Ganit Prakash - Class-X - Sphere
Let us work out 12 Solid Geometry
📘 Let us work out 12 Solutions
SphereGiven:
Radius of the sphere \( (r) \) = 10.5 cm
Whole surface area of a sphere is given by the formula \( 4\pi r^2 \).
\[\begin{array}{l} \text{Whole surface area} = 4 \times \frac{22}{7} \times (10.5)^2 \\ = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \\ = 22 \times 3 \times 21 \\ = 66 \times 21 \\ = 1386 \text{ sq cm.} \end{array}\]Therefore, the whole surface area of the sphere is 1386 sq cm.
Given:
Total cost of making the leather ball = ₹ 431.20
Rate of making = ₹ 17.50 per sq cm.
Total surface area of the ball \( = \frac{\text{Total Cost}}{\text{Rate}} \)
\[\begin{array}{l} \text{Surface Area} = \frac{431.20}{17.50} \\ = \frac{43120}{1750} \\ = 24.64 \text{ sq cm.} \end{array}\]Let the radius of the ball be \( r \) cm.
\[\begin{array}{l} 4\pi r^2 = 24.64 \\ \Rightarrow 4 \times \frac{22}{7} \times r^2 = 24.64 \\ \Rightarrow r^2 = \frac{24.64 \times 7}{4 \times 22} \\ \Rightarrow r^2 = \frac{172.48}{88} \\ \Rightarrow r^2 = 1.96 \\ \Rightarrow r = \sqrt{1.96} = 1.4 \text{ cm.} \end{array}\]Length of diameter = \( 2r = 2 \times 1.4 = \) 2.8 cm.
Given:
Diameter of the shotput ball = 7 cm
Radius \( (r) \) = \( \frac{7}{2} \) cm = 3.5 cm
The amount of iron in the ball is equal to its volume. The volume of a sphere is \( \frac{4}{3}\pi r^3 \).
\[\begin{array}{l} \text{Volume} = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 \\ = \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \\ = \frac{11 \times 49}{3} \\ = \frac{539}{3} \\ = 179\frac{2}{3} \text{ cubic cm.} \end{array}\]Therefore, there is \( 179\frac{2}{3} \) cubic cm of iron in the ball.
Given:
Diameter of the solid sphere = 28 cm
Radius \( (r) \) = \( \frac{28}{2} \) = 14 cm
According to Archimedes' principle, the volume of water displaced by a completely immersed solid is equal to the volume of the solid itself.
Volume of the sphere = \( \frac{4}{3}\pi r^3 \)
\[\begin{array}{l} = \frac{4}{3} \times \frac{22}{7} \times (14)^3 \\ = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 \\ = \frac{4 \times 22 \times 2 \times 196}{3} \\ = \frac{88 \times 392}{3} \\ = \frac{34496}{3} \\ = 11498\frac{2}{3} \text{ cubic cm.} \end{array}\]Therefore, the volume of water displaced is \( 11498\frac{2}{3} \) cubic cm.
Given:
Initial radius of the spherical balloon \( (r_1) \) = 7 cm
Final radius of the spherical balloon \( (r_2) \) = 21 cm
The surface area of a sphere is given by the formula \( 4\pi r^2 \).
Ratio of the surface areas in the two cases:
\[\begin{array}{l} = \frac{\text{Initial Surface Area}}{\text{Final Surface Area}} \\ = \frac{4\pi (r_1)^2}{4\pi (r_2)^2} \\ = \frac{(r_1)^2}{(r_2)^2} \\ = \left(\frac{7}{21}\right)^2 \\ = \left(\frac{1}{3}\right)^2 \\ = \frac{1}{9} \end{array}\]Therefore, the ratio of the surface areas of the balloon is 1:9.
Given:
Area of the sheet required to make the hemispherical bowl = \( 127\frac{2}{7} \) sq cm.
Since it is a bowl, it is open at the top. The area of the sheet is equal to its curved surface area.
Curved surface area of a hemisphere = \( 2\pi r^2 \)
Let the radius of the bowl be \( r \) cm.
\[\begin{array}{l} 2\pi r^2 = 127\frac{2}{7} \\ \Rightarrow 2 \times \frac{22}{7} \times r^2 = \frac{127 \times 7 + 2}{7} \\ \Rightarrow \frac{44}{7} \times r^2 = \frac{889 + 2}{7} \\ \Rightarrow \frac{44}{7} \times r^2 = \frac{891}{7} \\ \Rightarrow 44r^2 = 891 \\ \Rightarrow r^2 = \frac{891}{44} \\ \Rightarrow r^2 = \frac{81}{4} \quad [\text{Dividing numerator and denominator by 11}] \\ \Rightarrow r = \sqrt{\frac{81}{4}} \\ \Rightarrow r = \frac{9}{2} = 4.5 \text{ cm.} \end{array}\]Length of diameter = \( 2r = 2 \times 4.5 \) = 9 cm.
Given:
Radius of the solid spherical ball \( (r) \) = 2.1 cm
Part 1: Amount of iron (Volume)
The amount of iron in the ball is its volume, given by \( \frac{4}{3}\pi r^3 \).
\[\begin{array}{l} \text{Volume} = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 \\ = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10} \\ = 4 \times 22 \times \frac{1}{10} \times \frac{21}{10} \times \frac{21}{10} \\ = \frac{88 \times 441}{1000} \\ = \frac{38808}{1000} \\ = 38.808 \text{ cubic cm.} \end{array}\]Therefore, there is 38.808 cubic cm of iron in the ball.
Part 2: Curved Surface Area
For a solid sphere, the curved surface area is the same as its total surface area, given by \( 4\pi r^2 \).
\[\begin{array}{l} \text{Curved Surface Area} = 4 \times \frac{22}{7} \times (2.1)^2 \\ = 4 \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \\ = 4 \times 22 \times \frac{3}{10} \times \frac{21}{10} \\ = \frac{88 \times 63}{100} \\ = \frac{5544}{100} \\ = 55.44 \text{ sq cm.} \end{array}\]Therefore, the curved surface area of the iron ball is 55.44 sq cm.
Given:
Diameter of the large solid sphere = 14 cm
Radius of the large solid sphere \( (R) \) = \( \frac{14}{2} \) = 7 cm
Radius of each small sphere \( (r) \) = 3.5 cm = \( \frac{7}{2} \) cm
Volume of the large sphere = \( \frac{4}{3}\pi R^3 \)
Volume of one small sphere = \( \frac{4}{3}\pi r^3 \)
Let the number of small spheres made be \( n \).
\[\begin{array}{l} n = \frac{\text{Volume of large sphere}}{\text{Volume of one small sphere}} \\ = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} \\ = \left(\frac{R}{r}\right)^3 \\ = \left(\frac{7}{3.5}\right)^3 \\ = (2)^3 \\ = 8 \end{array}\]Therefore, 8 small spheres can be made.
Given:
Radii of the three small spheres are \( r_1 \) = 3 cm, \( r_2 \) = 4 cm, and \( r_3 \) = 5 cm.
Let the radius of the newly formed large sphere be \( R \) cm.
According to the problem, the total volume of the three small spheres is equal to the volume of the large sphere.
\[\begin{array}{l} \text{Volume of large sphere} = \text{Sum of volumes of 3 small spheres} \\ \frac{4}{3}\pi R^3 = \frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3 \\ \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (r_1^3 + r_2^3 + r_3^3) \end{array}\]Cancelling \( \frac{4}{3}\pi \) from both sides:
\[\begin{array}{l} R^3 = 3^3 + 4^3 + 5^3 \\ R^3 = 27 + 64 + 125 \\ R^3 = 216 \\ R^3 = 6^3 \\ R = 6 \text{ cm.} \end{array}\]Therefore, the length of the radius of the large sphere is 6 cm.
Given:
Diameter of the base of the hemispherical tomb = 42 dm
Radius \( (r) \) = \( \frac{42}{2} \) = 21 dm
Since the rate is given in square metres, we convert the radius into metres:
\( r = \frac{21}{10} \) m = 2.1 m
The upper surface to be coloured is the curved surface area (CSA) of the hemisphere.
Curved Surface Area \( = 2\pi r^2 \)
\[\begin{array}{l} = 2 \times \frac{22}{7} \times (2.1)^2 \text{ sq. m.} \\ = 2 \times \frac{22}{7} \times 2.1 \times 2.1 \\ = 44 \times 0.3 \times 2.1 \\ = 13.2 \times 2.1 \\ = 27.72 \text{ sq. m.} \end{array}\]Cost of colouring per square metre = ₹ 35
Total Cost = \( 27.72 \times 35 \)
\[\begin{array}{l} = \text{₹ } 970.20 \end{array}\]Therefore, the total cost of colouring the upper surface is ₹ 970.20.
Given:
Diameter of the first hollow sphere \( (D_1) \) = 21 cm
Radius of the first sphere \( (r_1) \) = \( \frac{21}{2} \) cm
Diameter of the second hollow sphere \( (D_2) \) = 17.5 cm
Radius of the second sphere \( (r_2) \) = \( \frac{17.5}{2} \) cm = \( \frac{35}{4} \) cm
The area of the sheet of metal required for a hollow sphere is equal to its total surface area, which is \( 4\pi r^2 \).
Ratio of the areas of sheets required:
\[\begin{array}{l} = \frac{\text{Surface Area of 1st Sphere}}{\text{Surface Area of 2nd Sphere}} \\ = \frac{4\pi (r_1)^2}{4\pi (r_2)^2} \\ = \left(\frac{r_1}{r_2}\right)^2 \\ = \left(\frac{\frac{21}{2}}{\frac{17.5}{2}}\right)^2 \\ = \left(\frac{21}{17.5}\right)^2 \\ = \left(\frac{210}{175}\right)^2 \end{array}\]Simplifying the fraction \( \frac{210}{175} \) by dividing numerator and denominator by 35:
\[\begin{array}{l} = \left(\frac{6}{5}\right)^2 \\ = \frac{36}{25} \end{array}\]Therefore, the ratio of the areas of the sheets of metal is 36 : 25.
Given:
Let the radius of the original sphere be \( R \).
Let the radius of the new (remaining) sphere be \( r \).
According to the problem, the curved surface area of the new sphere is half of the original one.
\[\begin{array}{l} 4\pi r^2 = \frac{1}{2} \times 4\pi R^2 \\ \Rightarrow r^2 = \frac{R^2}{2} \\ \Rightarrow r = \frac{R}{\sqrt{2}} \end{array}\]Let the volume of the original sphere be \( V_1 \) and the remaining sphere be \( V_2 \).
\[\begin{array}{l} V_1 = \frac{4}{3}\pi R^3 \\ V_2 = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{R}{\sqrt{2}}\right)^3 = \frac{4}{3}\pi \frac{R^3}{2\sqrt{2}} \end{array}\]The volume of the portion cut off \( = V_1 - V_2 \)
\[\begin{array}{l} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi \frac{R^3}{2\sqrt{2}} \\ = \frac{4}{3}\pi R^3 \left(1 - \frac{1}{2\sqrt{2}}\right) \\ = \frac{4}{3}\pi R^3 \left(\frac{2\sqrt{2} - 1}{2\sqrt{2}}\right) \end{array}\]Now, we find the ratio of the volumes of the portion cut off and the remaining portion (\( V_2 \)):
\[\begin{array}{l} \text{Ratio} = \frac{V_1 - V_2}{V_2} \\ = \frac{\frac{4}{3}\pi R^3 \left(\frac{2\sqrt{2} - 1}{2\sqrt{2}}\right)}{\frac{4}{3}\pi \frac{R^3}{2\sqrt{2}}} \\ = \frac{2\sqrt{2} - 1}{1} \end{array}\]Therefore, the ratio is \( (2\sqrt{2} - 1) : 1 \).
Given:
Radius of the globe \( (R) \) = 14 cm
Radius of each circular hole \( (r) \) = 0.7 cm
First, calculate the total curved surface area of the globe without holes:
\[\begin{array}{l} \text{Total CSA} = 4\pi R^2 \\ = 4 \times \frac{22}{7} \times (14)^2 \\ = 4 \times \frac{22}{7} \times 196 \\ = 4 \times 22 \times 28 \\ = 2464 \text{ sq. cm.} \end{array}\]Now, calculate the area of the two circular holes that are removed:
\[\begin{array}{l} \text{Area of 2 holes} = 2 \times \pi r^2 \\ = 2 \times \frac{22}{7} \times (0.7)^2 \\ = 2 \times \frac{22}{7} \times 0.49 \\ = 2 \times 22 \times 0.07 \\ = 3.08 \text{ sq. cm.} \end{array}\]The area of the metal sheet surrounding its curved surface is the Total CSA minus the area of the holes:
\[\begin{array}{l} \text{Area of metal sheet} = 2464 - 3.08 \\ = 2460.92 \text{ sq. cm.} \end{array}\]Therefore, the area of the metal sheet is 2460.92 sq. cm.
Given:
Radius of the large solid sphere \( (R) \) = 8 cm
Radius of each small marble \( (r) \) = 1 cm
Let the number of marbles formed be \( n \).
According to the problem, the total volume of all marbles is equal to the volume of the large solid sphere.
\[\begin{array}{l} n = \frac{\text{Volume of large solid sphere}}{\text{Volume of one marble}} \\ = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} \\ = \left(\frac{R}{r}\right)^3 \\ = \left(\frac{8}{1}\right)^3 \\ = (8)^3 \\ = 512 \end{array}\]Therefore, 512 marbles can be formed.
📘 15. Very short answer type question (V.S.A.)
(A) M.C.Q.Explanation:
Radius of the sphere = \( 2r \)
Volume of a sphere = \( \frac{4}{3}\pi (\text{radius})^3 \)
\[\begin{array}{l} \text{Volume} = \frac{4}{3}\pi (2r)^3 \\ = \frac{4}{3}\pi (8r^3) \\ = \frac{32\pi r^3}{3} \text{ cubic units.} \end{array}\]Correct Option: (a) \( \frac{32\pi r^3}{3} \) cubic unit
Explanation:
Let the radii of the two spheres be \( r_1 \) and \( r_2 \).
Ratio of their volumes:
\[\begin{array}{l} \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{1}{8} \\ \Rightarrow \left(\frac{r_1}{r_2}\right)^3 = \left(\frac{1}{2}\right)^3 \\ \Rightarrow \frac{r_1}{r_2} = \frac{1}{2} \end{array}\]Now, ratio of their curved surface areas:
\[\begin{array}{l} \frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \\ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \end{array}\]Correct Option: (b) 1:4
Explanation:
Radius of the solid hemisphere \( (r) \) = 7 cm
Whole surface area of a solid hemisphere = \( 3\pi r^2 \)
\[\begin{array}{l} \text{Whole surface area} = 3 \times \pi \times (7)^2 \\ = 3 \times \pi \times 49 \\ = 147\pi \text{ sq cm.} \end{array}\]Correct Option: (c) 147\pi sq cm.
Explanation:
Let the radii of the two spheres be \( r_1 \) and \( r_2 \).
Ratio of their curved surface areas:
\[\begin{array}{l} \frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{16}{9} \\ \Rightarrow \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{3}\right)^2 \\ \Rightarrow \frac{r_1}{r_2} = \frac{4}{3} \end{array}\]Now, ratio of their volumes:
\[\begin{array}{l} \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3 \\ = \left(\frac{4}{3}\right)^3 = \frac{64}{27} \end{array}\]Correct Option: (a) 64:27
Explanation:
Let the radius of the sphere be \( r \).
Curved surface area = \( 4\pi r^2 \)
Volume = \( \frac{4}{3}\pi r^3 \)
According to the condition: Area = 3 \( \times \) Volume
\[\begin{array}{l} 4\pi r^2 = 3 \times \left(\frac{4}{3}\pi r^3\right) \\ \Rightarrow 4\pi r^2 = 4\pi r^3 \\ \Rightarrow 1 = r \quad [\text{Dividing both sides by } 4\pi r^2] \\ \Rightarrow r = 1 \text{ unit} \end{array}\]Correct Option: (a) 1 unit
📘 15. Very short answer type question (V.S.A.)
(B) True or FalseIf we double the length of radius of a solid sphere, the volume of sphere will be doubled.
Explanation:
Let the original radius be \( r \). Original Volume \( (V_1) = \frac{4}{3}\pi r^3 \).
If the radius is doubled, new radius = \( 2r \).
New Volume \( (V_2) \):
\[\begin{array}{l} V_2 = \frac{4}{3}\pi (2r)^3 \\ = \frac{4}{3}\pi (8r^3) \\ = 8 \times \left(\frac{4}{3}\pi r^3\right) \\ = 8 V_1 \end{array}\]The volume becomes 8 times the original volume, not doubled.
Answer: False
If the ratio of curved surface areas of two hemispheres is 4:9, the ratio of their lengths of radii is 2:3
Explanation:
Let the radii of the two hemispheres be \( r_1 \) and \( r_2 \).
Curved surface area of a hemisphere = \( 2\pi r^2 \).
Ratio of their curved surface areas:
\[\begin{array}{l} \frac{2\pi r_1^2}{2\pi r_2^2} = \frac{4}{9} \\ \Rightarrow \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{2}{3}\right)^2 \\ \Rightarrow \frac{r_1}{r_2} = \frac{2}{3} \end{array}\]The ratio of their radii is indeed 2:3.
Answer: True
📘 15. Very short answer type question (V.S.A.)
(C) Fill in the blanksExplanation:
A sphere has only one continuous curved surface.
Answer: sphere
Explanation:
A solid hemisphere has two surfaces: one curved surface and one flat circular base.
Answer: 2
Explanation:
Whole surface area of a solid hemisphere = \( 3\pi (\text{Radius})^2 \)
Given Radius = \( 2r \)
\[\begin{array}{l} \text{Whole surface area} = 3\pi (2r)^2 \\ = 3\pi (4r^2) \\ = 12\pi r^2 \text{ sq units.} \end{array}\]Therefore, the coefficient before \( \pi r^2 \) is 12.
Answer: 12
📘 16. Short answer type (S.A.)
SphereExplanation:
Let the radius of the solid hemisphere be \( r \) units.
Volume of the solid hemisphere = \( \frac{2}{3}\pi r^3 \)
Whole surface area of the solid hemisphere = \( 3\pi r^2 \)
According to the condition:
\[\begin{array}{l} \frac{2}{3}\pi r^3 = 3\pi r^2 \\ \Rightarrow \frac{2}{3}r = 3 \quad [\text{Dividing both sides by } \pi r^2 \text{ as } r \neq 0] \\ \Rightarrow 2r = 9 \\ \Rightarrow r = \frac{9}{2} = 4.5 \end{array}\]Therefore, the length of the radius of the hemisphere is 4.5 units.
Explanation:
For the solid right circular cylinder:
Height \( (h) \) = 12 cm
Diameter = 12 cm \(\Rightarrow\) Radius \( (r_c) \) = 6 cm
Total surface area of the solid cylinder = \( 2\pi r_c(h + r_c) \)
\[\begin{array}{l} = 2 \times \pi \times 6(12 + 6) \\ = 12\pi \times 18 \\ = 216\pi \text{ sq. cm.} \end{array}\]Let the radius of the solid sphere be \( r_s \) cm.
Curved surface area of the sphere = \( 4\pi r_s^2 \)
According to the condition:
\[\begin{array}{l} 4\pi r_s^2 = 216\pi \\ \Rightarrow 4r_s^2 = 216 \\ \Rightarrow r_s^2 = \frac{216}{4} \\ \Rightarrow r_s^2 = 54 \\ \Rightarrow r_s = \sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6} \end{array}\]Therefore, the length of the radius of the sphere is \( 3\sqrt{6} \) cm.
Explanation:
Let the radius of the solid hemisphere be \( r_1 \) and the radius of the solid sphere be \( r_2 \).
Whole surface area of the solid hemisphere = \( 3\pi r_1^2 \)
Curved surface area of the solid sphere = \( 4\pi r_2^2 \)
According to the condition:
\[\begin{array}{l} 3\pi r_1^2 = 4\pi r_2^2 \\ \Rightarrow \frac{r_1^2}{r_2^2} = \frac{4}{3} \\ \Rightarrow \left(\frac{r_1}{r_2}\right)^2 = \frac{4}{3} \\ \Rightarrow \frac{r_1}{r_2} = \frac{\sqrt{4}}{\sqrt{3}} \\ \Rightarrow \frac{r_1}{r_2} = \frac{2}{\sqrt{3}} \end{array}\]Therefore, the ratio of the lengths of radius of hemisphere and sphere is \( 2 : \sqrt{3} \).
Explanation:
Let the radius of the solid sphere be \( r \).
Curved surface area, \( S = 4\pi r^2 \)
Volume, \( V = \frac{4}{3}\pi r^3 \)
Now, evaluating the expression \( \frac{S^3}{V^2} \):
\[\begin{array}{l} \frac{S^3}{V^2} = \frac{(4\pi r^2)^3}{\left(\frac{4}{3}\pi r^3\right)^2} \\ = \frac{64\pi^3 r^6}{\frac{16}{9}\pi^2 r^6} \\ = 64\pi^3 \times \frac{9}{16\pi^2} \\ = 4\pi \times 9 \\ = 36\pi \end{array}\]Therefore, the value of \( \frac{S^3}{V^2} \) is \( 36\pi \).
Explanation:
Let the original radius of the sphere be \( r \).
Original curved surface area, \( A_1 = 4\pi r^2 \)
New radius \( = r + 50\% \text{ of } r = r + 0.5r = 1.5r = \frac{3}{2}r \)
New curved surface area, \( A_2 \):
\[\begin{array}{l} A_2 = 4\pi \left(\frac{3}{2}r\right)^2 \\ = 4\pi \left(\frac{9}{4}r^2\right) \\ = 9\pi r^2 \end{array}\]Increase in curved surface area \( = A_2 - A_1 = 9\pi r^2 - 4\pi r^2 = 5\pi r^2 \)
Percentage increase:
\[\begin{array}{l} = \frac{\text{Increase in Area}}{\text{Original Area}} \times 100\% \\ = \frac{5\pi r^2}{4\pi r^2} \times 100\% \\ = \frac{5}{4} \times 100\% \\ = 5 \times 25\% \\ = 125\% \end{array}\]Therefore, its curved surface area will be increased by 125%.
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