Math, asked by Satan2002, 1 year ago

How to do this question?

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Answered by Rajdeep11111

Volume of cone = n x Volume of 1 lead shot
Where n is the number of lead shots.

Volume of total lead shots = Volume of water displaced.
So Volume of 1 lead shot x n = Volume of water displaced.

Volume of water displaced = 2/5 of volume of cone.

Volume of cone = πr²h/3
= π x 2.5 x 2.5 x 11 / 3
= 6.25 x 3.14 x 11 / 3
= 19.63 x 11/3
= 215.93/3 cm³
= 71.98cm³

Now, volume of water displaced = 2 x 71.98/5
= 143.96/5
= 28.79cm³

Volume of 1 lead shot = 4/3πr³
= 4 x 3.14 x 0.25 x 0.25 x 0.25 /3
= 0.2/3 cm³
= 0.067 cm³

Volume of 1 lead shot x n = Volume of water displaced.
or 0.067 x n = 28.79
=> n = 429.70
or 430 lead shots are dropped.

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Answered by acsahjosemon40

Answer:

Volume of cone = n x Volume of 1 lead shot

Where n is the number of lead shots.

Volume of total lead shots = Volume of water displaced.

So Volume of 1 lead shot x n = Volume of water displaced.

Volume of water displaced = 2/5 of volume of cone.

Volume of cone = πr²h/3

= π x 2.5 x 2.5 x 11 / 3

= 6.25 x 3.14 x 11 / 3

= 19.63 x 11/3

= 215.93/3 cm³

= 71.98cm³

Now, volume of water displaced = 2 x 71.98/5

= 143.96/5

= 28.79cm³

Volume of 1 lead shot = 4/3πr³

= 4 x 3.14 x 0.25 x 0.25 x 0.25 /3

= 0.2/3 cm³

= 0.067 cm³

Volume of 1 lead shot x n = Volume of water displaced.

or 0.067 x n = 28.79

=> n = 429.70

or 430 lead shots are dropped.

Hope this helps uh!

Stay blessed ✨

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