A `0.1 kg` mass is suspended from a wire of negligible mass. The length of the wire is `1 m` and its cross - sectional area is `4.9xx10^(-7) m^(2)`. If the mass is pulled a little in the vertically downward direction and released , it performs `SHM` with angular frequency `140 rad s^(-1)`. If the young's modulus of the material of the wire is ` pxx10^(9) Nm^(-2)`, find the value of `p`.

Options

A. 3
B. 4
C. 5
D. 7

Answer

B

Explanation

The mass attached to the end of the wire, executes a S.H.M. for which `omega=sqrt((g)/(l)) therefore l=(g)/(omega^(2))`
where l is the extension produced in the wire.
But `Y=(MgL)/(pi r^(2)(l))=(MgL)/(pi r^(2)g//omega^(2))=(ML omega^(2))/(A)`
`therefore Y=(0.1xx1xx140xx140)/(4.9xx10^(-7))=4xx10^(9)`
But `Y=nxx10^(9) therefore n=4`