A uniform round tube of diameter 6.32 cm has two piston tops which are balanced

Question

A uniform round tube of diameter 6.32 cm has two piston tops which are balanced at exactly the same height due to a mass (m) on the left side and mass of 15.3 kg on the right side. Oil of density 720 kg/m^3 fills the tube and a reduced pressure of 0.85 ATM exists around the 15.3 kg mass initially.

1. What is the pressure at the bottom of the oil filled tube if the height of the pistons is initially 27.9 cm above the bottom?

2. What mass (m) is required to balanced the system?

3. If the 0.85 ATM enclosure is opened to the air, by how much does the mass move ( assuming the piston is able to move up/down within the tube freely and that the piston top stays within the tube)?

Please help and explain your answers. Will rate high for correct answers!

Explanation / Answer

1)pressure at bottom = 15.3*9.8)/area +rho*g*h = 50530.41 pa = 0.4988 atm

h= 0.279 m

2)1.013*10^5+mg/area =0.85*1.013*10^5 +(15.3*9.8/area)

m = 10.435 Kg

3)for the pressure at bottom to remain same,

(15.3*9.8)/area +rho*g*0.279 = (10.435*9.8/area)+(rho*g*h2

h2 = 1.8298 m

so, height moved = 1.8298-0.279 = 1.551 m = 155.1 cm

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