Attachment:
image 1.jpg [ 32.27 KiB | Viewed 1345 times ]
Consider the image shown above.
Let the height of the cone be "
h" and "
R" be the radius of the cone. When the bottle is rested on its flat base, the volume of water is given by
Volume of the cone - volume of empty space above the water level
\(\frac{\pi}{3} R^2 h\) - \(\frac{\pi}{3} r_1^2 (8)\)
Now by similar triangle concept:
\(\frac{r_1}{8} = \frac{R}{h}\)
\(r_1 = \frac{8R}{h}\)
Sub \(r_1\) value in the above equation, we get
volume of water = \(\frac{\pi}{3} R^2 h - \frac{\pi}{3} (\frac{8R}{h})^2 (8)\)
\(\frac{\pi}{3} R^2 (h - \frac{512}{h^2})\) ......................................... (1)
Now consider the cone is tilted upside down. Volume of water is
\(\frac{\pi}{3} (r_2)^2 (h-2)\)
Again by similar triangle concept:
\(\frac{r_2}{h-2} = \frac{R}{h}\)
\(r_2 = \frac{(h-2) R}{h}\)
Substituting \(r_2\) in the above equation of volume of water when cone is tilted,
volume of water = \(\frac{\pi}{3} (\frac{(h-2) R}{h})^2 (h-2)\)
\(\frac{\pi}{3} R^2 \frac{(h-2)^3}{h^2}\) ..................................... (2)
Since volume of water in both the cases is equal, (1) = (2)
\(\frac{\pi}{3} R^2 (h - \frac{512}{h^2})\) = \(\frac{\pi}{3} R^2 \frac{(h-2)^3}{h^2}\)
\(h - \frac{512}{h^2} = \frac{(h-2)^3}{h^2}\)
\(h^3 - 512 = (h-2)^3\)
\(h^3 - 512 = h^3 - 6h^2 + 12h - 8\)
\(6h^2 - 12h - 504 = 0\)
\(h^2 - 2h - 84 = 0\)
Since \(h > 0 , h \approx {10 cm}\)
OPTION :
C
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Regards,
Chaitanya
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