JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 15)
A particle of mass ' $m$ ' and charge ' $q$ ' is fastened to one end ' $A$ ' of a massless string having equilibrium length $l$, whose other end is fixed at point ' $O$ '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the $x$-axis is
_28th_January_Morning_Shift_en_15_1.png)
Explanation
_28th_January_Morning_Shift_en_15_2.png)
In $$\Delta OMA,{{OM} \over {OA}} = \cos 60^\circ = {1 \over 2}$$
$$ \Rightarrow OM = {l \over 2}$$ (as $$OA = l$$)
$$MN = ON - OM = l - {l \over 2} = {l \over 2}$$
Using work $-$ energy theorem,
$${W_{all}} = \Delta k$$
$$ \Rightarrow {W_c} = {k_f} - {k_i}$$
$$ \Rightarrow qE{l \over 2} = {1 \over 2}m{v^2} - o$$ (As $${v_i} = o$$ and $$w = FS$$)
$$ \Rightarrow {v^2} = {{qEl} \over m}$$
$$ \Rightarrow v = \sqrt {{{qEl} \over m}} $$
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