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

Notes: Important formuals

Parameters of a Sphere: Radius of the Sphere ( $r$ )

Volume of the sphere $= \frac{4}{3} \pi r^3$

Surface area of a sphere $= 4 \pi r^2$

Question 1: The surface area of a sphere is $2464 \ cm^2$ , find its volume.

Answer:

Surface area of Sphere $2464 \ cm^2$

Therefore $4 \pi r^2 = 2464$

$\Rightarrow r^2 = \frac{2464 \times 7}{4 \times 22} = 196$

Therefore $r = 14 \ cm$

Volume $= \frac{4}{3} \pi r ^3 = \frac{4}{3} \pi (14) ^3 = 11498.66 \ cm^3$

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Question 2: The volume of a sphere is $38808 \ cm^3$ ; find its diameter and the surface area.

Answer:

Volume of Sphere $= 38808 \ cm^3$

Therefore $\frac{4}{3} \pi r^3 = 38808$

$r^3 =$ $\frac{3 \times 38808 \times 7}{4 \times 22}$ $= 9261 \Rightarrow r = 21 \ cm$

Surface area $= 4 \pi (21)^2 = 4 \times \frac{22}{7} \times (21)^2 = 5544 \ cm^2$

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Question 3: A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?

Answer:

Let the radius of the original ball be $r_1$ and the volume be $V_1$

Similarly, let the radius of the smaller ball be $r_2$ and the volume be $V_2$

Given $2 r_2 = r_1$

Therefore the number of smaller balls $= \frac{V_1}{V_2}$

$= \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}$ $= (\frac{r_1}{r_2})^3 = 8$

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Question 4: How many balls each of radius $1 \ cm$ can be made by melting a bigger ball whose diameter is $8 \ cm$ .

Answer:

Diameter of the bigger ball $= 8 cm$

Therefore the Radius of the bigger ball $= 4 \ cm$

Radius of the smaller ball $= 1 \ cm$

Therefore No of Smaller balls $=$ $\frac{\frac{4}{3} \pi (4)^3}{\frac{4}{3} \pi (1)^3}$ $= 4^3 = 64$

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Question 5: Eight metallic spheres; each of radius $2 \ mm$ , are melted and cast into a single sphere. Calculate the radius of the new sphere.

Answer:

Number of metallic spheres $= 8$

Radius of the metalic sphere $= 2 \ mm$

Let the radius of the new sphere $= r$

Therefore $8 \times \frac{4}{3} \pi (2)^3c= \frac{4}{3} \pi (r)^3$

$8 \times 8 = r^3 \Rightarrow r = 4 \ mm$

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Question 6: The volume of one sphere is $27$ times that of another sphere. Calculate the ratio of their: (i) radii (ii) surface areas

Answer:

Let the volume of the 1st sphere be $V_1$ and its radius be $r_1$ . Similarly, Let the volume of the 2nd sphere be $V_2$ and its radius be $r_2$ .

Given $V_1 = 27 V_2$

(i) Therefore $\frac{4}{3} \pi (r_1)^3 = 27 \times \frac{4}{3} \pi (r_2)^3$

$\Rightarrow r_1 = 3r_2 \Rightarrow r_1:r_2 = 3:1$

(ii) Let the surface area of the two spheres be $S_1$ and $S_2$ respectively.

Therefore $\frac{S_1}{S_2} = \frac{4 \pi (r_1)^2}{4 \pi (r_2)^2} = (\frac{r_1}{r_2})^2$ $= 9$

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Question 7: If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in its volume, what is the diameter of the sphere?

Answer:

Let the radius of the sphere $= r$

Given, Surface area = Volume of the sphere (only in terms of numerical value)

$\frac{4}{3} \pi (r)^3 = 4 \pi (r)^2$

$\Rightarrow r = 3 \ cm$

Therefore Diameter $= 6 \ cm$

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Question 8: A solid metal sphere is cut through its center into $2$ equal parts. If the diameter of the sphere is $3.5 \ cm$ , find the total surface area of each part correct to two decimal places.

Answer:

Diameter of the sphere $= 3.5 \ cm$

Radius $= \frac{3.5}{2} = 1.75 \ cm$

Total surface area $= \frac{1}{2} (4 \pi r^2) + \pi r^2 = 3 \pi r^2 = 3 \times \frac{22}{7} \times (1.75)^2 = 28.875 \ cm^2$

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Question 9: The internal and external diameters of a hollow hemispherical vessel are $21 \ cm$ and $28 \ cm$ respectively. Find: (i) internal curved surface area, (ii) external curved surface area, (iii) total surface area, (iv) volume of material of the vessel

Answer:

Internal Radius $(r) = \frac{21}{2} = 10.5 \ cm$

External Radius $(R) = \frac{28}{2} = 14 \ cm$

(i) Internal surface area $= \frac{1}{2} . 4 \pi r^2 = \frac{1}{2} . 4 \pi (10.5)^2 = 693 \ cm^2$

(ii) External surface area $= \frac{1}{2} . 4 \pi (14)^2 = 1232 \ cm^2$

(iii) Total surface area $= 2 \pi (14)^2 + 2 \pi (10.5)^2 + \pi (14^2-10.5^2) = 1925 + 269.5 = 2194.5 \ cm^2$

(iv) Volume of Material $= \frac{1}{2} \times \frac{4}{3} \pi (14^3 - 10.5^3) = 3323.83 \ cm^3$

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Question 10: A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.

Answer:

Let the radius of the sphere be $r_1$ and hemisphere be $r_2$

Surface area of sphere $= 4 \pi (r_1)^2$

Surface are of hemisphere $= 2 \pi (r_2)^2 + \pi r_2^2$

Volume of sphere $= \frac{4}{3} \pi (r_1)^3$

Volume of hemisphere $= \frac{2}{3} \pi (r_2)^3$

Therefore the ratios of their volume $= 2. (\frac{r_1}{r_2})^3$ … … … (i)

It is given that:

$4 \pi (r_1)^2 = 2 \pi (r_2)^2 + \pi (r_2)^2$

$\Rightarrow 4(r_1)^2 = 3(r_2)^2 \ or \ \frac{r_1}{r_2} = \frac{\sqrt{3}}{2}$

Substituting this in (i) we get

Ratios of their volume $= 2. (\frac{\sqrt{3}}{2})^3 = \frac{3 \sqrt{3}}{4}$

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Question 11: Metallic spheres of radii $6 \ cm, 8 \ cm$ and $10 cm$ respectively are melted and recasted into a single solid sphere. Taking $\pi = \frac{22}{7}$ , find the surface area of solid sphere formed.

Answer:

Let the radius of the new sphere $= r$

Volume of the new sphere = total volume of the three spheres. Therefore

$\frac{4}{3} \pi (r)^3 = \frac{4}{3} \pi (6)^3 + \frac{4}{3} \pi (8)^3 + \frac{4}{3} \pi (10)^3$

$\Rightarrow r^3 = 6^3 + 8^3 + 10^3 = 1728$

Therefore $r = 12 \ cm$

Surface area $= 4 \pi (12)^2 = 1810.28 \ cm^2$

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Question 12: The surface area of a solid sphere is increased by $21\%$ without changing its shape. Find the percentage increase in its: (i) radius (ii) volume.