Q149 / 225 · JEE Main 2004 · Mathematics

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

  • A
    $\frac{1}{8} \mathrm{~m} / \mathrm{s}$
  • B
    $\frac{1}{4}(\sqrt{3}-1) \mathrm{m} / \mathrm{s}$
  • C
    $\frac{1}{4} \mathrm{~m} / \mathrm{s}$
  • Correct answer
    D
    $\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$

Solution

Component along $\mathrm{OB}=\frac{\frac{1}{4} \sin 30^{\circ}}{\sin \left(45^{\circ}+30^{\circ}\right)}=\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}$.

Hence, the correct answer is option: 'D'.

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