Wednesday, January 25, 2017

CBSE Class 9 - Maths - Surface Areas and Volumes (NCERT Ex 13.4) (#cbseclass9notes)

Surface Areas and Volumes - Spheres
(NCERT Ex 13.4)

CBSE Class 9 - Maths - Surface Areas and Volumes (NCERT Ex 13.4) (#cbseclass9notes)


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).
CBSE Class 9 - Maths - Surface Areas and Volumes (NCERT Ex 13.4) (#cbseclass9notes)

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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