Question
Question: The radius of the sphere is \[\left( {4.6 \pm 0.1} \right)cm\]. The percentage error in its volume i...
The radius of the sphere is (4.6±0.1)cm. The percentage error in its volume is:
A. 4.60.1×100%
B. 3×4.60.1×100%
C. (4.60.1×100)3%
D. 31×4.60.1×100%
Solution
Complete step by step answer:
Error is the difference that we find when the difference between the actual(true) value and measured(observed) value of a given physical quantity is called an error.
Error= actual(true) value- measured(observed) value.
Under the errors there are 2 types of errors:
- systematic error and
- random errors
Percentage error is given by
Percentage error= Relative error ×100%
δa%=aˉΔaˉ×100%
Where Δaˉ=mean absolute error and aˉ=arithmetic mean.
The percentage error again has many types :
-Error in addition
-Error is subtraction
-Error in division
-And Error in multiplication.
In the above mentioned question radius has been given and volume is to be found, hence we use error in multiplication. Given: Radius(r) =(4.6±0.1)cm
V=34πr3
⇒V=34×π×(4.6)3=406.4cm3
Now in error in multiplication :if X and Y are 2 individual quantities to be multiplied and is the resultant product like Z=XY, then the maximum possible error is given by
ZΔZ=XΔX+YΔY
And percentage error is given by:
ZΔZ×100%=XΔX×100%+YΔY×100%
Where ZΔZ,XΔXandYΔY are relative errors in X,Y and Z.
Hence applying the above way we get:
VΔV=3(rΔr)×100%
∴VΔV=3×(4.60.1)×100%
Hence the correct option is option B.
Note: Percentage error should always be mentioned in %. We should not leave it, otherwise it is relative error.it can also be written as ΔV=V×3×(4.60.1)×100% but we have written as required in the options. Also remember that when physical quantities are multiplied the maximum error in the resultant physical quantity is equal to the sum of percentage errors of individual quantities.