Question

The dimensions of a solid cylinder is measured as given Mass \(= 19.42 \pm 0.02 \,kg\) Diameter \(= 20.20 \pm 0.02 \,cm\) Length \(= 10.10 \pm 0.02 \,cm\) Find out % error in density.

• A. \(0.5%\)
• B. \(0.3%\)
• C. \(0.4%\)
• D. \(0.7%\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When calculating a derived physical quantity, the maximum relative (or percentage) error is the sum of the relative errors of the individual measured quantities, each multiplied by the absolute value of its power in the formula.
Step 2: Key Formula or Approach:
The density \(\rho\) of a cylinder is: \[ \rho = \frac{M}{V} = \frac{M}{\pi r^2 \ell} = \frac{4M}{\pi d^2 \ell} \] The relative error formula is: \[ \frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta d}{d} + \frac{\Delta \ell}{\ell} \] Step 3: Detailed Explanation:
Substitute the given measurements and their absolute errors: \[ \frac{\Delta \rho}{\rho} = \frac{0.02}{19.42} + 2\left(\frac{0.02}{20.20}\right) + \frac{0.02}{10.10} \] \[ \frac{\Delta \rho}{\rho} = \frac{0.02}{19.42} + \frac{0.04}{20.20} + \frac{0.02}{10.10} \] Convert the relative error to a percentage by multiplying by 100: \[ % \text{ Error} = \left( \frac{0.02}{19.42} + \frac{0.04}{20.20} + \frac{0.02}{10.10} \right) \times 100% \] \[ % \text{ Error} \approx (0.00103 + 0.00198 + 0.00198) \times 100% \] \[ % \text{ Error} \approx 0.00499 \times 100% \approx 0.5% \] Step 4: Final Answer:
The percentage error in density is \(0.5%\).