Measurement Of Length - Distance

Measurements of length and distance are done in many ways. Did you know that you can use the average human body as means to measure? For example, the foot (which is literally the length measurement of an average human foot) is around 25 – 30 cm. This unit of measurement is still in use nowadays. We also use units like inches and yards which are still in use but they are not the standard units of length measurements.

A common way of measuring distances is the transit – time method. A signal is sent to a reference point and received back. The time taken for the signal to the reference point and back is measured. The speed of the travelling signal is already known. Using the distance-speed-time relation, the distance is calculated by,

\(2d\) = \(t~\times~v\)

where,

\(d\)is the distance from starting point to reference point

\(t\) is the time taken for the signal to be received back

\(v\) is the speed of the signal

Once\(t\) is calculated, \(2d\) can be calculated from this value. Transit-time measurement can be used to measure large distances to an extent. An interferometer is used to measure shorter distances.

How to find Accurate Measurements?

Check out the image given below for the measurement of length and thickness.

Length Measurement

Do you think that it is possible to measure the thickness of this ID card using the simple ruler available in your geometry boxes? The least count of those rulers is 1mm. What if the thickness of such an object lies between 1mm and 2mm. You can never get an accurate measurement.

To tackle this problem you can club say \(10\) or \(20\) of such ID cards and measure the collective thickness using a ruler. Say you record the measurement of \(20\) such ID cards to be \(3.4 cm\)

Total thickness = \(3.4 cm\)

No of ID cards = \(20\)

Thickness of single ID card = \(\frac{3.4}{20}\) = \(0.17 cm\) = \(1.7 mm\)

Other techniques of measuring lengths include electron microscope or diffraction measurements to compute the length of extremely small objects.

To learn more about measuring lengths and distances join us here @BYJU’S


Practise This Question

If radius of the sphere is (5.3 ± 0.1)cm. Then percentage error in its volume will be