(B) Let the surface area be $S = 4\pi r^2$, where $r$ is the radius.Differentiating with respect to time $t$, we get $\frac{dS}{dt} = 8\pi r \frac{dr}

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(B) Let the surface area be $S = 4\pi r^2$, where $r$ is the radius.Differentiating with respect to time $t$, we get $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.Given $\frac{dS}{dt} = 4\pi \text{ cm}^2/\text{s}$.When $S = 16\pi \text{ cm}^2$, we have $4\pi r^2 = 16\pi$, which implies $r^2 = 4$, so $r = 2 \text{ cm}$.Substituting these values into the derivative equation: $4\pi = 8\pi(2) \frac{dr}{dt}$.$4\pi = 16\pi \frac{dr}{dt}$.$\frac{dr}{dt} = \frac{4\pi}{16\pi} = 0.25 \text{ cm/s}$.

Class 12 Mathematics Application of Derivatives Rate of Change of Quantities

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The surface area of a spherical ball is increasing at the rate of $4\pi \text{ cm}^2/\text{s}$. The rate at which the radius is increasing when the surface area is $16\pi \text{ cm}^2$ is (in $\text{ cm/s}$)

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A
$0.5$
B
$0.25$
C
$0.125$
D
$1$

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The surface area of a spherical ball is increasing at the rate of $4\pi \text{ cm}^2/\text{s}$. The rate at which the radius is increasing when the surface area is $16\pi \text{ cm}^2$ is (in $\text{ cm/s}$)

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