We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be $2.63 \;s , 2.56 \;s , 2.42\; s , 2.71 \;s$ and $2.80 \;s$. Calculate the absolute errors, relative error or percentage error.
Answer The mean perlod of oscillation of the pendulum
$T \;=\frac{(2.63+2.56+2.42+2.71+2.80) \,s}{5}$
$\quad=\frac{13.12}{5} \;s$
$=2.624\, s $
$=2.62 \,s$
As the periods are measured to a resolution of $0.01 \,s ,$ all times are to the second decimal; it is proper to put this mean perlod also to the second decimal.
The errors in the measurements are
$2.63 \,s -2.62 \,s =0.01 \,s$
$2.56 \,s-2.62 \,s=-0.06 \,s$
$2.42\, s -2.62\, s =-0.20 \,s$
$2.71 \,s -2.62\, s =0.09 \,s$
$2.80\, s-2.62\, s=0.18\, s$
The arthmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
$ \Delta T_{\text {mean}} =[(0.01+0.06+0.20+0.09+0.18) \,s ] / 5 $
$=0.54 \,s / 5 $
$=0.11 \,s $
$T=2.6 \pm 0.1 \,s$
$\delta a=\frac{0.1}{2.6} \times 100=4 \%$
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Can error be completely eliminated ?
A certain body weighs $22.42\;g$ and has a measured volume of $4.7 \;cc .$ The possible error in the measurement of mass and volume are $0.01\; gm$ and $0.1 \;cc .$
Then maximum error in the density will be
A sliver wire has mass $(0.6 \pm 0.006) \; g$, radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$
Error in volume of a sphere is $6\%$. Error in its radius will be .......... $\%$