Get Answers to all your Questions

header-bg qa

ABC is a plane lamina of the shape of an equilateral triangle. D, E are midpoints of AB, AC and G is the centroid of the lamina. Moment of inertia of lamina about an axis passing through G and perpendicular to the plane ABC I_{0}. If the 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 _____________
Option: 1 11
Option: 2 9
Option: 3 8
Option: 4 7

Answers (1)


Let the side of the triangle is a and mass is m

\begin{aligned} &\text { MOI of plate ABC about centroid }\\ &\mathrm{I}_{0}=\frac{\mathrm{m}}{3}\left(\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)^{2} \times 3\right)=\frac{\mathrm{ma}^{2}}{12} \end{aligned}

Triangle ADE is also an equilateral triangle of side  \frac{a}{2}.

Let the moment of inertia of triangular plate ADE about its centroid (G') is I_1 and mass is  m_1

\begin{array}{l} \mathrm{m}_{1}=\frac{\mathrm{m}}{\frac{\sqrt{3} \mathrm{a}^{2}}{4}} \times \frac{\sqrt{3}}{4}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{m}}{4} \\ \\ \mathrm{I}_{1}=\frac{\mathrm{m}_{1}}{12}\left(\frac{\mathrm{a}}{2}\right)^{2}=\frac{\mathrm{m}}{4 \times 12} \frac{\mathrm{a}^{2}}{4}=\frac{\mathrm{ma}^{2}}{192} \end{array}

\text {from figure distance } \mathrm{GG}^{\prime}=\frac{\mathrm{a}}{\sqrt{3}}-\frac{\mathrm{a}}{2 \sqrt{3}}=\frac{\mathrm{a}}{2 \sqrt{3}}

\begin{aligned} &\text { so MOI of part ADE about centroid G is }\\ &\mathrm{I}_{2}=\mathrm{I}_{1}+\mathrm{m}_{1}\left(\frac{\mathrm{a}}{2 \sqrt{3}}\right)^{2}=\frac{\mathrm{ma}^{2}}{192}+\frac{\mathrm{m}}{4} \cdot \frac{\mathrm{a}^{2}}{12}=\frac{5 \mathrm{ma}^{2}}{192} \end{aligned}

now MOI of remaining part

\begin{aligned} &I_{re}=\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 & N=11 \end{aligned}

Posted by


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE