Zero error of an instrument introduces

A packet contains silver powder of mass $20.23 \,g \pm 0.01 \,g$. Some of the powder of mass $5.75 \,g \pm 0.01 \,g$ is taken out from it. The mass of the powder left back is ................

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 $X$ is related to four measurable quantities $a,\, b,\, c$ and $d$ as follows $X = a^2b^3c^{\frac {5}{2}}d^{-2}$. The percentange error in the measurement of $a,\, b,\, c$ and $d$ are $1\,\%$, $2\,\%$, $3\,\%$ and $4\,\%$ respectively. What is the percentage error in quantity $X$ ? If the value of $X$ calculated on the basis of the above relation is $2.763$, to what value should you round off the result.

Durring Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale. The $20^{\text {th }}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \ kg$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2} m$ and $3.25 \times 10^{-2} m$ of the main scale but now the $45^{\text {th }}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 m$. and its cross-sectional area is $8 \times 10^{-7} m ^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} m$. The maximum percentage error in the Young's modulus of the wire is