Maths question from Naib Tehsildar exam, 2018 by JKSSB
A right circular cone is 27 cm high and the radius of its base is 2 cm. The cone is melted and recast into a sphere
The diameter of the sphere is
Last updated Jun 24, 2026
Correct Answer:
Option B —
6 cm
When one geometric shape is melted and recast into another, the volume remains the same. To find the diameter of the sphere, we must equate the volume of the cone to the volume of the sphere.
1. Formulas Needed
Volume of a Cone: V (cone) = 1\3pir^2h
Volume of a Sphere: $V_{sphere} = 4\3\pi R^3
2. Step-by-Step Calculation
Step 1: Calculate the volume of the cone Given:
Radius (r) = 2 cm, Height (h) = 27 cm V_{cone} = 1\3 times pi times (2)^2 times 27 V_{cone} = pi times 4 times 9
V _cone = 36 pi cm^3
Step 2: Equate to the volume of the sphere.
Let the radius of the sphere be R
36\pi = 4\3pi R^3
Step 3: Solve for R
Cancel pi from both sides:
36 = 4\3 R^3
Multiply by 3 and divide by 4
R^3 = 36 times 3\4}
R^3 = 9 times 3 R^3 = 27
R = sqrt[3]{27} = 3 cm
Step 4: Find the diameter
The diameter is twice the radius:
Diameter = 2 times R = 2 times 3 = 6 cm
Final Result:
The diameter of the sphere is 6 cm
Answer verified by Quintessence Classes faculty — Karan Nagar, Srinagar.
