the error in measurement of radius of a sphere is 0.5%. find the error in permissible error in the surface area?
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Volume of a sphere, V =4/3 π r³ .
The error in the measurement of r i.e. ∆r/r = 2%,
Therefore
∆V/V = 3 ( ∆r/r) = 3× 2% = 6%
So error in volume of the sphere due to error in measured value of r = 6%.
The error in the measurement of r i.e. ∆r/r = 2%,
Therefore
∆V/V = 3 ( ∆r/r) = 3× 2% = 6%
So error in volume of the sphere due to error in measured value of r = 6%.
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