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Question
A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate
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Solution
Ball of salt is spherical
∴ Volume of ball, V = `4/3 pi"r"^3`
Where r = radius of the ball
As per the question, `"dV"/"dt" oo "S"`
Where S = surface area of the ball
⇒ `"d"/"dt" (4/3 pi"r"^3) oo 4pi"r"^2` .....[∵ S = 4πr2]
⇒ `4/3 pi * 3"r"^2 * "dr"/"dt" oo 4pi"r"^2`
⇒ `4pi"r"^2 * "dr"/"dt" = "K" * 4pi"r"^2` ......(K = Constant of proportionality)
⇒ `"dr"/"dt" = "K" * 4pi"r"^2`
∴ `"dr"/"dt" = "K" * 1` = K
Hence, the radius of the ball is decreasing at constant rate.
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