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A uniform non conducting rod of mass m and length l , with density \lambda is hinged at the mid point at origin so that it can rotate in horizontal plane. \overset {\rightarrow }E is parallel to x-axis in the entire region. Calculate time period of oscillation.

  • Option 1)

    2\pi \sqrt{\frac{m}{3\lambda}}

  • Option 2)

    2\pi \sqrt{\frac{m}{3\lambda E }}

  • Option 3)

    2\pi \sqrt{\frac{m}{\lambda}}

  • Option 4)

    2\pi \sqrt{\frac{m\lambda}{3E}}

 

Answers (1)

best_answer

As we learned

Line Charge -

Electric field and Potential due to a charged straight wire length and charge density \lambda.

- wherein

 

 

 

\tau = \int_{o}^{\frac{l}{2}} d\tau _{1} + \int_{o}^{\frac{l}{2}} d\tau _{2} = 2\int_{o}^{\frac{l}{2}} E \lambda dx (sin \Theta x)= \frac{E\lambda }{4}l^{2}sin \Theta

\frac{Ml^{2}}{12}\alpha = - \frac{E\lambda }{4}l^{2} \Theta \Rightarrow T = 2\pi \sqrt{\frac{m}{3E\lambda }}

 

 


Option 1)

2\pi \sqrt{\frac{m}{3\lambda}}

Option 2)

2\pi \sqrt{\frac{m}{3\lambda E }}

Option 3)

2\pi \sqrt{\frac{m}{\lambda}}

Option 4)

2\pi \sqrt{\frac{m\lambda}{3E}}

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Avinash

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