Water is flowing continuously from a tap having an internal diameter 8×10^-3 m. The water velocity as it leaves the tap is 0.4 m/s. The diameter of the water stream at a distance 2×10^-1 m below the tap is close to:
A) 5.0×10^-3m
B) 7.5×10^-3m
C) 9.6×10^-3m
D) 3.6×10^-3m
C is the correct answer
Nope, Not C.
see the solution here, number 16: http://prernaclasses.com/AIEEE2011PhyQuestionsSolutions.pdf
To determine the diameter of the water stream at a distance 2×10^-1 m below the tap, we can make use of the principle of conservation of mass.
According to this principle, the volume flow rate of water at any point in the stream should remain constant. Mathematically, this can be expressed as:
A1 * v1 = A2 * v2
Where A1 and v1 represent the cross-sectional area and velocity of the water stream at the tap, and A2 and v2 represent the cross-sectional area and velocity at a distance 2×10^-1 m below the tap.
We are given that the internal diameter of the tap is 8×10^-3 m, so we can calculate the cross-sectional area at the tap using the formula:
A1 = π * (d1/2)^2
Where d1 is the diameter of the tap. Plugging in the value, we have:
A1 = π * (8×10^-3/2)^2
≈ 0.0201 m^2
We are also given the velocity at the tap, which is 0.4 m/s.
Now, we can rearrange the conservation of mass equation to solve for A2:
A2 = (A1 * v1) / v2
Plugging in the values we have:
A2 = (0.0201 * 0.4) / v2
≈ 0.00804 / v2
From the given options, we need to find the diameter of the water stream. The diameter can be found using the formula:
d2 = 2 * √(A2 / π)
Plugging in the value of A2, we have:
d2 = 2 * √(0.00804 / π * v2)
To compare this value with the given options, we need to calculate v2 first. The distance 2×10^-1 m below the tap will be the same as the distance traveled by the water stream. We can calculate the time taken using the formula:
t = d / v
Plugging in the values, we have:
t = (2×10^-1) / 0.4
= 5×10^-1 s
Now, we can calculate v2 using the formula:
v2 = d2 / t
Plugging in the values, we have:
v2 = d2 / (5×10^-1)
= (2 * √(0.00804 / π * v2)) / (5×10^-1)
At this point, we need to use an iterative method or approximation techniques to solve for v2. This involves repeatedly substituting the expression for v2 on the right side of the equation until we get a value that converges. This process can be computationally intensive, especially without initial approximation values.
However, since we know that the answer should be close to one of the given options, we can use a trial and error approach. By substituting one option at a time into the equation for v2, we can check which option gives us a value of v2 that matches the known value.
Let's try each option:
A) d2 = 5.0×10^-3 m
v2 = (2 * √(0.00804 / π * 5.0×10^-3)) / (5×10^-1)
≈ 0.136 m/s (approximately)
B) d2 = 7.5×10^-3 m
v2 = (2 * √(0.00804 / π * 7.5×10^-3)) / (5×10^-1)
≈ 0.111 m/s (approximately)
C) d2 = 9.6×10^-3 m
v2 = (2 * √(0.00804 / π * 9.6×10^-3)) / (5×10^-1)
≈ 0.096 m/s (approximately)
D) d2 = 3.6×10^-3 m
v2 = (2 * √(0.00804 / π * 3.6×10^-3)) / (5×10^-1)
≈ 0.205 m/s (approximately)
Comparing the calculated values of v2 with the given value of 0.4 m/s, we can see that option D) gives the closest value to the known velocity. Therefore, the diameter of the water stream at a distance 2×10^-1 m below the tap is close to 3.6×10^-3 m, which is option D).