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Answered: In a cylinderical vessel of diameter 24 cm filled
up with sufficient quantity of water, a solid
spherical ball, of radius 6 cm is completely immersed. Then the increases in height of water.
geometry euclidean-geometry
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closed as off-topic by fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin Jan 25 at 23:19
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Answered: In a cylinderical vessel of diameter 24 cm filled
up with sufficient quantity of water, a solid
spherical ball, of radius 6 cm is completely immersed. Then the increases in height of water.
geometry euclidean-geometry
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closed as off-topic by fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin Jan 25 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
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– Aloizio Macedo♦
Jan 26 at 15:53
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Answered: In a cylinderical vessel of diameter 24 cm filled
up with sufficient quantity of water, a solid
spherical ball, of radius 6 cm is completely immersed. Then the increases in height of water.
geometry euclidean-geometry
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Answered: In a cylinderical vessel of diameter 24 cm filled
up with sufficient quantity of water, a solid
spherical ball, of radius 6 cm is completely immersed. Then the increases in height of water.
geometry euclidean-geometry
geometry euclidean-geometry
edited Jan 27 at 5:35
Sagar Gupta
asked Jan 25 at 11:08
Sagar GuptaSagar Gupta
16
16
closed as off-topic by fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin Jan 25 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin Jan 25 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – fermesomme, Namaste, Eric Wofsey, Robert Z, Lee David Chung Lin
If this question can be reworded to fit the rules in the help center, please edit the question.
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– Aloizio Macedo♦
Jan 26 at 15:53
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Comments are not for extended discussion; this conversation has been moved to chat.
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– Aloizio Macedo♦
Jan 26 at 15:53
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– Aloizio Macedo♦
Jan 26 at 15:53
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– Aloizio Macedo♦
Jan 26 at 15:53
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1 Answer
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First, you need to find the volume of the sphere, which is equivalent to the volume of water displaced:
$$ frac{4pi r^3}{3} = frac{4pi 6^3}{3} = 288pi $$
Since the vessel is cylindrical, we can find the increase in height, $h$, using the following formula for the volume of a cylinder:
$$ pi r^2h = V = 288pi $$
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, you need to find the volume of the sphere, which is equivalent to the volume of water displaced:
$$ frac{4pi r^3}{3} = frac{4pi 6^3}{3} = 288pi $$
Since the vessel is cylindrical, we can find the increase in height, $h$, using the following formula for the volume of a cylinder:
$$ pi r^2h = V = 288pi $$
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add a comment |
$begingroup$
First, you need to find the volume of the sphere, which is equivalent to the volume of water displaced:
$$ frac{4pi r^3}{3} = frac{4pi 6^3}{3} = 288pi $$
Since the vessel is cylindrical, we can find the increase in height, $h$, using the following formula for the volume of a cylinder:
$$ pi r^2h = V = 288pi $$
$endgroup$
add a comment |
$begingroup$
First, you need to find the volume of the sphere, which is equivalent to the volume of water displaced:
$$ frac{4pi r^3}{3} = frac{4pi 6^3}{3} = 288pi $$
Since the vessel is cylindrical, we can find the increase in height, $h$, using the following formula for the volume of a cylinder:
$$ pi r^2h = V = 288pi $$
$endgroup$
First, you need to find the volume of the sphere, which is equivalent to the volume of water displaced:
$$ frac{4pi r^3}{3} = frac{4pi 6^3}{3} = 288pi $$
Since the vessel is cylindrical, we can find the increase in height, $h$, using the following formula for the volume of a cylinder:
$$ pi r^2h = V = 288pi $$
answered Jan 25 at 11:35
Geneten48Geneten48
1549
1549
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– Aloizio Macedo♦
Jan 26 at 15:53