Water is flowing at the rate of 5 km/hr through a cylindrical pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Determine the time in which the level of water in the tank will rise by 7 cm.

### Concept & Flow Rate The volume of water accumulating in the tank over time $t$ equals the volume of water flowing out of the pipe over that same time. Treat the flow speed as the length of a cylinder formed per unit of time. $$ \text{Volume}_{\text{tank}} = \text{Area}_{\text{pipe}} \times \text{Speed} \times \text{Time} $$ ### Step-by-Step Solution * Target volume in tank = $L \times B \times H = 50 \text{ m} \times 44 \text{ m} \times 0.07 \text{ m}$ (converting 7 cm to meters). * Volume required = $2200 \times 0.07 = 154 \text{ m}^3$. * Pipe diameter = 14 cm, so $r = 7 \text{ cm} = 0.07 \text{ m}$. * Flow speed = 5 km/hr = 5000 m/hr. * Volume flowing per hour = $\pi \times r^2 \times \text{speed} = \frac{22}{7} \times (0.07)^2 \times 5000$. * Volume per hour = $\frac{22}{7} \times 0.0049 \times 5000 = 22 \times 0.0007 \times 5000 = 22 \times 3.5 = 77 \text{ m}^3/\text{hr}$. * Time required = $\frac{\text{Total Volume}}{\text{Volume per hour}} = \frac{154}{77} = 2 \text{ hours}$. ### Exam Strategy & Shortcut Align all units to meters before computing. Write the full equation to cross-cancel: $\frac{22}{7} \times \frac{7}{100} \times \frac{7}{100} \times 5000 \times t = 50 \times 44 \times \frac{7}{100}$. Cancel the $\frac{7}{100}$ on both sides. $22 \times \frac{1}{100} \times 5000 \times t = 50 \times 44$. $1100 t = 2200 \Rightarrow t = 2$. ### Common Pitfall Failing to convert the pipe radius (cm), water level rise (cm), and flow rate (km) into a single, uniform unit like meters. ### Final Answer Therefore, the correct answer is **2 hours**.