$A$ wire of length $2L$ is fixed between two walls. When a weight $W = mg$ is applied at its midpoint,it sags by a distance $x$ $(x << L)$. Then $m$ is equal to:

$A$ sphere of mass $2 \,kg$ and diameter $4.5 \,cm$ is attached to the lower end of a steel wire of $2 \,m$ length and area of cross-section $0.24 \times 10^{-6} \,m^2$. The wire is suspended from a $205 \,cm$ high ceiling of a room. When the system is made to oscillate as a simple pendulum, the sphere just grazes the floor at its lowest position. Find the velocity of the sphere at the lowest position. (Young's modulus of steel $= 2 \times 10^{11} \,Nm^{-2}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,ms^{-1}$)

$A$ uniform metallic wire is elongated by $0.04\, m$ when subjected to a linear force $F$. The elongation,if its length and diameter are doubled and subjected to the same force,will be ..... $cm$.

Two steel wires of same length $L$ but radii $r$ and $2r$ are connected together end to end and tied to a wall as shown. The force $F$ stretches the combination by $10 \ mm$. How far does the junction point $A$ move (in $mm$)?

What force is required to increase the length of a wire of diameter $4 \, mm$ and Young's modulus $9 \times 10^{10} \, N/m^2$ by $0.1\%$?

The Young's modulus of a rubber string $8\, cm$ long and density $1.5\, kg/m^3$ is $5 \times 10^8\, N/m^2$. If it is suspended from the ceiling in a room,the increase in length due to its own weight will be: