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The radius of the base of a cone is increasing at the rate of 0.03 cm/sec and the altitude is always in the ratio 3 : 4 with the radius. Then the rate of change in the total surface area of the cone when r = 14 cms is

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1.89 $π$ sq.cm/sec

48 $π$ sq.cm/sec

24 $π$ sq.cm/sec

none

answer is A.

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Detailed Solution

$\frac{dr}{dt} = 0.03 \frac{h}{r} = \frac{3}{4} h = \frac{3}{4} r$ Question Image

$l^{2} = r^{2} + h^{2} = r^{2} + \frac{9}{16} r^{2} = \frac{25}{16} r^{2} l = \frac{5}{4} r S = {πr}^{2} + πrl = {πr}^{2} + \frac{5 π}{4} r^{2} = \frac{9 π}{4} r^{2} \frac{ds}{dt} = \frac{9 π}{2} r \frac{dr}{dt} = 63 π \times 0.03 = 1.89 π$

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