- Home
- Standard 11
- Physics
हम एक सरल लोलक का दोलन-काल ज्ञात करते हैं। प्रयोग के क्रमिक मापनों में लिए गए पाठ्यांक हैं $: 2.63, s , 2.56\, s , 2.42\, s , 2.71\, s$ एवं $2.80\, s$ । निरपेक्ष त्रुटि, सापेक्ष त्रुटि एवं प्रतिशत त्रुटि परिकलित कीजिए।
Solution
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 $
$T=2.6 \pm 0.1 \,s$
$\delta a=\frac{0.1}{2.6} \times 100=4 \%$