A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a string of length L. Time period of pendulum is T_0. When parallel plates are charged, the time period changes to . The ratio \frac{T}{}T_0 is equal to

  • Option 1)

    \left ( \frac{g + \frac{qE}{m}}{g} \right )^{\frac{1}{2}}

  • Option 2)

    \left ( \frac{g} {g + \frac{qE}{m}}\right )^{\frac{3}{2}}

  • Option 3)

    \left ( \frac{g} {g + \frac{qE}{m}}\right )^{\frac{1}{2}}

  • Option 4)

    None of these

 

Answers (1)
V Vakul

As we have learnt

 

when electric field applied downward direction -

{T}'=2\pi \sqrt{\frac{l}{g+\frac{qE}{m}}}

{T}'< T

- wherein

 Net downward force mg' =mg + QE
\RightarrowEffect acceleration \Rightarrow g' = \left (g + \frac{QE}{m} \right )
Hence time period  T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{l}{g + \frac{QE}{m}}}

 


Option 1)

\left ( \frac{g + \frac{qE}{m}}{g} \right )^{\frac{1}{2}}

Option 2)

\left ( \frac{g} {g + \frac{qE}{m}}\right )^{\frac{3}{2}}

Option 3)

\left ( \frac{g} {g + \frac{qE}{m}}\right )^{\frac{1}{2}}

Option 4)

None of these

Exams
Articles
Questions