\begin{array}{1 1}(A)\; 1 m^3/h & (B)\; 0.1 m^3/h \\(C)\; 1.1 m^3/h & (D)\; 0.5 m^3/h \end{array}

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- Volume=$\pi r^2 h$

Step 1:

Let $h$ be the height of the cylindrical tank at any instant .

Volume of the cylindrical tank $=\pi r^2 h$

$\qquad\qquad\qquad\qquad\quad\quad\;\;=\pi(10)^2h$

$V=100\pi h$

Rate of change of volume=$\large\frac{dV}{dt}$

$\large\frac{dV}{dt}=$$100\pi \large\frac{dh}{dt}$------(1)

The tank is filled at the rate of $314$cubic feet per minute .

(i.e)$\large\frac{dV}{dt}$$=314$

Step 2:

From (1)

$314=100\pi \large\frac{dh}{dt}$

$\large\frac{dh}{dt}=\frac{314}{100\pi}$

$\qquad=\large\frac{314}{100\times 3.14}$$=1$

Hence the depth of the tank changes at $1$ cubic ft/minute

Hence part (A) is the correct answer.

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