The velocity, density, and pressure of a certain fluid are related by

Question

Tough and Tricky questions: Word Problems.

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

Kudos for a correct  solution .

DangerPenguin · Answer

5v^2 for v=10, 5(10)^2 = 5(100) = 500. So now we have 500+P = c. Let's use c=1000. This means that P would be 500.

5v^2 for v=5, 5(25) = 125. So now we have 125 + P = c and since we used c=1000 above we have 125 + P =1000 or P=875.

875-500 = 375 so we know that a decrease in velocity from 10m to 5m would raise the pressure by 375 pascals.

Answer C!

Ankur9 · Answer

Lets say V i  as the initial velocity and V f  as final velocity.
similarly P i  as the initial pressure and P f  as final pressure.
Given: V i  = 10m/s 
V f  = 5m/s
therefore 5(10)^2 + P i  = 5(5)^2 + P f 
P f  - P i  = 375.

akumar5 · Answer

as mentioned 5v^2+p=c
 p=c-5v^2

initial v=10, p=c-500........1
finial v=5, p=c-125..........2

so difference of equations 1 & 2 will give value of pressure rise =375 pascal

Answer C

PareshGmat · Answer

Answer = C = 375

Initial

P = C - 5*10^2 = C - 500

After velocity decrease:

P = C - 5*5^2 = C - 125

Increase = C - 125 - (C - 500) = 500 - 125 = 375

Answer = C

Bunuel · Answer

Official Solution:

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
 
This problem requires the use of a special given equation,  5v^2 + P = c . We are told that the velocity decreases from 10 meters per second to 5 meters per second. The key to this problem is to recognize that we have a "before" situation and an "after" situation. In the "before" situation, we have a certain velocity and pressure; in the "after" situation, we have a different velocity and pressure. What remains constant between the two situations is the constant  c . Thus, we should set up the given equation for both situations.
 
To distinguish the situations, let's use subscripts on the variables.  P_1  and  v_1  will indicate "before," while  P_2  and  v_2  will indicate "after."
 
Before:  5v_1^2 + P_1 = c 
  5(10)^2 + P_1 = c 
  500 + P_1 = c 
 
After:  5v_2^2 + P_2 = c 
  5(5)^2 + P_2 = c 
  125 + P_2 = c 
 
Now, we cannot solve for  c , but we can set the left sides of these two equations equal to each other, because they are both equal to  c .
  500 + P_1 = 125 + P_2 
 
Again, we cannot solve for either pressure, but we do not need to. What we need to find is the increase in pressure - in other words, how much the pressure rises by. As an expression, the increase in pressure is simply  P_2 - P_1 . Thus, we rearrange the equation to solve for this difference.
  500 - 125 = P_2 - P_1 
  375 = P_2 - P_1 
 
Answer: C.

Abhishek009 · Answer

{5(10)^2 + P } - {5(5)^2 + P } = 500 - 125 => 375 , Answer will be (C)

bumpbot · Answer

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The velocity, density, and pressure of a certain fluid are related by

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