Question

The radius of a sphere is measured to be (7.50 ± 0.85) cm. Suppose the percentage error in its volume is x. The value of x, to the nearest integer, is __________.

Hint:

The relative error in a calculated quantity is the sum of relative errors of components multiplied by their respective powers.

Answer 1

Correct Answer: 34

The formula for the volume of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). To find the percentage error in the volume, we start with the relative error expression. The relative error in the volume \( \frac{\Delta V}{V} \) is related to the relative error in the radius \( \frac{\Delta r}{r} \). When the radius \( r \) of a sphere is measured as \( r = 7.50 \pm 0.85 \) cm, the relative error in the radius is \( \frac{\Delta r}{r} = \frac{0.85}{7.50} \).

Since \( V \propto r^3 \), the percentage error in the volume is three times the percentage error in the radius. Therefore, the percentage error in the volume is calculated as:

\[ \text{Percentage Error in Volume} = 3 \times \left(\frac{\Delta r}{r} \times 100\%\right) \]

Substituting the given values, we have:

\[ \text{Percentage Error in Volume} = 3 \times \left(\frac{0.85}{7.50} \times 100\%\right) \]

Performing the calculations:

\[ = 3 \times (11.33\%) = 33.99\% \]

Rounding to the nearest integer, we get \( x = 34\% \).

Thus, the value of \( x \), the percentage error in the volume, is 34, which clearly falls within the given range of 34 to 34.