To solve the given problem, we need to determine the percentage errors in both the radius and the volume of a spherical body with a given radius and its associated error.
1. Determine the percentage error in radius:
- The radius \( r \) of the spherical body is given as \( 10 \) cm.
- The error in the radius \( \Delta r \) is given as \( 0.5 \) cm.
- The formula to calculate the percentage error in the radius is:
[tex]\[
\text{Percentage error in radius} = \left( \frac{\Delta r}{r} \right) \times 100
\][/tex]
Plugging in the given values:
[tex]\[
\text{Percentage error in radius} = \left( \frac{0.5}{10} \right) \times 100 = 5\%
\][/tex]
2. Determine the percentage error in volume:
- The volume \( V \) of a sphere is given by the formula:
[tex]\[
V = \frac{4}{3} \pi r^3
\][/tex]
- The percentage error in volume can be approximated by the expression:
[tex]\[
\frac{\Delta V}{V} \approx 3 \left( \frac{\Delta r}{r} \right)
\][/tex]
- This is because the volume of a sphere is proportional to the cube of the radius.
- Therefore, the percentage error in volume is approximately three times the percentage error in radius.
Using the previously calculated percentage error in radius:
[tex]\[
\text{Percentage error in volume} \approx 3 \times 5\% = 15\%
\][/tex]
3. Evaluate the given options:
- Option (1): The percentage error in radius is 5% -> True
- Option (2): The percentage error in volume is 10% -> False
- Option (3): The percentage error in volume is 15% -> True
- Option (4): Both (1) & (3) -> True
Thus, the correct options are:
(1), (3), and (4).
