JEE MAIN - Physics (2016 (Offline) - No. 26)
A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :
92 $$ \pm $$ 1.8 s
92 $$ \pm $$ 3 s
92 $$ \pm $$ 2 s
92 $$ \pm $$ 5.0 s
Explanation
Here t1 = 90 s, t2 = 91 s, t3 = 95 s, t4 = 92 s
Mean(t) = $${{{t_1} + {t_2} + {t_3} + {t_4}} \over 4}$$
= $${{90 + 91 + 95 + 92} \over 4}$$ = 92 s
Now mean deviation
= $${{2 + 1 + 3 + 0} \over 4}$$ = 1.5 s
Since least count of clock is one second, so reported mean time
= (92 $$ \pm $$ 2) s
Mean(t) = $${{{t_1} + {t_2} + {t_3} + {t_4}} \over 4}$$
= $${{90 + 91 + 95 + 92} \over 4}$$ = 92 s
Now mean deviation
= $${{2 + 1 + 3 + 0} \over 4}$$ = 1.5 s
Since least count of clock is one second, so reported mean time
= (92 $$ \pm $$ 2) s
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