Book Name |
: Ganit Prakash |

Subject |
: Mathematics |

Class |
: 10 (Madhyamik) |

Publisher |
: Prof Nabanita Chatterjee |

Chapter Name |
: Sphere (12th Chapter) |

**Application 1. **

**If the diameter of a ball is 42 cm, let us calculate how much leather is required for making the ball.**

**Solution:**

Diameter of the ball = 42 cm. \quad \therefore Radius of the ball = \frac{42}{2} cm. = 21 cm

\therefore Whole surface area of the ball = 4 ×(21)^{2} sq cm.

= 4 \pi × \frac{22}{7} × 21 \quad 21 sq cm

= 5544 sq cm,

\therefore The ball contains 5544 sq cm leather.

**Application 2. **

**The diameter of a sphere made of iron sheet is 14 cm; let us calculate late what is the cost of colouring the sphere at Rs. 2.50 per square cm. [Let me do myself]**

**Solution:**

Radius of the sphere (r) = \frac{14}{2} = 7 cm.

Area of the curved surface = 4 \pi r^{2}

= 4 × \frac{22}{7} × 7 × 7 sq cm. = 616 sq cm

Cost of colouring the sphere at Rs. 2.50 per sq cm.

= Rs. 2.50616 = Rs. 1540.

**Application 3. **

**The length of diameter of a solid spherical ball is 14 cm. Let us write by calculating how much stone is there in the spherical solid ball.**

**Solution:**

The length of radius of the solid spherical ball of stone = \frac{14}{2} cm = 7 cm

\therefore Solid spherical ball of stone contains stone of volume = \frac{4}{3} × \frac{22}{7} × 7^{3} cubic cm = 1437 \frac{1}{3} cu cm

**Application 4. **

**My spherical stone ball having the radius 0.7 dcm length is d pletely immersed into the water of the reservoir. Let us write by calculating much water will be overflowed from the reservoir. [Let me do it myself]**

**Solution:**

Radius of the spherical stone ball (r) = 0.7 dcm = \frac{7}{10} dcm.

Volume of the stone ball = \frac{4}{3} r^{3}

= \frac{4}{3} × \frac{22}{7} ×\left(\frac{7}{10}\right)^{3} \mathrm{cu} \mathrm{dcm} = \frac{4}{3} × \frac{22}{7} × \frac{7 × 7 × 7}{10 × 10 × 10} \mathrm{cu} \mathrm{dcm}

= 1.44 cu dcm.

**Application 6. **

**If the length of diameter of a hemispherical solid object is 14 cm, let ss write by calculating its whole surface area.**

**Solution:**

Whole surface area = 3 × \frac{22}{7} × 14 × 14 sq units = 1848 sq cm.

**Application 7. **

**If it requires 173.25 sq cm of sheet to make a hemispherical bowl them let us write by calculating the length of diameter of fore part of the bowl. [Let me do it myself]**

**[Answer hints : As the bowl is not a solid object, curved surface area of sheet will only be required.]**

**Solution:**

Let the diameter of the hemispherical bowl = z \pi \mathrm{rm}. \quad \therefore \text{ Radius } = \mathrm{r} \mathrm{cm}.

Area of on road surface = 2 \pi \mathrm{r}^{2}

\therefore 2 \pi r^{2} = 173.25 \quad \text { or, } 2 × \frac{22}{7} r^{2} = \frac{17325}{100} \\

\therefore r^{2} - \frac{17325}{100} × \frac{7}{22 × 2} = \frac{1575 × 7}{400} \\

\therefore r^{\circ} = \sqrt{\frac{1575 × 7}{400}} = \frac{105}{20} = 5.25

\therefore Radius = 5.25 cm & Diameter = 10.5 cm.

**Application 8. **

**If the length of radius of a solid hemispherical object is 14 cm, let us calculate the ratio of their volumes.**

**Solution:**

Stone contained in hemispherical paper weight is = \frac{2}{3} × \frac{22}{7} × 14 × 14 × 14 cubic units.

= 17248 cu cm.

**Application 10. **

**If the ratio of lengths of radii of two spheres is 1 : 2, let us write by calculating the ratio of their whole surface areas. [Let me do it myself]**

**Solution:**

Let the radius of the two spheres are r_{1} \text{ unit } \ 2 r_{1} unit respectively.

Total surface area of the 1st sphere = 4 \pi \mathrm{r}_{1}^{2}

Total surface area of the 2nd sphere = 4 \pi\left(2 r_{1}\right)^{2} = 4 \pi.4 r_{1}{ }^{2} = 16 \pi r_{1}{ }^{2}

\therefore Ratio of total surface area of the 1st sphere : total surface area of the 2nd sphere = 4 \pi r_{1}^{2}: 16 \pi r_{1}^{2} = 1: 4.

**Application 12. **

**If two spheres with the radii of 1 cm and 6 cm lengths are melted and a hollow sphere with the thickness of 1 cm is made, let us write by calculating the outer curved surface area of the hollow sphere.**

**Solution:**

Let the length of outer radius of the sphere is r cm.

\therefore The length of inner radius of that sphere is = (r – 1) cm.

By condition, \frac{4}{3} \pi r^{3} - \frac{4}{3} \pi(r - 1)^{3} = \frac{4}{3} \pi(1)^{3} + \frac{4}{3} \pi(6)^{3}

or, \frac{4}{3} \pi\left\{r^{3} - (r - 1)^{3}\right\} = \frac{4}{3} \pi(1 + 216)

or, r^{3} - r^{3} + 3 r^{2} – 3 r + 1 = 217

or, 3r^{2} – 3r – 216 = 0

or, r^{2} – r – 7^{2} = 0

or, r^{2} – 9r + 8r – 72 = 0 or, r(r – 9) + 8(r – 9) = 0

or, (r – 9)(r + 8) = 0

Either, r – 9 = 0 \quad \therefore r = 9

or, r + 8 = 0 \quad \therefore r = – 8

\therefore Since r \neq – 8, as r is the length of radius, i.e., it can not be negative. \therefore r = 9

\therefore The length of outer radius of the new hollow sphere is 9 cm.

\therefore Outer surface area = 4 × \frac{22}{7} × 9 × 9 sq cm = 1018 \frac{2}{7} sq cm

## LET US WORK OUT – 12

**Question 1**

**If the length of radius of a sphere is 10.5 cm, let us write by calculating the whole surface area of the sphere.**

**Solution:**

Radius = 10.5 cm

Total surface area = 4 \pi \mathrm{r}^{2} = 4 × \frac{22}{7} ×(10.5)^{2} sq cm

= 4 × \frac{22}{7} × 10.510 .5 sq cm = 1386 sq cm

**Question 2**

**If the cost of making a leather ball is Rs. 431.20 at Rs. 17.50 per square cm, let us write by calculating the length of diameter of the ball.**

**Solution:**

Surface area of the leather ball = 4 \pi r^{2}(r = radius)

According to the problem,

4 \pi r^{2} × 17.50 = 431.20

\therefore 4 \pi r^{2} = \frac{431.20}{17.50} = 24.64

or, 4 × \frac{22}{7} × r^{2} = 24.64

r^{2} = \frac{24.64 × 7}{2 × 22} = 1.96

\therefore \text{ r }= \sqrt{1.96} = 1.4

\therefore Diameter = 2 × 1.4 cm = 2.8 cm.

**Question 3**

**If the length of diameter of the ball used for playing shotput in our school is 7 cm, let us write by calculating how many cubic cm of iron is there in the ball.**

**Solution:**

Radius of the ball / r = \frac{7}{2} cm (as Diameter = 7 cm)

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

= \frac{4}{3} × \frac{22}{7} × \frac{7}{2} × \frac{7}{2} × \frac{7}{2} = \frac{539}{2} cu cm

= 179\frac{2}{3}cu cm

\therefore \text{ 179 } \frac{2}{3} cu cm iron is in the ball.

**Question 4**

**If the length of diameter of a solid sphere is 28 cm and it is completely immersed into the water, let us calculate the volume of water displaced by the sphere.**

**Solution:**

Diameter = 28 cm.

Radius (r) = \frac{28}{2} = 14 cm

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

= \frac{4}{3} × \frac{22}{7} × 14 × 14 × 14 cu cm

= 11498.67 cu cm.

\therefore Volume of displaced water = 11498.67 cu cm.

**Question 5**

**The length of radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it; let us find the ratio of surface areas of the balloon in two cases.**

**Solution:**

1st case radius (r) = 7 cm,

2nd case radius r_{2} = 21 cm.

\therefore Area of total surface of the 1st ball = 4 \pi \mathrm{r}_{1}{ }^{2} = 4 \pi(7)^{2} sq cm.

Area of total surface of the 2nd ball = 4 \pi \mathrm{r}_{2}^{2} = 4 \pi(21)^{2} sq cm.

\therefore Ratio of their total surface areas

= 4 \pi(7)^{2}: 4 \pi(21 .)^{2}

= 49 : 441 = 1 : 9.

**Question 6**

**127 \frac{2}{7} sq cm of sheet is required to make a hemispherical bowl. Let us write by calculating the length of diameter of the forepart of the bowl.**

**Solution:**

Surface area of the hemispherical bowl = 127 \frac{2}{7} sq cm = \frac{891}{7} sq cm.

If the radius = r cm.

\therefore 2 \pi r^{2} = \frac{891}{7}

2 × \frac{22}{7} r^{2} = \frac{891}{7}

\therefore r^{2} = \frac{891}{7} × \frac{7}{22 × 2} = \frac{81}{4} \quad \therefore \text{ r } = \frac{9}{2} cm.

\therefore Diameter = 2 × \frac{9}{2} = 9 cm.

**Question 7**

**The length of radius of a solid spherical ball is 2.1 cm; let us write by calculating how much cubic cm iron is there and let us find the curved surface area of the iron ball.**

**Solution:**

Radius of the spherical ball = 2.1 cm.

Volume of the sphere = \frac{4}{3} \pi ×(2.1)^{3} cu cm.

= \frac{4}{3} × \frac{22}{7} \quad 2.1 × 2.1 × 2.1 cu cm . = 38.81 cu cm.

Total surface area = 4 \pi r^{2} = 4 × \frac{22}{7} × (2.1)^{2} sq cm.

= 4 × \frac{22}{7} × 2.1 × 2.1 sq cm. = 55.44 sq cm.

**Question 8**

**The length of diameter of solid sphere of lead is 14 cm. If the sphere is melted, le us write by calculating how many spheres with length of 3.5 cm radius can be made.**

**Solution:**

Diameter of the solid sphere = 14 cm.

\therefore Radius (r) of the solid sphere = \frac{14}{2} \pi = 7 cm.

Volume of lead in the solid sphere = \frac{4}{3} \pi(7)^{3} cu cm.

Radius of the small sphere = 3.5 cm :

\therefore Volume of small sphere = \frac{4}{3} \pi(3.5)^{3} cu cm.

\therefore No. of small sphere = \frac{\frac{4}{3} \pi × 7 × 7 × 7}{\frac{4}{3} \pi × 3.5 × 3.5 × 3.5} = 2 × 2 × 2 = 8

**Question 9**

**Three spheres made of copper having the lengths of 3 cm, 4 cm and 5 cm radii are melted and a large sphere is made. Let us write by calculating the length of radius of the large sphere.**

**Solution:**

Volume of 1st sphere with radius 3 cm = \frac{4}{3} \pi(3)^{3} cu cm.

Volume of 2nd sphere with radius 4 cm = \frac{4}{3} \pi(4)^{3} cu cm

& Volume of 3rd sphere with radius 5 cm = \frac{4}{3} \pi(5)^{3} cu cm

Total volume of 3 small spheres = \frac{4}{3} \pi(3)^{3}+\frac{4}{3} \pi(4)^{3}+\frac{4}{3} \pi(5)^{3}

= \frac{4}{3} \pi\left(3^{3}+4^{3}+5^{3}\right) cu cm

= \frac{4}{3} \pi(27+64+125) cu cm

= \frac{4}{3} \pi × 216 cu cm

Now, let the radius of the new big sphere = R cm.

\therefore \text{ Its volume } = \frac{4}{3} \pi \mathrm{R}^{3}

According to the problem,

\frac{4}{3} \pi \mathrm{R}^{3} = \frac{4}{3} \pi × 216 \quad \therefore \mathrm{R}^{3} = 216

\therefore \text{ R } = \sqrt[3]{216} = 6 cm.

**Question 10**

**The length of diameter of base of a hemispherical tomb is 42 dm. Let us write by calculating the cost of colouring the upper surface of the tomb at the rate of Rs. 35 per square metre.**

**Solution:**

Diameter of base of a hemispherical tomb = 42 dm.

\therefore Radius of base (r) = \frac{42}{2} = 21 dm

Surface area = 2 \pi \mathrm{r}^{2} = 2 × \frac{22}{7} ×(21)^{2} sq dm

= 2 × \frac{22}{7} × 21 × 21 sq dm. = 2772 sq dm. = 27.72 sqm.

Total cost for colouring the upper surface of the tomb = Rs. 35 × 27.72 = Rs. 970.20.

**Question 11**

**Two hollow spheres with lengths of diameter 21 cm and 17.5 cm respectively are made from the sheets of the same metal. Let us calculate the Ratio of the area of sheets of metal required to make the two spheres.**

**Solution:**

Radius of the 1st sphere = \frac{21}{2} cm and radius of 2nd sphere = \frac{17.5}{2} cm.

Total surface area of the 2nd sphere = 4 \pi\left(\frac{21}{2}\right)^{2} sq cm.

Total surface area of the 2nd sphere = 4 \pi\left(\frac{17.5}{2}\right)^{2} sq cm.

\therefore Ratio of metal sheet to make the two spheres

= 4 \pi\left(\frac{21}{2}\right)^{2}: 4 \pi\left(\frac{17.5}{2}\right)^{2} \\

= \frac{21 × 21}{2 × 2}: \frac{17.5 × 17.5}{2 × 2} \\

= \frac{441}{4}: \frac{5359.375}{4} = \frac{441}{4} \frac{4}{306.25} = 36: 25

**Question 12**

**The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere is half of the previous one. Let us calculate the ratio of the volumes of the portion cut off and the remaining portion of the sphere.**

**Solution:**

Let the radius of old sphere = R cm and the radius of new sphere = r cm.

Surface area of the old sphere = 4 \pi \mathrm{R}^{2} sq cm.

Surface area of the new sphere = 4 \pi r^{2}

According to the question, \frac{1}{2} 4 \pi \mathrm{R}^{2} = 4 \pi \mathrm{r}^{2}

\therefore R = 2 r^{2} \quad\therefore R = \sqrt{2} r.

Volume of the old sphere = \frac{4}{3} \pi \mathrm{R}^{3} = \frac{4}{3} \pi(\sqrt{2} r)^{3} cu cm.

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

\therefore Volume of the remaining sphere = \frac{4}{3} \pi(\sqrt{2} r)^{3}-\frac{4}{3} \pi r^{3} = \frac{4}{3} \pi r^{3}(2 \sqrt{2}-1)

\therefore Ratio of the volumes of the portion cut off and the remaining portions of sphere

= \frac{4}{3} \pi r^{3}(2 \sqrt{2}-1): \frac{4}{3} \pi r^{3} = (\sqrt{2}-1) : 1

**Question 13**

**On the curved surface of the axis of a globe with the length of 14 cm radius, two circular holes are made, each of which has the length of radius 0.7 cm. Let us calcu. late the area of metal sheet surrounding its curved surface.**

**Solution:**

Radius of the globe = 14 cm

Surface area of the globe = 4π × 14 × 14 = 4 × \frac{22}{7} × 14 × 14 = 2464 sq cm.

Radius of each hole = 0.7 cm.

\therefore Area of two holes = 2 × \pi(0.7)^{2} = 2 × \frac{22}{7} × \frac{7}{10} × \frac{7}{10} = \frac{308}{100} = 3.08 sq cm.

\therefore Area of the metal sheet surrounding its curved surface = (2464-3.08) sq cm = 2460.92 sq cm.

**Question 14**

**Let us write by calculating how many marbles with lengths of 1 cm radius may be formed by melting solid sphere of iron having 8 cm length of radius.**

**Solution:**

Radius of the solid sphere = 8 cm.

Volume of the solid sphere = \frac{4}{3} (8) 3 cu cm.

Radius of the small sphere = 1 cm.

Volume of each small sphere = \frac{4}{3}(1) 3 cu cm.

\therefore No. of small solid spheres may be formed from the big solid sphere

= \frac{\frac{4}{3} \pi(8)^{3}}{\frac{4}{3} \pi(1)^{3}} = \frac{8 × 8 × 8}{1 × 1 × 1} = 512.

## Very short answer type question (V.S.A.)

** M.C.A. :**

**Question 1**

**The volume of a solid sphere having the radius of 2r units length is**

**\frac{32 nr^{3}}{3} cubic unit****\frac{16 nr^{3}}{3} cubic unit****\frac{8 nr^{3}}{3} cubic unit****\frac{64 nr^{3}}{3} cubic unit**

**Solution:**

Volume of solid sphere having the, radius of 2r. unit

= \frac{4}{3} π(2r)^{3} = \frac{4 π × 2r × 2r × 2r}{3} = \frac{32 πr^{3}}{3} cu units

**Ans.** (a) \frac{32 nr^{3}}{3} cubic unit

**Question 2**

**If the ratio of the volumes of two solid spheres is 1 : 8, the ratio of their curved surface areas is**

**1 : 2****1 : 4****1 : 8****1 : 16**

**Solution:**

If the ratio of the volumes of two solid sphere = 1 : 8

Ratio of their curved surface area = 1 : 4

**Ans.** (b) 1 : 4

**Question 3**

**The whole surface area of a solid hemisphere with length 7 cm radius is**

**588 π \mathrm{sq} \mathrm{cm}.****392 π \mathrm{sq} \mathrm{cm}.****147 π sq \mathrm{cm}.****98 π \mathrm{sq} \mathrm{cm}.**

**Solution:**

Whole surface area of a solid hemisphere of radius 7 cm = 3π (7)^{2} sq cm = 147π sq cm.

**Ans.** (c) 147π sq cm.

**Question 4**

**If the ratio of curved surface areas of two solid spheres is 16 : 9, the ratio of their volumes is**

**64 : 27****4 : 3****27 : 64****3 : 4**

**Solution:**

If the ratio of curved surface areas of two solid sphere is 16 : 9.

\thereforeratio of their volumes = 64: 27

**Ans.** (a) 64 : 27

**Question 5**

**If numerical value of curved surface area of a solid sphere is three times of its volume, the length of radius is**

**1 unit****2 unit****3 unit****4 unit**

**Ans. **(a) 1 unit

** Let us write whether the following statements are true or false :**

**Question 1**

**If we double the length of radius of a solid sphere, the volume of sphere will be doubled.**

**Solution:**

FALSE

**Question 2**

**If the ratio of curved surface areas of two hemispheres is 4 : 9, the ratio of their lengths of radii is 2 : 3.**

**Solution:**

TRUE

** Let us fill in the blanks :**

**Question 1**

**The name of solid which is composed of only one surface is __________________.**

**Solution:**

Sphere.

**Question 2**

**The number of curved surfaces of a solid hemisphere is __________________________.**

**Solution:**

One.

**Question 3**

**Solid hemisphere is 2r units, its whole surface area is 2r units, its whole surface area is _____________________ r ^{2} sq units.**

**Solution:**

12π r^{2}.

**Short answer type question (S.A.) :**

**Question 1**

**The numerical values of volume and whole surface area of a solid hemisphere are equal; let us write the length of radius of the hemisphere.**

**Solution:**

Let the radius of the hemisphere = r unit.

Volume of the hemisphere = \frac{2}{3} πr^{3} cu unit and total surface area = 3 πr^{2} sq unit.

According to the problem,

\frac{2}{3} πr^{3} = 3 πr^{2}

or, 2r = 9 \quad \therefore \text{ r }= \frac{9}{2} = 4.5 unit.

\therefore radius of the hemisphere = 4.5 unit. \thereforeradius of the hemisphere = 4.5 unit.

**Question 2**

**The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of the cylinder are 12 cm. Let us write the length of radius of the sphere.**

**Solution:**

Let the radius of solid sphere = r cm.

Surface area of solid sphere = 4πr^{2} sq cm^{2}

Again, height of the cylinder = 12cm.

Radius of the cylinder = 6cm.

Curved surface area of the cylinder = 2 π rh = 2 π × 6 × 12 sq cm = 144 π sq cm.

According to the problem,

4 πr^{2} = 144 π \quad \therefore { r }^{2} = 36, \quad or, r = 6

\therefore radius of the sphere = 6 cm.

**Question 3**

**Whole surface area of a solid hemisphere is equal to the curved surface area of the solid sphere. Let us write the ratio of lengths of radii of the hemisphere and the sphere.**

**Solution:**

Let the radius of the solid hemisphere = r_{1} unit

& the radius of the sphere = r_{2} unit

Total surface area of the hemisphere = 3 πr_{1}^{2} sq unit

& surface area of the sphere = 4 πr_{2}{ }^{2} sq unit

According to the question,

3 πr_{1}^{2} = 4 πr_{2}^{2}

or, \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{4}{3} \text{ or }, \frac{r_{1}}{r_{2}} = \frac{\sqrt{4}}{\sqrt{3}} = \frac{2}{\sqrt{3}}

**Question 4**

**If curved surface area of a solid sphere is S and volume is V, let us write the value of \frac{\mathrm{S}^{3}}{\mathrm{~V}^{3}}. [not putting the value of π ]**

**Solution:**

Let the radius of the sphere = r units

Surface area of sphere = S = 4 πr^{2}

Volume of sphere = V = \frac{4}{3} πr^{3}

\frac{S^{3}}{V^{2}} = \frac{\left(4r^{2}\right)^{3}}{\left(\frac{4}{3} πr^{3}\right)^{2}} = \frac{64 π^{3}r^{6}}{\frac{16}{9} π^{2}r^{6}} = \frac{64 × 9 π^{3}}{16 π^{2}} = 36 π

**Question 5**

**If the length of radius of a sphere is increased by 50%, let us write how much percent will be increased of its curved surface area.**

**Solution:**

Let the radius of the sphere = r unit

\therefore Surface area = 4 πr^{2} sq. unit

If the radius of the sphere is increased by 50%, new radius will be

r +\frac{50}{100}r = r + \frac{r}{2} = \frac{3r}{2} unit.

\therefore Surface area will be = 4 π\left(\frac{3r}{2}\right)^{2} = 9 πr^{2}

Area increased = 9 πr^{2}-4 πr^{2} = 5r^{2}

Increased percentage = \frac{5 πr^{2}}{4 πr^{2}} × 100% = 125%