The radius of a sphere is measured as $ (10 \pm 0.02\% ) $ cm. the error in the measurement of its volume is
(A) $ 25.1cc $
(B) $ 25.12cc $
(C) $ 2.51cc $
(D) $ 251.2cc $
Answer
624.9k+ views
Hint : When quantities are multiplied with one another, the errors present in each of the quantities will be added to one another. The volume of the sphere should be found before trying to find the error in the volume.
Formula used: In this solution we will be using the following formula;
$ PE = \dfrac{E}{{AV}} \times 100\% $ where $ PE $ is the percentage error of a measurement, $ E $ is the error of the measurement, and $ AV $ is the actual value.
$ V = \dfrac{4}{3}\pi {r^3} $ where $ V $ is the volume of a sphere, and $ r $ is the radius of the sphere.
Complete step by step answer:
The radius of a sphere is measured to be $ (10 \pm 0.02\% ) $ cm and the error of the volume is asked to be determined.
To do this, we shall calculate first the actual value of the volume (i.e. the value taken to be the volume).
This is calculated from
$ V = \dfrac{4}{3}\pi {r^3} $ where $ V $ is the volume of a sphere, and $ r $ is the radius of the sphere.
Hence,
$ V = \dfrac{4}{3}\pi {\left( {10} \right)^3} = 4189c{m^3} $ . This is also written as 4189 cc.
Now, the error in the radius, in percentage form, was given as $ 0.02\% $
Hence, the percentage error in the volume is $ 3 \times 0.02\% = 0.06\% $
Hence, the actual error in the volume would be given from
$ PE = \dfrac{E}{{AV}} \times 100\% $ , where $ PE $ is the percentage error of a measurement, $ E $ is the error of the measurement, and $ AV $ is the actual value, as
$ E = PE \times AV $ where $ PE $ is in percentage.
Hence,
$ E = \dfrac{{0.06}}{{100}} \times 4189 = 2.51cc $
Thus, the correct option is C.
Note:
For clarity, the percentage error of the volume was given as thrice its radius because when quantities are multiplied with one another, the errors present in each of the quantities will be added to one another. In this case the quantity was the radius, and
$ {r^3} = r \times r \times r $ then the error is
$ {E_r} = r + r + r = 3r $
The constant does not affect the percentage error.
Formula used: In this solution we will be using the following formula;
$ PE = \dfrac{E}{{AV}} \times 100\% $ where $ PE $ is the percentage error of a measurement, $ E $ is the error of the measurement, and $ AV $ is the actual value.
$ V = \dfrac{4}{3}\pi {r^3} $ where $ V $ is the volume of a sphere, and $ r $ is the radius of the sphere.
Complete step by step answer:
The radius of a sphere is measured to be $ (10 \pm 0.02\% ) $ cm and the error of the volume is asked to be determined.
To do this, we shall calculate first the actual value of the volume (i.e. the value taken to be the volume).
This is calculated from
$ V = \dfrac{4}{3}\pi {r^3} $ where $ V $ is the volume of a sphere, and $ r $ is the radius of the sphere.
Hence,
$ V = \dfrac{4}{3}\pi {\left( {10} \right)^3} = 4189c{m^3} $ . This is also written as 4189 cc.
Now, the error in the radius, in percentage form, was given as $ 0.02\% $
Hence, the percentage error in the volume is $ 3 \times 0.02\% = 0.06\% $
Hence, the actual error in the volume would be given from
$ PE = \dfrac{E}{{AV}} \times 100\% $ , where $ PE $ is the percentage error of a measurement, $ E $ is the error of the measurement, and $ AV $ is the actual value, as
$ E = PE \times AV $ where $ PE $ is in percentage.
Hence,
$ E = \dfrac{{0.06}}{{100}} \times 4189 = 2.51cc $
Thus, the correct option is C.
Note:
For clarity, the percentage error of the volume was given as thrice its radius because when quantities are multiplied with one another, the errors present in each of the quantities will be added to one another. In this case the quantity was the radius, and
$ {r^3} = r \times r \times r $ then the error is
$ {E_r} = r + r + r = 3r $
The constant does not affect the percentage error.
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