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Assuming the length of the day uniformly increases by 0.001 seconds per century. Calculate the net effect on the measure of time over 20 centuries.
(a) 3.2 hours
(b) 2.1 hours
(c) 2.4 hours
(d) 5 hours
Answer
473.1k+ views
Hint: First, we will find the total time difference between the start of the 1st century and the end of the 20th century. Next, we will find the average increase in the length of the day by taking the average of the initial time difference and the final time difference. Finally, we will be multiplying the average length of the day to the total number of days to find the net total time difference and then by converting from seconds to hours.
Complete step-by-step solution
Here it is given that the length of the day increases by 0.001 second or 1 millisecond per century. We have to find the time difference after 20 centuries.
We need to find the total difference in time between the beginning of the first century and the end of the 20th century initially.
Total difference = $\left( \dfrac{0.001\text{s}}{1\,\text{century}} \right)\cdot \left( \dfrac{20\,\text{centuries}}{1} \right)=0.02\text{s}$
Since the time difference is increasing uniformly, let us now find the average increase in the length of the day between the first century and the end of the 20th century.
Average $=\dfrac{{{D}_{i}}+{{D}_{f}}}{2}$, where ${{D}_{i}}$ = initial time difference, ${{D}_{f}}$ = final time difference
Average $=\dfrac{0+0.02}{2}$
$\begin{align}
& =\dfrac{0.02}{2} \\
& =0.01 \\
\end{align}$
Therefore, the average increase in the length of the day is 0.01s
Now, finally let us find the total time difference, which is all the time displacement that occurs throughout the 20 centuries added up is equal to T.
Here, we get
$\text{T}\,\text{=}\,{{D}_{avg}}\cdot n$, where ${{D}_{avg}}$ is the average increase in the length of the day and $n$ is the total number of days.
$\begin{align}
& \text{T}\,\text{=}\,\left( \dfrac{0.01\text{s}}{\text{1 day}} \right)\cdot \left( \dfrac{365.25}{1\,\text{year}} \right)\cdot \left( \dfrac{2000\,\text{years}}{1} \right) \\
& =7305
\end{align}$
We got the total time difference as 7305 seconds, but the question has been asked in hours, therefore we will divide total time difference (seconds) by $60\times 60$ to get the obtained time difference in hours.
We get,
$\begin{align}
& \text{T =}\dfrac{7305}{60\times 60} \\
& =\dfrac{7305}{3600} \\
& =2.029
\end{align}$
The closest number to the obtained answer is 2.1 hours.
Hence, the net effect on the measure of time over 20 centuries is 2.1 hours
Note: In this question, in the final step, we took the total number of days in a year is 365.25 as an average because in every leap year we have 366 days and we know that in 1 century there are 100 years, therefore, for 20 centuries there will be 2000 years. Also, note that 1ms = 0.001s.
Complete step-by-step solution
Here it is given that the length of the day increases by 0.001 second or 1 millisecond per century. We have to find the time difference after 20 centuries.
We need to find the total difference in time between the beginning of the first century and the end of the 20th century initially.
Total difference = $\left( \dfrac{0.001\text{s}}{1\,\text{century}} \right)\cdot \left( \dfrac{20\,\text{centuries}}{1} \right)=0.02\text{s}$
Since the time difference is increasing uniformly, let us now find the average increase in the length of the day between the first century and the end of the 20th century.
Average $=\dfrac{{{D}_{i}}+{{D}_{f}}}{2}$, where ${{D}_{i}}$ = initial time difference, ${{D}_{f}}$ = final time difference
Average $=\dfrac{0+0.02}{2}$
$\begin{align}
& =\dfrac{0.02}{2} \\
& =0.01 \\
\end{align}$
Therefore, the average increase in the length of the day is 0.01s
Now, finally let us find the total time difference, which is all the time displacement that occurs throughout the 20 centuries added up is equal to T.
Here, we get
$\text{T}\,\text{=}\,{{D}_{avg}}\cdot n$, where ${{D}_{avg}}$ is the average increase in the length of the day and $n$ is the total number of days.
$\begin{align}
& \text{T}\,\text{=}\,\left( \dfrac{0.01\text{s}}{\text{1 day}} \right)\cdot \left( \dfrac{365.25}{1\,\text{year}} \right)\cdot \left( \dfrac{2000\,\text{years}}{1} \right) \\
& =7305
\end{align}$
We got the total time difference as 7305 seconds, but the question has been asked in hours, therefore we will divide total time difference (seconds) by $60\times 60$ to get the obtained time difference in hours.
We get,
$\begin{align}
& \text{T =}\dfrac{7305}{60\times 60} \\
& =\dfrac{7305}{3600} \\
& =2.029
\end{align}$
The closest number to the obtained answer is 2.1 hours.
Hence, the net effect on the measure of time over 20 centuries is 2.1 hours
Note: In this question, in the final step, we took the total number of days in a year is 365.25 as an average because in every leap year we have 366 days and we know that in 1 century there are 100 years, therefore, for 20 centuries there will be 2000 years. Also, note that 1ms = 0.001s.
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