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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2020
  • Hard
  • Numerical

\(ABC\) is a plane lamina of the shape of an equilateral triagnle. \(D,\) \(E\) are mid points of \(AB\), \(AC\) and \(G\) is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through \(G\) and perpendicular to the plane \(ABC\) is \(I _{0} .\) If part \(ADE\) is removed, the moment of inertia of the remaining part about the same axis is \(\frac{ NI _{0}}{16}\) where \(N\) is an integer. Value of \(N\) is

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Answer & Solution

Correct Answer

(B) \(11\)

Step-by-step Solution

Detailed explanation

Let side of triangle is a and mass is \(m\)

\(MOI\) of plate \(ABC\) about centroid

\(I_{0}=\frac{m}{3}\left(\left(\frac{a}{2 \sqrt{3}}\right)^{2} \times 3\right)=\frac{m a^{2}}{12}\)

triangle \(ADE\) is also an equilateral triangle of side \(a / 2\)

Let moment of inertia of triangular plate \(ADE\) about it's centroid \((G')\) is \(I _{1}\) and \(mass\) is \(m _{1}\)

\(m _{1}=\frac{ m }{\frac{\sqrt{3} a ^{2}}{4}} \times \frac{\sqrt{3}}{4}\left(\frac{ a }{2}\right)^{2}=\frac{ m }{4}\)

\(I _{1}=\frac{ m _{1}}{12}\left(\frac{ a }{2}\right)^{2}=\frac{ m }{4 \times 12} \frac{ a ^{2}}{4}=\frac{ ma ^{2}}{192}\)

distance \(G G^{\prime}=\frac{a}{\sqrt{3}}-\frac{a}{2 \sqrt{3}}=\frac{a}{2 \sqrt{3}}\)

so \(MOI\) of part \(ADE\) about centroid \(G\) is

\(I _{2}= I _{1}+ m _{1}\left(\frac{ a }{2 \sqrt{3}}\right)^{2}=\frac{ ma ^{2}}{192}+\frac{ m }{4} \cdot \frac{ a ^{2}}{12}\)

\(=\frac{5 ma ^{2}}{192}\)

now \(MOI\) of remaining part

\(=\frac{m a^{2}}{12}-\frac{5 m a^{2}}{192}=\frac{11 m a^{2}}{12 \times 16}=\frac{11 I_{0}}{16}\)

\(\Rightarrow \quad N=11\)

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