Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm22.0 cm2. The area of cross section at the outlet is 10 mm210 mm2. If the piston is pushed at a speed of 4 cm s−14 cm s-1, the speed of outgoing f
Question
Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm22.0 cm2. The area of cross section at the outlet is 10 mm210 mm2. If the piston is pushed at a speed of 4 cm s−14 cm s-1, the speed of outgoing f
Solution
To find the speed of the outgoing liquid, we can use the principle of continuity, which states that the volume flow rate of a fluid remains constant along a pipe or tube.
The volume flow rate can be calculated using the equation:
Q = A1 * v1 = A2 * v2
Where Q is the volume flow rate, A1 and A2 are the cross-sectional areas at the inlet and outlet respectively, and v1 and v2 are the velocities at the inlet and outlet respectively.
Given that the area of cross section at the inlet (A1) is 2.0 cm^2 and the area of cross section at the outlet (A2) is 10 mm^2, we need to convert the units to be consistent.
1 cm^2 = 100 mm^2
So, A1 = 2.0 cm^2 = 2.0 * 100 mm^2 = 200 mm^2
Now, we can substitute the values into the equation:
Q = 200 mm^2 * v1 = 10 mm^2 * v2
Since the piston is pushed at a speed of 4 cm/s, we can substitute v1 = 4 cm/s into the equation:
200 mm^2 * 4 cm/s = 10 mm^2 * v2
Simplifying the equation:
800 mm^2 cm/s = 10 mm^2 * v2
Dividing both sides by 10 mm^2:
80 cm/s = v2
Therefore, the speed of the outgoing liquid is 80 cm/s.
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