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(C) For an uncharged particle in projectile motion,the range is $L = \frac{u^2 \sin 2	heta}{g} \quad \dots(i)$For a charged particle,the effective ac

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$L/3$

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(C) For an uncharged particle in projectile motion,the range is $L = \frac{u^2 \sin 2	heta}{g} \quad \dots(i)$For a charged particle,the effective acceleration becomes $g' = g + \frac{qE}{m}$.Given that the range for a particle with charge $q$ is $L/2$,we have:$\frac{L}{2} = \frac{u^2 \sin 2	heta}{g + \frac{qE}{m}}$Substituting $u^2 \sin 2	heta = Lg$ from equation $(i)$:$\frac{L}{2} = \frac{Lg}{g + \frac{qE}{m}}$$\Rightarrow g + \frac{qE}{m} = 2g \Rightarrow \frac{qE}{m} = g \quad \dots(ii)$Now,for a particle of mass $m$ and charge $2q$,the effective acceleration is $g'' = g + \frac{2qE}{m}$.Using equation $(ii)$,$g'' = g + 2g = 3g$.The new range $R$ is:$R = \frac{u^2 \sin 2	heta}{3g} = \frac{1}{3} \left( \frac{u^2 \sin 2	heta}{g} ight) = \frac{L}{3}$.

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