Let's solve the given problem step-by-step.
1. Dimensions of the Tank:
- Width = 0.5 meters
- Length = 2.0 meters
- Height = 4.0 meters
2. Rate of Droplets and their Weight:
- The tap drops water at a rate of 2 drops per second.
- Each drop weighs 0.05 grams.
3. Convert Drop Weight to Kilograms:
- Since 1 gram = 0.001 kilograms,
[tex]\[
\text{Drop weight in kilograms} = 0.05 \, \text{grams} \times 0.001 \, \text{kg/gram} = 0.00005 \, \text{kg}
\][/tex]
4. Volume of the Tank:
- The volume of the tank (although not directly needed to find the time) can be calculated as:
[tex]\[
\text{Volume} = \text{Width} \times \text{Length} \times \text{Height} = 0.5 \, \text{m} \times 2.0 \, \text{m} \times 4.0 \, \text{m} = 4.0 \, \text{m}^3
\][/tex]
5. Mass of One Drop of Water:
- As calculated, the mass of one drop is:
[tex]\[
\text{Mass of one drop} = 0.00005 \, \text{kg}
\][/tex]
6. Rate of Water Mass Addition to the Tank:
- The total mass added to the tank per second is the rate of drops times the mass of each drop.
[tex]\[
\text{Mass per second} = 2 \, \text{drops/second} \times 0.00005 \, \text{kg/drop} = 0.0001 \, \text{kg/second}
\][/tex]
7. Weight Added to the Tank Per Second:
- Weight is mass times gravitational acceleration [tex]\( g \)[/tex] (given as 10 m/s²):
[tex]\[
\text{Weight per second} = 0.0001 \, \text{kg/second} \times 10 \, \text{m/s}^2 = 0.001 \, \text{N/second}
\][/tex]
8. Bottom Area of the Tank:
- The bottom surface area of the tank is:
[tex]\[
\text{Area} = \text{Width} \times \text{Length} = 0.5 \, \text{m} \times 2.0 \, \text{m} = 1.0 \, \text{m}^2
\][/tex]
9. Required Force to Achieve the Desired Pressure:
- Given the required pressure is 200 Pascals (Pa), using the relationship [tex]\( \text{Pressure} = \frac{\text{Force}}{\text{Area}} \)[/tex], we rearrange it to find the required force:
[tex]\[
\text{Force required} = \text{Pressure} \times \text{Area} = 200 \, \text{Pa} \times 1.0 \, \text{m}^2 = 200 \, \text{N}
\][/tex]
10. Time Required to Reach the Desired Pressure:
- The total weight (force) added per second is 0.001 N/second. To accumulate the required force of 200 N:
[tex]\[
\text{Time required} = \frac{\text{Force required}}{\text{Weight per second}} = \frac{200 \, \text{N}}{0.001 \, \text{N/second}} = 200,000 \, \text{seconds}
\][/tex]
Therefore, it will take 200,000 seconds for the pressure at the bottom of the tank to reach 200 Pascals.
