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

Concept: Density of a cylinder \[ \rho = \frac{Mass}{Volume} \] Volume of cylinder \[ V = \pi r^2 l = \frac{\pi d^2 l}{4} \] Thus \[ \rho = \frac{4M}{\pi d^2 l} \] For error propagation \[ \frac{d\rho}{\rho} = \frac{dM}{M} + \frac{dl}{l} + \frac{2\,dd}{d} \] Step 1: Substitute given errors.} \[ \frac{d\rho}{\rho} = \left( \frac{0.02}{19.42} + \frac{0.02}{10.10} + \frac{2\times0.02}{20.20} \right) \]
Step 2: Calculate percentage error.} \[ \frac{d\rho}{\rho} = (0.00103 + 0.00198 + 0.00198) \] \[ \frac{d\rho}{\rho} \approx 0.005 \] \[ %\text{ error} = 0.005 \times 100 \] \[ %\text{ error} = 0.5% \] Final Result \[ %\text{ error in density} = 0.5% \]