If $Q= \frac{X^n}{Y^m}$ and $\Delta X$ is absolute error in the measurement of $X,$ $\Delta Y$ is absolute error in the measurement of $Y,$ then absolute error $\Delta Q$ in $Q$ is
$\Delta Q = \pm \left( {n\frac{{\Delta X}}{X} + m\frac{{\Delta Y}}{Y}} \right)$
$\Delta Q = \pm \left( {n\frac{{\Delta X}}{X} + m\frac{{\Delta Y}}{Y}} \right)Q$
$\Delta Q = \pm \left( {n\frac{{\Delta X}}{X} - m\frac{{\Delta Y}}{Y}} \right)Q$
$\Delta Q = \pm \left( {n\frac{{\Delta X}}{X} - m\frac{{\Delta Y}}{Y}} \right)$
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
Explain uncertainty or error in given measurement by suitable example.
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 ($2$)
In an experiment to determine the acceleration due to gravity $g$, the formula used for the time period of a periodic motion is $T=2 \pi \sqrt{\frac{7(R-r)}{5 g}}$. The values of $R$ and $r$ are measured to be $(60 \pm 1) \mathrm{mm}$ and $(10 \pm 1) \mathrm{mm}$, respectively. In five successive measurements, the time period is found to be $0.52 \mathrm{~s}, 0.56 \mathrm{~s}, 0.57 \mathrm{~s}, 0.54 \mathrm{~s}$ and $0.59 \mathrm{~s}$. The least count of the watch used for the measurement of time period is $0.01 \mathrm{~s}$. Which of the following statement($s$) is(are) true?
($A$) The error in the measurement of $r$ is $10 \%$
($B$) The error in the measurement of $T$ is $3.57 \%$
($C$) The error in the measurement of $T$ is $2 \%$
($D$) The error in the determined value of $g$ is $11 \%$
The acceleration due to gravity is measured on the surface of earth by using a simple pendulum. If $\alpha$ and $\beta$ are relative errors in the measurement of length and time period respectively, then percentage error in the measurement of acceleration due to gravity is