4. A water tank is a cuboid with dimensions 40 cm by 90 cm by 70 cm . The tank is initially empty and is then filled with the water which flows from a pipe of cross-sectional area \( 1.2 \mathrm{~cm}^{2} \) at a speed of \( 3.5 \mathrm{~m} / \mathrm{s} \). Calculate the time, in minutes, that it takes to fill the tanks completely.

To determine the time required to fill the cuboid-shaped water tank completely, we'll follow these steps: 1. **Calculate the Volume of the Tank**: The tank is a cuboid with dimensions: - Length (\( L \)) = 40 cm - Width (\( W \)) = 90 cm - Height (\( H \)) = 70 cm \[ \text{Volume} = L \times W \times H = 40\, \text{cm} \times 90\, \text{cm} \times 70\, \text{cm} = 252{,}000\, \text{cm}^3 \] Convert cubic centimeters to cubic meters (since flow rate is in meters per second): \[ 1\, \text{m}^3 = 1{,}000{,}000\, \text{cm}^3 \implies 252{,}000\, \text{cm}^3 = 0.252\, \text{m}^3 \] 2. **Determine the Flow Rate of Water from the Pipe**: The flow rate (\( Q \)) can be calculated using the formula: \[ Q = \text{Area} \times \text{Speed} \] Given: - Cross-sectional area of the pipe (\( A \)) = \( 1.2\, \text{cm}^2 \) - Speed of water (\( v \)) = \( 3.5\, \text{m/s} \) Convert area to square meters: \[ 1\, \text{cm}^2 = 0.0001\, \text{m}^2 \implies 1.2\, \text{cm}^2 = 0.00012\, \text{m}^2 \] Now, calculate the flow rate: \[ Q = 0.00012\, \text{m}^2 \times 3.5\, \text{m/s} = 0.00042\, \text{m}^3/\text{s} \] 3. **Calculate the Time to Fill the Tank**: The time (\( t \)) required to fill the tank is: \[ t = \frac{\text{Volume}}{\text{Flow Rate}} = \frac{0.252\, \text{m}^3}{0.00042\, \text{m}^3/\text{s}} = 600\, \text{seconds} \] Convert seconds to minutes: \[ 600\, \text{seconds} \div 60 = 10\, \text{minutes} \] **Answer:** It takes 10 minutes to fill the tank completely.