JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2017, 2019
- Hard
- MCQ
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.
Answer & Solution
Correct Answer
(A) \(I = \frac{{15}}{{16}}{I_0}\)
Step-by-step Solution
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 = \(I - \frac{1}{{16}}\)
\( = \frac{{151}}{{16}}\)
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