KVPY SB SX Exam 2011 Question Paper

Question : 97 of 120

Marks: +1, -0

At a distance l from a uniformly charged long wire, a charged particle is thrown radially outward with a velocity u in the direction perpendicular to the wire. When the particle reaches a distance 2l from the wire its speed is found to be

√2u

. The magnitude of the velocity, when it is a distance 4l away from the wire, is (ignore gravity)

$√3u$
$2u$
$2√2u$
$4u$

Validate

Solution:

energy conservation at

A \& B

{qV}_{ {A}}+\;{1}/{2} {mu}^{2}= {q} {V}_{ {B}}+\;{1}/{2} {m} × 2 {u}^{2}

{q}[ {V}_{ {A}}- {V}_{ {B}}]=\;{1}/{2} {mu}^{2}

{q} × \;{λ}/{2 π E_{0}} \ln 2=\;{1}/{2} {mu}^{2}

energy conservation at

A \& C

{qV}_{ {A}}+\;{1}/{2} {mu}^{2}= {qV}_{ {C}}+\;{1}/{2} {mv}^{2}

{q}[ {V}_{ {A}}- {V}_{ {C}}]+\;{1}/{2} {mu}^{2}=\;{1}/{2} {mv}^{2}

\;{ {q} λ}/{2 π E_{0}} \ln 4+\;{1}/{2} {mu}^{2}=\;{1}/{2} {mv}^{2}

\;{2 {q} λ}/{2 π E_{0}} \ln 2+\;{1}/{2} {mu}^{2}=\;{1}/{2} {mv}^{2}

{mu}^{2}+\;{1}/{2} {mu}^{2}=\;{1}/{2} {mv}^{2}

\;{3}/{2} {u}^{2}=\;{1}/{2} {v}^{2} ⇒ {v}=√{3} {u}

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