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(D) For two particles to collide,their positions at time $t$ must be equal: $r_1 + v_1 t = r_2 + v_2 t$.Rearranging the equation: $r_1 - r_2 = (v_2 -

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$8$

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(D) For two particles to collide,their positions at time $t$ must be equal: $r_1 + v_1 t = r_2 + v_2 t$.Rearranging the equation: $r_1 - r_2 = (v_2 - v_1) t$.Given $r_1 = (3 \hat{i} + 5 \hat{j})$ and $r_2 = (-5 \hat{i} - 3 \hat{j})$,the relative position vector is $r_1 - r_2 = (3 - (-5)) \hat{i} + (5 - (-3)) \hat{j} = (8 \hat{i} + 8 \hat{j}) 	ext{ m}$.The relative velocity is $v_2 - v_1 = (a \hat{i} + 7 \hat{j}) - (4 \hat{i} + 3 \hat{j}) = (a - 4) \hat{i} + 4 \hat{j}$.Substituting these into the collision condition with $t = 2 	ext{ s}$:$8 \hat{i} + 8 \hat{j} = ((a - 4) \hat{i} + 4 \hat{j}) 	imes 2$.Dividing by $2$: $4 \hat{i} + 4 \hat{j} = (a - 4) \hat{i} + 4 \hat{j}$.Comparing the coefficients of $\hat{i}$: $4 = a - 4$,which gives $a = 8$.

Column-$I$	Column-$II$
$(1)$ Velocity of $A$ relative to $B$ if $A$ and $B$ are moving perpendicular to each other	$(a)$ $\vec{v}_{rm} = \vec{v}_r + \vec{v}_m$
$(2)$ Velocity of rain drops relative to a man	$(b)$ $\vec{v}_{AB} = \vec{v}_A + \vec{v}_B$
	$(c)$ $v_{AB} = \sqrt{v_A^2 + v_B^2}$

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