Question 274: Units , Dimensions and Error Analysis

Question

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with
a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the
main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a
relative error of 2 %, the relative percentage error in the density is

Accepted Answer

Least count of screw gauge
$$ = {{Pitch} \over {No.\,of\,divisions\,on\,the\,circular\,scale}}$$
$$ = {{0.5\,mm} \over {50}}$$ = 0.01 mm
Diameter of ball, $$D = MSR + CSR 	imes LC$$
= 2.5 mm + (20)(0.01 mm) = 2.7 mm
Density, $$ho  = {{Mass} \over {Volume}} = {M \over {{{4\pi } \over 3}{{\left( {{D \over 2}} ight)}^3}}}$$
The relative error in the density is
$${{\Delta ho } \over ho } = {{\Delta M} \over M} + {{3\Delta D} \over D}$$
The relative percentage in the density is
$${{\Delta ho } \over ho } 	imes 100 = \left[ {{{\Delta M} \over M} + {{3\Delta D} \over D}} ight] 	imes 100$$
$$ = 2\%  + {{3 	imes 0.01} \over {2.7}} 	imes 100 = 2\%  + 1.11\%  = 3.11\% $$

Answer

0.9 %

Answer

2.4 %

Answer

3.1 %

Answer

4.2 %

Units , Dimensions and Error Analysis

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