Question #3755

An object of irregular geometric shape is lowered into a beaker of water. Find the volume of the
shape and the error of measurement.

Expert's answer

Lets measure radius of breaker beforehand, suppose it has radius R. Let’s also take measurements of heights of water before and after lowering our object. Suppose heights difference is h, then let’s consider it’s error δh as half of the lowest graduation of your measuring device. Radius measurement error δR is also the lowest graduation of your measuring device.

Volume of the shape we get by

V=πR^{2} h

And error will be

δ=√((δR)^{2}+(δH)^{2} )

Thus

V=πR^{2} h ± √((δR)^{2}+(δH)^{2} )

Volume of the shape we get by

V=πR

And error will be

δ=√((δR)

Thus

V=πR

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