Question

Are evenly spaced specific gravity markings on the cylinder of a hydrometer equal distances apart? In other words, is the depth \(d\) to which the cylinder is submerged linearly related to the density \(\rho\) of the fluid? To answer this question, assume that the cylinder has radius \(r\) and mass \(m .\) Find an expression for \(d\) in terms of \(\rho, r,\) and \(m\) and see if \(d\) is a linear function of \(\rho\).

Short answer

Answer: No, the depth to which the cylinder is submerged is not linearly related to the density of the fluid. Instead, they have an inverse relationship, as the expression for depth is \(d = \frac{m}{\pi \times r^2 \times \rho}\), which does not conform to the linear form \(d = a\rho + b\).

Step-by-step solution

Write the equation for buoyant force

The buoyant force experienced by the cylinder will be equal to the weight of the fluid it displaces. Mathematically, the buoyant force \(F_b\) can be represented as: $$F_b = V_{displaced} \times \rho \times g$$ where \(V_{displaced}\) is the volume of the fluid displaced by the cylinder, \(\rho\) is the density of the fluid, and \(g\) is the gravitational acceleration.

Calculate the volume of displaced fluid

The volume of the displaced fluid is the volume of the submerged part of the cylinder, which can be represented as: $$V_{displaced} = \pi \times r^2 \times d$$ where \(r\) is the radius of the cylinder and \(d\) is the depth to which it is submerged.

Calculate the weight of the cylinder

The weight of the cylinder, \(W\), can be calculated using the mass \(m\) and gravitational acceleration \(g\) as follows: $$W = m \times g$$

Apply the principle of buoyancy

According to the principle of buoyancy, the buoyant force is equal to the weight of the object submerged in the fluid. Therefore, we equate \(F_b\) and \(W\) and substitute the expressions from Steps 1, 2, and 3: $$(\pi \times r^2 \times d) \times \rho \times g = m \times g$$

Isolate \(d\)

The goal is to find an expression for \(d\) in terms of \(\rho, r,\) and \(m\). Divide both sides of the equation by \(\pi \times r^2 \times \rho \times g\) to isolate \(d\): $$d = \frac{m \times g}{\pi \times r^2 \times \rho \times g}$$ Since \(g\) appears in both the numerator and denominator, we can cancel it out: $$d = \frac{m}{\pi \times r^2 \times \rho}$$

Check for linearity

To determine if \(d\) is a linear function of \(\rho\), we examine the expression we derived for \(d\) in terms of \(\rho\). If \(d\) and \(\rho\) have a linear relationship, then their equation should take the form \(d = a\rho + b\), where \(a\) and \(b\) are constants. Looking back at the derived equation: $$d = \frac{m}{\pi \times r^2 \times \rho}$$ We observe that \(d\) and \(\rho\) do not have a linear relationship. They have an inverse relationship, as \(\rho\) is in the denominator, and the equation does not conform to the linear form \(d = a\rho + b\). Therefore, the depth \(d\) to which the cylinder is submerged is not linearly related to the density \(\rho\) of the fluid.