A sphere of mass of m = 1.69 kg is first placed directly on the plate of an electronic scale. The scale shows 16.58 N as the weight of the object. A large beaker containing a liquid with a density of 783 kg/m^3 is then placed on the scale and the scale is tared (i.e. zeroed to this weight as a reference point, it has a reading of zero with the beaker on the scale). The sphere, hung by a thin string, is lowered into the liquid and submerges below the surface. The scale reads 2.19 N with the sphere not touching the beaker. Calculate the density of the sphere.

Question

A sphere of mass of m = 1.69 kg is first placed directly on the plate of an electronic scale. The scale shows 16.58 N as the weight of the object. A large beaker containing a liquid with a density of 783 kg/m^3 is then placed on the scale and the scale is tared (i.e. zeroed to this weight as a reference point, it has a reading of zero with the beaker on the scale). The sphere, hung by a thin string, is lowered into the liquid and submerges below the surface. The scale reads 2.19 N with the sphere not touching the beaker. Calculate the density of the sphere.

My work so far:

F_buoyant = 16.58N - 2.19N = 14.39

F_buoyant = (Rho_liquid)(g)(V_liquid displaced)

14.39N = (783kg/m^3)(9.81m/s^2)(V_liquid displaced)

V_liquid displaced = 0.001873398 m^3 = 0.00187 m^3

So the density of the sphere:

rho_sphere = m_sphere/V_sphere

V_sphere = V_liquid displaced = 0.00187 m^3

rho_sphere = 1.69kg/0.00187m^3
= 902.1040911 kg/m^3
= 902 kg/m^3

... which is what I got for the final answer but apparently it is incorrect. If there is anything, what did I do wrong?

Answer 1

Your calculations appear to be correct, so it's possible that there may be a mistake elsewhere. However, I'll go through the process again step by step to make sure we didn't miss anything.

Let's start with the given information:
- Mass of the sphere, m = 1.69 kg
- Weight of the sphere on the scale, Fsph = 16.58 N
- Weight of the sphere in the liquid, Fsph_liquid = 2.19 N
- Density of the liquid, ρliquid = 783 kg/m^3

First, let's calculate the buoyant force on the sphere:
Fbuoyant = Fsph - Fsph_liquid
Fbuoyant = 16.58 N - 2.19 N
Fbuoyant = 14.39 N

Now, using the buoyant force equation:
Fbuoyant = ρliquid * g * Vliquid_displaced

Since the scale was tared with the beaker on it, the reading of the scale is the weight of the liquid displaced by the sphere. Therefore, Vliquid_displaced = reading of the scale.

Substituting the values into the equation:
14.39 N = 783 kg/m^3 * 9.81 m/s^2 * Vliquid_displaced

Solving for Vliquid_displaced:
Vliquid_displaced = 14.39 N / (783 kg/m^3 * 9.81 m/s^2)
Vliquid_displaced ≈ 0.00187 m^3

Now, to calculate the density of the sphere:
ρsphere = msphere / Vsphere

We know the mass of the sphere, msphere = 1.69 kg.
The volume of the sphere is the same as the volume of the liquid displaced, Vsphere = Vliquid_displaced ≈ 0.00187 m^3.

Substituting the values into the equation:
ρsphere = 1.69 kg / 0.00187 m^3
ρsphere ≈ 902.1 kg/m^3, which can be rounded to 902 kg/m^3.

So, based on these calculations, the density of the sphere is approximately 902 kg/m^3. If your answer matches these calculations, then it is correct. If not, please double-check the conversions and calculations to ensure accuracy.