Water is flowing at the rate of 12 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 33 m wide. Find the time in which the level of water in the tank will rise by 28 cm.

The volume of water flowing into the tank is equal to the volume of water that accumulates in the tank. The pipe is cylindrical. Diameter = 14 cm, so radius $r = 7$ cm. Area of cross-section of the pipe = $\pi r^2 = \pi (7)^2 = 49\pi$ cm$^2$. Speed of water flow = 12 km/hour. Convert to cm/hour: 12 km/ hour = $12 \times 1000 \times 100$ cm/hour = 1,200,000 cm/hour. Volume of water flowing out of the pipe per hour (Volume Flow Rate) = Area of cross-section $\times$ Speed Volume Flow Rate = $49\pi \text{ cm}^2 \times 1,200,000 \text{ cm/ hour}$. The rectangular tank has length $L = 60$ m = 6000 cm and width $W = 33$ m = 3300 cm. The desired rise in water level is $H = 28$ cm. The volume of water required in the tank = $L \times W \times H = 6000 \times 3300 \times 28$ cm$^3$. Let the time taken be $t$ hours. Volume of water needed = Volume Flow Rate $\times$ Time $6000 \times 3300 \times 28 = (49\pi \times 1,200,000) \times t$. Using $\pi = 22/7$: $6000 \times 3300 \times 28 = (49 \times 22/7 \times 1,200,000) \times t$. $6000 \times 3300 \times 28 = (7 \times 22 \times 1,200,000) \times t$. $6000 \times 3300 \times 28 = (154 \times 1,200,000) \times t$. $t = \frac{6000 \times 3300 \times 28}{154 \times 1,200,000}$. Simplify the expression: $t = \frac{6000 \times 3300 \times 28}{154 \times 1200000} = \frac{(6 \times 1000) \times (33 \times 100) \times 28}{154 \times (12 \times 100000)} = \frac{6 \times 33 \times 28 \times 100000}{154 \times 12 \times 100000}$. Cancel $100000$: $t = \frac{6 \times 33 \times 28}{154 \times 12}$. $t = \frac{6 \times (3 \times 11) \times (4 \times 7)}{(14 \times 11) \times (3 \times 4)} = \frac{6 \times 3 \times 11 \times 4 \times 7}{14 \times 11 \times 3 \times 4}$. Cancel $11 \times 3 \times 4$: $t = \frac{6 \times 7}{14} = \frac{42}{14} = 3$. The time taken is 3 hours.