A soft synthetic rubber ball of radius 9 cm is submerged to 7.5 m beneath the surface of a lake.
a) If the bulk modulus of the rubber compound is Brubber = 9 x106Pa , what is the diameter of the ball at that depth?
(b) A similar ball of equal radius, but made of thin rubber inflated with air, is also submerged to the same depth. The bulk modulus of air is Bair = 1.2 x105Pa . What is the radius of this ball at that depth? (Ignore any effects of temperature.)
To solve these problems, we need to use the concept of pressure and the bulk modulus of the materials involved.
a) To find the diameter of the rubber ball at a certain depth, we can use the equation:
Diameter = 2 * radius
First, let's calculate the pressure at a depth of 7.5 m beneath the surface of the lake. The pressure at a depth h in a fluid is given by the equation:
Pressure = ρ * g * h
Where:
ρ is the density of the fluid (in this case, water),
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
h is the depth.
Given that the density of water is approximately 1000 kg/m^3, the pressure at a depth of 7.5 m is:
Pressure = 1000 kg/m^3 * 9.8 m/s^2 * 7.5 m
Now, let's use the bulk modulus of the rubber compound (Brubber = 9 x 10^6 Pa) to find the change in volume of the ball due to the increased pressure at that depth.
The formula for the change in volume of an object under pressure is given by:
ΔV = -V * (ΔP / B)
Where:
ΔV is the change in volume,
V is the original volume,
ΔP is the change in pressure, and
B is the bulk modulus.
In this case, the change in pressure (ΔP) would be the pressure at the given depth minus the atmospheric pressure (which we can consider negligible for this problem).
Now, we can calculate the change in volume using the formula above. Suppose the original volume of the ball is V, and the change in volume is ΔV, then:
ΔV = -V * (Pressure / Brubber)
Since we are interested in the change in diameter (which is related to the change in volume), we can rewrite ΔV in terms of diameter (ΔD) by using the relation ΔV = π * (ΔD)^3 / 6.
With these equations, we can solve for ΔD and then find the new diameter of the ball:
ΔD = (6 * ΔV / π)^(1/3)
Finally, we can calculate the new diameter by adding the change in diameter to the original diameter.
b) To find the radius of the ball made of thin rubber inflated with air, we can use a similar approach.
Using the bulk modulus of air (Bair = 1.2 x 10^5 Pa), we can calculate the change in volume of the ball under pressure using the formula:
ΔV = -V * (Pressure / Bair)
Again, since we are interested in the change in radius (Δr), we can rewrite ΔV in terms of radius (Δr) by using the relation ΔV = (4/3) * π * (Δr)^3.
After finding Δr, we can calculate the new radius by adding the change in radius to the original radius.
Note: In both problems, it is important to convert the units to be consistent (e.g., meters for depth and radii, Pascal for pressure, etc.), and to use the appropriate formulas and formulas based on the given information.