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A sliver wire has mass $(0.6 \pm 0.006) \; g$, radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$
$4$
$3$
$6$
$7$
Solution
$M=(0.6 \pm 0.006) \; g$
$r=(0.5 \pm 0.005) \; mm$
$l=(4 \pm 0.04) \; cm$
$\rho=\frac{m}{V}$
$\Rightarrow \frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$
(Volume of cylinder $=\pi r^{2} l$ )
$=\frac{0.006}{0.6}+\frac{2 \times 0.005}{0.5}+\frac{0.04}{4}$
$100 \times \frac{\Delta \rho}{\rho}=4 \times 10^{-2} \times 100$
$\frac{\Delta \rho}{\rho} \times 100=4 \%$
Similar Questions
Students $I$, $II$ and $III$ perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum.
They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.
Least count for length $=0.1 \mathrm{~cm}$
Least count for time $=0.1 \mathrm{~s}$
| Student | Length of the pendulum $(cm)$ | Number of oscillations $(n)$ | Total time for $(n)$ oscillations $(s)$ | Time period $(s)$ |
| $I.$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
| $II.$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
| $III.$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
If $\mathrm{E}_{\mathrm{I}}, \mathrm{E}_{\text {II }}$ and $\mathrm{E}_{\text {III }}$ are the percentage errors in g, i.e., $\left(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100\right)$ for students $\mathrm{I}, \mathrm{II}$ and III, respectively,
