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 $$1s,$$ then the reported mean time should be-
A.
$$92 \pm 1.8\,s$$
B.
$$92 \pm 3\,s$$
C.
$$92 \pm 2\,s$$
D.
$$92 \pm 5.0\,s$$
Answer :
$$92 \pm 1.8\,s$$
Solution :
$$\eqalign{
& \Delta T = \frac{{\left| {\Delta {T_1}} \right| + \left| {\Delta {T_2}} \right| + \left| {\Delta {T_3}} \right| + \left| {\Delta {T_4}} \right|}}{4} \cr
& = \frac{{2 + 1 + 3 + 0}}{4} \cr
& = 1.5 \cr} $$
As the resolution of measuring clock is 1.5 therefore the mean time should be $$92 \pm 1.5$$
Releted MCQ Question on Basic Physics >> Unit and Measurement
Releted Question 1
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Pressure depends on distance as, $$P = \frac{\alpha }{\beta }exp\left( { - \frac{{\alpha z}}{{k\theta }}} \right),$$ where $$\alpha ,$$ $$\beta $$ are constants, $$z$$ is distance, $$k$$ is Boltzman’s constant and $$\theta $$ is temperature. The dimension of $$\beta $$ are-