A) 2
B) 1
C) \[\sqrt{2}\]
D) \[\sqrt{3}\]
E) \[\sqrt{3}+\sqrt{2}\]
Correct Answer: C
Solution :
In\[\Delta OBC,\] \[\sin \theta =\frac{BC}{OC}\] \[\Rightarrow \] \[r=l\sin \theta \] ?. (i) and \[\cos \theta =\frac{OB}{OC}\] \[\Rightarrow \] \[h=l\cos \theta \] ...(ii) Now, \[V=\frac{1}{3}\pi {{r}^{2}}h\] \[=\frac{1}{3}\pi {{l}^{2}}{{\sin }^{2}}\theta .l\cos \theta \] \[\Rightarrow \] \[V=\frac{\pi {{l}^{3}}}{3}{{\sin }^{2}}\theta \cos \theta \] \[\therefore \]\[\frac{dV}{d\theta }=\frac{\pi {{l}^{3}}}{3}[2\sin \theta co{{s}^{2}}\theta +{{\sin }^{2}}\theta -(\sin \theta )]\] \[=\frac{\pi {{l}^{3}}}{3}\sin \theta (2co{{s}^{2}}\theta -{{\sin }^{2}}\theta )\] and\[\frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{\pi {{l}^{3}}}{3}\cos \theta (2co{{s}^{2}}\theta -{{\sin }^{2}}\theta )\] \[+\frac{\pi {{l}^{3}}}{3}\sin \theta (2co{{s}^{2}}\theta -{{\sin }^{2}}\theta )=0\] On putting\[\frac{dV}{d\theta }=0,\]for maxima or minima \[\therefore \]\[\frac{\pi {{l}^{3}}}{3}\sin \theta (2{{\cos }^{2}}\theta -{{\sin }^{2}}\theta )=0\] \[\Rightarrow \] \[\tan \theta =\sqrt{2}\] At\[\theta ={{\tan }^{-1}}\sqrt{2}\] \[\frac{{{d}^{2}}V}{d{{\theta }^{2}}}=-ve<0\] \[\therefore \]Volume is maximum at\[\tan \theta =\sqrt{2}\].You need to login to perform this action.
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