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A velocity $\frac{1}{4} \mathrm{~m} / \mathrm{s}$ is resolved into two components along $\mathrm{OA}$ and $\mathrm{OB}$ making angles $30^{\circ}$ and $45^{\circ}$ respectively with the given velocity. Then the component along $\mathrm{OB}$ is
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$\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$
$\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$
Component along $\mathrm{OB}=\frac{\frac{1}{4} \sin 30^{\circ}}{\sin \left(45^{\circ}+30^{\circ}\right)}=\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$.
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