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Home » 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.
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.
CBSE | Class 10 | Exercise 13.2 | Maths | NCERT | Surface Areas And Volumes
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.

For the cone,

Range = 5 cm,

Tallness = 8 cm

Too,

Range of circle = 0.5 cm

The outline will resemble

Ncert solutions class 10 chapter 13-16

It is realized that,

Volume of cone = volume of water in the cone

    \[=\text{ }\pi r^2h\text{ }=\text{ }\left( 200/3 \right)\pi \text{ }cm^3\]

Presently,

Absolute volume of water overflown=

    \[\left( {\scriptscriptstyle 1\!/\!{ }_4} \right)\times \left( 200/3 \right)\text{ }\pi \text{ }=\left( 50/3 \right)\pi \]

Volume of lead shot

=\frac{4}{3}\pi r^3

=\frac{1}{6} \pi

Presently,

The quantity of lead shots = Total Volume of Water over flown/Volume of Lead shot

    \[=\text{ }\left( 50/3 \right)\pi /(\frac{1}{6})\pi \]

    \[=\text{ }\left( 50/3 \right)\times 6\text{ }=\text{ }100\]

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