Explain uncertainty or error in given measurement by suitable example.
Explain uncertainty or error in given measurement by suitable example.
$(1)$ Length and breadth of a thin rectangular plate is $l=16.2 \mathrm{~cm}$ and breadth $b=10.1 \mathrm{~cm}$ Least count of meterscale is $0.1 \mathrm{~cm}$ hence absolute error in measurement should be $0.1 \mathrm{~cm}$.
$l=(16.2 \pm 0.1) \mathrm{cm}$
$b=(10.1 \pm 0.1) \mathrm{cm}$
$\%$ error in measurement of length,
$l=(16.2 \pm 0.6 \%) \mathrm{cm}$
$\%$ error in measurement of breadth,
$b=(10.1 \pm 1 \%) \mathrm{cm}$
Area of rectangular plate,
$\mathrm{A} =l b$
$=16.2 \times 10.1=163.62 \mathrm{~cm}^{2}$
$\%$ error in $\mathrm{A}$ :
$=\frac{\Delta \mathrm{A}}{\mathrm{A}} \times 100=\frac{\Delta l}{l} \times 100+\frac{\Delta b}{b} \times 100$
$=(0.6 \%+1 \%)=1.6 \%$
$\Delta \mathrm{A} =\frac{1.6 \times 163.62}{100}=2.6$
Area : $\mathrm{A}=(163.62 \pm 2.6) \mathrm{cm}^{2}$
Here, minimum significant digit are $3$ hence area should be represented as, $\mathrm{A} \approx 163 \pm 3 \mathrm{~cm}^{2}$
$\therefore$ Error in measurement of area of plate is $3 \mathrm{~cm}^{2}$.
$(2)$ If a set of experimental data is specified to ' $n$ ' significant figures, a result obtained by combining the data will also be valid to ' $n$ ' significant figures.
However, if data is subtracted the number of significant figures can be reduced. For example,
$12.9 \mathrm{~g}-7.06 \mathrm{~g}=5.84 \mathrm{~g}$
Here, there are $3$ significant digits. But in subtraction digit after decimal point are considered hence this will be represented as $5.8 \mathrm{~g}$.
Similar Questions
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 $......\,\%$
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 $......\,\%$
- [JEE MAIN 2022]
What is accuracy in measurement ? Accuracy depend on which factors ?
What is accuracy in measurement ? Accuracy depend on which factors ?
The radius ( $\mathrm{r})$, length $(l)$ and resistance $(\mathrm{R})$ of a metal wire was measured in the laboratory as
$\mathrm{r}=(0.35 \pm 0.05) \mathrm{cm}$
$\mathrm{R}=(100 \pm 10) \mathrm{ohm}$
$l=(15 \pm 0.2) \mathrm{cm}$
The percentage error in resistivity of the material of the wire is :
$\mathrm{r}=(0.35 \pm 0.05) \mathrm{cm}$
$\mathrm{R}=(100 \pm 10) \mathrm{ohm}$
$l=(15 \pm 0.2) \mathrm{cm}$
The percentage error in resistivity of the material of the wire is :
- [JEE MAIN 2024]
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $z=x / y$. If the errors in $x, y$ and $z$ are $\Delta x, \Delta y$ and $\Delta z$, respectively, then
$z \pm \Delta z=\frac{x \pm \Delta x}{y \pm \Delta y}=\frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1} .$
The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in $\Delta y / y$, is $1 \mp(\Delta y / y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z=z\left(\frac{\Delta x}{x}+\frac{\Delta y}{y}\right)$.
The above derivation makes the assumption that $\Delta x / x \ll<1, \Delta y / y \ll<1$. Therefore, the higher powers of these quantities are neglected.
($1$) Consider the ratio $r =\frac{(1- a )}{(1+ a )}$ to be determined by measuring a dimensionless quantity a.
If the error in the measurement of $a$ is $\Delta a (\Delta a / a \ll<1)$, then what is the error $\Delta r$ in
$(A)$ $\frac{\Delta a }{(1+ a )^2}$ $(B)$ $\frac{2 \Delta a }{(1+ a )^2}$ $(C)$ $\frac{2 \Delta a}{\left(1-a^2\right)}$ $(D)$ $\frac{2 a \Delta a}{\left(1-a^2\right)}$
($2$) In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm$ 40 nuclei decayed in the first $1.0 s$. For $|x|<1$, In $(1+x)=x$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $s ^{-1}$, is
$(A) 0.04$ $(B) 0.03$ $(C) 0.02$ $(D) 0.01$
Give the answer or quetion ($1$) and ($2$)
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $z=x / y$. If the errors in $x, y$ and $z$ are $\Delta x, \Delta y$ and $\Delta z$, respectively, then
$z \pm \Delta z=\frac{x \pm \Delta x}{y \pm \Delta y}=\frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1} .$
The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in $\Delta y / y$, is $1 \mp(\Delta y / y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z=z\left(\frac{\Delta x}{x}+\frac{\Delta y}{y}\right)$.
The above derivation makes the assumption that $\Delta x / x \ll<1, \Delta y / y \ll<1$. Therefore, the higher powers of these quantities are neglected.
($1$) Consider the ratio $r =\frac{(1- a )}{(1+ a )}$ to be determined by measuring a dimensionless quantity a.
If the error in the measurement of $a$ is $\Delta a (\Delta a / a \ll<1)$, then what is the error $\Delta r$ in
$(A)$ $\frac{\Delta a }{(1+ a )^2}$ $(B)$ $\frac{2 \Delta a }{(1+ a )^2}$ $(C)$ $\frac{2 \Delta a}{\left(1-a^2\right)}$ $(D)$ $\frac{2 a \Delta a}{\left(1-a^2\right)}$
($2$) In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm$ 40 nuclei decayed in the first $1.0 s$. For $|x|<1$, In $(1+x)=x$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $s ^{-1}$, is
$(A) 0.04$ $(B) 0.03$ $(C) 0.02$ $(D) 0.01$
Give the answer or quetion ($1$) and ($2$)
- [IIT 2018]
If radius of the sphere is $(5.3 \pm 0.1)\;cm$. Then percentage error in its volume will be
If radius of the sphere is $(5.3 \pm 0.1)\;cm$. Then percentage error in its volume will be