Question 1: A cone of height $\displaystyle 15 \text{ cm }$ and diameter $\displaystyle 7 \text{ cm }$ is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

Cone: Height $\displaystyle = 15 \text{ cm } ,$ Diameter $\displaystyle = 7 \text{ cm }$

Hemisphere: Radius $\displaystyle = 3.5 \text{ cm }$

Total volume = volume of the cone + volume of the hemisphere

$\displaystyle = \frac{1}{3} \pi \times (3.5)^2 \times 15 + \frac{1}{2} \times \frac{4}{3} \pi \times (3.5)^3$

$\displaystyle = 192.5 + 89.833 = 282.33 \text{ cm}^3$

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Question 2: A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of a hemisphere. The radius of the base of the cone is $\displaystyle 3.5 meters$ and its volume is two-thirds of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal.

Cone: Height $\displaystyle = h \text{ cm } ,$ Diameter $\displaystyle = 7 \text{ cm }$

Hemisphere: Radius $\displaystyle = 3.5 \text{ cm }$

Therefore

$\displaystyle \frac{1}{3} \pi \times (3.5)^2 \times h = \frac{2}{3} \times \frac{1}{2} \times \frac{4}{3} \pi \times (3.5)^3$

$\displaystyle h = 2 \times \frac{4}{3} \times 3.5 = 4.67$

$\displaystyle \text{Total surface area of the solid } = \pi r l + \frac{1}{2} \times 4 \pi r^2$

$\displaystyle = 3.14 \times (3.5) \times \sqrt{3.5^2+4.67^2} + 2 \times 3.14 \times (3.5)^2$

$\displaystyle = 64.137 + 76.93 = 141.17 m^2$

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Question 3: From a rectangular solid of metal $\displaystyle 42 \text{ cm }$ by $\displaystyle 30 \text{ cm }$ by $\displaystyle 20 \text{ cm } ,$ a conical cavity of diameter $\displaystyle 14$ cm and depth $\displaystyle 24 \text{ cm }$ is drilled out. Find:

(i) the surface area of remaining solid,

(ii) the volume of remaining solid,

(iii) the weight of the material drilled out if it weighs $\displaystyle 7 gm / \text{ cm}^3 .$

Rectangular solid: $\displaystyle 42 \text{ cm }$ by $\displaystyle 30 \text{ cm }$ by $\displaystyle 20 \text{ cm }$

(i) Surface area of the solid = surface are of the rectangular solid – surface are of the base of the cone +curved surface ares of the cone

$\displaystyle = 2 (lb+bh+ hw) - \pi r^2 + \pi r l$

$\displaystyle = 2 (42 \times 30 + 30 \times 20 + 20 \times 42) - \frac{22}{7} \times (7)^2 + \frac{22}{7} \times 7 \times \sqrt{7^2+24^2}$

$\displaystyle = 5400 - 154 + 550 = 5796 \text{ cm}^2$

(ii) Volume = Volume of the solid – Volume of the cone

$\displaystyle = 42 \times 30 \times 20 - \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 24$

$\displaystyle = 25200 - 1232 = 23968 \text{ cm}^3$

(iii) Weight of the material drilled $\displaystyle = 1232 \times 7 = 8624 \text{ gms } = 8.624 \text{ kg }$

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Question 4: A cubical block of side $\displaystyle 7 \text{ cm }$ is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.

Box: Side $\displaystyle = 7 \text{ cm }$

Hemisphere: Radius $\displaystyle = 3.5 \text{ cm }$

Total surface area = surface area of the box – surface area of one side + surface area of the hemisphere

$\displaystyle = 6 (7)^ - (7)^ + \frac{1}{2} \times 4 \pi \times (3.5)^2$

$\displaystyle = 245 + 77 = 322 \text{ cm}^2$

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

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

Lead shot: Radius $\displaystyle = 0.5 \text{ cm } ,$ number of shots $\displaystyle = n$

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

$\displaystyle \Rightarrow n = \frac{5^2 \times 8}{4 \times 4 \times (0.5)^3} = 100$

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Question 6: A hemispherical bowl has a negligible thickness and the length of its circumference is $\displaystyle 198 \text{ cm }.$ Find the capacity of the bowl.

Circumference $\displaystyle = 198 \text{ cm }$

$\displaystyle \Rightarrow 2 \pi r = 198$

$\displaystyle \Rightarrow r = \frac{198}{2 \pi} = 31.5 \text{ cm }$

$\displaystyle \text{Therefore Volume of the bowl } = \frac{1}{2} \times \frac{4}{3} \pi \times (31.5)^3 = 65488.5 \text{ cm}^3$

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Question 7: Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius $\displaystyle r \text{ cm } .$

Sphere: Radius $\displaystyle = r \text{ cm }$

Cone: Radius $\displaystyle = r ,$ Height $\displaystyle = r$

$\displaystyle \text{The maximum volume of the cone } = \frac{1}{3} \pi \times (r)^2 \times r = \frac{1}{3} \pi r^3$

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Question 8: The radii of the bases of two solid right circular cones of same height are $\displaystyle r_1 ,$ and $\displaystyle r_2 .$ respectively. The cones are melted and recast into a solid sphere of radius R. Find the height of each cone in terms of $\displaystyle r_1, r_2 ,$ and $\displaystyle R .$

Cones: Radius $\displaystyle = r_1 ,$ Radius $\displaystyle = r_2 ,$ Height $\displaystyle = h$

Sphere: Radius $\displaystyle = R$

Therefore

$\displaystyle \frac{1}{3} \pi \times (r_1)^2 \times h + \frac{1}{3} \pi \times (r_1)^2 \times h = \frac{4}{3} \pi \times (R)^3$

$\displaystyle \Rightarrow (r_1^2+r_2^2)h = 4R^3$

$\displaystyle \Rightarrow h = \frac{4R^3}{(r_1^2+r_2^2)}$

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Question 9: A solid metallic hemisphere of diameter $\displaystyle 28 \text{ cm }$ is melted and recast into a number of identical solid cones, each of diameter $\displaystyle 14 \text{ cm }$ and height $\displaystyle 8 \text{ cm } .$ Find the number of cones so formed.

Hemisphere: Radius $\displaystyle = 14 \text{ cm }$

Cones: Radius $\displaystyle = 7 \text{ cm } ,$ Height $\displaystyle = 8 \text{ cm } ,$ Number of cones $\displaystyle = n$

Therefore $\displaystyle n \times \frac{1}{3} \pi \times (7)^2 \times 8 = \frac{2}{3} \pi \times (14)^3$

$\displaystyle \Rightarrow n = \frac{2 \times 14^3}{7^2 \times 8} = 14$

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Question 10: A cone and a hemisphere have the same base and the same height. Find the ratio between their volumes.

Cone: Radius $\displaystyle = r ,$ Height $\displaystyle = h$

Hemisphere: Radius $\displaystyle = r$

$\displaystyle \text{Ratio of their volumes } = \frac{\text{ Volume of Cone}}{\text{ Volume of Hemisphere } }$

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