A) \[\frac{1+\tan \alpha }{1-\tan \alpha }\]
B) \[\frac{1+\sin \alpha }{1-\cos \alpha }\]
C) \[\frac{1+\sin \alpha }{1-\sin \alpha }\]
D) \[\frac{1+\cos \alpha }{1-\cos \alpha }\]
Correct Answer: A
Solution :
Pressure at 0 level is same from both sides. \[{{d}_{1}}(1-sin\alpha )={{d}_{1}}(1-cos\alpha )+{{d}_{2}}[cos\alpha +sin\alpha ]\] \[\Rightarrow \]\[\frac{{{d}_{1}}}{{{d}_{2}}}=\frac{\sin \alpha +\cos \alpha }{\cos \alpha -\sin \alpha }\]You need to login to perform this action.
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