$z \pm \Delta z=\frac{x \pm \Delta x}{y \pm \Delta y}=\frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1} .$

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 ($2$)

diameter of capillary, $D= 1.25 \times 10^{-2}\; m$

rise of water, $h=1.45 \times 10^{-2}\; m $ 

Using $g= 9.80 \;m/s^2$ and the simplified relation $T = \frac{{rhg}}{2}\times 10^3 N/m$ , the possible error in surface tension is ........... $\%$