Question

A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5
cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of
radius 0.5cm are dropped into the vessel, of the water flows out. Find the number of
4
lead shots dropped into the vessel.
,
1:a metallic sphere of dia​

Answer 1

Answer:

I don't know what is answer of question. Sorry.

Please mark me as brainly.

Answer 2

Question:

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer:

$\text{Volume of Cone}=\pi\times r^2\times\dfrac{h}{3} = \dfrac{22}{7} \times5^2\times\dfrac{8}{3} =\dfrac{4400}{21} \: cm^3$

$\text{Volume of water flows}= \dfrac{1}{4} \ \text{of volume of water in cone} = \dfrac{1}{4} \times\dfrac{4400}{21} = \dfrac{1100}{21} \ cm^3$

$\\ \text{Radius of lead shot}= R \ cm$

$\text{Volume of lead shot (Sphere)}=\dfrac{4}{3}\times\pi\times R^3 = \dfrac{4}{3}\times \dfrac{22}{7} \times\bigg(\dfrac{5}{10}\bigg)^3 = \dfrac{11}{21} \ cm^3$

$\\ \text{Required number of slots} =\dfrac{\text{Volume of water flows}}{\text{Volume of one lead shot}} = \dfrac{1100}{21} \div \dfrac{11}{21}= \dfrac{1100}{21} \times \dfrac{21}{11} = 100$

$100 \ \ \ \sf{Answer}$

$@rohangupta0424$