There is a vertically suspended spring, mass system. When a block of mass $10 \mathrm{kg}$ is suspended from the lower end of the spring, it is stretched by $20 \mathrm{cm}$ under the load of a block at the equilibrium position. When an upward speed of$4 \mathrm{m} {\mathrm{s}}^{-1}$ is imparted to the block by giving a sharp impulse from below, how much high will it rise from the equilibrium position. 

The figure shows a hollow cube of side $a$ and volume $V$. There is a small chamber of volume $\frac{V}{4}$ in the cube as shown. This chamber is completely filled by $m \mathrm{kg}$ of water. Water leaks through a hole $H$ and spreads in the whole cube. Then, the work done by gravity in this process, assuming that the complete water finally lies at the bottom of the cube, is,

From what minimum height $h$ in metre must the system be released when the spring is in its natural length as shown in the figure so that after a perfectly inelastic collision $(e=0)$ of block $B$ with ground, $B$ may be lifted off ground.$(\text{Take,} k=40 \mathrm{N} {\mathrm{m}}^{-1}\text{,} g=10 \mathrm{m} {\mathrm{s}}^{-2}\text{,} m_{\mathrm{A}}=2 \mathrm{kg}\text{,} m_{\mathrm{B}}=4 \mathrm{kg}\text{.})$

A $3 \mathrm{kg}$ uniform rod rotates in a vertical plane about a smooth pivot at $B$. A spring of constant $k=300 {\mathrm{Nm}}^{-1}$ and of unstretched length $100 \mathrm{mm}$ is attached to the rod as shown. Knowing that in the position shown the rod has an angular velocity of $4 \mathrm{rad} {\mathrm{s}}^{-1}$ clockwise, determine the angular velocity of the rod after it has rotated through. $[g=10 \mathrm{m} {\mathrm{s}}^{-2} ]$

(a) $90º$        (b) $180º$