Chapter 13 Surface Areas and Volumes Ex 13.8 Solutions
Question - 1 : - Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
(Assume ╧А =22/7)
Answer - 1 : -
(i) Radius of sphere,r = 7 cm
Using, Volume ofsphere = (4/3) ╧Аr3
= (4/3)├Ч(22/7)├Ч73
= 4312/3
Hence, volume of thesphere is 4312/3 cm3
(ii) Radius of sphere,r = 0.63 m
Using, volume ofsphere = (4/3) ╧Аr3
= (4/3)├Ч(22/7)├Ч0.633
= 1.0478
Hence, volume of thesphere is 1.05 m3┬а(approx).
Question - 2 : - Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
(Assume ╧А =22/7)
Answer - 2 : -
(i) Diameter = 28 cm
Radius, r = 28/2 cm =14cm
Volume of the solidspherical ball = (4/3) ╧Аr3
Volume of the ball =(4/3)├Ч(22/7)├Ч143┬а= 34496/3
Hence, volume of theball is 34496/3 cm3
(ii) Diameter = 0.21 m
Radius of the ball=0.21/2 m= 0.105 m
Volume of the ball =(4/3 )╧Аr3
Volume of the ball =(4/3)├Ч (22/7)├Ч0.1053┬аm3
Hence, volume of theball = 0.004851 m3
Question - 3 : - The diameter of a metallic ball is 4.2cm. What is the mass of the ball,if the density of the metal is 8.9 g per cm3? (Assume ╧А=22/7)
Answer - 3 : -
Given,
Diameter of a metallicball = 4.2 cm
Radius(r) of themetallic ball, r = 4.2/2 cm = 2.1 cm
Volume formula = 4/3╧Аr3
Volume of the metallicball = (4/3)├Ч(22/7)├Ч2.1 cm3
Volume of the metallicball = 38.808 cm3
Now, usingrelationship between, density, mass and volume,
Density = Mass/Volume
Mass = Density ├Чvolume
= (8.9├Ч38.808) g
= 345.3912 g
Mass of the ball is 345.39g (approx).
Question - 4 : - The diameter of the moon is approximately one-fourth of the diameter ofthe earth. What fraction of the volume of the earth is the volume of the moon?
Answer - 4 : -
Let the diameter ofearth be тАЬdтАЭ. Therefore, the radius of earth will be will be d/2
Diameter of moon willbe d/4 and the radius of moon will be d/8
Find the volume of themoon :
Volume of the moon =(4/3) ╧Аr3┬а= (4/3) ╧А (d/8)3┬а= 4/3╧А(d3/512)
Find the volume of theearth :
Volume of the earth =(4/3) ╧Аr3= (4/3) ╧А (d/2)3┬а= 4/3╧А(d3/8)
Fraction of the volumeof the earth is the volume of the moon
Answer: Volume of moonis of the 1/64 volume of earth.
Question - 5 : - How many litres of milk can a hemispherical bowl of diameter 10.5cm hold?(Assume ╧А = 22/7)
Answer - 5 : -
Diameter ofhemispherical bowl = 10.5 cm
Radius ofhemispherical bowl, r = 10.5/2 cm = 5.25 cm
Formula for volume ofthe hemispherical bowl = (2/3) ╧Аr3
Volume of thehemispherical bowl = (2/3)├Ч(22/7)├Ч5.253┬а= 303.1875
Volume of thehemispherical bowl is 303.1875 cm3
Capacity of the bowl =(303.1875)/1000 L = 0.303 litres(approx.)
Therefore,hemispherical bowl can hold 0.303 litres of milk.
Question - 6 : - A hemi spherical tank is made up of an ironsheet 1cm thick. If the inner radius is 1 m, then find the volume of the ironused to make the tank. (Assume ╧А = 22/7)
Answer - 6 : -
Inner Radius of thetank, (r ) = 1m
Outer Radius (R ) =1.01m
Volume of the ironused in the tank = (2/3) ╧А(R3тАУ r3)
Put values,
Volume of the ironused in the hemispherical tank = (2/3)├Ч(22/7)├Ч(1.013тАУ 13)= 0.06348
So, volume of the ironused in the hemispherical tank is 0.06348 m3.
Question - 7 : - Find the volume of a sphere whose surface area is 154 cm2.(Assume ╧А = 22/7)
Answer - 7 : -
Let r be the radius ofa sphere.
Surface area of sphere= 4╧Аr2
4╧Аr2┬а=154 cm2┬а(given)
r2┬а=(154├Ч7)/(4 ├Ч22)
r = 7/2
Radius is 7/2 cm
Now,
Volume of the sphere =(4/3) ╧Аr3
Question - 8 : - A dome of a building is in the form of a hemi sphere. From inside, it waswhite-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20per square meter, find the
(i) inside surface area of the dome (ii) volume of the air inside thedome
(Assume ╧А = 22/7)
Answer - 8 : -
(i) Cost ofwhite-washing the dome from inside = Rs 4989.60
Cost of white-washing1m2┬аarea = Rs 20
CSA of the inner sideof dome = 498.96/2 m2 ┬а= 249.48 m2
(ii) Let the innerradius of the hemispherical dome be r.
CSA of inner side ofdome = 249.48 m2┬а(from (i))
Formula to find CSA ofa hemi sphere = 2╧Аr2
2╧Аr2┬а=249.48
2├Ч(22/7)├Чr2┬а=249.48
r2┬а=(249.48├Ч7)/(2├Ч22)
r2┬а=39.69
r = 6.3
So, radius is 6.3 m
Volume of air insidethe dome = Volume of hemispherical dome
Using formula, volumeof the hemisphere = 2/3 ╧Аr3
=(2/3)├Ч(22/7)├Ч6.3├Ч6.3├Ч6.3
= 523.908
= 523.9(approx.)
Volume of airinside the dome is 523.9 m3.
Question - 9 : - Twenty-seven solid iron spheres, each of radius r and surface area S aremelted to form a sphere with surface area SтАЩ. Find the
(i) radius rтАЩ of the new sphere,
(ii) ratio of Sand SтАЩ.
Answer - 9 : -
Volume of the solidsphere = (4/3)╧Аr3
Volume of twenty sevensolid sphere = 27├Ч(4/3)╧Аr3┬а= 36 ╧А r3
(i) New solid ironsphere radius = rтАЩ
Volume of this newsphere = (4/3)╧А(rтАЩ)3
(4/3)╧А(rтАЩ)3┬а=36 ╧А r3
(rтАЩ)3┬а=27r3
rтАЩ= 3r
Radius of new spherewill be 3r (thrice the radius of original sphere)
(ii) Surface area ofiron sphere of radius r, S =4╧Аr2
Surface area of ironsphere of radius rтАЩ= 4╧А (rтАЩ)2
Now
S/SтАЩ = (4╧Аr2)/(4╧А (rтАЩ)2)
S/SтАЩ = r2/(3rтАЩ)2┬а=1/9
The ratio of S and SтАЩ is 1:9.
Question - 10 : - A capsule of medicine is in the shape of a sphere of diameter 3.5mm. Howmuch medicine (in mm3) is needed to fill this capsule? (Assume ╧А =22/7)
Answer - 10 : -
Diameter of capsule =3.5 mm
Radius of capsule, sayr = diameter/ 2 = (3.5/2) mm = 1.75mm
Volume of sphericalcapsule = 4/3 ╧Аr3
Volume of sphericalcapsule = (4/3)├Ч(22/7)├Ч(1.75)3┬а= 22.458
Answer: The volume ofthe spherical capsule is 22.46 mm3.