A steel ball of `50mm` radius is kept on a hole of concrete. The diameter of the hole is `0.05mm` less then the diameter of the steel ball at `20^(@)C`. Coefficient of volume expansion of the steel ball is `3.2x 10^(-6)//^(@)C` . At what temperature the ball will be inside the hole ?

Options

A. `70^(@)C`
B. `30^(@)C`
C. `55^(@)C`
D. `62^(@)C`

Answer

A

Explanation

`D_(S)-D_(H)=0.05` mm
Where `D_(S)` is diameter of ball and `D_(H)` is diameter of concrete hole.
As we increase temperature
`D_(H)gtD_(S)` for bal to enter hole
`D_(H)(1+12xx10^(-6)DeltaT-D_(S)(1+(3.2)/(3)xx10^(-6)DeltaT)gt0`
`(12xx10^(-4)-(3.2)/(3)xx10^(-4))DeltaTgt0.05`