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(D) The volume of coffee in the cylindrical pot is given by $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height of the coffee level.Given th

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$\frac{16}{9\pi}$

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(D) The volume of coffee in the cylindrical pot is given by $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height of the coffee level.Given the diameter of the pot is $15 \, cm$,the radius $r = \frac{15}{2} \, cm$.Since the pot is cylindrical,the radius $r$ remains constant as the level rises,so $\frac{dr}{dt} = 0$.Differentiating the volume with respect to time $t$,we get:$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.We are given $\frac{dV}{dt} = 100 \, cm^3/min$.Substituting the values:$100 = \pi \left( \frac{15}{2} ight)^2 \frac{dh}{dt}$$100 = \pi \left( \frac{225}{4} ight) \frac{dh}{dt}$$\frac{dh}{dt} = \frac{100 	imes 4}{225 \pi} = \frac{400}{225 \pi} = \frac{16}{9 \pi} \, cm/min$.Thus,the rate at which the level in the pot is rising is $\frac{16}{9 \pi} \, cm/min$.

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