question_answer

    A velocity\[\frac{1}{4}\]m/s is resolved into two components along OA and OB making angles \[30{}^\circ \]and\[45{}^\circ ,\]respectively with the given velocity. Then, the component along OB is

A)  \[\frac{1}{4}m/s\]                          

B)  \[\frac{1}{4}(\sqrt{3}-1)m/s\]

C)  \[\frac{1}{8}m/s\]                          

D)  \[\frac{1}{8}(\sqrt{6}-\sqrt{2})m/s\]

Correct Answer: D

Solution :

                Using\[\frac{w}{\sin 30{}^\circ }=\frac{w}{\sin 105{}^\circ }\] \[\Rightarrow \]               \[w=\frac{v\sin 30{}^\circ }{\sin 105{}^\circ }\]                 \[=\frac{v.\frac{1}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\frac{\sqrt{2}v}{\sqrt{3}+1}\]                 \[=\frac{\sqrt{2}v(\sqrt{3}-1)}{3-1}=\frac{v}{\sqrt{2}}(\sqrt{3}-1)\]                 \[=\frac{1}{4\sqrt{2}}(\sqrt{3}-1)=\frac{1}{8}(\sqrt{6}-\sqrt{2})m/s\]