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timama [110]
3 years ago
5

A cylindrical rod with length (1.7 m) and radius (2 cm) is fixed from one end and a mass of (20 kg) is attached to the other fre

e end. The elongation in the rod is (0.5 mm). If this cylinder is exposed to an excess pressure of (180 atm.) above the atmospheric pressure its volume decreases by (1.2 %). Find Poisson’s ratio for the rod material.
Physics
1 answer:
belka [17]3 years ago
3 0

Answer:

0.44

Explanation:

Poisson's ratio is:

ν = (3K − E) / 6K

where K is the bulk modulus and E is Young's modulus.

Young's modulus is:

E = FL / (AΔL)

where F is the force, L is the initial length, A is the cross sectional area, and ΔL is the change in length.

E = (20 kg × 9.8 m/s²) (1.7 m) / (π (0.02 m)² × 0.0005 m)

E = 0.530×10⁹ Pa

Bulk modulus is:

K = -ΔP / (ΔV/V)

where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.

K = -(180 atm × 101325 Pa/atm) / (-0.012)

K = 1.52×10⁹ Pa

Therefore, the Poisson's ratio is:

ν = (3(1.52×10⁹ Pa) − 0.530×10⁹ Pa) / 6(1.52×10⁹ Pa)

ν = (3(1.52) − 0.530) / 6(1.52)

ν = 0.442

Rounded to 2 significant figures, the Poisson's ratio is 0.44.

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47.4 m

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3 years ago
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