Numerical Problems Based on Measurement of Time for Class 11 Physics

Total life span of the man, $T = 60\text{ years}$.

We convert this time into seconds (considering $365.25\text{ days}$ in a year to account for leap years):

Time taken for one heart beat, $t = 0.8\text{ s}$.

Total number of heart beats ($N$) is the total time divided by the time per beat:

Total time elapsed, $T = 70\text{ years}$.

Converting this time into seconds:

The difference (error) in the clocks over this entire duration is $\Delta T = 0.2\text{ s}$.

The accuracy in measuring a time interval of $1\text{ s}$ is the error per second:

Thus, the accuracy of the standard atomic clock is approximately $1\text{ s}$ in $10^{10}\text{ s}$.

Given the actual age of the universe $T_u = 10^{10}\text{ years}$ and the actual time mankind has existed $T_m = 10^6\text{ years}$.

The fraction of the universe’s life that mankind has existed is:

If the scaled age of the universe is exactly $1\text{ day}$, we first convert $1\text{ day}$ into seconds:

The scaled time for mankind’s existence is the same fraction of this new universe age: