A solid cone is 25 cm. high and the radius of its base is 50cm. It is melted and recast into a solid sphere. Determine the surface area of the sphere. (Use = 22/7)
Answers
Given :-
♦ A solid cone is 25 cm. high and the radius of its base is 50cm.
♦ It is melted and recast into a solid sphere.
To find :-
♦ The Surface Area of the sphere.
Solution :-
Given that
♦ Radius of the base of a cone (r) = 50 cm
♦ Height of the cone (h) = 25 cm
We know that
♦ Volume of a cone = (1/3) πr²h cubic units
Volume of the cone = (22/7)×(50)²×25 cm³
=> V =(1/3)× (22/7)×50×50×25 cm³
=> V = (22×50×50×25)/(3×7) cm³
=> V = (22×50×50×25)/21 cm³
=> V = 1375000/21 cm³
♦Volume of the cone = 1375000/21 cm³
We know that
Let the radius of the sphere be r units
♦ Volume of a sphere = (4/3)πr³ cubic units
As we know,
" If a solid is melted and recast into another solid then the volume of the first solid is equal to the volume of the resultant solid " .
Given that
The Cone is melted and recast into a Sphere.
Volume of the sphere = Volume of the cone
=> (4/3)πr³ = 1375000/21
=> (4/3)×(22/7)×r³ = 1375000/21
=> (88/21)×r³ = 1375000/21
=> r³ = (1375000/31)×(21/88)
=> r³ = (1375000×21)/(21×88)
=> r³ = 1375000/88
=> r³ = 15625
=> r = 25 cm
♦ Radius of the Sphere = 25 cm
We know that
♦ Surface Area of a sphere is 4πr² sq.units
The Surface Area of the Sphere
=> TSA = 4×(22/7)×(25)² cm²
=> TSA = (4×22×25×25)/7 cm²
=> TSA = 55000/7 cm²
=> TSA = 7857.14 cm² (approximately)
Answer :-
♦ The Surface Area of the Sphere is 7857.14 cm²
Used Concept :-
→ If a solid is melted and recast into another solid then the volume of the first solid is equal to the volume of the resultant solid .
Used formulae:-
→ Volume of a cone = (1/3)πr²h cubic units
→ Surface Area of a sphere is 4πr² sq.units
→ Volume of a sphere = (4/3)πr³ cubic units
- r = radius
- h = height
- π = 22/7
