The vertical extension in a light spring by a weight of $1\, kg$ suspended from the wire is $9.8\, cm$. The period of oscillation
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(c) $mg = kx$
==> $\frac{m}{k} = \frac{x}{g}$
==> $T = 2\pi \sqrt {\frac{m}{k}} = 2\pi \sqrt {\frac{x}{g}} $
$ = 2\pi \sqrt {\frac{{9.8 \times {{10}^{ - 2}}}}{{9.8}}} = \frac{{2\pi }}{{10}}\,sec$
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