Question

The percentage error in the calculated volume of a sphere, if there is \(2%\) error in its diameter measurement, is _____.

• A. \(1\)
• B. \(2\)
• C. \(6\)
• D. \(8\)

Answer 1

The Correct Option is C

Step 1: Understanding the Question
We are asked to find the percentage error in the volume of a sphere given the percentage error in its diameter. This is a classic application of error propagation using calculus.
Step 2: Key Formula or Approach
1. The volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). 2. The diameter is \( D = 2r \), so the radius is \( r = D/2 \). We can express the volume in terms of the diameter. 3. For a quantity \( Q \) that depends on a variable \( x \) as \( Q = kx^n \), the relative error is given by \( \frac{\Delta Q}{Q} = n \frac{\Delta x}{x} \). 4. The percentage error is the relative error multiplied by 100.
Step 3: Detailed Explanation
First, let's write the volume \(V\) of the sphere in terms of its diameter \(D\). \[ V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi \frac{D^3}{8} = \frac{\pi}{6}D^3 \] To find the relationship between the errors, we can use logarithmic differentiation. \[ \ln(V) = \ln\left(\frac{\pi}{6}D^3\right) = \ln\left(\frac{\pi}{6}\right) + 3\ln(D) \] Differentiating with respect to the variables, for small changes (errors), we get: \[ \frac{dV}{V} = 0 + 3 \frac{dD}{D} \] For small finite errors, we can approximate this as: \[ \frac{\Delta V}{V} \approx 3 \frac{\Delta D}{D} \] This equation relates the relative error in volume to the relative error in diameter. To find the percentage error, we multiply both sides by 100: \[ \left(\frac{\Delta V}{V} \times 100%\right) = 3 \times \left(\frac{\Delta D}{D} \times 100%\right) \] We are given that the percentage error in the diameter measurement is 2%. \[ % \text{ Error in V} = 3 \times (2%) = 6% \] Step 4: Final Answer
The percentage error in the calculated volume is 6.