A solid metal cylinder of 12 cm height and 7 cm radius is melted and recast into two cones in the proportion of 1 : 2 (volume), keeping the height 12 cm. What would be the percentage change in the flat surface area before and after
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 50%.
Volume of the Cylinder:
Volume of a cylinder = πr²h
Given, r = 7 cm, h = 12 cm
So, Volume = π × 7² × 12 = 588π cm³
Volume of the Two Cones:
The cones are in the ratio 1:2.
So, their volumes are 1/3 × 588π/3 = 196π cm³ and 2/3 × 588π/3 = 392π cm³.
Radius of the Cones:
Volume of a cone = 1/3 × πr²h
For the smaller cone: 196π = 1/3 × πr² × 12
Solving for r, we get r₁ = 7 cm
For the larger cone: 392π = 1/3 × πr² × 12
Solving for r, we get r₂ = 7√2 cm
Flat Surface Area of the Cylinder:
The flat surface area of the cylinder is the sum of the areas of the two bases.
So, the initial flat surface area = 2 × π × 7² = 98π cm²
Flat Surface Area of the Two Cones:
The flat surface area of each cone is the area of its base.
So, the total flat surface area of the two cones = π × 7² + π × (7√2)² = 147π cm²
Percentage Change in Flat Surface Area:
Percentage change = [(Final area - Initial area) / Initial area] × 100
Percentage change = [(147π - 98π) / 98π] × 100 = 50%
Therefore, the percentage change in the flat surface area is 50%.