A cylindrical drum of radius 7 cm and height 2 m is being kept in a vertical position filled with milk. If the milk is leaking at 14 cm³/sec from its lower base, then the rate of decrease in the level of milk is:

[Take $\pi = \frac{22}{7}$]

Given information:

Radius of cylindrical drum: $r = 7$ cm

Height of drum: $2$ m

Rate of milk leaking: $14$ cm³/sec

$\pi = \frac{22}{7}$

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The drum is kept vertically (standing upright). The radius stays constant at $7$ cm, while only the height $h$ of milk changes as milk leaks out.

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Volume of milk in cylinder:

$V = \pi r^2 h$

Substituting $r = 7$ cm:

$V = \pi \times (7)^2 \times h$

$V = \pi \times 49 \times h$

$V = 49\pi h$

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When milk leaks, both volume and height decrease together. Differentiating with respect to time:

$\frac{dV}{dt} = 49\pi \times \frac{dh}{dt}$

Where $\frac{dV}{dt} = -14$ cm³/sec (negative because decreasing) and $\frac{dh}{dt}$ is the rate of change of height.

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Substituting the values:

$-14 = 49\pi \times \frac{dh}{dt}$

Using $\pi = \frac{22}{7}$:

$-14 = 49 \times \frac{22}{7} \times \frac{dh}{dt}$

$-14 = \frac{49 \times 22}{7} \times \frac{dh}{dt}$

$-14 = 7 \times 22 \times \frac{dh}{dt}$

$-14 = 154 \times \frac{dh}{dt}$

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Solving for $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{-14}{154}$

$\frac{dh}{dt} = \frac{-1}{11}$ cm/sec

The negative sign indicates the level is decreasing.

Therefore, the rate of decrease in milk level is $\frac{1}{11}$ cm/sec.