$$, downward) and the tension in the string ($$T$$, along the string). For the bob to have centripetal acceleration, the string must tilt away from the vertical (outward from the centre), so that the horizontal component of tension provides the centripetal force.

Next,

Let $$\theta$$ be the angle the string makes with the vertical.

Vertical equilibrium (no vertical acceleration):

$$ T\cos\theta = mg \quad \cdots (1) $$

Horizontal direction (centripetal acceleration $$a_c = v^2/R$$):

$$ T\sin\theta = \frac{mv^2}{R} \quad \cdots (2) $$

From here,

$$ \frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/R}{mg} $$

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

Continuing,

$$ \tan\theta = \frac{(20)^2}{40 \times 10} = \frac{400}{400} = 1 $$

$$ \theta = \tan^{-1}(1) = 45° = \frac{\pi}{4} $$

The string makes an angle of $$\dfrac{\pi}{4}$$ with the vertical.

The correct answer is Option 3: $$\dfrac{\pi}{4}$$.

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.

Lets write down equation for both

For 4 Kg

$$Mg\sin\theta_1\ -T=Ma$$

$$40\sin60^{\circ\ }\ -T=4a$$

For 1 Kg

$$T-mg\sin\theta_2\ =ma$$

$$T-10\sin30^{\circ\ }\ =a$$

Now add 2 equations

$$40\sin60^{\circ\ }-10\sin30^{\circ\ }\ =5a$$

$$40\times\ \frac{\sqrt{\ 3}}{2}-10\times\ \frac{1}{2}\ =5a$$

$$20\ \sqrt{\ 3}-5\ =5a$$

$$4\sqrt{\ 3}-1=a$$

Now substitute a in any equation

$$T-10\sin30^{\circ\ }\ =4\sqrt{\ 3}-1$$

$$T-5\ =4\sqrt{\ 3}-1$$

$$T=4\sqrt{\ 3}+4\ =\ 4\left(\sqrt{\ 3}+1\right)$$

A garden roller of mass 70 kg is pushed with a force of 200 N at 30 degrees below the horizontal. We need to find the normal reaction.

To begin,

Three forces act on the roller:

1. Weight $$W = mg = 70 \times 10 = 700$$ N (acting downward)

2. Applied force $$F = 200$$ N at 30 degrees below the horizontal (has both horizontal and vertical components)

3. Normal reaction $$N$$ from the ground (acting upward)

Next,

Since the force is applied at 30 degrees below the horizontal:

Horizontal component: $$F_x = F\cos 30° = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3}$$ N (forward)

Vertical component: $$F_y = F\sin 30° = 200 \times \frac{1}{2} = 100$$ N (downward, since the force is below the horizontal)

From here,

Since the roller moves horizontally (no vertical acceleration), the net vertical force must be zero. The upward force (normal reaction) must balance all downward forces (weight + vertical component of applied force):

$$ N = mg + F\sin 30° $$

$$ N = 700 + 100 = 800\;\text{N} $$

Note: The vertical component of the applied force acts downward because the force is directed below the horizontal. This increases the normal reaction beyond just the weight of the roller. If the force were directed above the horizontal, the vertical component would reduce the normal reaction.

The normal reaction on the roller is 800 N.

The correct answer is Option 3: 800 N.

To compare impulses, remember one key idea:

Impulse = area under the F-t graph

Now evaluate each figure based on its shape and dimensions:

For graph a

This is a triangle with $$base\ 1s\ and\ hieght\ 0.5N$$.
Area (impulse) =

$$\frac{1}{2}\times\ 1\times\ 0.5=0.25N$$

For graph B

This is a rectangle with constant force $$0.5N\ \ for\ a\ time\ of\ 2s$$.
Area (impulse) =

$$0.5\times\ 2\ =1N$$

For graph C

Triangle with base 1 s and height 0.75 N
Area =

$$\frac{1}{2}\times\ 1\times\ 0.75=0.375N$$

For graph D

Triangle with base 2 s and height 0.5 
Area =$$\frac{1}{2}\times\ 2\times\ 0.5=0.5N$$

Therefore from this we can conclude that fig(b) has the highest impulse

The assertion states that an electric fan continues to rotate for some time after the current is switched off, and this is correct because when the current to an electric fan is switched off, the fan does not stop immediately but continues to rotate and gradually slows down over several seconds, a commonly observed phenomenon. According to Newton’s First Law of Motion (the Law of Inertia), an object in motion continues in its state of motion unless acted upon by an external unbalanced force, and for rotational motion the analogous principle states that a rotating body continues to rotate unless acted upon by an external torque. The fan blades, motor rotor, and other rotating parts have mass and therefore possess rotational inertia (moment of inertia $$I$$). When the current is switched off, there is no longer a driving torque from the motor; however, the fan has angular momentum given by $$L = I\omega$$ where $$I$$ is the moment of inertia of the rotating parts and $$\omega$$ is the angular velocity. This angular momentum cannot vanish instantaneously, so the fan gradually decelerates due to frictional forces (air resistance and bearing friction) which provide a retarding torque $$\tau_{friction}$$, resulting in an angular deceleration $$\alpha = \frac{\tau_{friction}}{I}$$. Since the frictional torque is relatively small compared to the moment of inertia, it takes considerable time for the fan to come to rest.

The reason suggests that the fan continues to rotate due to inertia of motion, which is correct since it identifies that the fan’s continued rotation is due to rotational inertia, representing a direct application of Newton’s First Law extended to rotational motion, and thus the reason explains the assertion because the rotating fan tends to maintain its state of rotation and only external torques (friction, air resistance) can bring it to rest.

The correct answer is Option 3: Both A and R are correct and R is the correct explanation of A.

Given , 

A momentum-time graph is given,

p-t

momentum p=mv

as m is constant , graph is directly proportional to v

so the slope of the graph is the acceleration 

$$\ F=ma$$

So higher the slope higher the Force

Observe the graph, order of slopes is 

c>b>a

Therefore maximum force is in c and minimum force is in b

We need to find the coefficient of kinetic friction for a $$45°$$ rough incline, given that the time to slide down is $$n$$ times that on a smooth incline.

Acceleration on a smooth 45° incline.

$$a_1 = g\sin 45° = \frac{g}{\sqrt{2}}$$

Acceleration on a rough 45° incline.

$$a_2 = g(\sin 45° - \mu_k \cos 45°) = \frac{g}{\sqrt{2}}(1 - \mu_k)$$

Next, relate the times using kinematic equations.

For the same distance $$s$$ starting from rest:

$$s = \frac{1}{2}a_1 t_1^2 = \frac{1}{2}a_2 t_2^2$$

Given $$t_2 = n \cdot t_1$$:

$$a_1 t_1^2 = a_2 (nt_1)^2 = a_2 n^2 t_1^2$$

$$a_1 = n^2 a_2$$

Next, solve for $$\mu_k$$.

$$\frac{g}{\sqrt{2}} = n^2 \cdot \frac{g}{\sqrt{2}}(1 - \mu_k)$$

$$1 = n^2(1 - \mu_k)$$

$$\mu_k = 1 - \frac{1}{n^2}$$

The coefficient of kinetic friction is $$1 - \frac{1}{n^2}$$.

The correct answer is Option 4: $$1 - \frac{1}{n^2}$$.

Each ball has mass $$m$$ and moves with speed $$v$$. After striking the wall normally, it bounces back with the same speed.

Change in momentum for each ball:

$$ \Delta p = mv - (-mv) = 2mv $$

Total change in momentum for 100 balls:

$$ \Delta p_{\text{total}} = 100 \times 2mv = 200mv $$

The total force exerted on the wall (by Newton's second law):

$$ F = \frac{\Delta p_{\text{total}}}{t} = \frac{200mv}{t} $$

We have a block of mass $$m = 5$$ kg with an applied force $$F = 30$$ N that travels a distance $$s = 50$$ m in time $$t = 10$$ s, starting from rest (with $$g = 10$$ m/s$$^2$$).

Since the block starts from rest, using $$s = ut + \frac{1}{2}at^2$$ gives

$$50 = 0 + \frac{1}{2} \times a \times (10)^2 = 50a$$

so $$a = 1$$ m/s$$^2$$.

Now applying Newton's second law along the surface (the friction force opposes motion):

$$F - \mu_k mg = ma$$

$$30 - \mu_k \times 5 \times 10 = 5 \times 1$$

$$30 - 50\mu_k = 5$$

$$50\mu_k = 25$$

So $$\mu_k = 0.50$$. Hence, the correct answer is 0.50.

Find the coefficient of kinetic friction for a block sliding down a $$30°$$ incline with acceleration $$\frac{g}{4}$$.

$$ma = mg\sin 30° - \mu_k mg\cos 30°$$

$$\frac{g}{4} = g\sin 30° - \mu_k g\cos 30°$$

$$\frac{1}{4} = \frac{1}{2} - \mu_k \cdot \frac{\sqrt{3}}{2}$$

$$\mu_k \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$

$$\mu_k = \frac{1}{4} \cdot \frac{2}{\sqrt{3}} = \frac{1}{2\sqrt{3}}$$

The correct answer is Option B: $$\frac{1}{2\sqrt{3}}$$.

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.

A body of mass 500 g moves along the x-axis with velocity $$v = 10\sqrt{x}$$ m/s. We need to find the force acting on it.

First, we state the given quantities.

Mass: $$m = 500$$ g $$= 0.5$$ kg

Velocity as a function of displacement: $$v = 10\sqrt{x} = 10x^{1/2}$$ m/s

Next, we find the acceleration using the chain rule.

Since velocity is given as a function of position (not time), we use the relation:

$$ a = v\frac{dv}{dx} $$

This follows from the chain rule: $$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$$.

First, find $$\frac{dv}{dx}$$:

$$ \frac{dv}{dx} = \frac{d}{dx}(10x^{1/2}) = 10 \times \frac{1}{2}x^{-1/2} = \frac{5}{\sqrt{x}} $$

Now compute the acceleration:

$$ a = v \cdot \frac{dv}{dx} = 10\sqrt{x} \times \frac{5}{\sqrt{x}} = 50\;\text{m/s}^2 $$

Notice that the $$\sqrt{x}$$ terms cancel, giving a constant acceleration independent of position.

From this, we apply Newton's second law.

$$ F = ma = 0.5 \times 50 = 25\;\text{N} $$

The force acting on the body is 25 N.

The correct answer is Option 2: 25 N.

We have a bullet of mass $$m = 0.1$$ kg travelling at $$u = 400$$ m/s that gets embedded in a block of mass $$M = 3.9$$ kg. The combined system then slides a distance $$s = 20$$ m before coming to rest.

Since the bullet gets embedded in the block, this is a perfectly inelastic collision. By conservation of momentum:

$$mu = (m + M)v$$

$$0.1 \times 400 = (0.1 + 3.9) \times v$$

$$40 = 4v$$

$$v = 10 \text{ m/s}$$

Now, after the collision, the block-bullet system decelerates due to friction and comes to rest. Using the kinematic equation $$v^2 = u^2 - 2as$$ with final velocity zero:

$$0 = (10)^2 - 2a(20)$$

$$a = \frac{100}{40} = 2.5 \text{ m/s}^2$$

Since friction is the only horizontal force causing this deceleration, we have $$a = \mu g$$. So:

$$2.5 = \mu \times 10$$

$$\mu = 0.25$$

Hence, the correct answer is Option 4.

Using conservation of momentum for a perfectly inelastic collision:

$$mv + 2m(0) = (m + 2m)v'$$

$$mv = 3mv'$$

$$v' = \frac{v}{3}$$

The combined mass moves with velocity $$\frac{v}{3}$$.

The correct answer is Option 1.

Let mass $$=m,angle, θ=45°$$

Normal reaction:
$$N=mg\cosθ$$

Friction:
$$f=μN=μmg\cosθ$$

Now case 1: just moving up the incline

Friction acts down the plane

$$F₁=mg\sinθ+μmg\cosθ$$

Now case 2: just preventing sliding down

Friction acts up the plane

$$F₂=mg\sinθ−μmg\cosθ.$$

Given:

$$F_1=2F_2$$

Substitute:

$$mg\sinθ+μmg\cosθ=2(mg\sinθ−μmg\cosθ)$$

Cancel mg:

$$\sinθ+μ\cosθ=2\sinθ−2μ\cosθ$$

Bring terms together:

$$\sinθ−2\sinθ=−2μ\cosθ−μ\cosθ$$
$$−\sinθ=−3μ\cosθ$$

So:

$$μ=\frac{(\sinθ)}{(3\cosθ)}$$
$$μ=\frac{(\tanθ)}{3}$$

Now$$θ=45°,\tan45=1$$

So:

$$μ=\frac{1}{3}$$

Final answer:

$$\frac{1}{3}$$

Statement I: An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable.

Statement II: Force exerted by the floor on a person is more than their weight when the elevator goes down with increasing speed.

Evaluation of Statement I:

When the elevator moves with uniform speed (zero acceleration), by Newton's second law: $$T - Mg = 0$$, so $$T = Mg$$. The tension in the cable exactly balances the weight. Statement I is true.

Evaluation of Statement II:

"Going down with increasing speed" means the elevator has a downward acceleration ($$a$$ directed downward).

For a person of mass $$m$$ inside: $$mg - N = ma$$ (taking downward as positive).

$$N = m(g - a)$$

Since $$a > 0$$ (downward acceleration), $$N < mg$$. The normal force is less than the person's weight, not more. Statement II is false.

Statement I is true but Statement II is false, which corresponds to Option 2.

$$\vec{r} = 10t\hat{i} + 15t^2\hat{j} + 7\hat{k}$$

Velocity: $$\vec{v} = \frac{d\vec{r}}{dt} = 10\hat{i} + 30t\hat{j}$$

Acceleration: $$\vec{a} = \frac{d\vec{v}}{dt} = 30\hat{j}$$

Since $$\vec{F} = m\vec{a}$$, the force is along the positive y-axis direction ($$30m\hat{j}$$).

The direction of net force is along the positive y-axis.

A ball of mass $$200$$ g rests on a vertical post of height $$20$$ m. A bullet of mass $$10$$ g hits it horizontally. After collision, both travel independently.

Both the ball and bullet undergo projectile motion from height $$h = 20$$ m with zero initial vertical velocity.

$$h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{10}} = 2$$ s

Ball lands at 30 m: $$v_{\text{ball}} = \frac{30}{2} = 15$$ m/s

Bullet lands at 120 m: $$v_{\text{bullet}} = \frac{120}{2} = 60$$ m/s

Before collision: only bullet has momentum.

$$m_{\text{bullet}} \times u = m_{\text{ball}} \times v_{\text{ball}} + m_{\text{bullet}} \times v_{\text{bullet}}$$

$$0.01 \times u = 0.2 \times 15 + 0.01 \times 60$$

$$0.01u = 3 + 0.6 = 3.6$$

$$u = 360$$ m/s

The initial velocity of the bullet is $$360$$ m/s.

The correct answer is Option 4: $$360$$ m/s.

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.

Given: mass $$m = 500$$ g = 0.5 kg, velocity $$\vec{v} = (2t\hat{i} + 3t^2\hat{j})$$ m/s.

Acceleration:

$$ \vec{a} = \frac{d\vec{v}}{dt} = (2\hat{i} + 6t\hat{j})\;\text{m/s}^2 $$

At $$t = 1$$ s: $$\vec{a} = (2\hat{i} + 6\hat{j})$$ m/s².

Force:

$$ \vec{F} = m\vec{a} = 0.5(2\hat{i} + 6\hat{j}) = (\hat{i} + 3\hat{j})\;\text{N} $$

Comparing with $$(\hat{i} + x\hat{j})$$: $$x = 3$$.

The correct answer is $$x = 3$$.

The force acting on the particle is $$\vec{F}= -k\left(x\hat{i}+y\hat{j}\right)$$ with $$k=1$$ and the mass is $$m=1\;\text{kg}$$.

Hence the equations of motion in the $$x$$ and $$y$$ directions are
$$m\frac{d^2x}{dt^2}= -x,\qquad m\frac{d^2y}{dt^2}= -y$$
or, since $$m=1$$,
$$\frac{d^2x}{dt^2}= -x,\qquad \frac{d^2y}{dt^2}= -y.$$

Notice that the force always points along $$-\,(x\hat{i}+y\hat{j})$$, i.e. along the radius vector $$\vec{r}_{xy}=x\hat{i}+y\hat{j}$$ lying in the $$xy$$-plane. Therefore the torque about the origin is

$$\vec{\tau}= \vec{r}\times\vec{F}= (x\hat{i}+y\hat{j})\times\bigl(-(x\hat{i}+y\hat{j})\bigr)=\vec{0},$$

which implies that the angular momentum about the origin is conserved. The component of angular momentum along the $$z$$-axis is

$$L_z = (\vec{r}\times\vec{p})_z = x p_y - y p_x = m(xv_y - yv_x).$$

Because $$m=1\;\text{kg}$$, the conserved quantity simplifies to

$$xv_y - yv_x = \text{constant}.$$

Compute this constant using the initial data at $$t=0$$:
$$x_0 = \frac{1}{\sqrt{2}},\quad y_0 = \sqrt{2},\quad v_{x0} = -\sqrt{2},\quad v_{y0} = \sqrt{2}.$$

$$\begin{aligned} x_0v_{y0} - y_0v_{x0} &= \left(\frac{1}{\sqrt{2}}\right)(\sqrt{2}) - (\sqrt{2})(-\sqrt{2}) \\ &= 1 - (-2) \\ &= 3\;\text{m}^2\text{s}^{-1}. \end{aligned}$$

The motion in the $$z$$-direction is unaffected by the force, so $$z(t) = z_0 + v_{z0}t$$. With $$z_0 = 0$$ and $$v_{z0}= \dfrac{2}{\pi}\;\text{m s}^{-1}$$, the condition $$z=0.5\;\text{m}$$ gives $$t = \dfrac{\pi}{4}\;\text{s}$$, but the conserved quantity $$xv_y - yv_x$$ remains the same at every instant.

Therefore, when $$z = 0.5\;\text{m}$$, the value of $$(xv_y - yv_x)$$ is still

$$3\;\text{m}^2\text{s}^{-1}.$$

Concept:
For an inextensible string,

$$\sum_{ }^{ }T⃗⋅a⃗=0$$

i.e., sum of components of accelerations along tension = 0.

Applying to system (counting segments):

$$-4Ta_1-2Ta_2-Ta_3-Ta_4=0$$

Cancel T:

$$-4a_1-2a_2-a_3-a_4=0$$

Final Answer:

$$-4a_1-2a_2-a_3-a_4=0$$

Consider the frame of reference of the $$100\ kg$$ block. Since this block is accelerating towards the right with acceleration a, this frame is non-inertial, and hence a pseudo force ma acts on each mass towards the left.

For the $$10\ kg$$ block (which moves left relative to the big block with acceleration $$2 \text{ m s}^{-2}$$), the net acceleration in this non-inertial frame is $$2 \text{ m s}^{-2}$$ towards the left. Applying Newton’s second law along the horizontal direction,

$$10a-T=10\times2\quad$$

where 10a is the pseudo force and T is the tension acting opposite to it.

For the 20 kg hanging mass, which moves upward with acceleration  $$2 \text{ m s}^{-2}$$ applying Newton’s second law in vertical direction,

$$T−20g=20\times2⇒T−200=40$$

Adding equations (1) and (2),

$$10a−200=60⇒10a=260⇒a=26m/s^2$$

Substituting back into (2),

$$T=240N$$

Now, considering the horizontal motion of the entire system, the 100 kg block and the 20kg mass together move with acceleration a. The only external horizontal force is F, and tension T opposes motion. Hence,

$$F-T=(100+20)a$$

$$F−240=120\times26⇒F=3360N$$

Two masses $$M_1$$ and $$M_2$$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. This is known as an Atwood machine. We need to find the ratio $$\dfrac{a_1}{a_2}$$.

Consider two masses $$M_1$$ and $$M_2$$ (where $$M_2 > M_1$$) connected by a string over a frictionless pulley. Let the tension in the string be $$T$$ and the acceleration of the system be $$a$$.

For mass $$M_2$$ moving downward, the net force gives $$M_2 g - T = M_2 a$$. Since mass $$M_1$$ moves upward with the same magnitude of acceleration, we have $$T - M_1 g = M_1 a$$. Adding these equations yields $$M_2 g - M_1 g = (M_1 + M_2) a$$ and therefore $$a = \dfrac{(M_2 - M_1)}{(M_1 + M_2)} \cdot g$$.

Substituting $$M_2 = 2M_1$$ gives $$a_1 = \dfrac{(2M_1 - M_1)}{(M_1 + 2M_1)} \cdot g = \dfrac{M_1}{3M_1} \cdot g = \dfrac{g}{3}$$.

Next, substituting $$M_2 = 3M_1$$ gives $$a_2 = \dfrac{(3M_1 - M_1)}{(M_1 + 3M_1)} \cdot g = \dfrac{2M_1}{4M_1} \cdot g = \dfrac{g}{2}$$.

From this, $$\dfrac{a_1}{a_2} = \dfrac{g/3}{g/2} = \dfrac{g}{3} \times \dfrac{2}{g} = \dfrac{2}{3}$$.

The correct answer is Option B: $$\dfrac{2}{3}$$.

Let us start by considering the forces on the 4M block

There is an external Force pulling it which is $$F=2Mg$$ and $$T\left(tension\ of\ the\ rope\right)$$

we know that here masses M and 4M have the same acceleration , lets take it as 'a'

now writing equation for the 4M block

$$2Mg\ -\ T\ =4Ma$$

and for the M block 

Forces at $$T\left(tension\ of\ the\ rope\right)$$ and $$F=Mg$$(due to gravity)

now writing equation for M block

$$T-Mg=Ma$$

now on adding these both equations

$$2Mg-Mg=5Ma$$

$$Mg=5Ma$$

$$\frac{g}{5}=a$$

Now substituting value of a in equation 1

$$2Mg\ -\ T\ =4M\times\ \frac{g}{5}$$

$$2Mg\ -\frac{4Mg}{5}\ =T$$

$$\frac{\left(10Mg\ -4Mg\right)}{5}\ =T$$

$$\frac{6Mg}{5}\ =T$$

Therefore x=6

We begin by noting that a mass of $$10$$ kg hangs from a rope of length $$5$$ m, and that a horizontal force of $$30$$ N is applied at the midpoint of the rope. Let $$T_1$$ denote the tension in the upper half and $$T_2$$ the tension in the lower half, while $$\alpha$$ and $$\beta$$ are the angles these halves make with the vertical.

Since there is no horizontal force below the midpoint, the lower half supports only the weight of the mass, so it hangs vertically and thus $$\beta = 0$$. Consequently, $$T_2 = mg = 10 \times 10 = 100$$ N.

At the midpoint, three forces are in equilibrium: the tension $$T_1$$ along the upper rope at an angle $$\alpha$$ to the vertical, the downward tension $$T_2 = 100$$ N, and the applied horizontal force $$F = 30$$ N. From horizontal equilibrium we have $$T_1 \sin\alpha = 30$$ ... (i), and from vertical equilibrium we have $$T_1 \cos\alpha = T_2 = 100$$ ... (ii).

Dividing equation (i) by equation (ii) gives $$\tan\alpha = \frac{30}{100} = \frac{3}{10} = 3 \times 10^{-1}$$, so $$\alpha = \tan^{-1}(3 \times 10^{-1})$$.

Comparing this expression for $$\alpha$$ with the given form $$\alpha = \tan^{-1}(x \times 10^{-1})$$ shows that $$x = 3$$.

The answer is $$\boxed{3}$$.

Let the total length of the chain be $$L = 6$$ m and the coefficient of static friction be $$\mu = 0.5$$.

Let $$x$$ be the length of the chain hanging over the edge. Then $$(L - x) = (6 - x)$$ is the length on the table.

Let the mass per unit length of the chain be $$\lambda = \frac{m}{L}$$.

Weight of the hanging part (pulling the chain down):

$$W_{hang} = \lambda x g = \frac{m x g}{6}$$

Normal force on the table part:

$$N = \lambda (6 - x) g = \frac{m(6 - x)g}{6}$$

Maximum static friction force:

$$f = \mu N = 0.5 \times \frac{m(6 - x)g}{6}$$

At the maximum hanging length, the system is on the verge of sliding, so:

$$W_{hang} = f$$

$$\frac{m x g}{6} = 0.5 \times \frac{m(6 - x)g}{6}$$

Cancelling common terms:

$$x = 0.5(6 - x)$$

$$x = 3 - 0.5x$$

$$1.5x = 3$$

$$x = 2 \text{ m}$$

The maximum length of the chain hanging from the table is 2 m.

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.

Given: Mass $$m = 100$$ g $$= 0.1$$ kg, Force $$\vec{F} = (10\hat{i} + 5\hat{j})$$ N, initial velocity = 0 (starts from rest), $$t = 2$$ s.

Find the acceleration: Using Newton's second law $$\vec{F} = m\vec{a}$$:

$$\vec{a} = \frac{\vec{F}}{m} = \frac{10\hat{i} + 5\hat{j}}{0.1} = (100\hat{i} + 50\hat{j}) \text{ m/s}^2$$

Find the position at $$t = 2$$ s: Since the object starts from rest ($$\vec{u} = 0$$), using $$\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2$$:

$$\vec{s} = \frac{1}{2}(100\hat{i} + 50\hat{j})(2)^2$$

$$\vec{s} = \frac{1}{2}(100\hat{i} + 50\hat{j})(4)$$

$$\vec{s} = (200\hat{i} + 100\hat{j}) \text{ m}$$

So $$a = 200$$ and $$b = 100$$.

Find $$\frac{a}{b}$$: $$\frac{a}{b} = \frac{200}{100} = 2$$

The answer is 2.

Step 1: Acceleration of system
When the 6th ball just leaves the table:

• Number of hanging balls = 6
• Driving force = 6mg N
• Total mass of system = 10m

$$a=\frac{\text{net force}}{\text{total mass}}=\ \frac{\ 6mg}{10m}=\ \frac{\ 3g}{5}$$

​Step 2: Tension between 7th and 8th ball

Consider the last 3 balls (8th, 9th, 10th) lying on the table:

• Mass of these 3 balls = 3m
• They move with same acceleration $$a=\frac{3g}{5}$$

$$T=(\text{mass})\times(\text{acceleration})$$

 $$T=3m\cdot\frac{3g}{5}$$

$$substitute\ m=2kg,g=10m/s^2:$$

$$T=3\times2\times\ \frac{\ 3\times\ 10}{5}​=36N$$

Final Answer:

36 N

We need to find the impulse imparted to the ball when the batsman hits it back without changing its speed.

Since the mass of the ball is $$m = 0.4$$ kg and its speed is $$v = 15$$ m/s, and the ball is hit back in the opposite direction with the same speed, we define a direction convention to set up our calculations.

Let the initial direction of the ball towards the batsman be negative, so $$v_i = -15$$ m/s. After being hit back, the ball moves in the opposite direction, which we take as positive: $$v_f = +15$$ m/s.

Now the impulse can be calculated as the change in momentum: Impulse = Change in momentum = $$m(v_f - v_i)$$, which gives $$0.4 \times (15 - (-15))$$, or equivalently $$0.4 \times 30$$, resulting in $$12$$ N s.

The impulse imparted to the ball is 12 N s.

Based on the provided image, we can accurately solve for x by balancing the forces acting on the block along the inclined plane.

1. Forces and Components

To keep the block stationary on a smooth incline, the net force along the plane must be zero.

• Component of Weight (mg): Acts down the incline as $$mg \sin(60^\circ)$$.
• Component of Horizontal Force (F): Acts up the incline as $$F \cos(60^\circ)$$.

2. Equilibrium Equation

For the block to remain stationary:

$$F \cos(60^\circ) = mg \sin(60^\circ)$$

$$F = mg \tan(60^\circ)$$

3. Numerical Calculation

Given:

• Mass (m) = $$200\text{ g} = 0.2\text{ kg}$$
• $$g = 10\text{ m/s}^2$$
• Weight (m
• g) = $$0.2 \times 10 = 2\text{ N}$$
• Angle ($$\theta$$) = $$60^\circ$$
• Force ($$F$$) = $$\sqrt{x}\text{ N}$$

Substituting the values:

$$\sqrt{x} = 2 \times \tan(60^\circ)$$

$$\sqrt{x} = 2 \times \sqrt{3}$$

Squaring both sides to find $x$:

$$x = (2\sqrt{3})^2$$$$x = 4 \times 3$$

$$\boxed{x = 12}$$

We apply the principle of conservation of linear momentum to solve this problem. The mass of the man is $$m_1 = 60$$ kg, the mass of the trolley is $$m_2 = 120$$ kg, the initial velocity of the trolley is $$0$$ m/s (stationary), and the final velocity of the (man + trolley) system is $$v_f = 2$$ m/s. Let the velocity of the running man be $$v$$ m/s.

Total momentum before equals total momentum after. Substituting into this gives:

$$m_1 \times v + m_2 \times 0 = (m_1 + m_2) \times v_f$$

$$60 \times v + 120 \times 0 = (60 + 120) \times 2$$

$$60v = 180 \times 2$$

$$60v = 360$$

$$v = \frac{360}{60}$$

$$v = 6 \text{ m/s}$$

Hence, the velocity of the running man when he jumps into the car is 6 m/s.

Let the projectile be launched with speed $$v$$ and angle of projection $$\theta$$ from a horizontal ground.

Horizontal component of velocity: $$v_x = v\cos\theta$$
Vertical component of velocity: $$v_y = v\sin\theta$$

Time to reach the highest point (region with acceleration $$g$$)
At the top, the vertical velocity becomes zero. Using the first equation of motion,

$$0 = v_y - gt_{\text{up}} \;\; \Longrightarrow \;\; t_{\text{up}} = \frac{v\sin\theta}{g}$$

Height of the highest point
Using $$v_y^{2} - u_y^{2} = 2g(h_{\text{max}} - 0)$$ with $$v_y = 0$$ and $$u_y = v\sin\theta$$:

$$0 - (v\sin\theta)^2 = -2gh_{\text{max}}$$
$$\Rightarrow h_{\text{max}} = \frac{v^{2}\sin^{2}\theta}{2g}$$

Descent in the region where acceleration is $$g' = \dfrac{g}{0.81}$$
From rest at the top, the projectile falls the height $$h_{\text{max}}$$ under acceleration $$g'$$. Using $$h = \tfrac12 g' t_{\text{down}}^{2}$$:

$$\frac{v^{2}\sin^{2}\theta}{2g} = \frac12 g' t_{\text{down}}^{2}$$
$$\Longrightarrow t_{\text{down}} = \frac{v\sin\theta}{\sqrt{g\,g'}}$$

Total time of flight in the two-gravity situation

$$T' = t_{\text{up}} + t_{\text{down}} = \frac{v\sin\theta}{g} + \frac{v\sin\theta}{\sqrt{g\,g'}} = v\sin\theta\left(\frac1g + \frac1{\sqrt{g\,g'}}\right)$$

Horizontal range in the new situation

$$d' = v_x\,T' = v\cos\theta \, v\sin\theta\left(\frac1g + \frac1{\sqrt{g\,g'}}\right)$$

Original range (uniform gravity $$g$$)

$$d = v\cos\theta \times \frac{2v\sin\theta}{g} = \frac{v^{2}\sin2\theta}{g}$$

Ratio of the two ranges

$$n = \frac{d'}{d} = \frac{v\cos\theta\,v\sin\theta\left(\dfrac1g + \dfrac1{\sqrt{g\,g'}}\right)}{v\cos\theta \times \dfrac{2v\sin\theta}{g}} = \frac{\left(\dfrac1g + \dfrac1{\sqrt{g\,g'}}\right)}{\dfrac{2}{g}} = \frac12\left(1 + \sqrt{\frac{g}{g'}}\right)$$

Substituting $$g' = \dfrac{g}{0.81} \;\Longrightarrow\; \frac{g}{g'} = 0.81$$

$$\sqrt{\frac{g}{g'}} = \sqrt{0.81} = 0.9$$

Therefore, $$n = \frac12\bigl(1 + 0.9\bigr) = \frac{1.9}{2} = 0.95$$

The new range is $$d' = 0.95\,d$$. Hence, the required value of $$n$$ is 0.95.

A bag is dropped on a conveyor belt moving at 2 m/s. We need to find the distance the bag travels on the belt during slipping.

The bag is dropped gently (initial velocity = 0) onto a belt moving at $$v = 2$$ m/s, and friction accelerates the bag until it reaches belt speed. The friction force on the bag is $$\mu m g$$, so the acceleration of the bag is $$a = \mu g = 0.4 \times 10 = 4 \text{ m/s}^2$$. The bag reaches the belt speed when $$v = a t$$, which gives $$2 = 4t \implies t = 0.5 \text{ s}$$.

During this time, the distance moved by the bag in the ground frame is $$s_{\text{bag}} = \frac{1}{2} a t^2 = \frac{1}{2} \times 4 \times 0.25 = 0.5 \text{ m}$$, and the belt moves $$s_{\text{belt}} = v \times t = 2 \times 0.5 = 1 \text{ m}$$. Therefore, the distance the bag travels relative to the belt while slipping is $$s_{\text{belt}} - s_{\text{bag}} = 1 - 0.5 = 0.5 \text{ m}$$.

The correct answer is Option B: $$0.5 \text{ m}$$.

A ball of mass $$0.15 \text{ kg}$$ hits the wall at $$12 \text{ m s}^{-1}$$ and bounces back with the same speed under a force of $$100 \text{ N}$$, so we need to find the contact time.

Taking the initial direction of motion as positive, the initial momentum is $$p_i = m \times v = 0.15 \times 12 = 1.8 \text{ kg m s}^{-1}$$. Next, the final momentum after bouncing back is $$p_f = m \times (-v) = 0.15 \times (-12) = -1.8 \text{ kg m s}^{-1}$$. From this, the change in momentum is $$\Delta p = p_f - p_i = -1.8 - 1.8 = -3.6 \text{ kg m s}^{-1}$$.

Since $$F \cdot \Delta t = |\Delta p|$$ and the magnitude of the change in momentum is $$|\Delta p| = 3.6 \text{ kg m s}^{-1}$$, substituting gives $$\Delta t = \dfrac{|\Delta p|}{F} = \dfrac{3.6}{100} = 0.036 \text{ s}$$.

The correct answer is Option B: $$0.036 \text{ s}$$.

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.

For the beaker to revolve with the disc, the friction must provide the necessary centripetal force.

The centripetal force required for circular motion at distance $$R$$ from the center is:

$$F_c = m\omega^2 R$$

The maximum static friction available is:

$$f_{max} = \mu m g$$

For the beaker to not slide, friction must be sufficient:

$$m\omega^2 R \leq \mu m g$$

Cancelling $$m$$ from both sides:

$$\omega^2 R \leq \mu g$$

$$R \leq \frac{\mu g}{\omega^2}$$

Hence, the correct answer is Option B.

A monkey of mass $$50 \text{ kg}$$ climbs on a rope that can withstand a maximum tension of $$T_{max} = 350 \text{ N}$$. We need to check if the rope breaks in either case.

We start by determining the tension when the monkey climbs down with acceleration $$a = 4 \text{ m/s}^2$$, taking downward as positive:

$$mg - T = ma$$

$$T = m(g - a) = 50(10 - 4) = 50 \times 6 = 300 \text{ N}$$

Since $$T = 300 \text{ N} \lt 350 \text{ N}$$, the rope does not break while climbing down.

Next, when the monkey climbs up with acceleration $$a = 5 \text{ m/s}^2$$, the net upward force must produce this acceleration:

$$T - mg = ma$$

$$T = m(g + a) = 50(10 + 5) = 50 \times 15 = 750 \text{ N}$$

Since $$T = 750 \text{ N} \gt 350 \text{ N}$$, the rope breaks while climbing upward.

Now we evaluate the options:

Option A: $$T = 700 \text{ N}$$ while climbing upward — Incorrect, $$T = 750 \text{ N}$$.

Option B: $$T = 350 \text{ N}$$ while going downward — Incorrect, $$T = 300 \text{ N}$$.

Option C: Rope will break while climbing upward — Correct, since $$750 \text{ N} \gt 350 \text{ N}$$.

Option D: Rope will break while going downward — Incorrect, since $$300 \text{ N} \lt 350 \text{ N}$$.

Therefore, the correct answer is Option C: Rope will break while climbing upward.

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.

$$

We are given: mass $$m = 10$$ kg, initial velocity $$u = 9.8$$ m/s, coefficient of friction $$\mu = 0.5$$, $$g = 9.8$$ m/s$$^2$$.

Calculate the deceleration due to friction: the frictional force is $$f = \mu mg$$, so the deceleration is:

$$ a = \mu g = 0.5 \times 9.8 = 4.9 \text{ m/s}^2 $$

Use the kinematic equation to find the distance: using $$v^2 = u^2 - 2as$$ where $$v = 0$$ (block comes to rest):

$$ 0 = u^2 - 2as $$

$$ s = \frac{u^2}{2a} = \frac{(9.8)^2}{2 \times 4.9} = \frac{96.04}{9.8} = 9.8 \text{ m} $$

Therefore, the correct answer is Option A.

A block of mass $$M$$ inside a box descends vertically with acceleration $$a$$. The block exerts a force equal to one-fourth of its weight on the floor.

When the system descends with acceleration $$a$$, the apparent weight (normal force $$N$$) is:

$$N = M(g - a)$$

The block exerts a force $$N = \frac{Mg}{4}$$ on the floor (by Newton's third law, the floor exerts the same force on the block).

$$M(g - a) = \frac{Mg}{4}$$

$$g - a = \frac{g}{4}$$

$$a = g - \frac{g}{4} = \frac{3g}{4}$$

Hence, the correct answer is Option B: $$\dfrac{3g}{4}$$.

A block of mass $$M$$ slides down a rough inclined plane with constant velocity, so its acceleration is zero and the net force on the block is zero. The forces acting on the block are the weight $$Mg$$ acting vertically downward, the normal reaction $$N$$ perpendicular to the inclined surface, and the friction force $$f$$ along the inclined surface (opposing motion and directed up the incline). The contact force between the block and the inclined surface is the resultant of the normal reaction and the friction force, and for equilibrium this contact force must balance the weight exactly, giving $$|\text{Contact force}| = Mg$$.

Verification: Along the incline one has $$f = Mg\sin\theta$$ and perpendicular to the incline $$N = Mg\cos\theta$$. Thus the magnitude of the contact force is $$\sqrt{N^2 + f^2} = \sqrt{M^2g^2\cos^2\theta + M^2g^2\sin^2\theta} = Mg\sqrt{\cos^2\theta + \sin^2\theta} = Mg\,. $$

Answer: Option A: $$Mg$$

Given:$$m=2kg,\ M=8kg,\ μ=0.5,\ g=9.8m/s^2$$

Step 1: Maximum acceleration (no slipping)

Maximum acceleration comes from maximum static friction acting on the top block:

$$f_{\max}=\mu N=\mu mg\ $$

This friction is the only horizontal force accelerating the top block:

$$⁡f_{\max}=ma_{\max}$$

$$μmg=ma_{\max}\ ​⇒\ a_{\max}​=μg=0.5\times9.8=4.9m/s^2$$

This ensures the top block does not slip over the bottom block.

Friction provides acceleration to top block:

$$a_{\max}​=μg=0.5\times9.8=4.9m/s^2$$

Step 2: Treat system as one body
Total mass:

m+M=2+8=10 kg

Step 3: Maximum force

$$F_{\max}​=(m+M)a_{\max}​=10\times4.9=49N$$

Final Answer:

49 N

An object of mass 5 kg moving at 25 m/s hits a wall and comes to rest. We compare two scenarios in which it stops in 3 s and in 5 s. The impulse imparted to the object is given by the change in momentum: $$m \times \Delta v$$. In both cases the initial momentum is the same and the final momentum is zero, so the impulse is

$$J = m(v_f - v_i) = 5(0 - 25) = -125 \text{ kg m/s}$$

and its magnitude is 125 N·s.

The average force over the stopping time is calculated as

$$\frac{\text{Impulse}}{\Delta t}$$

When the object stops in 3 s, the average force is

$$F_1 = \frac{125}{3} = 41.67$$ N,

whereas stopping in 5 s yields

$$F_2 = \frac{125}{5} = 25$$ N.

Thus, although the impulse is the same in both cases, the average force differs due to the different time intervals. Therefore, the correct answer is Option B: impulse will be the same for both cases but the average force will be different.

Now here we have 2 bodies of masses $$m_1$$ and $$m_2$$

Let us consider the forces acting on these 

On $$m_2$$ the forces acting are $$T\ and\ F=\ m_2g$$

As they are in opposite directions and for the system to be in equilibrium, they have to be equal

$$T\ =\ m_2g$$

coming to the first body of mass $$m_1$$

Forces along the direction of incline are $$T\ and\ F_{ }=\ m_1g\sin\theta\ $$

These are also equal as the system is in equilibrium

So , $$T\ =m_1g\sin\theta\ $$

We know  , $$T\ =\ m_2g$$

After Substituting T

$$m_1g\sin\theta\ =\ m_2g$$

$$5g\sin\theta\ =\ 3g$$

$$5\sin\theta\ =\ 3$$

$$\sin\theta\ =\ \frac{3}{5}$$

Now coming to Forces which are  perpendicular to the incline

 $$N\ and\ m_1g\cos\theta\ $$

Even they have to equal

so , $$N\ =\ m_1g\cos\theta\ $$

As $$\sin\theta\ =\frac{3}{5}\ then\ \cos\ \theta\ =\frac{4}{5}\ $$

Then $$N\ =\ m_1g\times\ \frac{4}{5}$$

$$N\ =5\times\ g\times\ \frac{4}{5}$$

$$N\ =5\times\ 10\times\ \frac{4}{5}$$

$$N\ =\ 10\times\ 4$$

$$N\ =\ 40N$$

This is the force exerted by the inclined plane on the body of mass $$m_1$$

A block takes 2 s to slide down a frictionless incline of $$30^\circ$$ and length $$l$$, inside a lift going up with uniform velocity. We need to find the time when the incline angle changes to $$45^\circ$$. Since the lift moves with uniform velocity (zero acceleration), there is no pseudo force, and the effective gravitational acceleration remains $$g$$. For a frictionless incline at angle $$\theta$$, the component of gravitational acceleration along the incline is $$a = g\sin\theta$$. The block starts from rest and slides a distance $$l$$, giving $$l = \tfrac{1}{2}(g\sin 30^\circ)\,t_1^2 = \tfrac{1}{2}\cdot g\cdot \tfrac{1}{2}\cdot (2)^2 = g\;. $$ Therefore: $$l = g\quad\text{(i)}$$

When the incline angle becomes $$45^\circ$$, the block slides the same length $$l$$, so $$l = \tfrac{1}{2}(g\sin 45^\circ)\,t_2^2 = \tfrac{1}{2}\cdot g\cdot \tfrac{1}{\sqrt{2}}\cdot t_2^2\;. $$ Using equation (i), we set $$g = \tfrac{g}{2\sqrt{2}}\;t_2^2\,, $$ which leads to $$1 = \tfrac{t_2^2}{2\sqrt{2}}\quad\Longrightarrow\quad t_2^2 = 2\sqrt{2} = 2\times1.414 = 2.828\;. $$ Hence $$t_2 = \sqrt{2.828}\approx1.68\text{ s}\;. $$

We can also express this as $$t_2 = 2^{3/4} = (8)^{1/4} \approx 1.68\text{ s}\;. $$

Verification using the ratio method: From the two cases: $$\frac{t_2^2}{t_1^2} = \frac{\sin 30^\circ}{\sin 45^\circ} = \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\;, $$ so $$t_2^2 = \frac{t_1^2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\quad\Longrightarrow\quad t_2 = \sqrt{2\sqrt{2}}\approx1.68\text{ s}\;. $$

Answer: Option C: 1.68 s

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.

Two billiard balls of mass $$m = 0.05 \text{ kg}$$ each, moving in opposite directions at $$10 \text{ m s}^{-1}$$, collide and rebound with the same speed. Contact time $$t = 0.005 \text{ s}$$.

Calculate the change in momentum of one ball.

Taking the initial direction of one ball as positive, its velocity changes from $$+10 \text{ m s}^{-1}$$ to $$-10 \text{ m s}^{-1}$$:

Calculate the force using the impulse-momentum theorem.

The correct answer is Option B: $$200 \text{ N}$$.

We have three forces $$\vec F_1$$, $$\vec F_2$$ and $$\vec F_3$$. The statement tells us that these three vectors are represented by the three sides of a triangle and that they are taken in the same cyclic order, i.e. the head of $$\vec F_1$$ is joined to the tail of $$\vec F_2$$, the head of $$\vec F_2$$ is joined to the tail of $$\vec F_3$$ and finally the head of $$\vec F_3$$ returns to the tail of $$\vec F_1$$, thereby closing the triangle.

Whenever three vectors form the three sides of a closed triangle in the same order, we can write, by simple head-to-tail addition, the vector equation

$$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0.$$

Rearranging this equation, we obtain

$$\vec F_1 + \vec F_2 = -\vec F_3.$$

Now, the condition for equilibrium of concurrent forces is that the resultant force must be zero. This condition is expressed algebraically as

$$\vec F_{\text{net}} = \sum \vec F = \vec 0.$$

Comparing this general equilibrium condition with the relation $$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0,$$ we see that the resultant of the three forces vanishes. Hence there is no unbalanced force, so the system is in equilibrium. Since we can imagine these three forces acting at the same point (because only their magnitudes and directions matter for the triangle law), they are described as concurrent forces. Therefore, Statement I is true.

Next, let us examine Statement II. The condition for translatory (translational) equilibrium is exactly the same: the vector sum of all external forces acting on the body must be zero, i.e.

$$\sum \vec F = \vec 0.$$

We have already established that if three forces can be arranged to form a closed triangle in the same order, then

$$\vec F_1 + \vec F_2 + \vec F_3 = \vec 0.$$

This directly satisfies the translational equilibrium condition. No net force means no linear acceleration of the body. Hence Statement II is also true.

Both statements are individually correct, and Statement II is merely a restatement of the vector equilibrium condition without going into the concurrency detail. Consequently, both statements are true.

Hence, the correct answer is Option 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 apply the law of conservation of momentum. Initially, both the bullet and gun are at rest, so the total momentum is zero.

The bullet has mass $$m = 4 \text{ g} = 0.004 \text{ kg}$$ and muzzle speed $$v = 50 \text{ ms}^{-1}$$.

The impulse imparted to the gun equals the change in momentum of the bullet (Newton's third law). The magnitude of impulse is: $$J = mv = 0.004 \times 50 = 0.2 \text{ kg m s}^{-1}$$

By conservation of momentum, the gun (mass $$M = 4 \text{ kg}$$) recoils with velocity $$V$$ such that: $$MV = mv$$ $$V = \frac{mv}{M} = \frac{0.004 \times 50}{4} = \frac{0.2}{4} = 0.05 \text{ m s}^{-1}$$

Therefore, the impulse imparted to the gun is $$0.2 \text{ kg m s}^{-1}$$ and the velocity of recoil is $$0.05 \text{ m s}^{-1}$$.

To move the body with the minimum possible force, the force $$F$$ should be applied at an angle $$\theta$$ above the horizontal. Resolving forces: the horizontal component $$F\cos\theta$$ must overcome friction, and the vertical component $$F\sin\theta$$ reduces the normal reaction.

The normal reaction is $$N = mg - F\sin\theta$$ and the friction force is $$f = \mu N = \mu(mg - F\sin\theta)$$. For the body to just move, $$F\cos\theta = \mu(mg - F\sin\theta)$$, giving $$F = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$$.

To minimize $$F$$, we maximize the denominator $$\cos\theta + \mu\sin\theta$$. Taking the derivative and setting it to zero: $$-\sin\theta + \mu\cos\theta = 0$$, so $$\tan\theta = \mu = \frac{1}{\sqrt{3}}$$, which gives $$\theta = 30°$$.

The maximum value of the denominator is $$\cos 30° + \frac{1}{\sqrt{3}}\sin 30° = \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}} \times \frac{1}{2} = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{3 + 1}{2\sqrt{3}} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$$.

Therefore $$F = \frac{\frac{1}{\sqrt{3}} \times 1 \times 10}{\frac{2}{\sqrt{3}}} = \frac{10/\sqrt{3}}{2/\sqrt{3}} = \frac{10}{2} = 5$$ N.

The body of mass $$m = 2$$ kg starts from rest at the origin and moves under a constant force $$\vec{F} = 2\hat{i} + 3\hat{j} + 5\hat{k}$$ N.

The acceleration is $$\vec{a} = \frac{\vec{F}}{m} = \frac{2\hat{i} + 3\hat{j} + 5\hat{k}}{2} = 1\hat{i} + 1.5\hat{j} + 2.5\hat{k}$$ m/s$$^2$$.

Since the body starts from rest at the origin, the displacement at time $$t$$ is $$\vec{s} = \frac{1}{2}\vec{a}t^2$$.

At $$t = 4$$ s: $$\vec{s} = \frac{1}{2}(1\hat{i} + 1.5\hat{j} + 2.5\hat{k})(16) = 8\hat{i} + 12\hat{j} + 20\hat{k}$$.

The coordinates are $$(8, 12, 20)$$. Comparing with the given coordinates $$(8, b, 20)$$, we find $$b = 12$$.

We have two bodies. The first body has mass $$m_1 = 2\ \text{kg}$$ and is moving with initial speed $$u_1 = 4\ \text{m s}^{-1}$$. The second body has mass $$m_2$$ (unknown for now) and is initially at rest, so $$u_2 = 0$$.

Because the collision is perfectly elastic, two basic conservation laws can be applied:

1. Conservation of linear momentum $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2.$$

2. Conservation of kinetic energy $$\dfrac12 m_1 u_1^{2} + \dfrac12 m_2 u_2^{2} = \dfrac12 m_1 v_1^{2} + \dfrac12 m_2 v_2^{2}.$$

After the collision, the first body continues in the same direction but with one-fourth of its initial speed. Hence

$$v_1 = \dfrac{u_1}{4} = \dfrac{4}{4} = 1\ \text{m s}^{-1}.$$

Let the speed of the second body after collision be $$v_2$$ (to be found). We now substitute the known numbers into the two conservation equations.

Step 1: Momentum conservation $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$2 \times 4 + m_2 \times 0 = 2 \times 1 + m_2 v_2$$ $$8 = 2 + m_2 v_2.$$

Rearranging gives $$m_2 v_2 = 8 - 2 = 6. \quad -(1)$$

Step 2: Kinetic-energy conservation $$\dfrac12 m_1 u_1^{2} + 0 = \dfrac12 m_1 v_1^{2} + \dfrac12 m_2 v_2^{2}$$ $$\dfrac12 \times 2 \times 4^{2} = \dfrac12 \times 2 \times 1^{2} + \dfrac12 m_2 v_2^{2}$$ $$1 \times 16 = 1 \times 1 + \dfrac12 m_2 v_2^{2}$$ $$16 = 1 + \dfrac12 m_2 v_2^{2}.$$

Subtracting 1 from both sides, we obtain $$\dfrac12 m_2 v_2^{2} = 15$$ $$m_2 v_2^{2} = 30. \quad -(2)$$

Step 3: Solving for $$v_2$$ We divide equation (2) by equation (1): $$\dfrac{m_2 v_2^{2}}{m_2 v_2} = \dfrac{30}{6}$$ $$v_2 = 5\ \text{m s}^{-1}.$$

Step 4: Determining $$m_2$$ (only for completeness) Using equation (1): $$m_2 = \dfrac{6}{v_2} = \dfrac{6}{5} = 1.2\ \text{kg}.$$

Step 5: Velocity of the centre of mass The centre of mass (COM) velocity is constant throughout the motion. It can be found from the initial momenta: $$v_{\text{cm}} = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}.$$

Substituting the values, we get $$v_{\text{cm}} = \dfrac{2 \times 4 + 1.2 \times 0}{2 + 1.2} = \dfrac{8}{3.2}\ \text{m s}^{-1}.$$

Now, $$\dfrac{8}{3.2} = 2.5\ \text{m s}^{-1}.$$

The problem states that the COM speed can be expressed as $$\dfrac{x}{10}\ \text{m s}^{-1}$$. So we set

$$\dfrac{x}{10} = 2.5 \quad\Longrightarrow\quad x = 2.5 \times 10 = 25.$$

So, the answer is $$25$$.

Let the coefficient of friction be $$\mu = \frac{\sqrt{x}}{5}$$ and the incline angle be $$\theta = 30^\circ$$. During ascent, both the component of gravity along the plane and friction act downward along the slope, so the deceleration is $$a_{\text{up}} = g(\sin\theta + \mu\cos\theta)$$. During descent, gravity acts down the slope while friction acts upward, giving acceleration $$a_{\text{down}} = g(\sin\theta - \mu\cos\theta)$$.

If the body is launched with initial speed $$u$$, it decelerates uniformly to rest, so the time of ascent is $$t_a = \frac{u}{a_{\text{up}}}$$. The distance traveled up the slope is $$s = \frac{u^2}{2a_{\text{up}}}$$. Starting from rest, the body descends this same distance $$s$$ with acceleration $$a_{\text{down}}$$, so $$s = \frac{1}{2}a_{\text{down}}t_d^2$$, giving $$t_d = \sqrt{\frac{2s}{a_{\text{down}}}} = \frac{u}{\sqrt{a_{\text{up}} \cdot a_{\text{down}}}}$$.

We are given $$t_a = \frac{1}{2}t_d$$:

$$\frac{u}{a_{\text{up}}} = \frac{1}{2} \cdot \frac{u}{\sqrt{a_{\text{up}} \cdot a_{\text{down}}}}$$

$$2\sqrt{a_{\text{up}} \cdot a_{\text{down}}} = a_{\text{up}}$$

Squaring both sides: $$4 a_{\text{down}} = a_{\text{up}}$$

Substituting with $$\sin 30^\circ = \frac{1}{2}$$ and $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$:

$$4\left(\frac{1}{2} - \mu\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \mu\frac{\sqrt{3}}{2}$$

$$2 - 2\sqrt{3}\mu = \frac{1}{2} + \frac{\sqrt{3}}{2}\mu$$

$$\frac{3}{2} = 2\sqrt{3}\mu + \frac{\sqrt{3}}{2}\mu = \frac{5\sqrt{3}}{2}\mu$$

$$\mu = \frac{3}{2} \cdot \frac{2}{5\sqrt{3}} = \frac{3}{5\sqrt{3}} = \frac{\sqrt{3}}{5}$$

Since $$\mu = \frac{\sqrt{x}}{5}$$, we get $$\sqrt{x} = \sqrt{3}$$, so $$x = 3$$.

We consider the motion from the point of view of an observer sitting inside the car. Since the car is accelerating, its interior is a non-inertial frame, so in addition to the real forces we must introduce a pseudo (fictitious) force on every mass in the direction opposite to the car’s acceleration.

The given data are

$$a = 10 \text{ m s}^{-2}$$ (acceleration of the car upward along the plane) $$g = 10 \text{ m s}^{-2}$$ (acceleration due to gravity, vertically downward) $$\alpha = 30^\circ \quad$$ (inclination of the plane to the horizontal)

The real force on the bob is its weight $$\vec{W}=m\vec{g},$$ vertically downward.

The pseudo force is $$\vec{F}_p = -m\vec{a},$$ opposite to the car’s acceleration, i.e. down the plane. Because the car’s acceleration is parallel to the plane and up the slope, the pseudo force is parallel to the plane and down the slope, of magnitude $$|\vec{F}_p| = ma = 10m.$$

We now add these two vectors to obtain the effective force (or effective acceleration) that the bob “feels” in the non-inertial frame:

$$\vec{g}_{\text{eff}} = \vec{g} + \left(-\vec{a}\right).$$

To combine them, we express the pseudo force in components relative to the vertical:

• The direction “down the slope” makes an angle of $$\alpha = 30^\circ$$ below the horizontal. The angle between this direction and the vertical is therefore $$90^\circ - 30^\circ = 60^\circ.$$
• Hence the pseudo force of magnitude $$10m$$ has
   vertical component $$10m\cos 60^\circ,$$
   horizontal (transverse) component $$10m\sin 60^\circ.$$

Writing the magnitudes only (the factor m cancels everywhere),

Vertical component of $$\vec{g}_{\text{eff}}$$: $$g_{\text{vert}} = 10 + 10\cos 60^\circ = 10 + 10\left(\dfrac12\right) = 10 + 5 = 15.$$

Horizontal (lateral) component of $$\vec{g}_{\text{eff}}$$: $$g_{\text{horiz}} = 10\sin 60^\circ = 10\left(\dfrac{\sqrt3}{2}\right) = 5\sqrt3.$$

The string aligns itself along the direction of $$\vec{g}_{\text{eff}},$$ so the angle $$\theta$$ that the string makes with the true vertical is determined by

$$\tan\theta = \dfrac{g_{$$ horiz $$}}{g_{$$ vert $$}} = \dfrac{5\sqrt3}{15} = \dfrac{\sqrt3}{3} = \dfrac1{\sqrt3}.$$

We know the standard trigonometric value

$$\tan 30^\circ = \dfrac1{\sqrt3}.$$

Therefore,

$$\theta = 30^\circ.$$

So, the answer is $$30^\circ$$.

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$$^{-1}$$.

We have a wooden block of mass $$m = 0.5$$ kg pressed against a vertical rough wall by a horizontal force $$F$$. The coefficient of static friction is $$\mu_s = 0.2$$.

When the block is pushed horizontally against the wall, the normal reaction from the wall equals the applied force. So we have $$N = F$$.

The frictional force acts upward along the wall to prevent the block from sliding down. The maximum static friction is $$f = \mu_s N = \mu_s F$$.

For the block to remain stationary on the wall, the friction must balance the weight of the block. So $$f \geq mg$$, which gives us $$\mu_s F \geq mg$$.

Substituting the values, we get $$0.2 \times F \geq 0.5 \times 10$$.

$$0.2F \geq 5$$

$$F \geq \frac{5}{0.2} = 25 \text{ N}$$

The minimum horizontal force required is $$F = 25$$ N.

So, the answer is $$25$$.

We have a curved inclined plane whose vertical cross-section is given by $$y = \frac{x^2}{4}$$. The coefficient of friction is $$\mu = 0.5$$.

At any point on the curve, the slope of the tangent gives the angle of inclination $$\theta$$ with the horizontal. We find the slope by differentiating: $$\frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}$$.

Now, $$\tan\theta = \frac{dy}{dx} = \frac{x}{2}$$.

A stationary block will not slip as long as the component of gravity along the surface does not exceed the maximum static friction. This condition is $$\tan\theta \leq \mu$$.

Substituting, we get $$\frac{x}{2} \leq 0.5$$, which gives $$x \leq 1$$.

The maximum height is reached when $$x = 1$$. Substituting into the equation of the curve, $$y = \frac{1^2}{4} = \frac{1}{4} = 0.25$$ m.

Converting to centimetres, $$y = 0.25 \times 100 = 25$$ cm.

So, the answer is $$25$$.

We are given two situations for the same body moving down the same distance $$s$$ on two different 30° inclined planes, one smooth (no friction) and the other rough (with friction).

For any motion with uniform acceleration, the kinematic relation that connects distance travelled from rest, acceleration and time is stated first:

$$s=\dfrac12\,a\,t^2.$$

Motion on the smooth plane
The only component of gravity acting along the plane is $$g\sin\theta$$, where $$\theta=30^\circ$$. Therefore

$$a_{\text{smooth}}=g\sin\theta.$$

If the time taken is $$T$$, we substitute in the kinematic formula:

$$s=\dfrac12 \,a_{\text{smooth}}\,T^2 =\dfrac12 \,g\sin\theta\,T^2.$$

Motion on the rough plane
Let the coefficient of friction be $$\mu$$. The normal reaction is $$N=mg\cos\theta$$, so the frictional force is $$\mu N = \mu mg\cos\theta$$ acting up the plane. The net force down the plane is therefore

$$mg\sin\theta-\mu mg\cos\theta =m\bigl(g\sin\theta-\mu g\cos\theta\bigr).$$

Hence the acceleration on the rough plane is

$$a_{\text{rough}}=g\sin\theta-\mu g\cos\theta.$$

The time of descent now is given to be $$\alpha T$$, so the kinematic relation gives

$$s=\dfrac12 \,a_{\text{rough}}\,( \alpha T )^2 =\dfrac12 \bigl(g\sin\theta-\mu g\cos\theta \bigr)\alpha^2 T^2.$$

Equating the two expressions for the same distance $$s$$

$$\dfrac12 g\sin\theta\,T^2 =\dfrac12 \bigl(g\sin\theta-\mu g\cos\theta\bigr)\,\alpha^2 T^2.$$

The common factors $$\dfrac12\,g\,T^2$$ cancel, leaving

$$\sin\theta =\alpha^2\bigl(\sin\theta-\mu\cos\theta\bigr).$$

Expanding the right side, we have

$$\sin\theta =\alpha^2\sin\theta-\alpha^2\mu\cos\theta.$$

Rearranging to isolate $$\mu$$:

$$\alpha^2\mu\cos\theta =\alpha^2\sin\theta-\sin\theta =(\alpha^2-1)\sin\theta,$$

so

$$\mu=\dfrac{\sin\theta}{\cos\theta}\;\dfrac{\alpha^2-1}{\alpha^2}.$$

Because $$\dfrac{\sin\theta}{\cos\theta}=\tan\theta,$$ we write

$$\mu=\tan\theta\;\dfrac{\alpha^2-1}{\alpha^2}.$$

Given $$\theta=30^\circ,$$ we recall the standard trigonometric value

$$\tan30^\circ=\dfrac1{\sqrt3}.$$

Thus

$$\mu=\dfrac1{\sqrt3}\;\dfrac{\alpha^2-1}{\alpha^2}.$$

This matches the form stated in the question, namely

$$\mu=\dfrac1{\sqrt{x}}\;\dfrac{\alpha^2-1}{\alpha^2},$$

and by direct comparison we identify

$$x=3.$$

So, the answer is $$3$$.

We have a particle of mass $$M$$ that is initially at rest, so its initial velocity is zero. A time-dependent force acts on it along a fixed direction for the duration $$0 \le t \le 2T$$. The magnitude of the force is given by

$$F(t)=F_0\left[1-\left(\frac{t-T}{T}\right)^2\right].$$

According to Newton’s second law, the relation between force and acceleration is

$$F(t)=M\,a(t)=M\,\frac{dv}{dt}.$$

We can rewrite this as

$$\frac{dv}{dt}=\frac{F(t)}{M}.$$

To obtain the velocity after the force has acted for the full interval, we integrate both sides with respect to time from the initial instant $$t=0$$ to the final instant $$t=2T$$. Because the particle starts from rest, its initial velocity is zero. Hence

$$v(2T)-v(0)=\int_{0}^{2T}\frac{dv}{dt}\,dt=\int_{0}^{2T}\frac{F(t)}{M}\,dt.$$

Since $$v(0)=0$$, the velocity after time $$2T$$ is simply

$$v=\frac{1}{M}\int_{0}^{2T}F(t)\,dt.$$

Substituting the given expression for $$F(t)$$, we have

$$v=\frac{1}{M}\int_{0}^{2T}F_0\left[1-\left(\frac{t-T}{T}\right)^2\right]dt.$$

Now we factor out the constant $$F_0$$:

$$v=\frac{F_0}{M}\int_{0}^{2T}\left[1-\frac{(t-T)^2}{T^2}\right]dt.$$

To simplify the integral, let us change variables. Define

$$x=t-T \quad\Longrightarrow\quad t=x+T.$$

When $$t=0$$, we get $$x=-T$$; and when $$t=2T$$, we get $$x=+T$$. In terms of the new variable, the integral becomes

$$v=\frac{F_0}{M}\int_{-T}^{T}\left[1-\frac{x^2}{T^2}\right]dx.$$

We now carry out the integration term by term:

$$\int_{-T}^{T}1\,dx = \Bigl[x\Bigr]_{-T}^{T}=T-(-T)=2T,$$

$$\int_{-T}^{T}\frac{x^2}{T^2}\,dx = \frac{1}{T^2}\int_{-T}^{T}x^2\,dx = \frac{1}{T^2}\left[\frac{x^3}{3}\right]_{-T}^{T} = \frac{1}{T^2}\left(\frac{T^3}{3}-\frac{(-T)^3}{3}\right)=\frac{1}{T^2}\left(\frac{T^3}{3}+\frac{T^3}{3}\right)=\frac{2T}{3}.$$

Hence the complete integral evaluates to

$$\int_{-T}^{T}\left[1-\frac{x^2}{T^2}\right]dx \;=\; 2T-\frac{2T}{3}=\frac{4T}{3}.$$

Substituting this result back, we get

$$v=\frac{F_0}{M}\left(\frac{4T}{3}\right)=\frac{4F_0T}{3M}.$$

Therefore the speed of the particle after the force has acted for the time interval $$2T$$ is

$$v=\frac{4F_0T}{3M}.$$

Hence, the correct answer is Option C.

We consider the instantaneous motion of the rocket when its mass is still $$m = 1000\text{ kg}$$ and it has to acquire an upward acceleration of $$a = 20\ \text{m s}^{-2}$$. At that very instant the rocket ejects gases backward with a speed, relative to the rocket, of $$u = 500\ \text{m s}^{-1}$$.

According to the variable-mass (rocket) mechanics, the backward ejection of mass produces a forward thrust. The magnitude of this thrust is obtained from the basic relation

$$\text{Thrust} = \left(\dfrac{dm}{dt}\right)u,$$

where $$\dfrac{dm}{dt}$$ is the rate at which mass is being lost by the rocket (this rate is positive in magnitude although the actual change in mass is negative).

The forces acting on the rocket, all taken along the vertical line of motion, are:

1. Upward thrust $$\left(\dfrac{dm}{dt}\right)u$$,

2. Downward weight $$mg$$.

The net upward force must produce the required upward acceleration. Hence Newton’s second law gives

$$\left(\dfrac{dm}{dt}\right)u - mg = ma.$$

Now we substitute the known numerical values. We have

$$m = 1000\ \text{kg}, \qquad g = 10\ \text{m s}^{-2}, \qquad a = 20\ \text{m s}^{-2}, \qquad u = 500\ \text{m s}^{-1}.$$

Putting these into the above equation,

$$\left(\dfrac{dm}{dt}\right)(500) - (1000)(10) = (1000)(20).$$

First evaluate the right-hand side:

$$ (1000)(20) = 20000. $$

So we have

$$500\dfrac{dm}{dt} - 10000 = 20000.$$

Move the weight term to the right:

$$500\dfrac{dm}{dt} = 20000 + 10000 = 30000.$$

Solve for $$\dfrac{dm}{dt}$$ by dividing both sides by 500:

$$\dfrac{dm}{dt} = \dfrac{30000}{500}.$$

Carry out the division:

$$\dfrac{dm}{dt} = 60\ \text{kg s}^{-1}.$$

Thus the fuel must be burnt at a rate of $$60\ \text{kilograms per second}$$ to give the rocket the desired acceleration.

Hence, the correct answer is Option B.

We have a body of mass $$M$$ moving initially with speed $$V_0$$ along the positive $$x$$-axis. It collides elastically with a second body of mass $$m$$ that is at rest. After the collision the two bodies move with speeds $$u$$ (for mass $$M$$) and $$v$$ (for mass $$m$$) making angles $$\theta_1$$ and $$\theta_2$$ respectively with the original direction. The question states that $$\theta_1$$ and $$\theta_2$$ are equal, so we write $$\theta_1=\theta_2=\theta$$.

Because the collision is perfectly elastic, both linear momentum and kinetic energy are conserved. We now translate those words into equations.

Conservation of linear momentum in component form:

Along $$x$$:

$$M V_0 \;=\; M u\cos\theta + m v\cos\theta$$

Along $$y$$ (taking the upward direction for the angle of mass $$M$$ and the downward direction for mass $$m$$ so that the net initial $$y$$-momentum is zero):

$$0 \;=\; M u\sin\theta \;-\; m v\sin\theta$$

From the second equation we solve for $$u$$:

$$M u\sin\theta = m v\sin\theta \;\Longrightarrow\; u = \dfrac{m}{M}\,v.$$

Substituting this value of $$u$$ into the first momentum equation gives

$$M V_0 = M\bigl(\tfrac{m}{M}v\bigr)\cos\theta + m v\cos\theta = m v\cos\theta + m v\cos\theta = 2 m v\cos\theta.$$

Hence the speed of mass $$m$$ after collision is

$$v = \dfrac{M V_0}{2 m\cos\theta}.$$

Conservation of kinetic energy now supplies a second relation. The usual formula is $$\tfrac12 M V_0^2 = \tfrac12 M u^2 + \tfrac12 m v^2.$$ Removing the factor $$\tfrac12$$ gives

$$M V_0^2 = M u^2 + m v^2.$$

We already have $$u = \dfrac{m}{M}v,$$ so we substitute:

$$M V_0^2 = M\!\left(\dfrac{m}{M}v\right)^{\!2} + m v^2 = M\!\left(\dfrac{m^2}{M^2}v^2\right) + m v^2 = \dfrac{m^2}{M}\,v^2 + m v^2 = v^2\!\left(\dfrac{m^2}{M} + m\right).$$

Factor out $$m$$ to simplify the right-hand side:

$$M V_0^2 = v^2 \, m \!\left(\dfrac{m}{M} + 1\right) = v^2 \, m \!\left(\dfrac{m + M}{M}\right).$$

Multiplying both sides by $$M$$ removes the denominator:

$$M^2 V_0^2 = v^2\,m\,(m+M).$$

Now we substitute for $$v$$ from the momentum result:

$$v^2 = \left(\dfrac{M V_0}{2 m\cos\theta}\right)^{\!2} = \dfrac{M^2 V_0^2}{4 m^{\,2}\cos^{2}\theta}.$$

Putting this into the energy equation yields

$$M^2 V_0^2 \;=\; \dfrac{M^2 V_0^2}{4 m^{\,2}\cos^{2}\theta}\;\,m\,(m+M).$$

The factor $$M^2 V_0^2$$ appears on both sides, so it cancels out completely:

$$1 = \dfrac{m(m+M)}{4 m^{\,2}\cos^{2}\theta} = \dfrac{m+M}{4 m\cos^{2}\theta}.$$

Re-arranging this gives the squared cosine of the common angle:

$$\cos^{2}\theta = \dfrac{m+M}{4 m}.$$

We know from trigonometry that $$\cos^{2}\theta \le 1$$ for all real angles. Therefore

$$\dfrac{m+M}{4 m} \;\le\; 1.$$

Multiplying both sides by $$4m$$ we get

$$m + M \;\le\; 4 m.$$

Isolating $$M$$ leads directly to

$$M \;\le\; 3 m.$$

Thus the largest possible value of the ratio $$\dfrac{M}{m}$$ that still allows the two deflection angles to be equal is

$$\dfrac{M}{m} = 3.$$

Hence, the correct answer is Option A.

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.

Let the initial velocity of $$m_1$$ be $$u$$ and $$m_2$$ is at rest. After collision, both move with equal speeds $$v$$ in opposite directions. So $$m_1$$ moves with speed $$v$$ in the negative direction and $$m_2$$ moves with speed $$v$$ in the positive direction.

Applying conservation of momentum: $$m_1 u = m_2 v - m_1 v$$, which gives $$m_1 u = v(m_2 - m_1)$$.

For the collision to be physically valid, we also use the coefficient of restitution. For elastic collision, $$e = 1$$: $$e = \frac{v - (-v)}{u - 0} = \frac{2v}{u} = 1$$, giving $$u = 2v$$.

Substituting into the momentum equation: $$m_1(2v) = v(m_2 - m_1)$$, so $$2m_1 = m_2 - m_1$$, which gives $$m_2 = 3m_1$$.

Therefore $$m_2 : m_1 = 3 : 1$$.

We have a constant force $$\vec F = 40\hat i + 10\hat j \;{\rm N}$$ acting on a body whose mass is $$m = 5\;{\rm kg}$$.

According to Newton’s second law, the relation between force and acceleration is first stated as $$\vec F = m\,\vec a.$$

We now solve for the acceleration $$\vec a$$ by dividing the force vector by the mass:

$$\vec a = \frac{\vec F}{m} = \frac{40\hat i + 10\hat j}{5} = \left(\frac{40}{5}\right)\hat i + \left(\frac{10}{5}\right)\hat j = 8\hat i + 2\hat j\;{\rm m\,s^{-2}}.$$

The body starts from rest, so its initial velocity is

$$\vec v_0 = 0\hat i + 0\hat j\;{\rm m\,s^{-1}}.$$

For motion with constant acceleration, the position vector after time $$t$$ is given by the kinematic formula (which we state explicitly):

$$\vec r = \vec r_0 + \vec v_0\,t + \frac12\,\vec a\,t^{2}.$$

The body is released from the origin, so $$\vec r_0 = 0\hat i + 0\hat j$$. Because $$\vec v_0 = 0$$, the second term also vanishes. Therefore, the position vector simplifies to

$$\vec r = \frac12\,\vec a\,t^{2}.$$

Substituting the acceleration $$\vec a = 8\hat i + 2\hat j$$ and the time $$t = 10\;{\rm s}$$, we obtain

$$\vec r = \frac12\,(8\hat i + 2\hat j)\,(10)^{2} = \frac12\,(8\hat i + 2\hat j)\,(100).$$

Carrying out the multiplication, we get component-wise values:

$$\vec r = \left[\frac12 \times 8 \times 100\right]\hat i + \left[\frac12 \times 2 \times 100\right]\hat j = (4 \times 100)\hat i + (1 \times 100)\hat j = 400\hat i + 100\hat j\;{\rm m}.$$

Thus the particle’s position vector after 10 seconds is $$\left(400\hat i + 100\hat j\right)\,{\rm m}$$.

Hence, the correct answer is Option C.

We have a particle of mass $$m$$ moving only in the $$xy$$-plane, so its velocity may be written as $$\vec v = v_x \hat i + v_y \hat j$$ with $$v_x = \dfrac{dx}{dt}$$ and $$v_y = \dfrac{dy}{dt}$$. The force acting on it is given in the question as

$$\vec F = k\bigl(v_y\hat i + v_x\hat j\bigr).$$

By Newton’s second law $$\vec F = m\vec a$$, therefore the acceleration components satisfy

$$\begin{aligned} a_x &= \dfrac{F_x}{m} = \dfrac{k}{m}\,v_y,\\[4pt] a_y &= \dfrac{F_y}{m} = \dfrac{k}{m}\,v_x. \end{aligned}$$

So the two coupled differential equations for the velocity components are

$$\begin{aligned} \frac{dv_x}{dt} &= \frac{k}{m}\,v_y, \quad\quad (1)\\[6pt] \frac{dv_y}{dt} &= \frac{k}{m}\,v_x. \quad\quad (2) \end{aligned}$$

Now we examine each statement one by one, using these relations.

Testing the constancy of $$\vec v\times\vec a$$ (Option A).

Because motion is confined to the plane, both $$\vec v$$ and $$\vec a$$ have zero $$z$$-components, so their cross-product points along $$\hat k$$ and its magnitude equals the scalar expression

$$\vec v \times \vec a = (v_x\hat i + v_y\hat j)\times\bigl(a_x\hat i + a_y\hat j\bigr) = (v_x a_y - v_y a_x)\,\hat k.$$

Substituting $$a_x$$ and $$a_y$$ from above,

$$\begin{aligned} v_x a_y - v_y a_x &= v_x\left(\frac{k}{m}v_x\right) - v_y\left(\frac{k}{m}v_y\right)\\ &= \frac{k}{m}\left(v_x^2 - v_y^2\right). \end{aligned}$$

We differentiate this quantity with respect to time to see whether it changes:

$$\begin{aligned} \frac{d}{dt}\!\left(v_x a_y - v_y a_x\right) &=\frac{k}{m}\,\frac{d}{dt}\!\left(v_x^2 - v_y^2\right)\\[6pt] &=\frac{k}{m}\,\bigl(2v_x\dot v_x - 2v_y\dot v_y\bigr)\\[6pt] &=\frac{2k}{m}\,\Bigl(v_x\frac{dv_x}{dt} - v_y\frac{dv_y}{dt}\Bigr). \end{aligned}$$

Using Eqs. (1) and (2): $$\dot v_x = \dfrac{k}{m}v_y$$ and $$\dot v_y = \dfrac{k}{m}v_x$$, so

$$\begin{aligned} \frac{d}{dt}\!\left(v_x a_y - v_y a_x\right) &=\frac{2k}{m}\Bigl(v_x\left(\frac{k}{m}v_y\right) - v_y\left(\frac{k}{m}v_x\right)\Bigr)\\[6pt] &=\frac{2k}{m}\cdot\frac{k}{m}\,(v_x v_y - v_y v_x)\\[6pt] &=\frac{2k^2}{m^2}\,(0)=0. \end{aligned}$$

The time derivative is zero, so $$v_x a_y - v_y a_x$$, and hence $$\vec v\times\vec a$$, is a constant vector. Therefore statement A is correct.

Testing whether $$\vec F$$ could be a magnetic force (Option B).

A magnetic (Lorentz) force on a charge $$q$$ has the form $$\vec F = q\,\vec v\times\vec B$$, which is always perpendicular to $$\vec v$$, implying $$\vec v\cdot\vec F = 0$$. For the given force, however,

$$\vec v\cdot\vec F = v_x(kv_y) + v_y(kv_x) = 2k\,v_x v_y,$$

which is in general non-zero. Hence the given force cannot come purely from a magnetic field, so statement B is false.

Testing the constancy of kinetic energy (Option C).

Kinetic energy is $$K = \dfrac12 m v^2 = \dfrac12 m(v_x^2 + v_y^2).$$ Its time derivative is

$$\frac{dK}{dt} = m\,\vec v\cdot\vec a.$$

We have just computed $$\vec v\cdot\vec a = 2\frac{k}{m}v_x v_y$$, so

$$\frac{dK}{dt} = m\left(2\frac{k}{m}v_x v_y\right) = 2k\,v_x v_y,$$

which is not generally zero. Thus kinetic energy is not constant, and statement C is false.

Testing whether $$\vec v\cdot\vec a$$ itself is constant (Option D).

From above $$\vec v\cdot\vec a = 2\dfrac{k}{m}v_x v_y.$$ Differentiating using the product rule and Eqs. (1) and (2),

$$\begin{aligned} \frac{d}{dt}\!\bigl(\vec v\cdot\vec a\bigr) &=2\frac{k}{m}\Bigl(v_x\dot v_y + v_y\dot v_x\Bigr)\\[6pt] &=2\frac{k}{m}\Bigl(v_x\cdot\frac{k}{m}v_x + v_y\cdot\frac{k}{m}v_y\Bigr)\\[6pt] &=2\frac{k^2}{m^2}\bigl(v_x^2 + v_y^2\bigr), \end{aligned}$$

which is strictly positive (unless the particle is at rest). Therefore $$\vec v\cdot\vec a$$ is not constant, and statement D is false.

Only Option A stands up to scrutiny.

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 radius of the hemispherical ditch be $$R = 1\ \text{m}$$. We consider the insect when it is at some point on the inner surface that is a vertical height $$h$$ above the lowest point (the bottom).

Draw a radius from the centre of the hemisphere to the insect. Let the angle between this radius and the vertical diameter be $$\theta$$. With this definition, the lowest point corresponds to $$\theta = 0$$ and any higher point corresponds to a positive $$\theta$$.

From simple geometry of a circle, the vertical height of the point above the bottom is related to $$\theta$$ by

$$h = R - R\cos\theta.$$

Since $$R = 1\ \text{m}$$, this becomes

$$h = 1 - \cos\theta.$$

Now we analyse the forces acting on the insect when it is momentarily at rest and on the verge of slipping downwards. The forces are:

1. The weight $$mg$$ acting vertically downward.
2. The normal reaction $$N$$ acting radially inward along the radius.
3. The static friction force $$f$$ acting up the surface (because it opposes the impending downward slide).

We resolve the weight into two components with respect to the surface:

• Normal component: $$mg\cos\theta$$ (directed along the normal, toward the centre).
• Tangential component: $$mg\sin\theta$$ (directed down the surface).

By Newton’s second law in the normal direction, because there is no motion perpendicular to the surface, the normal reaction must balance the normal component of weight, so

$$N = mg\cos\theta.$$

The condition for impending slip is that the tangential component of weight equals the maximum static friction. Stating the friction formula first, the maximum static friction is

$$f_{\text{max}} = \mu N,$$

where $$\mu = 0.75$$ is the coefficient of static friction. Setting this equal to the tangential component, we get

$$mg\sin\theta = \mu N = \mu\,mg\cos\theta.$$

Dividing both sides by $$mg\cos\theta$$ yields

$$\frac{\sin\theta}{\cos\theta} = \mu \quad\Longrightarrow\quad \tan\theta = \mu.$$

Substituting $$\mu = 0.75$$, we have

$$\tan\theta = 0.75 = \frac34.$$

We can associate $$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac34$$ with a right-angled triangle whose opposite side is 3, adjacent side is 4, and hypotenuse is 5. Hence

$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac45 = 0.80.$$

Substituting this value in the height expression $$h = 1 - \cos\theta$$ gives

$$h = 1 - 0.80 = 0.20\ \text{m}.$$

Hence, the correct answer is Option A.

We have a block that is projected up an inclined plane making an angle $$\theta = 30^{\circ}$$ with the horizontal. The initial velocity is $$v_0$$. While the block slides, kinetic friction acts opposite to its motion, and the coefficient of kinetic friction is $$\mu_k$$ (we shall write it simply as $$\mu$$). The block first moves upward, stops momentarily, and then slides back down to its starting point with speed $$\dfrac{v_0}{2}$$.

Let the upward direction along the plane be positive. The forces along the plane are the component of gravity $$mg\sin\theta$$ (down the plane) and the kinetic friction force $$\mu mg\cos\theta$$ (also down the plane because motion is upward). Hence, during the upward motion the net acceleration is

$$a_1 = -\bigl(g\sin\theta + \mu g\cos\theta\bigr).$$

The block starts with speed $$v_0$$ and comes to rest after travelling some distance $$s$$ up the plane, so we use the constant-acceleration relation (first state the formula):

For motion with initial speed $$u$$, final speed $$v$$, acceleration $$a$$ and displacement $$s$$, the kinematic equation is $$v^{2}=u^{2}+2as$$. Applying it upward,

$$0^{2}=v_0^{2}+2\,a_1\,s \;\Longrightarrow\; 0=v_0^{2}+2\bigl(-g\sin\theta-\mu g\cos\theta\bigr)s.$$

Re-arranging,

$$v_0^{2}=2\bigl(g\sin\theta+\mu g\cos\theta\bigr)s,$$

so

$$s=\dfrac{v_0^{2}}{2\bigl(g\sin\theta+\mu g\cos\theta\bigr)}. \quad -(1)$$

Now the block slides back down. While it moves downward, gravity’s component $$mg\sin\theta$$ is down the plane (this time along the direction of motion) but friction $$\mu mg\cos\theta$$ is still up the plane, opposing the motion. Thus the net acceleration downward is

$$a_2=g\sin\theta-\mu g\cos\theta.$$

The block starts this return journey from rest (at the top) and gains speed, reaching $$\dfrac{v_0}{2}$$ after descending the same distance $$s$$. Again using $$v^{2}=u^{2}+2as$$ with $$u=0,\;v=\dfrac{v_0}{2},\;a=a_2,\;s=s$$, we get

$$\Bigl(\dfrac{v_0}{2}\Bigr)^{2}=2\,a_2\,s,$$

that is,

$$\dfrac{v_0^{2}}{4}=2\bigl(g\sin\theta-\mu g\cos\theta\bigr)s. \quad -(2)$$

Substituting the value of $$s$$ from equation (1) into equation (2):

$$\dfrac{v_0^{2}}{4}=2\bigl(g\sin\theta-\mu g\cos\theta\bigr)\,\dfrac{v_0^{2}}{2\bigl(g\sin\theta+\mu g\cos\theta\bigr)}.$$

The factor $$v_0^{2}$$ cancels out, and one factor of 2 in numerator and denominator cancels as well, giving

$$\frac{1}{4}=\frac{g\sin\theta-\mu g\cos\theta}{g\sin\theta+\mu g\cos\theta}.$$

The acceleration due to gravity $$g$$ cancels on both numerator and denominator. Putting $$\sin30^{\circ}=\dfrac{1}{2}$$ and $$\cos30^{\circ}=\dfrac{\sqrt3}{2}$$, we have

$$\frac{1}{4}=\frac{\dfrac12-\mu\dfrac{\sqrt3}{2}}{\dfrac12+\mu\dfrac{\sqrt3}{2}}.$$

Multiplying numerator and denominator by 2 to clear the fractions gives

$$\frac{1}{4}=\frac{1-\mu\sqrt3}{1+\mu\sqrt3}.$$

Cross-multiplying,

$$(1+\mu\sqrt3)=4(1-\mu\sqrt3).$$

Expanding the right-hand side,

$$1+\mu\sqrt3=4-4\mu\sqrt3.$$

Now collect the terms containing $$\mu$$ on one side and constants on the other:

$$\mu\sqrt3+4\mu\sqrt3=4-1,$$

which simplifies to

$$5\mu\sqrt3=3.$$

Solving for $$\mu$$,

$$\mu=\frac{3}{5\sqrt3}=\frac{3}{5\times1.732} \approx \frac{3}{8.660} \approx 0.346.$$

The problem states that this value is close to $$\dfrac{I}{1000}$$, so $$I\approx346$$. The nearest integer to $$I$$ is therefore 346.

Hence, the correct answer is Option A: 346.

We have two bodies, both of mass $$m = 0.1\ \text{kg}$$.

The initial velocity of body A is $$\vec u_A = 3\hat i\ \text{m s}^{-1}$$ and that of body B is $$\vec u_B = 5\hat j\ \text{m s}^{-1}$$.

After an elastic collision, body A is given to move with $$\vec v_A = 4(\hat i + \hat j) = 4\hat i + 4\hat j\ \text{m s}^{-1}$$. Let the final velocity of body B be $$\vec v_B = v_{Bx}\hat i + v_{By}\hat j$$.

Because the collision is isolated, the total linear momentum is conserved. The law of conservation of momentum states $$m\vec u_A + m\vec u_B = m\vec v_A + m\vec v_B.$$

Substituting every known vector and cancelling the common mass $$m$$ on both sides gives

$$3\hat i + 5\hat j = (4\hat i + 4\hat j) + (v_{Bx}\hat i + v_{By}\hat j).$$

Now we equate the components separately:

For the $$\hat i$$ component: $$3 = 4 + v_{Bx}\ \Longrightarrow\ v_{Bx} = 3 - 4 = -1.$$

For the $$\hat j$$ component: $$5 = 4 + v_{By}\ \Longrightarrow\ v_{By} = 5 - 4 = 1.$$

So the velocity of body B after the collision is

$$\vec v_B = -1\hat i + 1\hat j\ \text{m s}^{-1}.$$

Next, we need the kinetic energy of body B after the collision. The kinetic-energy formula is $$K = \frac12 m v^2,$$ where $$v^2 = \vec v \cdot \vec v$$ is the square of the speed.

First compute $$v_B^2$$: $$v_B^2 = (-1)^2 + (1)^2 = 1 + 1 = 2.$$

Now substitute in the energy formula:

$$K_B = \frac12 (0.1)\,(2) = 0.1\ \text{J}.$$

This energy is written in the form $$\dfrac{x}{10}\ \text{J}$$, and since $$0.1\ \text{J} = \dfrac1{10}\ \text{J},$$ we identify

$$x = 1.$$

Hence, the correct answer is Option 1.