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प्रश्न
The volume of a sphere is increasing at 3 cm3/sec. The rate at which the radius increases when radius is 2 cm, is
पर्याय
\[\frac{3}{32\pi}cm/\sec\]
\[\frac{3}{16\pi}cm/\sec\]
\[\frac{3}{48\pi}cm/\sec\]
\[\frac{1}{24\pi}cm/\sec\]
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उत्तर
\[\frac{3}{16\pi}cm/\sec\]
\[\text { Let r be the radius and V be the volume of the sphere at any time t.Then },\]
\[V=\frac{4}{3} \pi r^3 \]
\[\Rightarrow\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}\]
\[\Rightarrow\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}\]
\[\Rightarrow\frac{dr}{dt}=\frac{3}{4\pi \left( 2 \right)^2}\]
\[\Rightarrow\frac{dr}{dt}=\frac{3}{16\pi}cm/sec\]
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