Question

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.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\huge\mathfrak{Bonjour\:dear!!!}$

☺☺

$\huge\mathfrak\blue{Question:}$

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.

$\huge\mathfrak\red{Solution:}$

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 , 4/3 × pi × (5)^3 m^3

Therefore, 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 3 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

Therefore, 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

= Rs. (2.8 × 4 × 22 × 36) / 7

= Rs. 1267.20

_____________________

$<marquee > HOPE IT HELPS !!!$

$<marquee > BE BRAINLY...$

❤❤❤