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Density and volume of a body are given as $(20 \pm 4)\text{ gm/cm}^3$ and $(10 \pm 1)\text{ cm}^3$ respectively. The absolute error in measurement of mass is

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$60\text{ gm}$

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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