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A particle A has a charge +q and particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electrical potential difference, the ratio of their speeds \frac{V_A}{V_B} will becomes

  • Option 1)

    2:1

  • Option 2)

    1:2

  • Option 3)

    1:4

  • Option 4)

    4:1

 

Answers (1)

As we have learnt,

 

when Charged Particle at rest in uniform field -

Velocity - 

\dpi{100} v=\frac{QEt}{m}=\sqrt{\frac{2Q\Delta V}{m}}

 

 

- wherein

\Delta V= Potential difference.

 

 We know that kinetic energy K = \frac{1}{2}mv^2 = QV. Since m and V are same so, v^2 \propto Q\Rightarrow \frac{V_A}{V_B} = \sqrt{\frac{Q_A}{Q_B}} = \sqrt{\frac{q}{4q}} = \frac{1}{}2

 


Option 1)

2:1

Option 2)

1:2

Option 3)

1:4

Option 4)

4:1

Posted by

Vakul

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