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(B) Let the position of $Q$ be at the origin $(0, 0)$ and $P$ be at $(0, 10)$.The velocity of $P$ is $\vec{v}_P = -10 \hat{i} \ kmph$.The velocity of

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$0.5$

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(B) Let the position of $Q$ be at the origin $(0, 0)$ and $P$ be at $(0, 10)$.The velocity of $P$ is $\vec{v}_P = -10 \hat{i} \ kmph$.The velocity of $Q$ is $\vec{v}_Q = 10 \hat{j} \ kmph$.The relative velocity of $P$ with respect to $Q$ is $\vec{v}_{PQ} = \vec{v}_P - \vec{v}_Q = -10 \hat{i} - 10 \hat{j} \ kmph$.The magnitude of relative velocity is $|\vec{v}_{PQ}| = \sqrt{(-10)^2 + (-10)^2} = 10\sqrt{2} \ kmph$.The angle made by the relative velocity vector with the $y$-axis is $45^\circ$.The minimum separation occurs when the position vector of $P$ relative to $Q$ is perpendicular to the relative velocity vector $\vec{v}_{PQ}$.The time taken is $t = \frac{d \cos 45^\circ}{|\vec{v}_{PQ}|} = \frac{10 	imes \frac{1}{\sqrt{2}}}{10\sqrt{2}} = \frac{1}{2} = 0.5 \ hr$.

There are two particles $P$ and $Q$ separated by a distance of $10 \ km$ and moving with velocities of $10 \ kmph$ as shown in the figure. The time elapsed before they reach their minimum separation is.........$hr$.

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