Surface Areas and Volumes - Spheres

(NCERT Ex 13.4)

**Q1: Find the surface area of a sphere of radius**

**(i) 10.5 cm**

**(ii) 5.6 cm**

**(iii) 14 cm**

Answer:

(i) r = 10.5 cm

Surface area of the sphere = 4πr²

= 4 × (22/7) × (10.5)²

=

**1386 cm² (Answer)**

(ii) r = 5.6 cm

Surface area of the sphere = 4πr²

= 4 × (22/7) × (5.6)²

=

**394.24 cm² (Answer)**

(iii) r = 14 cm

Surface area of the sphere = 4πr²

= 4 × (22/7) × 14 × 14

=

**2464 cm² (Answer)**

**Q2: Find the surface area of sphere of a diameter :**

**(i) 14 cm**

**(ii) 21 cm**

**(iii) 3.5 m**

Answer:

(i) r = 14/2 = 7cm

Surface area of the sphere = 4πr²

= 4 × (22/7) × 7²

=

**616 cm²**

(ii) r = 21/2 cm

Surface Area = 4πr²

= 4 × (22/7) × (21/2)²

= 22 × 22 × 3 × 3

=

**1386 cm²**

(iii) r = 3.5/2m

Surface Area = 4πr²

= 4 × (22/7) × (35/20)²

=

**38.5 m²**

**Q3: Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)**

Answer: Given, r = 10 cm

Total surface area of the hemisphere = 3πr²

= 3 × 3.14 × (10)² cm²

= 3 × 3.14 × 100 cm² =

**942 cm² (Answer)**

**Q4: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.**

Answer: For r = 7 cm

Surface Area = 4πr²

= 4 × π × 7 × 7 cm²

For R = 14 cm

Surface Area = 4πr²

= 4 × π × 14 × 14 cm²

∴ Required Ratio is

= | 4×π×7×7 |

4 × π × 14 × 14 |

= 1/4 =

**1: 4 (Answer)**

**Q5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm².**

Answer:

Give r = 10.5 /2 = 5.25 cm

Inner surface area of the bowl = 2πr²

= 2 × (22/7) × (5.25)²

= 44 × 0.75 × 5.25

= 173.25 cm²

Cost of tin plating 100 cm² = ₹ 16

Cost of tin plating 173.25 = (16/100) × 173.25

=

**₹ 27.72 (Answer)**

**Q6: Find the radius of a sphere whose surface area is 154 cm².**

Answer: Surface area of the sphere = 4πr²

⇒ 154 = 4 × (22/7) × r²

⇒ r² = (154 × 7) / (4 × 22) = 7 × 7 / 4

⇒ r = 7/2 =

**3.5 cm (Answer)**

**Q7: The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.**

Answer: Let diameter of the earth = 2r

Then radius of the earth = r

∴ Diameter of the moon = 2r/4 = r/2

∴ Radius of the moon = r/4

surface area of the moon = 4π(r/4)²

= πr²/4

Surface area of the earth = 4πr²

∴ Required ratio = (πr²/4) ÷ (4πr²)

= 1/16

=

**1 : 16**

**Q8: A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.**

Answer: Inner radius of the bowl (r) = 5 cm

Thickness of the steel = 0.25 cm

∴ Outer radius of the bowl (R) = (5 + 0.25) cm = 5.25 cm

Outer curved surface area of the bowl = 2πR²

= 2 × (22/7) × (5.25)²

=

**173.25 cm² (Answer)**

**Q9: A right circular cylinder just encloses a sphere of radius r (see Figure). Find**

**(i) surface area of the sphere,**

**(ii) curved surface area of the cylinder,**

**(iii) ratio of the areas obtained in (i) and (ii).**

Answer: Given, radius of the sphere = r

Radius of the cylinder = r

And, height of the cylinder = 2r

(i) Surface area of the sphere =

**4πr² (Answer)**

(ii) Curved surface area of the cylinder = 2πrh

2π × r × 2r =

**4πr² (Answer)**

(iii) Required Ratio = 4πr² / 4πr² = 1/1

=

**1:1 (Answer)**

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