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In the measurment of the period of a simple pendulum the readings turn out to be 

(1) 2.63 s  (2) 2.56 s (3) 2.42 s (4) 2.71 s  (5) 2.80 s 

Calculate % error in the measurment 

Option: 1

\pm 4


Option: 2

\pm 3


Option: 3

\pm 5


Option: 4

\pm 2


Answers (1)

best_answer

As we have learned 

percentage error =\dpi{100} \frac{\Delta\bar{ a}}{a_{m}}\times 100%

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T_{mean} = \frac{2.63+2.56+2.42+2.71+2.80}{5} = 2.624\\\\ T_{mean}\approx 2.62 \\\\ absolute\: \: \: error \\ \Delta T_1 = 2.62 - 2.63 = -0.01 s \: \: \: \Delta T_2 = 2.62 - 2.56 = + 0.06s\\\\\Delta T_3 = 2.62 - 2.42 = + 0.20 s \: \: \Delta T_4 = 2.62-2.71 = -0.09s\\\\ \Delta T_5 = 2.62 -2.80 = -0.18s

\Delta T _{mean} = \frac{\sum \left | \Delta T_1 \right |}{n} = \frac{0.54}{5} = 0.11 s

 so, percentage \ error=\pm \frac{\Delta T_{mean}}{T_{mean}}\times100 = \pm \frac{0.11}{2.62} \times 100 \ \approx \pm 4 \%

 

 

 

 

Posted by

sudhir kumar

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