Solveeit Logo

Question

Question: If the length of rod A is \(3.25 \pm 0.01cm\) and that of B is \(4.19 \pm 0.01cm\) then how much rod...

If the length of rod A is 3.25±0.01cm3.25 \pm 0.01cm and that of B is 4.19±0.01cm4.19 \pm 0.01cm then how much rod B is longer than rod A?

Explanation

Solution

We know that error is the difference between the measured value and the true value. It is an error while we take a measurement with a measuring instrument. It is unidirectional. Here apply the combination of errors in addition. Using this find how much rod B is longer than rod A.

Complete step by step answer:
Errors are categorized as two types: they are systematic error and random error.
Error can occur in the positive direction or negative direction. Errors may be because of either controllable or uncontrollable reasons.
By selecting an instrument with better resolution can reduce the error in the system. Also, we can say the difference between the measured value and the true value.
It is an error while we take a measurement with a measuring instrument.
The sum of absolute errors in the quantities is the final result when two quantities are added or subtracted.
We have to know that all the measurements will have both systematic and random errors.
Here the length of rod A is 3.25±0.01cm3.25 \pm 0.01cm and b is 4.19±0.01cm4.19 \pm 0.01cm
From the above statement it is clear that rod B is longer than that of rod A.
Then we apply rod B is longer than Rod A by
m=4.19±0.01(3.25±0.01)\Rightarrow m = 4.19 \pm 0.01 - \left( {3.25 \pm 0.01} \right)
Hence, we can write
m=(4.193.25)±(0.01+0.01)\Rightarrow m = \left( {4.19 - 3.25} \right) \pm \left( {0.01 + 0.01} \right)
On simplification we get
m=0.94±0.02cm\Rightarrow m = 0.94 \pm 0.02cm
Hence we get the required answer.

Note: While calculating errors in multiplication and division, the maximum relative error is the sum of relative errors of quantities that are multiplied or divided.
We have to know that all the measurements will have both systematic and random errors.
By selecting an instrument with better resolution can reduce the error in the system.