Q200 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
If the errors involved in the measurement of a side and the mass of a cube are $2\%$ and $4\%$ respectively, the maximum permissible error in the density of the material is
- A$2\%$
- B$6\%$
- Correct answerC$10\%$
- D$12\%$
Solution
The density of a cube is given by $\rho = \dfrac{m}{l^3}$
The maximum permissible percentage error in density is given by
$\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{\Delta m}{m} + 3 \dfrac{\Delta l}{l} \right) \times 100$
Given $\dfrac{\Delta m}{m} \times 100 = 4\%$ and $\dfrac{\Delta l}{l} \times 100 = 2\%$
$\dfrac{\Delta \rho}{\rho} \times 100 = 4\% + 3 \times 2\% = 10\%$
Answer: $10\%$
The maximum permissible percentage error in density is given by
$\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{\Delta m}{m} + 3 \dfrac{\Delta l}{l} \right) \times 100$
Given $\dfrac{\Delta m}{m} \times 100 = 4\%$ and $\dfrac{\Delta l}{l} \times 100 = 2\%$
$\dfrac{\Delta \rho}{\rho} \times 100 = 4\% + 3 \times 2\% = 10\%$
Answer: $10\%$
Hence, the correct answer is option: 'C'.