A) Disc of radius\[a\]
B) Ring of radius\[a\]
C) Square lamina of side\[2a\]
D) Pour rods of length \[2a\] making a square
Correct Answer: D
Solution :
For square of side, \[I=\frac{1}{2}m{{a}^{2}}\] For ring, \[I=m{{a}^{2}}\] For square of side \[2a\] \[=\frac{M}{12}{{[2a]}^{2}}+{{(2a)}^{2}}]=\frac{2}{3}M{{a}^{2}}\] For square of rod of length \[2a\] \[I=4\left[ M\frac{{{(2a)}^{2}}}{12}+M{{a}^{2}} \right]=\frac{16}{3}M{{a}^{2}}\] Hence, moment of inertia is maximum for square of four rods.
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