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yashikaaggarwal
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27 Jul 2020, 08:51
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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58% (02:59) correct
42%
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A hollow right circular cylinder of radius r and height 4r is standing vertically on a plane. If a solid right circular cone of radius 2r and height 6r is placed with its vertex down in the cylinder, then the volume of the portion of the cone outside the cylinder is:
A. 8/(3πr^3)
B. 2πr^3
C. 9/(8πr^3)
D. 7πr^3
E. 4/(3πr^3)
A. 8/(3πr^3)
B. 2πr^3
C. 9/(8πr^3)
D. 7πr^3
E. 4/(3πr^3)
27 Jul 2020, 11:55
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yashikaaggarwal
please see the image attached for solution.
trick here is: max radius of cone that can be put inside hallow cylinder is "r"
and corresponding height of cone inside the cylinder is "3r"
Attachments
IMG_20200728_002228.jpg [ 1.26 MiB | Viewed 4582 times ]
27 Jul 2020, 12:27
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The radius of the inverted cone is the same as that of the top of the cylinder = r. The cone looks as shown in the diagram.
By Theory of similarity, \(\frac{AF}{AG} \)= \(\frac{DE }{ BC}\)
\(\frac{x }{ 6r} = \frac{2r }{ 4r}\)
X = 3r
Volume of the cone with radius = 2r and h = 6r = \(\frac{1}{3} \pi * (2r)^2\) * 6r = 8\(\pi r^3\)
Volume of the cone with radius = r and h = 3r = \(\frac{1}{3} \pi * (r)^2\) * 3r = \(\pi r^3\)
Therefore the difference in volumes = 8\(\pi r^3\) -\(\pi r^3 \)= 7\(\pi r^3\)
Option D
Arun Kumar
By Theory of similarity, \(\frac{AF}{AG} \)= \(\frac{DE }{ BC}\)
\(\frac{x }{ 6r} = \frac{2r }{ 4r}\)
X = 3r
Volume of the cone with radius = 2r and h = 6r = \(\frac{1}{3} \pi * (2r)^2\) * 6r = 8\(\pi r^3\)
Volume of the cone with radius = r and h = 3r = \(\frac{1}{3} \pi * (r)^2\) * 3r = \(\pi r^3\)
Therefore the difference in volumes = 8\(\pi r^3\) -\(\pi r^3 \)= 7\(\pi r^3\)
Option D
Arun Kumar
Attachments
1.jpg [ 560.14 KiB | Viewed 4444 times ]
