question_answer

A closed vessel the inside of which is a circular cone of height h contains some water in it. When the cone is vertical with its vertex downwards, the water stands to a height \[\frac{h}{2}\]. To what height will it stand when the vessel is inverted?

A)  \[\frac{h}{2}\]                                 

B)  \[\frac{{{h}^{\frac{1}{3}}}}{2}\]

C)  \[{{7}^{\frac{1}{3}}}\frac{h}{2}\]             

D)         \[h\left( 1-\frac{{{7}^{\frac{1}{3}}}}{2} \right)\]  

Correct Answer: D

Solution :

  \[\frac{AD}{AB}=\frac{DE}{BC}\] \[\therefore \]    \[\frac{\frac{h}{2}}{h}=\frac{DE}{r}\] or            \[DE=\frac{r}{2}\] \[\therefore \]Volume of water \[=\frac{1}{3}\pi {{\left( \frac{r}{2} \right)}^{2}}\frac{h}{2}=\frac{\pi {{r}^{2}}h}{24}\] \[\therefore \] Volume of the remaining part of the cone                                 \[=\frac{1}{3}\pi {{r}^{2}}h-\frac{1}{24}\pi {{r}^{2}}h=\frac{7\pi \,{{r}^{2}}h}{24}\] When the vessel is inverted and the height of water be k                                                        then        \[AF=h-k\] \[\therefore \]  \[\frac{AF}{AB}=\frac{FG}{BC}\] or            \[\frac{h-k}{h}=\frac{{{r}_{1}}}{r}\] or            \[{{r}_{1}}=\frac{(h-h)r}{h}\] \[\therefore \] Vol. of the remaining part \[=\frac{1}{3}\,\pi r_{1}^{2}\,(h-k)\] or            \[\frac{7\pi {{r}^{2}}h}{24}=\frac{1}{3}\pi \frac{{{(h-k)}^{2}}{{r}^{2}}}{{{h}^{2}}}(h-k)\] or            \[{{(h-k)}^{{}}}=\frac{7}{8}{{h}^{3}}\] or            \[h-k=\frac{{{7}^{\frac{1}{3}}}h}{2}\] or            \[k=h-\frac{{{7}^{\frac{1}{3}}}h}{2}=h\left( 1-\frac{{{7}^{\frac{1}{3}}}}{2} \right)\]