A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8 \mathrm{~N} / \mathrm{m}^2$ and its radius is 4 mm , then maximum acceleration the lift can take is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}^2$.

(take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ and $\pi=3.14$ )

The two wires $A$ and $B$ of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire $A$ and wire $B$ is 20/11. When the joined wire is kept under certain tension the elongations in the wires $A$ and $B$ are equal. If the length of wire $A$ is 2.2 m , then the length of wire $B$ is

$\_\_\_\_$ m.

Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is $\_\_\_\_$ - ( $S$ is the surface tension of mercury).

The Young's modulus of steel wire of radius $r$ and length $L$ is $Y$.

If the radius $r$ and length $L$ of the wire are doubled then the value of $Y$