If radius of the sphere is (5.0 ± 0.2) cm, then the percentage error in its surface area is
Solution:
The radius of the sphere is given as r = (5.0 ± 0.2) cm.
This means the measured value of radius is R = 5.0 cm and the absolute error in radius is ΔR = 0.2 cm.
The percentage error in radius is given by:
Percentage error in R = (ΔR / R) × 100%
Percentage error in R = (0.2 / 5.0) × 100% = (2 / 50) × 100% = 4%
The surface area of a sphere is given by the formula A = 4πR².
To find the percentage error in the surface area, we use the rules of error propagation. If a quantity Q depends on another quantity X as Q = kXⁿ (where k is a constant), then the fractional error in Q is given by ΔQ/Q = n(ΔX/X).
In our case, A = 4πR². Here, 4π is a constant, and the area A depends on R².
So, the fractional error in A is:
ΔA / A = 2 × (ΔR / R)
To find the percentage error in A, we multiply by 100%:
Percentage error in A = 2 × (ΔR / R) × 100%
Percentage error in A = 2 × (Percentage error in R)
Substitute the calculated percentage error in R:
Percentage error in A = 2 × 4% = 8%
Thus, the percentage error in the surface area of the sphere is 8%.