Class 10: Cone and Sphere – Exercise 22(d) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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.