Login/Sign Up
Home » A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel. (2003)
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel. (2003)
CBSE | Class 10 | Exercise 1 | maths | Maths | Mensuration | ML Aggarwal
A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel. (2003)

Solution:

Given height of the cone, h = 11 cm

Radius of the cone, r = 2.5 cm

Volume of the cone = (1/3)r2h

= (1/3)×2.52×11

= (11/3)×6.25 cm3

When lead shots are dropped into vessel, (2/5) of water flows out.

Volume of water flown out = (2/5)×(11/3)×6.25

= (22/15)×6.25

= (137.5/15) cm3

Radius of sphere, R = 0.25 cm = ¼ cm

Volume of sphere = (4/3)R3

= (4/3)×(1/4)3

= /48 cm3

Number of lead shots dropped = Volume of water flown out/ Volume of sphere

= (137.5/15) ÷/48

= (137.5/15) ×48/

= 137.5×48/15

= 440

Hence the number of lead shots dropped is 440.

i

More Material