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 m to x = 5 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) ?

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

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

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

(d) \(\left(2 t + 3 t^{3}\right) W\)

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

(a)m₁²u₁+m₂²u₂-$\epsilon$=m₁²v₁+m₂²v₂

(b)$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²-$\epsilon$

(c)$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²-$\epsilon$=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²

(d)$\frac{1}{2}$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_o. It collides with the ground, loses 50% of its energy in a collision and rebounds to the same height. The initial velocity v_o is: (Take g = 10 ms^-2)

1. 14 ms^-1
2. 20 ms^-1
3. 28 ms^-1
4. 10 ms^-1

On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v/3. The second block's speed after the collision is:

(1)2√2v/3
 

(2)3v/4
 

(3)3v/√2
 

(4)√3v/2