A) \[\frac{1}{2}m{{\omega }^{2}}x\]
B) \[\frac{1}{2}m{{\omega }^{2}}\frac{{{x}^{2}}}{\ell }\]
C) \[\frac{1}{2}m{{\omega }^{2}}\ell \left( l-\frac{x}{l} \right)\]
D) \[\frac{1}{2}\frac{m{{\omega }^{2}}}{l}({{l}^{2}}-{{x}^{2}})\]
Correct Answer: D
Solution :
Mass of element of rod of length \[dr,\] \[dm=\left( \frac{m}{l} \right)dr\] \[dF=dm\times a=\left( \frac{m}{l} \right)dr\times {{\omega }^{2}}r\] \[\therefore \]Tension at P \[F=\int_{x}^{l}{\left( \frac{m}{l} \right)dr\times }\,{{\omega }^{2}}r\] \[=\frac{m}{l}{{\omega }^{2}}\left[ \frac{{{r}^{2}}}{2} \right]_{x}^{l}=\frac{1}{2}\frac{{{\omega }^{2}}}{l}({{l}^{2}}-{{x}^{2}})\] Hence, the correction option is [d].You need to login to perform this action.
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