Question

The bulk density of the chips was measured by weighing a batch of 195 liters of chips. The measurement result is 42.5kg. The volume was measured to within ± 2.5 liters. At what accuracy should the mass be measured to determine the bulk density to an accuracy of at least ± 3%?

Answer 1

By the definition, bulk density is

$$\rho=\frac{m}{V}$$

Let,

$$V_0=195\,liters\\[0.3cm] \Delta V=2.5\,liters\\[0.3cm] m_0=42.5\,kg$$

Then,

$$\rho=\overline{\rho}+\Delta\rho,\,where\\[0.3cm] \overline{\rho}=\frac{m_0}{V_0}$$

By the condition of the problem

$$\Delta\rho=0.03\cdot\overline{\rho}$$

As we know, to find the random error of indirect measurements, you should use the formulas:

$$F(x,y,z,\ldots)\longrightarrow\\[0.3cm] \Delta F=\pm\sqrt{\left(\frac{\partial F}{\partial x}\cdot\Delta x\right)^2+\left(\frac{\partial F}{\partial y}\cdot\Delta y\right)^2+\cdots}$$

In our case,

$$\rho=\frac{m}{V}\longrightarrow\\[0.3cm] \Delta\rho=\pm\sqrt{\left(\frac{\partial\rho}{\partial m}\cdot\Delta m\right)^2+\left(\frac{\partial\rho}{\partial V}\cdot\Delta V\right)^2}\longrightarrow\\[0.3cm] (0.03\cdot\overline{\rho})^2=\left(\frac{\Delta m}{V_0}\right)^2+\left(-\frac{m_0\Delta V}{V_0^2}\right)^2\longrightarrow\\[0.3cm] \left.\left(\frac{\Delta m}{V_0}\right)^2=\left(\frac{0.03m_0}{V_0}\right)^2-\left(-\frac{m_0\Delta V}{V_0^2}\right)^2\right|\times V_0^2\\[0.3cm] (\Delta m)^2=(0.03m_0)^2-\left(-\frac{m_0\Delta V}{V_0}\right)^2\longrightarrow\\[0.3cm] \Delta m=m_0\cdot\sqrt{0.0081-\left(\frac{\Delta V}{V_0}\right)^2}=42.5\cdot\sqrt{0.0081-\left(\frac{2.5}{195}\right)^2}\\[0.3cm] \Delta m\approx1.15\,(kg)$$

ANSWER

$$\Delta m\approx 1.15\,kg$$