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Exercise 13.8 - Chapter 13 Surface Areas & Volumes class 9 ncert solutions Maths - SaraNextGen [2024-2025]


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

Question 1:

Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3045/Chapter%2013_html_4110777b.gif

Answer:

(i) Radius of sphere = 7 cm

Volume of sphere = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3045/Chapter%2013_html_364858a3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3045/Chapter%2013_html_m573dad8a.gif

Therefore, the volume of the sphere is 1437https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3045/Chapter%2013_html_18a817bc.gif cm3.

(ii) Radius of sphere = 0.63 m

Volume of sphere = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3045/Chapter%2013_html_364858a3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3045/Chapter%2013_html_m58f178a9.gif

Therefore, the volume of the sphere is 1.05 m3 (approximately).

 

Question 2:

Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_4110777b.gif

Answer:

(i) Radius (r) of ball = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_m68f8168d.gif

Volume of ball = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_364858a3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_45e0ef1d.gif

Therefore, the volume of the sphere is https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_m6c2d21da.gifcm3.

(ii)Radius (r) of ball = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_m1b55a3d6.gif= 0.105 m

Volume of ball = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_364858a3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3047/Chapter%2013_html_39e6bcdc.gif

Therefore, the volume of the sphere is 0.004851 m3.

 

Question 3:

The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3050/Chapter%2013_html_4110777b.gif

Answer:

Radius (r) of metallic ball = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3050/Chapter%2013_html_m1c0d4079.gif

Volume of metallic ball = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3050/Chapter%2013_html_364858a3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3050/Chapter%2013_html_m6890667a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3050/Chapter%2013_html_m7a38cda6.gif

Mass = Density × Volume

= (8.9 × 38.808) g

= 345.3912 g

Hence, the mass of the ball is 345.39 g (approximately).

 

Question 4:

The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Answer:

Let the diameter of earth be d. Therefore, the radius of earth will be https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_m83168b7.gif.

Diameter of moon will be https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_m966fc41.gif and the radius of moon will be https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_m6c14caa6.gif.

Volume of moon = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_364858a3.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_762ac726.gif

Volume of earth = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_364858a3.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_3e6fad09.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_6f69902a.gif

Therefore, the volume of moon is https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3056/Chapter%2013_html_m7119943e.gifof the volume of earth.

Question 5:

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3057/Chapter%2013_html_4110777b.gif

Answer:

Radius (r) of hemispherical bowl = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3057/Chapter%2013_html_m1ff5e274.gif = 5.25 cm

Volume of hemispherical bowl = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3057/Chapter%2013_html_m27f0291.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3057/Chapter%2013_html_m48bda24f.gif

= 303.1875 cm3

Capacity of the bowl =https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3057/Chapter%2013_html_19b2ecd9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3057/Chapter%2013_html_3f2d1dcd.gif

Therefore, the volume of the hemispherical bowl is 0.303 litre.

 

Question 6:

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3060/Chapter%2013_html_4110777b.gif

Answer:

Inner radius (r1) of hemispherical tank = 1 m

Thickness of hemispherical tank = 1 cm = 0.01 m

Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3060/Chapter%2013_html_m3196ff15.gif

Question 7:

Find the volume of a sphere whose surface area is 154 cm2https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3063/Chapter%2013_html_4110777b.gif

Answer:

Let radius of sphere be r.

Surface area of sphere = 154 cm2

⇒ 4πr2 = 154 cm2

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3063/Chapter%2013_html_42c60133.gif

Volume of sphere = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3063/Chapter%2013_html_364858a3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3063/Chapter%2013_html_m6a9b264f.gif

Therefore, the volume of the sphere is https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3063/Chapter%2013_html_m63cba014.gif cm3.

 

Question 8:

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square meter, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome. https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3065/Chapter%2013_html_4110777b.gif

Answer:

(i) Cost of white-washing the dome from inside = Rs 498.96

Cost of white-washing 1 m2 area = Rs 2

Therefore, CSA of the inner side of dome = https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3065/Chapter%2013_html_2a193e66.gif

= 249.48 m2

(ii) Let the inner radius of the hemispherical dome be r.

CSA of inner side of dome = 249.48 m2

r2 = 249.48 m2

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3065/Chapter%2013_html_6e8eff8d.gif

⇒ r = 6.3 m

Volume of air inside the dome = Volume of hemispherical dome

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3065/Chapter%2013_html_m7127549f.gif

= 523.908 m3

= 523.9 m3 (approximately)

Therefore, the volume of air inside the dome is 523.9 m3.

Question 9:

Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the

(i) radius r‘ of the new sphere, (ii) ratio of S and S’.

Answer:

(i)Radius of 1 solid iron sphere = r

Volume of 1 solid iron sphere https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3068/Chapter%2013_html_1b4aae08.gif

Volume of 27 solid iron spheres https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3068/Chapter%2013_html_3f477544.gif

27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be r‘.

Volume of new solid iron sphere https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3068/Chapter%2013_html_m20b32e7f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3068/Chapter%2013_html_m62ef1779.gif

(ii) Surface area of 1 solid iron sphere of radius r = 4πr2

Surface area of iron sphere of radius r‘ = 4π (r‘)2

= 4 π (3r)2 = 36 πr2

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3068/Chapter%2013_html_71d679e4.gif

Question 10:

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3071/Chapter%2013_html_4110777b.gif

Answer:

Radius (r) of capsule https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3071/Chapter%2013_html_670f1b10.gif

Volume of spherical capsule https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3071/Chapter%2013_html_1b4aae08.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/9/7/110/3071/Chapter%2013_html_44a0576.gif

= 22.458 mm3

= 22.46 mm3 (approximately)

Therefore, the volume of the spherical capsule is 22.46 mm3.

Also Read : Exercise-13.9-Chapter-13-Surface-Areas-&-Volumes-class-9-ncert-solutions-Maths

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