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MathematicsMensuration
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A solid metallic sphere of radius 6 cm is melted and recast into 27 identical smaller spheres. What is the radius of each new smaller sphere?

A

1 cm

B

2 cm

C

3 cm

D

4 cm

Correct Answer

ЁЯУР MA тАв Math Concept & PrincipleMathematicsMensuration
Option B

2 cm

Quick Summary:

When a solid object is melted and recast into multiple identical smaller objects, the total volume remains constant due to the conservation of mass. By equating the volume of the large sphere to the sum of the volumes of the 27 smaller spheres, we can solve for the new radius.

ЁЯФмSCScience SolutionConcept & Principle
ЁЯТб Explanation

When a solid object is melted and recast into multiple identical smaller objects, the total volume remains constant due to the conservation of mass. By equating the volume of the large sphere to the sum of the volumes of the 27 smaller spheres, we can solve for the new radius.

ЁЯТб Everyday Analogy (Real-World Intuition)

Imagine melting a large ball of clay into 27 equal-sized tiny marbles; the total amount of clay stays exactly the same, just divided into smaller pieces.

ЁЯза Memory Mnemonic / Shortcut Aid

Volume stays the same when you melt and play a game!

ЁЯФв Key Formulas

V=43╧Аr3V = \frac{4}{3} \pi r^3V=34тАЛ╧Аr3 тАФ Volume of a sphere

R3=27r3R^3 = 27 r^3R3=27r3 тАФ Conservation of volume equation

тЪЩя╕П Working Principle

The volume of a sphere is given by the formula V=43╧Аr3V = \frac{4}{3} \pi r^3V=34тАЛ╧Аr3. Since the volume is conserved during the melting and recasting process, the relation is given by Vlarge=27imesVsmallV_{\text{large}} = 27 imes V_{\text{small}}VlargeтАЛ=27imesVsmallтАЛ, which simplifies to R3=27imesr3R^3 = 27 imes r^3R3=27imesr3.

ЁЯУМ Key Points
  • тЦ╕

    Volume remains invariant during melting and recasting.

  • тЦ╕

    The radius scales linearly with the cube root of the number of recast objects.

ЁЯЫая╕П Applications / Uses
  • тЦ╕

    Metal casting and foundry work

  • тЦ╕

    Manufacturing identical spherical components

ЁЯУД Additional Information
  • тЦ╕

    Radius of large sphere R=6R = 6R=6 cm.

  • тЦ╕

    Option B is 2 cm, which correctly satisfies 63=2166^3 = 21663=216 and 216/27=8216 / 27 = 8216/27=8, and the cube root of 8 is 2.

ЁЯУК Diagram / Illustration
Mensuration: Sphere Melting & Recasting1Given Data: Large Metallic SphereRadius of large sphere (R) = 6 cm, Recast into n = 27 identical smallspheres2Working Principle: Conservation of VolumeVolume of large sphere = 27 ├Ч Volume of small sphere (V = (4/3)╧Аr┬│)3Equation Setup & Substitution(4/3)╧АR┬│ = 27 ├Ч (4/3)╧Аr┬│ тЗТ R┬│ = 27 r┬│ тЗТ 6┬│ = 27 r┬│4Solve for Small Radius (r)216 = 27 r┬│ тЗТ r┬│ = 216 / 27 = 8 тЗТ r = тИЫ8 = 2 cmCorrect Option: B) 2 cm (Radius scales as R / тИЫn = 6 / 3 = 2 cm)
тЬЕ

B is correct тАФ The radius of each new smaller sphere is 2 cm based on the conservation of volume.

Core Concepts Used
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Conservation of Volume Mensuration Volume of Sphere
ЁЯТб EXAM TIP

Always equate volumes when dealing with melting and recasting problems in mensuration.

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