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

Answer:

Cone: Height = 15 \ cm , Diameter = 7 \ cm

Hemisphere: Radius = 3.5 \ cm

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

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

= 192.5 + 89.833 = 282.33 \ cm^3

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

Answer:

Cone: Height = h \ cm , Diameter = 7 \ cm

Hemisphere: Radius = 3.5 \ cm

Therefore

\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

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

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

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

= 64.137 + 76.93 = 141.17 \ m^2 

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Question 3: From a rectangular solid of metal 42 \ cm  by 30 \ cm  by 20 \ cm , a conical cavity of diameter 14 cm and depth latex 24 \ 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 7 gm / cm^3 .

Answer:

Rectangular solid: 42 \ cm  by 30 \ cm  by 20 \ 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

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

= 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}

= 5400 - 154 + 550 = 5796 \ cm^2

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

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

= 25200 - 1232 = 23968 \ cm^3

(iii) Weight of the material drilled = 1232 \times 7 = 8624 \ gms = 8.624 \ kg 

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

Answer:

Box: Side = 7 \ cm

Hemisphere: Radius = 3.5 \ cm

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

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

= 245 + 77 = 322 \ cm^2

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Question 5: 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 rim. 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.

Answer:

Cone: Height = 8 \ cm , Radius = 5 \ cm

Lead shot: Radius = 0.5 \ cm , number of shots = n

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

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

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Question 6: A hemi-spherical bowl has negligible thickness and the length of its circumference is 198 \ cm . Find the capacity of the bowl.

Answer:

Circumference = 198 \ cm

\Rightarrow 2 \pi r = 198 

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

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

Answer:

Sphere: Radius = r \ cm

Cone: Radius = r , Height = r

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 r_1 , and 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 r_1, r_2 , and R .

Answer:

Cones: Radius = r_1 , Radius = r_2 , Height = h

Sphere: Radius = R

Therefore

\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

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

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

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

Answer:

Hemisphere: Radius = 14 \ cm

Cones: Radius = 7 \ cm , Height = 8 \ cm , Number of cones = n

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

 \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.

Answer:

Cone: Radius  = r , Height  = h

Hemisphere: Radius  = r

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

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

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