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JKSSB · Written · 2018
Question from JKSSB Naib Tehsildar 2018 Written
Maths Jammu & Kashmir Services Selection Board Last updated May 9, 2026
JKSSB Written 2018

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

Answer & Explanation

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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.

About this question

JKSSB Naib Tehsildar 2018 Written

Details

Exam JKSSB
Recruitment Naib Tehsildar
Stage Written
Year 2018
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