Three solid spheres of brass whose radii are 3 m, 4 m and 5 m respectively are melted into a single solid sphere. Find the cost of gold polishing on this bigger sphere at the rate of Rs 2.80 per square metre.
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Volume of three spheres having radii 3 m, 4 m and 5 m respectively are 4/3 × pi × (3)^3 m^3 , 4/3 × pi × (4)^3 m^3 and 4/3 × pi × (5)^3 m^3
Volume of the new sphere = (4/3 × pi × (3)^3 + 4/3 × pi × (4)^3 + 4/3 × pi × (5)^3) m^3
= 4/3 × pi × (27 + 64 + 125) m^3 = 4/3 × pi × 216 m^3 ...... (i)
Now, let the radius of the new bigger sphere be R m
So, the volume of the new bigger sphere = 4/3 × pi × R^3 m^3
Since the bigger sphere is cast by melting above three solid spheres,
Therefore, volume of the bigger solid sphere = volume of 3 smaller solid spheres
Therefore, 4/3 × pi × R^3 = 4/3 × pi × 216 [by (i)]
Therefore, R = cube root 216 = cube root 6 × 6 × 6 = 6 m
Curved surface area of the bigger sphere = 4 × pi × R^2
= 4 × 22/7 × (6)^2 m^2
Cost of gold polishing = Rs 2.80 / m^2
Therefore, cost of gold polishing on 4 × 22 × 36 / 7 m^2 = Rs. 2.8 × 4 × 22 × 36 / 7
= Rs. 1267.20
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Volume of three spheres having radii 3 m, 4 m and 5 m respectively are 4/3 × pi × (3)^3 m^3 , 4/3 × pi × (4)^3 m^3 and 4/3 × pi × (5)^3 m^3
Volume of the new sphere = (4/3 × pi × (3)^3 + 4/3 × pi × (4)^3 + 4/3 × pi × (5)^3) m^3
= 4/3 × pi × (27 + 64 + 125) m^3 = 4/3 × pi × 216 m^3 ...... (i)
Now, let the radius of the new bigger sphere be R m
So, the volume of the new bigger sphere = 4/3 × pi × R^3 m^3
Since the bigger sphere is cast by melting above three solid spheres,
Therefore, volume of the bigger solid sphere = volume of 3 smaller solid spheres
Therefore, 4/3 × pi × R^3 = 4/3 × pi × 216 [by (i)]
Therefore, R = cube root 216 = cube root 6 × 6 × 6 = 6 m
Curved surface area of the bigger sphere = 4 × pi × R^2
= 4 × 22/7 × (6)^2 m^2
Cost of gold polishing = Rs 2.80 / m^2
Therefore, cost of gold polishing on 4 × 22 × 36 / 7 m^2 = Rs. 2.8 × 4 × 22 × 36 / 7
= Rs. 1267.20
_____________________
HOPE IT HELPS ❤❤
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