A) 6.54 cm
B) 3.64 cm
C) 1.77 cm
D) 0.88 cm
Correct Answer: C
Solution :
Here mass M = 50 g and radius R = 2.5 cm. Required moment of inertia of the disc is given by \[I=\frac{M{{R}^{2}}}{2}=M{{K}^{2}}\] So, \[{{K}^{2}}=\frac{{{R}^{2}}}{2}\] or \[K=\frac{R}{\sqrt{2}}=\frac{2.5}{\sqrt{2}}=\frac{2.5\sqrt{2}}{2}\] \[=1.767\,cm\] \[=1.77\,cm\]You need to login to perform this action.
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