JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 22)
In free space, a particle A of charge 1$$\mu $$C is held fixed at a point P. Another particle B of the same charge and
mass 4$$\mu $$g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P
is :
$$\left[ {Take\,{1 \over {4\pi { \in _0}}} = 9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right]$$
1.0 m/s
6.32 $$ \times $$ 104 m/s
2.0 $$ \times $$ 103 m/s
1.5 $$ \times $$ 102 m/s
Explanation
qA = 1 $$\mu $$c ; qB = 1 $$\mu $$c, mB = 4 Ć 10ā9 kg, rAB = 10ā3 m
$${1 \over 2}{M_B}{V^2} = k{q_A}{q_B}\left\{ {{1 \over {{{10}^{ - 13}}}} - {1 \over {9 \times {{10}^{ - 3}}}}} \right\}$$
$${1 \over 2}4 \times {10^{ - 9}}{V^2} = 9 \times {10^9} \times {10^{ - 6}} \times {8 \over 9} \times {10^3}$$
$${V^2} = {8 \over 2} \times {10^9} = 4 \times {10^9}$$
V = $$\sqrt {40} \times {10^4}\,m/s$$ = 6.32 $$ \times $$ 104 m/s
$${1 \over 2}{M_B}{V^2} = k{q_A}{q_B}\left\{ {{1 \over {{{10}^{ - 13}}}} - {1 \over {9 \times {{10}^{ - 3}}}}} \right\}$$
$${1 \over 2}4 \times {10^{ - 9}}{V^2} = 9 \times {10^9} \times {10^{ - 6}} \times {8 \over 9} \times {10^3}$$
$${V^2} = {8 \over 2} \times {10^9} = 4 \times {10^9}$$
V = $$\sqrt {40} \times {10^4}\,m/s$$ = 6.32 $$ \times $$ 104 m/s
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