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(D) The density $\rho$ of a cube is given by $\rho = \frac{m}{L^3}$,where $m$ is the mass and $L$ is the edge length.Given: $m = 10.00 \, kg$,$\Delta

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None of these

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(D) The density $\rho$ of a cube is given by $\rho = \frac{m}{L^3}$,where $m$ is the mass and $L$ is the edge length.Given: $m = 10.00 \, kg$,$\Delta m = 0.10 \, kg$,$L = 0.10 \, m$,$\Delta L = 0.01 \, m$.The measured density is $\rho = \frac{10.00}{(0.10)^3} = \frac{10.00}{0.001} = 10000 \, kg/m^3$.Since the relative error in $L$ is $\frac{\Delta L}{L} = \frac{0.01}{0.10} = 0.1$,which is not small,we calculate the maximum and minimum possible density values.$\rho_{max} = \frac{m + \Delta m}{(L - \Delta L)^3} = \frac{10.10}{(0.09)^3} = \frac{10.10}{0.000729} \approx 13854.6 \, kg/m^3$.$\rho_{min} = \frac{m - \Delta m}{(L + \Delta L)^3} = \frac{9.90}{(0.11)^3} = \frac{9.90}{0.001331} \approx 7438.0 \, kg/m^3$.The absolute error in density is $\Delta \rho = \frac{\rho_{max} - \rho_{min}}{2} = \frac{13854.6 - 7438.0}{2} = \frac{6416.6}{2} = 3208.3 \, kg/m^3$.Since this value is not among the given options,the correct choice is $D$.

In the density measurement of a cube,the mass and edge length are measured as $(10.00 \pm 0.10) \, kg$ and $(0.10 \pm 0.01) \, m$ respectively. The error in the measurement of density is

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