question_answer

A uniform rod of mass m and length \[l\] rotates in a horizontal plane with an angular velocity co about a vertical axis passing through one end. The tension in the rod at a distance \[x\]from the axis is

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].