A body moves from rest with a constant acceleration. Which one of the following graphs represents the variation of its kinetic energy K with the distance travelled x ?

1. 

2.

3. 

4. 

The diagrams represent the potential energy U as a function of the inter-atomic distance r. Which diagram corresponds to stable molecules found in nature.

1. 

2. 

3. 

4. 

The relationship between the force F and the position x of a body is as shown in the figure. The work done in displacing the body from \(x = 1\text{ m}\) to \(x = 5\text{ m}\) will be:

A particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function) 

1. 

2. 

3. 

4. 

A body of mass 1 kg begins to move under the action of a time dependent force \(F = 2 t\) \(\hat{i} + 3 t^{2}\ \hat{j}\) N, where \(\hat{i}\) and \(\hat{j}\) are unit vectors along X and Y axis, What power will be developed by the force at the time (t) ?

1. \(\left(2 t^{2} + 4 t^{4}\right) W\)         

2. \(\left(2 t^{3} + 3 t^{4}\right) W\)

3. \(\left(2 t^{3} + 3 t^{5}\right) W\)         

4. \(\left(2 t + 3 t^{3}\right) W\)

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. $\sqrt{2\mathrm{gR}}$

2. $\sqrt{3gR}$

3. $\sqrt{5gR}$

4. $\sqrt{gR}$
 

A block of mass \(10\) kg, moving in the \(x\text-\)direction with a constant speed of \(10\) ms^-1, is subjected to a retarding force \(F=0.1x \) J/m during its travel from \(x =20\) m to \(30\) m. Its final kinetic energy will be:

A particle of mass m is driven by a machine that delivers a constant power of k watts. If the particle starts from rest the force on the particle at time t is:

1. $\sqrt{\left(\frac{mk}{2}\right)}{t}^{-1/2}$
 

2. $\sqrt{\left(mk\right)}{t}^{-1/2}$
 

3. $\sqrt{\left(2mk\right)}{t}^{-1/2}$
 

4. $\frac{1}{2}\sqrt{\left(mk\right)}{t}^{-1/2}$

Two particles of masses m₁,m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy $\epsilon$. If final velocities of particles be v₁ and v₂, then we must have

1. m₁²u₁+m₂²u₂-$\epsilon$=m₁²v₁+m₂²v₂

2. $\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²-$\epsilon$

3. $\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²-$\epsilon$=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²

4. $\frac{1}{2}$m₁²u₁²+$\frac{1}{2}$m₂²u₂²+$\epsilon$=$\frac{1}{2}$m₁²v₁²+$\frac{1}{2}$m₂²v₂²
 

A ball is thrown vertically downwards from a height of \(20\) m with an initial velocity \(v_0\). It collides with the ground, loses \(50\%\) of its energy in a collision and rebounds to the same height. The initial velocity \(v_0\) is: (Take \(g = 10~\text{m/s}^2\))
1. \(14~\text{m/s}\)
2. \(20~\text{m/s}\)
3. \(28~\text{m/s}\)
4. \(10~\text{m/s}\)