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Q.

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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a

1.89π sq.cm/sec

b

48πsq.cm/sec

c

24 πsq.cm/sec

d

none

answer is A.

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Detailed Solution

drdt=0.03       hr=34                        h=34r  Question Image

l2=r2+h2=r2+916r2                   =2516r2 l=54r S=πr2+πrl=πr2+5π4r2=9π4r2 dsdt=9π2rdrdt=63π×0.03                         =1.89π

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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