The error in the measurement of the radius of a sphere is `0.5 %`. What is the permissible percentage error in the measurement of its (a) surface area and (b) volume ?

Explanation

To solve the problem of finding the permissible percentage error in the measurement of the surface area and volume of a sphere given the error in the measurement of its radius, we will follow these steps:

### Step 1: Understand the given data
The error in the measurement of the radius of the sphere is given as \(0.5\%\).

### Step 2: Recall the formulas for surface area and volume
- The surface area \(A\) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
- The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3}\pi r^3 \]

### Step 3: Determine the relationship between the percentage error in radius and surface area
The percentage error in a quantity that is a power of the radius can be calculated using the formula:
\[ \text{Percentage error in } A = n \times \text{Percentage error in } r \]
where \(n\) is the power of \(r\) in the formula for the quantity.

For surface area:
- Here, \(n = 2\) (since the radius is squared).
- Therefore, the percentage error in surface area is:
\[ \text{Percentage error in } A = 2 \times 0.5\% = 1\% \]

### Step 4: Determine the relationship between the percentage error in radius and volume
Using the same formula for volume:
- Here, \(n = 3\) (since the radius is cubed).
- Therefore, the percentage error in volume is:
\[ \text{Percentage error in } V = 3 \times 0.5\% = 1.5\% \]

### Step 5: Summarize the results
- The permissible percentage error in the measurement of the surface area is \(1\%\).
- The permissible percentage error in the measurement of the volume is \(1.5\%\).

### Final Answer:
(a) The permissible percentage error in the measurement of surface area is \(1\%\).
(b) The permissible percentage error in the measurement of volume is \(1.5\%\).

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