MCQOPTIONS
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| 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 \[\frac{ds}{dt}\propto r\]. | |