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 \[k=h-\frac{{{7}^{\frac{1}{3}}}h}{2}=h\left( 1-\frac{{{7}^{\frac{1}{3}}}}{2} \right)\]You need to login to perform this action.
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