A grinding mill manufactures spherical steel balls of various radii, ie. 5mm, 7mm, 10mm, 14mm, 16mm and so on. One day the mill manufactured 10 balls of 7mm radius and 20 balls of 3.5mm radius. The mill was having 18,000 mm3 steel. This 18,000 mm3 steel was melted and 10 balls of 7mm radius and 20
balls of 3.5 mm radius were made and the remaining steel was stored for future use.
Answer the following questions :
i) Find the volume of 10 balls each of radius 7 mm.
ii) Find the volume of 20 balls each of radius 3.5 mm.
iii) How much steel was kept for future use?
OR
Find the surface area of one ball of radius 7 mm.
Detailed Solution :
(i) Volume of 10 balls each of radius 7 mm
= 10×Volume of each ball of radius 7 mm
= 10×(4/3)πr³
= 10×(4/3)(22/7)(7)³
= 10×(4/3)22×49
= (40×22×49)/3
= 43120/3
= 14373.33 mm³ (approx)
(ii) Volume of 20 balls each of radius 3.5 mm
r=3.5 mm
r= 3.5/10 = 7/2 mm
Required Volume = 20×Volume of each ball of radius 7/2 mm
= 20×(4/3)πr³
= 20×(4/3)(22/7)(7/2)³
= 10×(1/3)(22)(49)
= (10×22×49)/3
= 10780/3
= 3593.33 mm³ (approx)
= 10×Volume of each ball of radius 7 mm
= 10×(4/3)πr³
= 10×(4/3)(22/7)(7)³
= 10×(4/3)22×49
= (40×22×49)/3
= 43120/3
= 14373.33 mm³ (approx)
(ii) Volume of 20 balls each of radius 3.5 mm
r=3.5 mm
r= 3.5/10 = 7/2 mm
Required Volume = 20×Volume of each ball of radius 7/2 mm
= 20×(4/3)πr³
= 20×(4/3)(22/7)(7/2)³
= 10×(1/3)(22)(49)
= (10×22×49)/3
= 10780/3
= 3593.33 mm³ (approx)
(iii) Total amount of steel = 18000 mm³
Amount of steel used = 14373.33 + 3593.33
Amount of steel used = 14373.33 + 3593.33
= 17966.66 mm³
Amount of steel kept for future use
= 18000-17966.66
= 33.34 mm³ (approx)
OR
Surface area of one ball of radius 7 mm
= 4πr²
= 4(22/7)(7)²
= 4×22×7
= 28×22
= 616 mm²