Please refer to the following lecture notes for the formulas used in this exercise: Notes

Question 1: A solid sphere of radius \displaystyle 15 \text{ cm } is melted and recast into solid right circular cones of radius \displaystyle 2.5 and height \displaystyle 8 \text{ cm } . Calculate the number of cones recast. [2013]

Answer:

Sphere: Radius \displaystyle = 15 \text{ cm }  

Cone: Radius \displaystyle = 2.5 \text{ cm } and Height \displaystyle = 8 \text{ cm }  

\displaystyle \text{Therefore number of cones re-casted } = \frac{\frac{4}{3} \times \pi \times (15)^3}{\frac{1}{3} \times \pi \times (2.5)^2 \times 8} = 270  

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Question 2: A hollow sphere of internal and external diameters \displaystyle 4 \text{ cm } and \displaystyle 8 \text{ cm } respectively is melted into a cone of base diameter \displaystyle 8 \text{ cm } . Find the height of the cone. [2002]

Answer:

Internal diameter \displaystyle = 4 \text{ cm } \Rightarrow Internal radius \displaystyle = 2 \text{ cm }  

External diameter \displaystyle = 8 \text{ cm } \Rightarrow External radius \displaystyle = 4 \text{ cm }  

Radius of the \displaystyle \text{ cone } = 4 \text{ cm }  

\displaystyle \therefore \frac{4}{3} \times \pi \times (3)^3 - \frac{4}{3} \times \pi \times (2)^3 = \frac{1}{3} \times \pi \times (4)^2 \times h  

\displaystyle \Rightarrow h = \frac{4 \times (64-8)}{16} = 14 \text{ cm }  

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Question 3: The radii of the internal and external surfaces of a metallic spherical shell are \displaystyle 3 \text{ cm } and \displaystyle 5 \text{ cm } respectively. It is melted and recast into a solid right circular cone of height \displaystyle 32 \text{ cm } . Find the diameter of the base of the cone.

Answer:

Internal radius \displaystyle = 3 \text{ cm }  

External radius \displaystyle = 5 \text{ cm }  

Height of the cone \displaystyle = 32 \text{ cm }  

\displaystyle \therefore \frac{4}{3} \times \pi \times (5)^3 - \frac{4}{3} \times \pi \times (5)^3 = \frac{1}{3} \times \pi \times (r)^2 \times 32  

\displaystyle \Rightarrow r^2 = \frac{4 \times (125-27)}{32} = 12.25 \text{ cm }  

Therefore \displaystyle r = 3.5 \text{ cm } and hence diameter \displaystyle = 7 \text{ cm }  

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Question 4: Total volume of three identical cones is the same as that of a bigger cone whose height is \displaystyle 9 \text{ cm } and diameter \displaystyle 40 \text{ cm } . Find the radius of the base of each smaller cone, if height of each is \displaystyle 108 \text{ cm } .

Answer:

Let the radius of the cone \displaystyle = r  

Height of the cone \displaystyle = 108 \text{ cm }  

Bigger cone: Height \displaystyle = 9 \text{ cm } and radius \displaystyle = 20 \text{ cm }  

\displaystyle \therefore 3 \times \frac{1}{3} \times \pi \times (r)^2 \times 108 = \frac{1}{3} \times \pi \times (20)^2 \times 9  

\displaystyle 3 r^2 \times 108 = 20^2 \times 9  

\displaystyle \Rightarrow r= \frac{20}{6} = 3\frac{1}{3}  

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Question 5: A solid rectangular block of metal \displaystyle 49 \text{ cm } by \displaystyle 44 \text{ cm } by \displaystyle 18 \text{ cm } is melted and formed into a solid sphere. Calculate the radius of the sphere.

Answer:

Dimension of the block \displaystyle = 49 cm \times 44 cm \times 18 \text{ cm }  

Let the radius of the sphere \displaystyle = r \text{ cm }  

\displaystyle \therefore 49 \times 44 \times 18 = \frac{4}{3} \times \pi \times (r)^3  

\displaystyle r^3 = \frac{49 \times 44 \times 18}{4 \times 22} = 21^3 \Rightarrow r = 21 \text{ cm }  

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Question 6: A hemispherical bowl of internal radius \displaystyle 9 \text{ cm } is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter \displaystyle 3 \text{ cm } and height \displaystyle 4 \text{ cm } . How many containers are necessary to empty the bowl?

Answer:

Bowl: Internal radius \displaystyle = 9 \text{ cm }  

Cone: radius \displaystyle = 1.5 \text{ cm } , Height \displaystyle = 4 \text{ cm }  

\displaystyle \therefore \frac{1}{2} \frac{4}{3} \times \pi \times (9)^3 = n \times \frac{1}{3} \times \pi \times (1.5)^2 \times 4  

\displaystyle \Rightarrow n = \frac{9 \times 9 \times 9 }{2 \times 1.5 \times 1.5} = 162  

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Question 7: A hemispherical bowl of diameter \displaystyle 7.2 \text{ cm } is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius \displaystyle 4.8 \text{ cm } . Find the height of the cone if it is completely filled. [2010]

Answer:

Hemisphere: Radius \displaystyle = 3.6 \text{ cm }  

Cone: Radius \displaystyle = 4.8 \text{ cm } , Height \displaystyle = h  

\displaystyle \therefore \frac{1}{2} \frac{4}{3} \times \pi \times (3.6)^3 = \frac{1}{3} \times \pi \times (4.8)^2 \times h  

\displaystyle \Rightarrow h = \frac{2 \times 3.6^3}{4.8^2} = 4.05 \text{ cm }  

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Question 8: A solid cone of radius \displaystyle 5 \text{ cm } and height \displaystyle 8 \text{ cm } is melted and made into small spheres of radius \displaystyle 0.5 \text{ cm } . Find the number of spheres formed. [2011]

Answer:

Cone: Radius \displaystyle = 5 \text{ cm } , Height \displaystyle = 8 \text{ cm }  

Sphere: Radius \displaystyle = 0.5 \text{ cm }  

\displaystyle n \times \frac{4}{3} \times \pi \times (0.5)^3 = \frac{1}{3} \times \pi \times (5)^2 \times 8  

\displaystyle \Rightarrow n = \frac{5^2 \times 8}{4 \times 0.5^3} = 400  

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Question 9: The total area of a solid metallic sphere is \displaystyle 1256  \text{ cm}^2 . It is melted and recast into solid right circular cones of radius \displaystyle 2.5 \text{ cm } and height \displaystyle 8 \text{ cm } . Calculate: (i) the radius of the solid sphere, (ii) the number of cones recast. [2000]

Answer:

Surface area \displaystyle = 1256  \text{ cm}^2  

(i) \displaystyle \therefore 4 \pi r^2 = 1256  

\displaystyle \Rightarrow r^2 = \frac{1256}{4 \times 3.14} = 100  

\displaystyle \Rightarrow r = 10 \text{ cm }  

(ii) Cone: Radius \displaystyle = 2.5 \text{ cm } , Height \displaystyle = 8 \text{ cm }  

\displaystyle \frac{4}{3} \times \pi \times (10)^3 = n \times \frac{1}{3} \times \pi \times (2.5)^2 \times 8  

\displaystyle \Rightarrow n = \frac{4 \times 10^3}{2.5^2 \times 8} = 80  

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Question 10: A solid metallic cone, with radius \displaystyle 6 \text{ cm } and height \displaystyle 10 \text{ cm } , is made of some heavy metal A. In order to reduce its weight, a conical hole is made in the cone as, shown and it is completely filled with a lighter metal B. The conical hole has a diameter of \displaystyle 6 \text{ cm } and depth \displaystyle 4 \text{ cm } . calculate the ratio of the volume of metal A to the volume of the metal B in the solid.

Answer:

Cone: Radius \displaystyle = 6 \text{ cm } , Height \displaystyle = 10 \text{ cm }  

Conical hole: Radius \displaystyle = 3 \text{ cm } , Height \displaystyle = 4 \text{ cm }  

\displaystyle \text{Volume of metal A } = \frac{1}{3} \times \pi \times (6)^2 \times 10 - \frac{1}{3} \times \pi \times (3)^2 \times 4 = 108 \pi  

\displaystyle \text{Volume of metal B } = \frac{1}{3} \times \pi \times (3)^2 \times 4 = \frac{36 \pi}{3}  

\displaystyle \text{Therefore Ratio } = \frac{108 \pi}{\frac{36 \pi}{3}} = 9:1  

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Question 11: A hollow sphere of internal and external \displaystyle 6 \text{ cm } and \displaystyle 8 \text{ cm } respectively is melted and recast into small cones of base radius \displaystyle 2 \text{ cm } and height \displaystyle 8 \text{ cm } . Find the number of cones. [2012]

Answer:

Sphere: Internal radius \displaystyle = 6 \text{ cm } , External radius \displaystyle = 8 \text{ cm }  

Cone: Radius \displaystyle = 2 \text{ cm } , Height \displaystyle = 8 \text{ cm }  

\displaystyle \frac{4}{3} \times \pi \times (8)^3 - \frac{4}{3} \times \pi \times (6)^3 = n \times \frac{1}{3} \times \pi \times (2)^2 \times 8  

\displaystyle \Rightarrow n = \frac{4 \times (8^3-6^3)}{2^2 \times 8} = 37  

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Question 12: The surface area of a solid metallic sphere is \displaystyle 2464  \text{ cm}^2 . It is melted and recast into solid right circular cones of radius \displaystyle 3.5 \text{ cm } and height \displaystyle 7 \text{ cm } . Calculate: (i) the radius of the sphere (ii) the number of cones recast. (Take \displaystyle \pi = \frac{22}{7} ) [2014]

Answer:

Surface area of sphere \displaystyle = 2464  \text{ cm}^2  

Cone: Radius \displaystyle = 3.5 \text{ cm } , Height \displaystyle = 7 \text{ cm }  

\displaystyle \text{(i) } 4 \pi r^2 = 2464 \Rightarrow r^2 = \frac{2464 \times 7}{4 \times 22} = 196  

Hence \displaystyle R = 14 \text{ cm }  

\displaystyle \text{(ii) } \frac{4}{3} \times \pi \times (14)^3 = n \times \frac{1}{3} \times \pi \times (3.5)^2 \times 7  

\displaystyle \Rightarrow n = \frac{4 \times 14^3}{3.5^2 \times 7} = 128