A point like particle of mass ‘m’ (very small) is projected in the vertically upward direction with velocity $V_{0}$ where already exist, a uniform horizontal electric field $\vec{E}$ . The field strength is such that $qE = \frac{3}{4} mg$ , where q is the charge on the particle. After what time the radius of curvature of the charged particle will be minimum.

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$\frac{25 V_{0}}{16 g}$

$\frac{8 V_{0}}{25 g}$

$\frac{V_{0}}{g}$

$\frac{16 V_{0}}{25 g}$

answer is C.

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Detailed Solution

When radius of curvature is minimum $\vec{v} is ⊥ to \vec{a}$

Velocity as function of time, $\vec{v} = (\frac{q E}{m} t) \hat{i} + (V_{0} - g t) \hat{j}$

Acceleration as function of time, $\vec{a} = (\frac{q E}{m}) \hat{i} + (- g) \hat{j}$

$\vec{v} . \vec{a} = 0$

$\Rightarrow t = \frac{16 V_{0}}{25 g}$

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