A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is $k^{\text{'}}$. If they are connected in parallel and force constant is $k^{\text{'}\text{'}},<mspace\; width="1em"></mspace>then<mspace\; width="1em"></mspace>k^{\text{'}}:k^{\text{'}\text{'}}$ is 

1. 1:6

2. 1:9

3. 1:11

4. 6:11

Consider a drop of rain water having mass \(1\) g falling from a height of \(1\) km. It hits the ground with a speed of \(50\) m/s.  Take \(g\) constant with a value of \(10\) m/$s^2$.  The work done by the 
(i) gravitational force and the (ii) resistive force of air is:

Two identical balls \(\mathrm{A}\) and \(\mathrm{B}\) having velocities of \(0.5~\text{m/s}\) and \(-0.3~\text{m/s}\) respectively collide elastically in one dimension. The velocities of \(\mathrm{B}\) and \(\mathrm{A}\) after the collision respectively will be:
1. \(-0.5 ~\text{m/s}~\text{and}~0.3~\text{m/s}\) 
2. \(0.5 ~\text{m/s}~\text{and}~-0.3~\text{m/s}\)
3. \(-0.3 ~\text{m/s}~\text{and}~0.5~\text{m/s}\)
4. \(0.3 ~\text{m/s}~\text{and}~0.5~\text{m/s}\)

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. $\sqrt{gR}$
 

Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a) and then by the same force (case b). The work done by external force, W_P and W_Q on the springs P and Q in case (a) and case (b) respectively are related as, 

1. W_P=W_Q ; W_P>W_Q

2. W_P=W_{Q ;} W_P=W_Q

3. W_P>W_{Q ;} W_Q>W_P

4. W_P<W_{Q ;} W_Q<W_P

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:

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. \(10~\text{m/s}\)