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.

Answer:

Let the initial Surface area be S_1 , Radius be (r_1) and Volume be V_1 .

Let the final Surface area be S_2 , Radius be (r_2) and Volume be V_2 .

Given S_1 \times 1.21 = S_2

Therefore 4 \pi (r_1)^2 \times 1.21 = 4 \pi (r_2)^2

\Rightarrow r_1 \times 1.1 = r_2

This means that there has been an increase of 10\% in the radius.

Taking ratios of the volumes we get = \frac{\frac{4}{3} \pi  (r_2)^3}{\frac{4}{3} \pi  (r_1)^3} = (\frac{r_2}{r_1})^3 = 1.1^3 = 1.331

This means that the volume has increased by 33.1\% 

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