Q127 / 155 · WBJEE 2026 · Physics
Density and volume of a body are given as $(20 \pm 4)$ gm/cm$^3$ and $(10 \pm 1)$ cm$^3$ respectively. The absolute error in measurement of mass is
- A$20$ gm
- B$30$ gm
- C$45$ gm
- Correct answerD$60$ gm
Solution
Mass $M = \rho \times V$
Given $\rho = 20$ gm/cm$^3$, $\Delta \rho = 4$ gm/cm$^3$
$V = 10$ cm$^3$, $\Delta V = 1$ cm$^3$
$M = 20 \times 10 = 200$ gm
The maximum fractional error in mass is given by
$\dfrac{\Delta M}{M} = \dfrac{\Delta \rho}{\rho} + \dfrac{\Delta V}{V}$
$\dfrac{\Delta M}{200} = \dfrac{4}{20} + \dfrac{1}{10}$
$\dfrac{\Delta M}{200} = \dfrac{4 + 2}{20} = \dfrac{6}{20}$
$\Delta M = 200 \times \dfrac{6}{20} = 60$ gm
The absolute error in measurement of mass is $60$ gm.
Answer: $60$ gm
Given $\rho = 20$ gm/cm$^3$, $\Delta \rho = 4$ gm/cm$^3$
$V = 10$ cm$^3$, $\Delta V = 1$ cm$^3$
$M = 20 \times 10 = 200$ gm
The maximum fractional error in mass is given by
$\dfrac{\Delta M}{M} = \dfrac{\Delta \rho}{\rho} + \dfrac{\Delta V}{V}$
$\dfrac{\Delta M}{200} = \dfrac{4}{20} + \dfrac{1}{10}$
$\dfrac{\Delta M}{200} = \dfrac{4 + 2}{20} = \dfrac{6}{20}$
$\Delta M = 200 \times \dfrac{6}{20} = 60$ gm
The absolute error in measurement of mass is $60$ gm.
Answer: $60$ gm
Hence, the correct answer is option: 'D'.