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An electron of mass m(e ) initially at r...

An electron of mass `m_(e )` initially at rest moves through a certain distance in a uniform electric field in time `t_(1)`. A proton of mass `m_(p)` also initially at rest takes time `t_(2)` to move through an equal distance in this uniform electric field.Neglecting the effect of gravity, the ratio of `t_(2)//t_(1)` is nearly equal to

A

1

B

`((m_(c))/(m_(p)))^(1//2)`

C

`((m_(p))/(m_(e)))^(1//2)`

D

`((m_(p))/(m_(e)))`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • An electron of mass M_e , initially at rest, moves through a certain distance in a uniform electric field in time t_1 . A proton of mass M_p also initially at rest, takes time t_2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t_2//t_1 is nearly equal to

    A
    `1`
    B
    `sqrt((M_p)/(M_e))`
    C
    `sqrt(M_e/M_p)`
    D
    `1836`

  • An electron of mass m_e starts from rest, moves a certain distance in uniform field and takes time t_1 . A proton of mass m_p also starts from rest in the same field and covers the same distande in time t_2 ,then t_1//t_2 is equal to

    A
    `sqrt(m_e/m_p)`
    B
    `sqrt(m_p/m_e)`
    C
    1
    D
    1836

  • An electron initially at rest falls a distance of 2 cm in a uniform electric field of magnitude 3 xx 10^(4) N C^(-1) . The time taken by the electron to fall to this distance is

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    `1.3 xx 10^(2) s`
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    D
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