One solid sphere, having mass 1kg & diameter 0.3m is suspended from a wire. If the twisting couple per unit twist for the wire is $6\times {10}^{-3}\mathrm{N}-\mathrm{m}/\mathrm{radian},$ then the time period of small oscillations

 $T=2\pi \sqrt{\frac{I}{C}}\mathrm{.........}\left(1\right)$
$I=\frac{2}{5}M{R}^2\mathrm{.......}\left(2\right)$
By equation (1) & (2)
$T=2\pi \sqrt{\frac{2M{R}^2}{5C}}$
Given,   $M=1kg$
$\begin{array}{l}R=0.15m\\ C=6\times {10}^{-3}N-m/rad.\end{array}$
T=?
$\Rightarrow T=6.28\sqrt{\frac{0.4\times 1\times {\left(0.15\right)}^2}{6\times {10}^{-3}}}$
$T=6.28\sqrt{\frac{9\times {10}^{-3}}{6\times {10}^{-3}}}=6.28\times \sqrt{1.5}$
T=7.7 sec.