A load of mass 100 gm increases the length of wire by 10 cm. If the system is kept in oscillation, its time period is `(g=10m//s^(2))`

Options

A. 0.314 s
B. 3.14 s
C. 0.628 s
D. 6.28 s

Answer

C

Explanation

`F= - Kx therefore |K| =|(F)/(x)|`
`therefore K=(100xx10^(-3))/(10xx10^(-2))=10 N//m`
and `T=2pi sqrt((m)/(K))= 2pi sqrt((100xx10^(-3))/(10))`
`=2pi sqrt(10^(-2))=2pi xx0.1=0.628 s`