A metal cylinder of length 1 m and radius 10 cm is stretched so that its length increases by 10%. If the Poisson ratio for this metal is 0.3, by what percentage does its volume increase?
A)0.0%.
B)3.7%.
C)3.0%.
D)1.0%.
E)-3.0%.

Question

Quizplus · Accepted Answer

The original height of the cylinder is 1 m and the original radius is 10 cm = 0.1 m.

After stretching, the new height is 1.1 m and the new radius is also 0.1 m (assuming uniform stretching).

Using the formula for the volume of a cylinder,

V = πr²h

The original volume is 
V = π(0.1)²(1) = 0.01π m³

The new volume is 
V' = π(0.1)²(1.1) ≈ 0.011 m³

The percentage increase in volume is 
((V' - V)/V) x 100% = ((0.011π - 0.01π)/0.01π) x 100% ≈ 3.7%

However, we also need to take into account the Poisson ratio, which relates the strain in the direction of stretching (longitudinal strain) to the strain in the perpendicular direction (lateral strain).

For a cylinder, the lateral strain is equal to the axial strain times the Poisson ratio.

The axial strain is the percentage increase in length, which is 10%, or 0.1.

So the lateral strain is (0.1)(0.3) = 0.03.

This means that the radial strain is 0.03, while the axial strain is 0.1.

Using the formula for the volume of a cylinder again, with the stretched values, we get

V'' = π(0.1 + 0.03)²(1.1 + 0.1) ≈ 0.011528 m³

The percentage increase in volume taking into account the Poisson ratio is

((V'' - V)/V) x 100% = ((0.011528π - 0.01π)/0.01π) x 100% ≈ 3.0%

Therefore, the answer is B, 3.7%.

A Metal Cylinder of Length 1 M and Radius 10