A hollow cylindrical column of steel supports a load of 20,000 kg. The inner and the outer radii of the column are 50 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of the column. Given, Young's modulus of steel `=2.0 xx 10^(11) Nm^(2)` and `g=10ms^(-2)`.

Options

A. `7 xx 10^(-16)`
B. `2.895 xx 10^(-10)`
C. `5.8 xx 10^(-12)`
D. `2.895 xx 10^(-6)`

Answer

D

Explanation

Here, `F=mg = 20,000 xx 10`
`= 2 xx 10^(5) N`.
Inner radius of the column,
`r_(1) = 50 cm = 0.50m`
Outer radius of the column,
`r_(2)=60 cm = 0.60 m`.
Area of cross - section of the column
`A=pi(r_(2)^(2)-r_(1)^(2)) =pi [(0.6)^(2)-(0.5)^(2)]`
`=0.11 pi m^(2)`
Now `Y=(F)/(A) xx (l)/(Delta l)`
or, `(Delta l)/(l) = F/(AY) = (2 xx 10^(5))/((0.11 pi) xx 2 xx 10^(11))`
`=2.895 xx 10^(-6)`.