A) \[\tan \theta \,=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\]
B) \[\tan \theta \,=\frac{1-\sqrt{2}}{\sqrt{2}(1+\sqrt{3})}\]
C) \[\tan \theta \,\frac{1-\sqrt{3}}{1+\sqrt{2}}\]
D) \[\tan \theta \,=\frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}\]
Correct Answer: D
Solution :
\[2mv'\,\sin \theta \,=\frac{mv}{\sqrt{2}}+\frac{mv\sqrt{3}}{2}\] \[3\,mv'\,\cos \theta \,=\frac{mv}{2}\,-\frac{mv}{\sqrt{2}}\] \[\sin \theta \,\frac{\frac{1}{\sqrt{2}}\,+\frac{\sqrt{3}}{2}}{\frac{1}{2}-\frac{1}{\sqrt{2}}}\] \[=\frac{\sqrt{2}+\sqrt{3}}{1-\sqrt{2}}\]
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