Question

A body has weight $22.42$ gm and has a measured volume of $4.7$ cc. The possible error in the measurement of mass and volume is $0.01$ gm and $0.1$ cc. The maximum error in the density will be
A. $22\% $
B. $2.2\% $
C. $0.22$%
D. $0.022\% $

Answer 1

Hint:We can find the maximum error in the density of the body by using error analysis of quantities and errors in a product and division. Also, we use the concept of relative or fractional error and percentage error.

Formulae used:
Relative / fractional error $ = \dfrac{{\Delta a}}{a}$
Percentage error % $ = \dfrac{{\Delta a}}{a} \times 100$
Maximum error in division $ = \left( {\dfrac{{\Delta a}}{a} + \dfrac{{\Delta b}}{b}} \right)$
Where, \[\Delta a,\Delta b\] - are the errors in quantities $a,b$respectively.

Complete step by step answer:
We are given that a body has weight $22.42$ gm and has a measured volume of $4.7$ cc and the possible error in the measurement of mass and volume is $0.01$ gm and $0.1$ cc, then,
$m = 22.42$ gm , \[\Delta m = 0.01\] gm
And $v = 4.7$ cc, $\Delta v = 0.1$ cc
Then, the relative error in measurement of mass and volume will be
\[\dfrac{{\Delta m}}{m} = \dfrac{{0.01}}{{22.42}} = \dfrac{1}{{2242}}\] and \[\dfrac{{\Delta v}}{v} = \dfrac{{0.1}}{{4.7}} = \dfrac{1}{{47}}\]
We know that, Density of the body is mass per volume of the body.
$d = \dfrac{m}{v}$

The error in the density will be
\[\dfrac{{\Delta d}}{d} = \dfrac{{\Delta m}}{m} - \dfrac{{\Delta v}}{v}\]
But, the maximum error in the density will be as
\[{\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }} = \dfrac{{\Delta m}}{m} + \dfrac{{\Delta v}}{v}\]
Percentage maximum error in measurement of density is
\[{\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = \left( {\dfrac{{\Delta m}}{m} + \dfrac{{\Delta v}}{v}} \right) \times 100\% \]
Substituting the values, we get
\[{\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = \left( {\dfrac{1}{{2242}} + \dfrac{1}{{47}}} \right) \times 100\% \]
\[\Rightarrow {\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = 0.0217 \times 100\% \]
\[\therefore {\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = 2.17\% \approx 2.2\% \]

Hence, option B is correct.

Note:We should identify the original quantity and the error of the measurement.The maximum value of fractional error in the division or product of two or more quantities is equal to the sum of the fractional errors in the individual quantities.