Two resistors of resistances $R_1 = (100 \pm 3) \,\Omega $ and $R_2 = (200 \pm 4)$ are connected in series. The maximm absolute error and percentage error in equivalent resistance of the series combination is
$7\, \Omega , 2.3 \%$
$1\, \Omega , 0.3 \%$
$3\, \Omega , 1 \%$
$4\, \Omega , 1.3 \%$
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
$\mathrm{z} \pm \Delta \mathrm{z}=\frac{\mathrm{x} \pm \Delta \mathrm{x}}{\mathrm{y} \pm \Delta \mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}}\left(1 \pm \frac{\Delta \mathrm{x}}{\mathrm{x}}\right)\left(1 \pm \frac{\Delta \mathrm{y}}{\mathrm{y}}\right)^{-1} .$
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 $\mathrm{z}$ will be $\Delta \mathrm{z}=\mathrm{z}\left(\frac{\Delta \mathrm{x}}{\mathrm{x}}+\frac{\Delta \mathrm{y}}{\mathrm{y}}\right)$.
The above derivation makes the assumption that $\Delta x / x<<1, \Delta \mathrm{y} / \mathrm{y} \ll<1$. Therefore, the higher powers of these quantities are neglected.
($1$) Consider the ratio $\mathrm{r}=\frac{(1-\mathrm{a})}{(1+\mathrm{a})}$ to be determined by measuring a dimensionless quantity a.
If the error in the measurement of $\mathrm{a}$ is $\Delta \mathrm{a}(\Delta \mathrm{a} / \mathrm{a} \ll<1)$, then what is the error $\Delta \mathrm{r}$ in
$(A)$ $\frac{\Delta \mathrm{a}}{(1+\mathrm{a})^2}$ $(B)$ $\frac{2 \Delta \mathrm{a}}{(1+\mathrm{a})^2}$ $(C)$ $\frac{2 \Delta \mathrm{a}}{\left(1-\mathrm{a}^2\right)}$ $(D)$ $\frac{2 \mathrm{a} \Delta \mathrm{a}}{\left(1-\mathrm{a}^2\right)}$
($2$) In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm$ $40$ nuclei decayed in the first $1.0 \mathrm{~s}$. For $|\mathrm{x}| \ll 1$, In $(1+\mathrm{x})=\mathrm{x}$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $\mathrm{s}^{-1}$, is
$(A) 0.04$ $(B) 0.03$ $(C) 0.02$ $(D) 0.01$
Give the answer quetion ($1$) and ($2$)
A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as follows: $P=\frac{a^{2} b^{2}}{(\sqrt{c} d)}$ The percentage errors of measurement in $a, b, c$ and $d$ are $1 \%, 3 \%, 4 \%$ and $2 \%$ respectively. What is the percentage error in the quantity $P$ ? If the value of $P$ calculated using the above relation turns out to be $3.763,$ to what value should you round off the result?
The most accurate reading of the length of a $6.28 \,cm$ long fibre is ............... $cm$
Out of absolute error, relative error and fractional error which has unit and which has no unit ?
A physical quantity is $A = P^2/Q^3.$ The percentage error in measurement of $P$ and $Q$ is $x$ and $y$ respectively. Maximum error in measurement of $A$ is