Given Measurement Uncertainties
The problem provides the following measurements:
- Density ($\rho$): $(20 \pm 4)\text{ gm/cm}^3$. This means the nominal density is $\rho = 20\text{ gm/cm}^3$ and the absolute error in density is $\Delta\rho = 4\text{ gm/cm}^3$.
- Volume ($V$): $(10 \pm 1)\text{ cm}^3$. This means the nominal volume is $V = 10\text{ cm}^3$ and the absolute error in volume is $\Delta V = 1\text{ cm}^3$.
Calculating Nominal Mass
Mass ($M$) is calculated using the formula $M = \rho \times V$. We first find the nominal mass using the nominal values of density and volume:
$ M = \rho \times V = (20\text{ gm/cm}^3) \times (10\text{ cm}^3) = 200\text{ gm} $
Error Propagation Rule for Multiplication
When two quantities are multiplied (like density and volume to find mass), their relative errors add up. The formula for the relative error in mass ($\Delta M / M$) is:
$ \frac{\Delta M}{M} = \frac{\Delta \rho}{\rho} + \frac{\Delta V}{V} $
Calculating Relative Errors
Calculate the relative error for density and volume:
- Relative error in density: $ \frac{\Delta \rho}{\rho} = \frac{4\text{ gm/cm}^3}{20\text{ gm/cm}^3} = 0.2 $
- Relative error in volume: $ \frac{\Delta V}{V} = \frac{1\text{ cm}^3}{10\text{ cm}^3} = 0.1 $
Calculating Absolute Error in Mass
Now, find the total relative error in mass:
$ \frac{\Delta M}{M} = 0.2 + 0.1 = 0.3 $
To find the absolute error in mass ($\Delta M$), multiply the nominal mass ($M$) by the total relative error:
$ \Delta M = M \times \left( \frac{\Delta M}{M} \right) = (200\text{ gm}) \times (0.3) = 60\text{ gm} $
Final Answer
The absolute error in the measurement of mass is 60 gm.
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(least count of length =0.1 cm
least count for time =0.1s )
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