A) 36.1 cm
B) 25.2 cm
C) 17.7 cm
D) zero
Correct Answer: A
Solution :
When rod is bent at its mid point to make angle \[{{90}^{o}}\] \[{{x}_{CM}}=\frac{m\times \frac{1}{4}+m\times \frac{1}{4}}{m\times m}=\frac{1}{8}m=12.5\,\,cm\] \[{{y}_{CM}}=\frac{m\times \frac{1}{4}+m\times \frac{1}{4}}{m+m}=\frac{1}{8}m=12.5\,\,cm\] The distance of the cantre of mass from the centre of the rod \[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\] \[=\sqrt{{{(12.5)}^{2}}+{{(12.5)}^{2}}}\] \[=\sqrt{312.5}-17.7\,cm\]
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