Question 240: Units , Dimensions and Error Analysis

Question

$	ext { A physical quantity } Q 	ext { is related to four observables } a, b, c, d 	ext { as follows : }$
$Q = \frac{ab^4}{cd}$
where, $\mathrm{a}=(60 \pm 3) \mathrm{Pa} ; \mathrm{b}=(20 \pm 0.1) \mathrm{m} ; \mathrm{c}=(40 \pm 0.2) \mathrm{Nsm}^{-2}$ and $\mathrm{d}=(50 \pm 0.1) \mathrm{m}$, then the percentage error in Q is $\frac{x}{1000}$, where $x=$ _________ .

Accepted Answer

Given, $Q = {{a{b^4}} \over {cd}}$
$a = (60 \pm 3)\,Pa \Rightarrow a = 60\,Pa,\,\Delta a = 3Pa$
$b = (20 \pm 0.1)\,m \Rightarrow b = 20m,\,\Delta b = 0.1\,m$
$c = (40 \pm 0.2)\,Ns{m^{ - 2}} \Rightarrow c = 40\,Ns{m^{ - 2}},\,\Delta c = 0.2\,Ns{m^{ - 2}}$
$d = (50 \pm 0.1)\,m \Rightarrow \,d = 50\,m,\,\Delta d = 0.1m$
As, $Q = {{a{b^4}} \over {cd}}$
by taking ln on both sides,
$\ln Q = \ln a + u\ln b - \ln c - \ln d$
Now, by differentiating,
${{dQ} \over Q} = {{da} \over a} + 4{{db} \over b} - {{dc} \over c} - {{dd} \over d}$
So, maximum fractional error in Q is given by,
${{\Delta Q} \over Q} = {{\Delta a} \over a} + 4{{\Delta b} \over b} + {{\Delta c} \over c} + {{\Delta d} \over d}$
$ \Rightarrow {{\Delta Q} \over Q} = {3 \over {60}} + 4\left( {{{0.1} \over {20}}} ight) + {{0.2} \over {40}} + {{0.1} \over {40}}$
$ = {1 \over {20}} + {1 \over {50}} + {1 \over {200}} + {1 \over {500}}$
$ \Rightarrow {{\Delta Q} \over Q} = {{50 + 20 + 5 + 2} \over {1000}} = {{77} \over {1000}}$
Hence the % error in $Q = {{\Delta Q} \over Q} 	imes 100\% $
$ = {{7700} \over {1000}}\%  = {x \over {1000}}$ (given)
So, $x = 7700$

Units , Dimensions and Error Analysis

Integer / Numeric Answer Type