JEE Advance - Physics (2018 - Paper 1 Offline - No. 15)
$$$z \pm \Delta z = {{x \pm \Delta x} \over {y \pm \Delta y}} = {x \over y}\left( {1 \pm {{\Delta x} \over x}} \right){\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}}.$$$
The series expansion for $${\left( {1 \pm {{\Delta y} \over y}} \right)^{ - 1}},$$ to first power in $$\Delta y/y.$$ is $$1 \pm \left( {\Delta y/y} \right).$$ The relative errors in independent variables are always added. So the error in $$z$$ will be
$$$\Delta z = z\left( {{{\Delta x} \over x} + {{\Delta y} \over y}} \right).$$$
The above derivation makes the assumption that $$\Delta x/x < < 1,$$ $$\Delta y/y < < 1.$$ Therefore, the higher powers of these quantities are neglected.
In an experiment the initial number of radioactive nuclei is $$3000.$$ It is found that $$1000 \pm 40$$ nuclei decayed in the first $$1.0s.$$ For $$\left| x \right| < < 1.$$ $$\ln \left( {1 + x} \right) = x$$ up to first power in $$x.$$ The error $$\Delta \lambda ,$$ in the determination of the decay constant $$\lambda ,$$ in $${s^{ - 1}},$$ is
Explanation
The law of radioactive decay gives number of undecayed nuclei N at time t as $$N = {N_0}{e^{ - \lambda t}}$$, where N0 is number of nuclei at t = 0 and $$\lambda$$ is the decay constant. The number of nuclei decayed till time t is given by $${N_d} = {N_0} - N$$. Thus, the decay constant is given by
$$\lambda = {1 \over t}\ln {{{N_0}} \over N} = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}}$$.
Since, N0 = 3000 and t = 1 s are given without any error, we assume that error in $$\lambda$$ is due to error in Nd only. Hence, we can write above equation as
$$\lambda + \Delta \lambda = {1 \over t}\ln {{{N_0}} \over {{N_0} - ({N_d} + \Delta {N_d})}}$$
$$ = {1 \over t}\ln {{{N_0}} \over {({N_0} - {N_d})(1 - \Delta {N_d}/({N_0} - {N_d}))}}$$
$$ = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}} - {1 \over t}\ln \left( {1 - {{\Delta {N_d}} \over {{N_0} - {N_d}}}} \right)$$
$$ \approx \lambda + {1 \over t}{{\Delta {N_d}} \over {{N_0} - {N_d}}}$$,
which gives $$\Delta \lambda = {{\Delta {N_d}} \over {t({N_0} - {N_d})}}$$. Substitute $$\Delta$$Nd = 40, t = 1 s, and Nd = 1000 to get $$\Delta$$$$\lambda$$ = 0.02.
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