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Question: A body has weight \(22.42\) gm and has a measured volume of \(4.7\) cc. The possible error in the me...

A body has weight 22.4222.42 gm and has a measured volume of 4.74.7 cc. The possible error in the measurement of mass and volume is 0.010.01 gm and 0.10.1 cc. The maximum error in the density will be
A. 22%22\%
B. 2.2%2.2\%
C. 0.220.22%
D. 0.022%0.022\%

Explanation

Solution

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 =Δaa = \dfrac{{\Delta a}}{a}
Percentage error % =Δaa×100 = \dfrac{{\Delta a}}{a} \times 100
Maximum error in division =(Δaa+Δbb) = \left( {\dfrac{{\Delta a}}{a} + \dfrac{{\Delta b}}{b}} \right)
Where, Δa,Δb\Delta a,\Delta b - are the errors in quantities a,ba,brespectively.

Complete step by step answer:
We are given that a body has weight 22.4222.42 gm and has a measured volume of 4.74.7 cc and the possible error in the measurement of mass and volume is 0.010.01 gm and 0.10.1 cc, then,
m=22.42m = 22.42 gm , Δm=0.01\Delta m = 0.01 gm
And v=4.7v = 4.7 cc, Δv=0.1\Delta v = 0.1 cc
Then, the relative error in measurement of mass and volume will be
Δmm=0.0122.42=12242\dfrac{{\Delta m}}{m} = \dfrac{{0.01}}{{22.42}} = \dfrac{1}{{2242}} and Δvv=0.14.7=147\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=mvd = \dfrac{m}{v}

The error in the density will be
Δdd=ΔmmΔvv\dfrac{{\Delta d}}{d} = \dfrac{{\Delta m}}{m} - \dfrac{{\Delta v}}{v}
But, the maximum error in the density will be as
(Δdd)max=Δmm+Δvv{\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }} = \dfrac{{\Delta m}}{m} + \dfrac{{\Delta v}}{v}
Percentage maximum error in measurement of density is
(Δdd)max%=(Δmm+Δvv)×100%{\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = \left( {\dfrac{{\Delta m}}{m} + \dfrac{{\Delta v}}{v}} \right) \times 100\%
Substituting the values, we get
(Δdd)max%=(12242+147)×100%{\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = \left( {\dfrac{1}{{2242}} + \dfrac{1}{{47}}} \right) \times 100\%
(Δdd)max%=0.0217×100%\Rightarrow {\left( {\dfrac{{\Delta d}}{d}} \right)_{\max }}\% = 0.0217 \times 100\%
(Δdd)max%=2.17%2.2%\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.