The length of a metallic wire is $l$. The tension in the wire is $T_1$ for length $l_1$ and the tension in the wire is $T_2$ for length $l_2$. Find the original length $l$.

Two wires $A$ and $B$ made of the same material and areas of cross-section in the ratio $1: 2$ are stretched by the same force. If the masses of the wires $A$ and $B$ are in the ratio $2: 3$, then the ratio of the elongations of the wires $A$ and $B$ is

$A$ wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $F$,its length increases by $5\,cm$. Another wire of the same material of length $4L$ and radius $4r$ is pulled by a force $4F$ under the same conditions. The increase in length of this wire is $....cm$.

If the given graph shows the load $(W)$ attached to and the elongation $(\Delta l)$ produced in a wire of length $1 \,m$ and area of cross-section $1 \,mm^2$, then the Young's modulus of the material of the wire is

Two wires each of radius $0.2\,cm$ and negligible mass, one made of steel and the other made of brass, are loaded as shown in the figure. The elongation of the steel wire is $.........\times 10^{-6}\,m$. [Young's modulus for steel $= 2 \times 10^{11}\,N/m^2$ and $g = 10\,m/s^2$]

The unit of Young's modulus is