Question

The MI of solid sphere about the tangent is 70 $kg{m}^2$. Its MI about any diameter is -

• A. 25 $kg{m}^2$
• B. 20 $kg{m}^2$
• C. 50 $kg{m}^2$
• D. 15 $kg{m}^2$

Answer 1

The correct option is B 20 $kg{m}^2$

$\begin{array}{c}\text{Given, moment of inertia of solid sphere about a tangent}\\ =70kg-m^2\end{array}$

$\text{By parallel axis theorem,}$

$\begin{array}{cc}{I}_{\text{tangent}}& =I_{\text{diameter}}+m{r}^2\\ \Rightarrow 70kg-m^2=& \frac{2}{5}m{r}^2+m{r}^2=\frac{7}{5}m{r}^2\\ \therefore m{r}^2=50kg\cdot m^2& \end{array}$

$\text{Hence,}{I}_{diameter}=\frac{2}{5}{mr}^2=(\frac{2}{5}\times 50)kg{m}^2=20kg-m^2$