Question

A vessel is in the form of an inverted cone. Its
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.
O00​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• 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.

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$\large\underline\blue{\bold{To \:  Find :-  }}$

• The number of lead shots dropped in the vessel.

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

${{ \boxed{\large{\bold\green{Volume_{(cone)}\: = \dfrac{1}{3} \:\pi r^2h}}}}}$

where,

• r = radius of cone
• h = height of cone

${{ \boxed{\large{\bold\green{Volume_{(sphere)}\: =\dfrac{4}{3} \:\pi r^3}}}}}$

where,

• r = radius of sphere

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$\large\underline\purple{\bold{Solution :- }}$

☆ Given Dimensions of Cone :-

¤ Radius of cone, r = 5 cm

¤ Height of cone, h = 8 cm

$\large{\bold\green{Volume_{(cone)}\: =\dfrac{1}{3} \:\pi r^2h}}$

$\tt\implies \:Volume_{(cone)} = \dfrac{1}{3} \times \dfrac{22}{7} \times {5}^{2} \times 8 \:$

$\tt\implies \:Volume_{(cone)} = \dfrac{4400}{21} \: {cm}^{3}$

$\tt \: Volume \: of \: water \: overflow \: = \dfrac{1}{4} \times Volume_{(cone)}$

$\tt \: Hence, Volume \: of \: water \: overflow \: = \dfrac{1}{4} \times \dfrac{4400}{21}$

$\tt\implies \:Volume \: of \: water \: overflow = \dfrac{1100}{21} \: {cm}^{3}$

Volume of 1 lead shot

☆Lead shot is in the form of sphere

☆ Radius of sphere, r = 0.5 cm

☆ Volume of 1 lead shot = Volume of sphere

$\tt\implies \:{{ {\large{\bold\green{Volume_{(lead \: shot)}\: =\dfrac{4}{3} \:\pi r^3}}}}}$

$\tt\implies \:Volume_{(lead \: shot)} = \dfrac{4}{3} \times \dfrac{22}{7} \times {(0.5)}^{3}$

$\tt\implies \:Volume_{(lead \: shot)} = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$

$\tt\implies \:Volume_{(lead \: shot)} = \dfrac{11}{21} \: {cm}^{ 3}$

☆ Now,

$\tt \ Number \: of \: lead \: shots \: = \dfrac{Volume_{(cone)}}{Volume_{(lead \: shot)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \:  = \dfrac{1100}{21} \div \dfrac{11}{21}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \:  = 100$

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More info:

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²