Q151 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
If the mass of the object is measured as $m = (7 \mp 0.14)$ kg and the density is measured as $\rho = (5 \mp 0.25)$ kg/m$^3$ respectively. The percentage error in the measurement of volume is
- A$3\%$
- B$5\%$
- Correct answerC$7\%$
- D$9\%$
Solution
Volume is given by the relation $V = \dfrac{m}{\rho}$
The maximum percentage error in volume is
$\dfrac{\Delta V}{V} \times 100 = \left( \dfrac{\Delta m}{m} + \dfrac{\Delta \rho}{\rho} \right) \times 100$
Substituting the given values:
$\dfrac{\Delta V}{V} \times 100 = \left( \dfrac{0.14}{7} + \dfrac{0.25}{5} \right) \times 100$
$\dfrac{\Delta V}{V} \times 100 = (0.02 + 0.05) \times 100$
$\dfrac{\Delta V}{V} \times 100 = 0.07 \times 100 = 7\%$
Answer: $7\%$
The maximum percentage error in volume is
$\dfrac{\Delta V}{V} \times 100 = \left( \dfrac{\Delta m}{m} + \dfrac{\Delta \rho}{\rho} \right) \times 100$
Substituting the given values:
$\dfrac{\Delta V}{V} \times 100 = \left( \dfrac{0.14}{7} + \dfrac{0.25}{5} \right) \times 100$
$\dfrac{\Delta V}{V} \times 100 = (0.02 + 0.05) \times 100$
$\dfrac{\Delta V}{V} \times 100 = 0.07 \times 100 = 7\%$
Answer: $7\%$
Hence, the correct answer is option: 'C'.