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The mass of the body is $10.000\,g$ and its volume is $10.00\,cm^3$. If the measured values are expressed upto the correct significant figures, the maximum error in the measurement of density is
$0.0011\,g\,cm^{-3}$
$0.001\,g\,cm^{-3}$
$0.0001\,g\,cm^{-3}$
$0.021\,g\,cm^{-3}$
Solution
$\frac{\Delta \rho}{\rho}= \frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{\Delta \mathrm{V}}{\mathrm{V}}=\left[\frac{0.001}{10.000}+\frac{0.01}{10.00}\right]=10^{-4}+10^{-3}$
$\quad=0.1 \times 10^{-3}+10^{-3} =1.1 \times 10^{-3}$
$\Delta \rho=1.1 \times 10^{-3} \times \frac{10.000}{10.00}$
$=1.1 \times 10^{-3}$
$=0.0011\,g\,cm^{-3}$
Similar Questions
A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \delta} .$ The value of $g$ is taken to be $9.8 \,{m} / {s}^{2}$, without any significant error, his observation are as following.
| Physical Quantity | Least count of the Equipment used for measurement | Observed value |
| Mass $({M})$ | $1\; {g}$ | $2\; {kg}$ |
| Length of bar $(L)$ | $1\; {mm}$ | $1 \;{m}$ |
| Breadth of bar $(b)$ | $0.1\; {mm}$ | $4\; {cm}$ |
| Thickness of bar $(d)$ | $0.01\; {mm}$ | $0.4 \;{cm}$ |
| Depression $(\delta)$ | $0.01\; {mm}$ | $5 \;{mm}$ |
Then the fractional error in the measurement of ${Y}$ is
