Four identical hollow cylindrical columns of mild steel support a big structure of mass $50,000 \;kg$. The inner and outer radii of each column are $30$ and $60\; cm$ respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Total force exerted, $F=M g=50000 \times 9.8 N$
Stress $=$ Force exerted on a single column $=\frac{50000 \times 9.8}{4}=122500 N$
Young's modulus, $Y=\frac{\text { Stress }}{\text { strain }}$
Strain $=\frac{\frac{F}{A}}{Y}$
Where,
Area, $A=\pi\left(R^{2}-r^{2}\right)=\pi\left((0.6)^{2}-(0.3)^{2}\right)$
Strain $=\frac{122500}{\pi\left[(0.6)^{2}-(0.3)^{2}\right] \times 2 \times 10^{11}}=7.22 \times 10^{-7}$
Hence, the compressional strain of each column is $7.22 \times 10^{-7}$
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 $\pi=3.2$.
The force required to stretch a wire of crosssection $1 cm ^{2}$ to double its length will be ........ $ \times 10^{7}\,N$
(Given Yong's modulus of the wire $=2 \times 10^{11}\,N / m ^{2}$ )
Two wires $A$ and $B$ are of same materials. Their lengths are in the ratio $1 : 2$ and diameters are in the ratio $2 : 1$ when stretched by force ${F_A}$ and ${F_B}$ respectively they get equal increase in their lengths. Then the ratio ${F_A}/{F_B}$ should be
Explain experimental determination of Young’s modulus.
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