$\frac{{100}}{{dt}} = \frac{{\sigma 4\pi {r^2}}}{{\frac{4}{3}\pi {r^3}\rho \,c\,J}}({200^4} - {0^4})$
==> $dt = \frac{{r\rho \,c\,J}}{{48\sigma }} \times {10^{ - 6}}s = \frac{{r\rho \,c}}{\sigma }.\frac{{4.2}}{{48}} \times {10^{ - 6}}$
$ = \frac{7}{{80}}\frac{{r\rho \,c}}{\sigma }\mu \,s\tilde --\frac{7}{{72}}\frac{{r\rho \,c}}{\sigma }\mu \,s$ [As $J = 4.2$ ]
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$\mathrm{z} \pm \Delta \mathrm{z}=\frac{\mathrm{x} \pm \Delta \mathrm{x}}{\mathrm{y} \pm \Delta \mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}}\left(1 \pm \frac{\Delta \mathrm{x}}{\mathrm{x}}\right)\left(1 \pm \frac{\Delta \mathrm{y}}{\mathrm{y}}\right)^{-1} .$
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 ($2$)
