A student measures time for 20 oscillations o | vedclass

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(A) Step $1$: Calculate the mean time $(T_{mean})$.$T_{mean} = \frac{30 + 32 + 35 + 35}{4} = \frac{132}{4} = 33 \ s$.Step $2$: Calculate the absolute

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$(33 \pm 2)$

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(A) Step $1$: Calculate the mean time $(T_{mean})$.$T_{mean} = \frac{30 + 32 + 35 + 35}{4} = \frac{132}{4} = 33 \ s$.Step $2$: Calculate the absolute errors $(\Delta T_i = |T_i - T_{mean}|)$.$\Delta T_1 = |30 - 33| = 3 \ s$$\Delta T_2 = |32 - 33| = 1 \ s$$\Delta T_3 = |35 - 33| = 2 \ s$$\Delta T_4 = |35 - 33| = 2 \ s$Step $3$: Calculate the mean absolute error $(\Delta T_{mean})$.$\Delta T_{mean} = \frac{3 + 1 + 2 + 2}{4} = \frac{8}{4} = 2 \ s$.Step $4$: The result is expressed as $T_{mean} \pm \Delta T_{mean}$.Thus,the correct mean time is $(33 \pm 2) \ s$.

$A$ student measures time for $20$ oscillations of a simple pendulum as $30 \ s, 32 \ s, 35 \ s$ and $35 \ s$. If the minimum division in the measuring clock is $1 \ s$,then the correct mean time (in seconds) is:

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