Problem 73
Question
There are atomic clocks capable of measuring time with an accuracy of 1 part in \(10^{11} .\) If two such clocks are operated with precision, then after running for \(5000 \mathrm{yr}\), these will record (a) a difference of nearly \(2 \mathrm{~s}\) (b) a difference of 1 day (c) a difference of \(10^{11} \mathrm{~s}\) (d) a difference of \(1 \mathrm{yr}\)
Step-by-Step Solution
Verified Answer
(a) a difference of nearly 2 seconds
1Step 1: Understand the Time Measurement Accuracy
The atomic clocks have an accuracy of 1 part in \(10^{11}\). This means that over a large time period, the error in measurement will only be \(1\) part out of \(10^{11}\) parts.
2Step 2: Calculate Total Seconds in 5000 Years
First, convert 5000 years into seconds. Each year has about \(365.25\) days (considering leap years), each day has \(24\) hours, each hour has \(3600\) seconds. So:\[5000 \times 365.25 \times 24 \times 3600 = 1.57788 \times 10^{11} \text{ seconds}\]
3Step 3: Determine Measurement Error Over 5000 Years
Given the clock's precision of 1 part in \(10^{11}\), the error after \(1.57788 \times 10^{11}\) seconds is:\[\frac{1.57788 \times 10^{11}}{10^{11}} = 1.57788 \text{ seconds}\]
4Step 4: Conclusion Based on Calculations
After calculating, the error or difference in time measured by the atomic clocks over \(5000\) years will be approximately \(1.57788\) seconds. So the correct choice is that the difference is nearly \(2\) seconds.
Key Concepts
Time Measurement AccuracyMeasurement Error CalculationAtomic Clock Precision
Time Measurement Accuracy
Time measurement accuracy is an essential aspect of any clock, especially for sophisticated devices like atomic clocks. These clocks operate with an incredible degree of precision. Time measurement accuracy, in this context, is expressed as 1 part in \(10^{11}\). This means for every \(10^{11}\) parts of time measured, only one part might be slightly off.
Such precision ensures that these clocks remain reliable over long periods, only accumulating minor errors.
Such precision ensures that these clocks remain reliable over long periods, only accumulating minor errors.
- This level of accuracy allows us to effectively gauge long intervals of time, making atomic clocks crucial in technological applications that rely on perfect timing.
- Atomic clocks are useful for global positioning systems (GPS), telecommunications, and scientific research.
Measurement Error Calculation
Calculating measurement error is critical to understanding how precise these timekeeping devices are over lengthy periods. For atomic clocks with an accuracy of 1 part in \(10^{11}\), it is significant to determine how much error builds over time.
In a scenario where such a clock operates for \(5000\) years, let's break it down:
1. **Determine Total Time in Seconds**: Convert \(5000\) years into seconds: - There are approximately \(365.25\) days per year (accounting for leap years). - Each day has \(24\) hours, and each hour has \(3600\) seconds. - Therefore, the total time in seconds is: \[5000 \times 365.25 \times 24 \times 3600 = 1.57788 \times 10^{11} \text{ seconds}\] 2. **Calculate Error Over Time**: Given the accuracy of 1 part in \(10^{11}\), calculate potential error: - The potential error is: \[\frac{1.57788 \times 10^{11}}{10^{11}} = 1.57788 \text{ seconds}\]The computed error indicates that even over \(5000\) years, these clocks remain staggeringly precise, with an expected error of only around \(1.57788\) seconds.
In a scenario where such a clock operates for \(5000\) years, let's break it down:
1. **Determine Total Time in Seconds**: Convert \(5000\) years into seconds: - There are approximately \(365.25\) days per year (accounting for leap years). - Each day has \(24\) hours, and each hour has \(3600\) seconds. - Therefore, the total time in seconds is: \[5000 \times 365.25 \times 24 \times 3600 = 1.57788 \times 10^{11} \text{ seconds}\] 2. **Calculate Error Over Time**: Given the accuracy of 1 part in \(10^{11}\), calculate potential error: - The potential error is: \[\frac{1.57788 \times 10^{11}}{10^{11}} = 1.57788 \text{ seconds}\]The computed error indicates that even over \(5000\) years, these clocks remain staggeringly precise, with an expected error of only around \(1.57788\) seconds.
Atomic Clock Precision
Atomic clock precision sets the standard for modern timekeeping. These clocks employ the oscillations of atoms, such as cesium or rubidium, to generate a consistent time interval.
The advantageous precision arises because atomic frequencies are far less affected by environmental changes than mechanical or quartz clocks. Each tick is generated by the oscillation of atoms at stable frequencies, maintaining a reliable measure of time.
The advantageous precision arises because atomic frequencies are far less affected by environmental changes than mechanical or quartz clocks. Each tick is generated by the oscillation of atoms at stable frequencies, maintaining a reliable measure of time.
- Atomic clocks remain the backbone of international timekeeping standards and scientific explorations.
- Their precision is why they serve vital roles in fields like astronomy and GPS technology.
Other exercises in this chapter
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If there is a positive error of \(50 \%\) in the measurement of speed of a body, then the error in the measurement of kinetic energy is (a) \(25 \%\) (b) \(50 \
View solution Problem 75
The radius of the sphere is \((4.3 \pm 0.1) \mathrm{cm}\). The percentage error in its volume is (a) \(\frac{0.1}{4.3} \times 100\) (b) \(3 \times \frac{0.1 \ti
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