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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?
1 cm
2 cm
3 cm
4 cm
2 cm
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.
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.
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.
Volume stays the same when you melt and play a game!
V=34тАЛ╧Аr3 тАФ Volume of a sphere
R3=27r3 тАФ Conservation of volume equation
The volume of a sphere is given by the formula V=34тАЛ╧Аr3. Since the volume is conserved during the melting and recasting process, the relation is given by VlargeтАЛ=27imesVsmallтАЛ, which simplifies to R3=27imesr3.
Volume remains invariant during melting and recasting.
The radius scales linearly with the cube root of the number of recast objects.
Metal casting and foundry work
Manufacturing identical spherical components
Radius of large sphere R=6 cm.
Option B is 2 cm, which correctly satisfies 63=216 and 216/27=8, and the cube root of 8 is 2.
B is correct тАФ The radius of each new smaller sphere is 2 cm based on the conservation of volume.
Always equate volumes when dealing with melting and recasting problems in mensuration.