How much force is required to produce an increase of $0.2\%$ in the length of a brass wire of diameter $0.6\, mm$ (Young’s modulus for brass = $0.9 \times {10^{11}}N/{m^2}$)

Overall changes in volume and radii of a uniform cylindrical steel wire are $0.2 \%$ and $0.002 \%$ respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is ($Y = 2.0 × 10^{11} Nm^{-2}$)

The temperature of a wire of length $1$ metre and area of cross-section $1\,c{m^2}$ is increased from $0°C$ to $100°C$. If the rod is not allowed to increase in length, the force required will be $(\alpha = {10^{ - 5}}/^\circ C$ and $Y = {10^{11}}\,N/{m^2})$

Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are $Y_{1}$ and $Y_{2}$. The combination behaves as a single wire then its Young's modulus is:

The area of cross section of a steel wire $(Y = 2.0 \times {10^{11}}N/{m^2})$ is $0.1\;c{m^2}$. The force required to double its length will be

A steel wire of diameter $0.5 mm$ and Young's modulus $2 \times 10^{11} N m ^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0 m$. A vernier scale with $10$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0 mm$, is attached. The $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2 kg$, the vernier scale division which coincides with a main scale division is. . . . Take $g =10 m s ^{-2}$ and $\pi=3.2$.