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If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
  • a)
    8%
  • b)
    2%
  • c)
    4%
  • d)
    6%
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
If the error in the measurement of radius of a sphere is 2%, then the...
Explanation:

To understand the error in the determination of the volume of a sphere, let's first understand the formula for calculating the volume of a sphere and how it is affected by the measurement of the radius.

The formula for the volume of a sphere is given by:

V = (4/3)πr³

where V is the volume and r is the radius of the sphere.

Error in the Measurement of Radius:
The error in the measurement of the radius is given as 2%. This means that the measured value of the radius will have an error of 2%.

Propagation of Errors:
When errors are involved in multiple measurements and calculations, the concept of error propagation is used. In this case, we need to determine how the error in the radius affects the determination of the volume.

Formula for Error Propagation:
The formula for propagating errors through a mathematical function is given by:

Δf = |df/dr| * Δr

where Δf is the error in the function, |df/dr| is the derivative of the function with respect to the variable, and Δr is the error in the variable.

Error in the Determination of Volume:
To find the error in the determination of the volume, we need to calculate the derivative of the volume formula with respect to the radius.

dV/dr = 4πr²

Now, we can substitute the values into the error propagation formula:

ΔV = |dV/dr| * Δr
= 4πr² * Δr

Since the error in the radius is given as 2%, we can substitute this value into the equation:

ΔV = 4πr² * 0.02

Simplifying further:

ΔV = 0.08πr²

Percentage Error:
To express the error as a percentage, we can divide the error by the actual value of the volume:

Percentage Error = (ΔV / V) * 100
= (0.08πr²) / ((4/3)πr³) * 100
= 0.08 * (3/4) * 100
= 6%

Therefore, the error in the determination of the volume of the sphere is 6%, which corresponds to option D.
Free Test
Community Answer
If the error in the measurement of radius of a sphere is 2%, then the...
Error in the determination of the volume = 3×2% = 6%
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