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The velocity, density and pressure of a certain fluid...

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by BTGmoderatorLU » Fri Dec 15, 2017 1:55 pm
The velocity, density and pressure of a certain fluid are related by the equation 5v^2 + P = c, where v is the velocity in meters per second, P is the pressure in Pascals, and c is a constant. If the velocity of this fluid decreases from 10 meters per second to 5 meters per second, by how many pascals does the pressure in the fluid rise?

A. 125
B. 250
C. 375
D. 500
E. 625

The OA is C.

I'm confused with this DS question. Please, can any expert assist me with it? Thanks in advanced.
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by [email protected] » Fri Dec 15, 2017 3:25 pm
Hi LUANDATO,

We're told that the Velocity, Density and Pressure of a certain fluid are related by the equation 5(V^2) + P = C, where V is the velocity in meters per second, P is the pressure in Pascals, and C is a constant. We're asked to find the increase in the number of pascals that the fluid will rise when the velocity of this fluid decreases from 10 meters per second to 5 meters per second. This question can essentially be solved by plugging in the values and tracking the results.

IF.... V=10, the equation becomes...
5(10^2) + P = C
500 + P = C

IF.... V=5, the equation becomes...
5(5^2) + P = C
125 + P = C

Since the overall value of 5(V^2) decreases by 375, the "P" has to increase by 375 to 'compensate' for the loss and get the total back up to the constant (C).

Final Answer: C

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