A uniform vertical glass tube open at the lower end and sealed at the lower end is lowered into sea water thus trappung the air int he tube . It is observed that when the tube is submerged to a depth 10 meters the sea water has entered the lower half of the tube . To what depth must the tube be lowered so that the sea water fill three quarters of the tube?
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To determine the depth to which the tube must be lowered so that the sea water fills three-quarters of the tube, let's break down the problem and find a solution step by step.
1. Let's consider the initial situation where the tube is submerged to a depth of 10 meters, and the sea water has entered the lower half of the tube. This means that the top half of the tube is still filled with air.
2. Now, we need to figure out the relationship between the depth of the sea water and the air in the tube. Since the tube is uniform, we can assume that the volume of air in the top half is equal to the volume of sea water in the bottom half.
3. Let's denote the total depth of the tube as "D" (which we need to find), and the depth at which the sea water fills three-quarters of the tube as "x".
4. Using the given information, we can set up a proportion to find the relationship between the air volume and the sea water volume:
Air volume / Sea water volume = top half depth / bottom half depth
The top half depth is D - x (since there are x meters of sea water in the bottom half). And the bottom half depth is x. Therefore, the proportion becomes:
Air volume / Sea water volume = (D - x) / x
5. We also know that initially (at a depth of 10 meters), the sea water fills the bottom half, which is 10 meters. So, we can set up another proportion using this relationship:
Air volume / Sea water volume = top half depth / bottom half depth
Air volume / 10 = (D - 10) / 10 => Air volume = 10(D - 10)
6. Equating the two expressions for the air volume, we get:
10(D - 10) = (D - x) * x
7. Expanding and simplifying the equation, we get a quadratic equation:
10D - 100 = Dx - x^2
8. Rearranging the equation, we get:
x^2 - Dx + 10D - 100 = 0
9. To solve this equation, we use the quadratic formula:
x = [-(-D) ± √(D^2 - 4(1)(10D - 100))] / 2(1)
10. Simplifying the equation further, we have:
x = [D ± √(D^2 - 40D + 400)] / 2
11. Since we're interested in the depth at which the sea water fills three-quarters of the tube, we can set up an equation and solve for D. Letting x = 0.75D:
0.75D = [D ± √(D^2 - 40D + 400)] / 2
12. Simplifying the equation, we get:
0.75D = D/2 ± √(D^2 - 40D + 400)/2
13. We solve for D by multiplying both sides of the equation by 2:
1.5D = D ± √(D^2 - 40D + 400)
14. Simplifying further, we eliminate the ± sign since we're considering positive values:
1.5D = D + √(D^2 - 40D + 400)
15. Rearranging the equation, we get:
0.5D = √(D^2 - 40D + 400)
16. Squaring both sides of the equation, we get:
0.25D^2 = D^2 - 40D + 400
17. Simplifying and rearranging, we have a quadratic equation:
0.75D^2 - 40D + 400 = 0
18. Solving this quadratic equation, we find the value for D. The positive solution will give the depth at which the sea water fills three-quarters of the tube.
19. Once we have the value of D, we can substitute it into the equation for x = 0.75D to find the actual depth at which the sea water fills three-quarters of the tube.
Please note that you'll need to solve the quadratic equation shown in step 18 to obtain the final answer.