An engine pump is used to pump a liquid of density ρ continuously through a pipe of cross-sectional area A. If the speed of flow of the liquid in the pipe is v, then the rate at which kinetic energy is being imparted to the liquid is

(1) $\frac{1}{2}A\rho v^3$

(2) $\frac{1}{2}A\rho v^2$

(3) $\frac{1}{2}A\rho v$

(4) $A\rho v$

A particle free to move along the x-axis has potential energy given by $U\left(x\right)=k[1-e^{-x^2}]$ for $-\infty \le x\le +\infty$, where k is a positive constant of appropriate dimensions. Then 

(1) At point away from the origin, the particle is in unstable equilibrium

(2) For any finite non-zero value of x, there is a force directed away from the origin

(3) If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

(4) For small displacements from x = 0, the motion is simple harmonic

The potential energy of a system is represented in the first figure. the force acting on the system will be represented by

(1)

(2)

(3)

(4)

The force acting on a body moving along x-axis varies with the position of the particle as shown in the fig.

The body is in stable equilibrium at

(1) x = x₁

(2) x = x₂

(3) both x₁ and x₂

(4) neither x₁ nor x₂

The potential energy of a particle varies with distance x as shown in the graph. The force acting on the particle is zero at

(1) C

(2) B

(3) B and C

(4) A and D

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)

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 20m with an initial velocity v_o . It collides with the ground, loses 50% of it's energy in collision and rebounds to the same height. The initial velocity v_o is (Take, g = 10 ms^-2)

(1) 14 ms^-1

(2) 20ms^-1

(3) 28ms^-1

(4) 10ms^-1

The potential energy of a particle in a force field is $U=\frac{A}{r^2}-\frac{B}{r},$where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is 

(a) B/2A                                     (b)2A/B

(c)A/B                                        (d)B/A

A car of mass m starts from rest and  accelerates so that the instantaneous power delivered to the car has a constant magnitude P_o. The instantaneous velocity of this car is proportional to

1. t₂P₀

2. t^1/2

3. t^–1/2

4. t / √m