Question Video: Finding the Difference between the Relative Errors of Two Measurements | Nagwa Question Video: Finding the Difference between the Relative Errors of Two Measurements | Nagwa

Question Video: Finding the Difference between the Relative Errors of Two Measurements Physics • First Year of Secondary School

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An experimental apparatus for measuring the acceleration due to gravity produces an absolute error of 0.05 m/s². The apparatus is used on two different planets. One planet has an accepted value for its gravitational acceleration of 3.72 m/s² and the other has an accepted value for its gravitational acceleration of 8.87 m/s². Find the difference between the percent relative errors of the measurements of gravitational acceleration made by the apparatus on the two planets. Give your answer to one decimal place.

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Video Transcript

An experimental apparatus for measuring the acceleration due to gravity produces an absolute error of 0.05 meters per second squared. The apparatus is used on two different planets. One planet has an accepted value for its gravitational acceleration of 3.72 meters per second squared, and the other has an accepted value for its gravitational acceleration of 8.87 meters per second squared. Find the difference between the percent relative errors of the measurements of gravitational acceleration made by the apparatus on the two planets. Give your answer to one decimal place.

There is a lot of information in this statement, so let’s zero in on what we actually need to do. We are told to find the difference between the percent relative errors of two different measurements. So the first thing that we’ll need to know is how to calculate a percent relative error. Relative error is one of the ways to quantify the difference between a measured value and the accepted value of the same quantity. In particular, relative error expresses this difference as a fraction of the accepted value. Now the actual numerical difference between the measured value and the accepted value of a given quantity is known as the absolute error.

So using this, we can rephrase the definition for relative error as the absolute error divided by the accepted value of the quantity. This will give us a fraction. And to convert that fraction into a percent, we simply multiply by 100. Looking back at our statement, we see that we are given the absolute error of the apparatus, which is the absolute error for every measurement that is taken. And additionally, we know the accepted values for the gravitational acceleration of each of our planets.

From our definition for relative error, we see that this is exactly the information that we need to calculate the relative error of measurements of gravitational acceleration taken by this apparatus on each planet. On the first planet, the accepted value for the gravitational acceleration is 3.72 meters per second squared and the absolute error of any measurement is 0.05 meters per second squared. So the relative error on planet one is 0.05 meters per second squared, the absolute error, divided by 3.72 meters per second squared, the accepted value for the gravitational acceleration.

We note that the numerator and denominator have the same units, meters per second squared. So the overall quotient is actually dimensionless. This is exactly correct. Relative error is a fraction that does not depend on the particular choice of units we choose for our measurement. It only depends on the relative sizes of the actual values. 0.05 divided by 3.72 is approximately 0.0134, which is 1.34 percent. Note that we are keeping two decimal places in this answer because our final answer is meant to be given to one decimal place. So if we keep two decimal places during the calculation, we can give our answer accurate to one decimal place.

On the second planet, the absolute error is again 0.05 meters per second squared. But now the accepted value for the gravitational acceleration is 8.87 meters per second squared. Once again, the units in the numerator are the same as the units in the denominator, exactly as they should be. And the result is a dimensionless fraction. 0.05 divided by 8.87 is approximately 0.0056, which is 0.56 percent.

Now that we have calculated the percent relative error for each measurement, we need to find the difference between these quantities. We will choose to perform the difference so that our result is positive. This is sensible because relative error is always a positive quantity, so it makes sense to have the difference between two relative errors be positive. The difference between the relative errors is 1.34 percent minus 0.56 percent, which is 0.78 percent. And rounding to one decimal place, we have that the difference between the percent relative errors of measurements taken of the gravitational acceleration on each of these two planets is 0.8 percent. From this answer, we see why relative error is such a useful quantity when describing measurements.

Although both measurements have the same absolute error, the net effect of this absolute error on the measurement is smaller on the second planet, where the gravitational acceleration is larger, than it is on the first planet, where the gravitational acceleration is smaller. To consider an extreme example, on a planet where the gravitational acceleration were 0.01 meters per second squared, our experimental apparatus would be completely useless because the absolute error of the apparatus is five times larger than the value of the quantity that we’re trying to measure. On such a planet, the relative error would be 500 percent.

By taking into account the relative scales of the absolute error of the measurement as well as the actual value we are trying to measure, the relative error gives us a good indicator of whether or not a measurement is reasonable.

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