1.

The rate of change of the surface area of a sphere of radius r when the radius is increasing at the rate of 2 cm/sec is proportional to [Karnataka CET 2003]

A.   \[\frac{1}{r}\]
B.   \[\frac{1}{{{r}^{2}}}\]
C.   \[\because \]Surface area \[s=4\pi {{r}^{2}}\] and \[\frac{dr}{dt}=2\] \ \[\frac{ds}{dt}=4\pi \times 2r\frac{dr}{dt}\] = \[8\pi r\times 2=16\pi r\]Þ  \[\frac{ds}{dt}\propto r\].
D.   \[{{r}^{2}}\]
Answer» C.   \[\because \]Surface area \[s=4\pi {{r}^{2}}\] and \[\frac{dr}{dt}=2\] \ \[\frac{ds}{dt}=4\pi \times 2r\frac{dr}{dt}\] = \[8\pi r\times 2=16\pi r\]Þ  \[\frac{ds}{dt}\propto r\].


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