Question 225: 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

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