The active element of a certain laser is made of a glass rod 30cm long and 1.5cm in diameter. The average coefficient of linear expansion of glass is 9 x 10^-6°C-1. If the temperature of the rod increases by 65°C, what is the increase in its volume?
Ok, most important is the change in length in a laser, but volume will be
coeffVolumeExpansion=appx 3*coeffLinearExpansion.
See the section isotropic materials:
http://en.wikipedia.org/wiki/Thermal_expansion
To find the increase in volume of the glass rod, we need to consider its linear expansion.
The formula for linear expansion is given by:
ΔL = α * L * ΔT
Where:
ΔL is the change in length,
α is the coefficient of linear expansion,
L is the original length, and
ΔT is the change in temperature.
In this case, we have:
L = 30 cm (the original length of the rod),
α = 9 x 10^-6 °C^-1 (the coefficient of linear expansion of glass), and
ΔT = 65 °C (the change in temperature).
Substituting these values into the formula, we can find the change in length:
ΔL = (9 x 10^-6 °C^-1) * (30 cm) * (65 °C)
ΔL = 0.01755 cm
Now, to find the change in volume, we need to consider the formula for volume:
V = π * r^2 * L
Where:
V is the volume,
π is a mathematical constant (approximately 3.14159),
r is the radius of the rod, and
L is the original length of the rod.
In this case, the radius of the rod is half its diameter, so:
r = 0.75 cm = 0.0075 m
The original volume of the rod is:
V = π * (0.0075 m)^2 * (0.30 m)
V = 0.00053085379 m^3
To find the change in volume, we can multiply the change in length by the cross-sectional area:
ΔV = ΔL * π * r^2
ΔV = (0.0001755 m) * π * (0.0075 m)^2
ΔV ≈ 7.843 x 10^-8 m^3
Therefore, the increase in volume of the glass rod is approximately 7.843 x 10^-8 cubic meters.