By melting two solid spheres of 1cm and 6cm a hollow spheres of 1cm thick is formed. Find the area of outer curve surface of new sphere.
Answers
Given :
By melting two solid spheres of 1cm and 6cm a hollow spheres of 1cm thick is formed.
To find:
Find the area of outer curve surface of new sphere.
Solution :
We need to remember some points before solving such problems or questionsRecasting, melting, reformed &
transformation, if these words are in any questions, then it means we have to find out volume of given dimension.
Volume of sphere → 4/3 Tr³
According to the given condition
Thickness of hollow sphere after melting = 1cm
Radius of the first sphere (x) = 1cm
Radius of the second sphere (y) = 6cm
Consider internal radius be xInternal radius (r) = x
* External radius (R)
→ internal radius + thickness = x + 1
* Volume of two sphere Volume of hollow
sphere
→ 4/3 π r³ + 4/3 πr³ = 4/3πR³ - 4/3πr³
→ 4/3TTX³ + 4/3TTу³ = 4/3π(R³ - r³)
4/3π(x³ + y³) = 4/3π(R³ - r³)
→ 4/3π(x³ + y³) = 4/3π{(x + 1)³ - x³}
Cancel 4/3TT & apply identity
(a + b)³ a³ + b³ + 3ab(a + b)
(6)³ + 1 = (x³ +1 + 3*x*1(x + 1)) - x³
→ 216 + 1 = (x³ + 1 + 3x(x + 1)} - x³
→216+1= x³ + 1 + 3x² + 3x - x³
→ 216 = 3x² + 3x + 1-1+x³ -x
→216 = 3x² + 3x
→ 3x² + 3x - 216 = 0
Take 3 as a common
→ 3(x² + x - 72) = 0
→x² + x - 72 = 0
Splitting middle term
→x² + 9x - 8x - 72 = 0
→x(x + 9) - 8(x + 9) = 0
→ (x + 9)(x - 8) = 0
•°• x = -9 or x = 8
* Length never be in negative
→ Take radius = 8cm = x= Internal radius
→ External radius = x + 1 = 9cm
Outer curved surface area of new sphere
→ 4TTR²
Put the value of external radius
→ 4TT(9)²
→ 4TT × 814 × 22/7 x 81
→ 88 × 81/7
→ 1018.28cm²
Hence, Outer curved surface area of new hollow sphere is 1018.28 cm²