The equilateral triangle \(ABC\) is cut from a thin solid sheet of wood. (See figure) \(D, E\) and \(F\) are the mid points of its sides as shown and \(G\) is the centre of the triangle. The moment of inertia of the triangle about an axis passing through \(G\) and perpendicular to the plane of the triangle is \(I_0.\) If the smaller triangle \(DEF\) is removed from \(ABC,\) the moment of inertia of the remaining figure about the same axis is \(I.\) Then

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(A) \(I = \frac{{15}}{{16}}{I_0}\)

Detailed explanation

\(I\, \propto \,m{\ell ^2}\,\,\,\,\,\,\left( {let\,\sigma = mass\,present\,area} \right)\)

\(\therefore \,\,\,{I_1}\, \propto \,{\ell ^4}\,\,\,\,\,\,\,...\left( 1 \right)\)

\(and\,\,{I_2}\, \propto \,{\left( {\frac{\ell }{2}} \right)^4}\,\,\,\,\,\,\,\,...\left( 2 \right)\)

\(So,\,{I_2} = \frac{I}{{16}}\)

Moment of inertia of remaining sheet = \(I - \frac{1}{{16}}\)

\( = \frac{{151}}{{16}}\)

Same subject

From JEE Mains