question_answer

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

B)        \[\frac{1}{4}(\sqrt{3}-1)m/s\]    

C) \[\frac{1}{4}m/s\]           

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

Correct Answer: D

Solution :

\[v=\frac{1}{4}m/{{s}^{2}}\] Component of v along OB \[=\frac{\frac{1}{4}\sin {{30}^{o}}}{\sin ({{30}^{o}}+{{45}^{o}})}=\frac{\frac{1}{4}\sin {{30}^{o}}}{\sin {{75}^{o}}}\] \[=\frac{\frac{1}{4}.\frac{1}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\frac{\sqrt{2}}{4(\sqrt{3}+1)}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\] \[=\frac{\sqrt{2}(\sqrt{3}-1)}{4(3-1)}\] \[=\frac{1}{8}(\sqrt{6}-\sqrt{2})m/s\]