Question
If the radius of a sphere is measured as 9 ~cm with an error of 0.03 ~cm , then find the approximating error in calculating its volume.
BITSAT
Mathematics
Application of Derivatives
2021
BITSAT 2021
BITSAT Mathematics Question (2021) — Solution
Options
- A. 2.46 cm ^ 3
- B. 8.62 cm ^ 3
- C. 9.72 cm ^ 3
- D. 7.6 cm ^ 3
Answer
C. 9.72 cm ^ 3
Step-by-step solution
Let r be the radius of the sphere and r be the error in measuring the radius. Then, r=9 ~cm and r=0.03 ~cm Let V be the volume of the sphere. Then, array l V = 4 3 r^ 3 \\ dV dr =4 r^ 2 \\ ( dV dr )_ r=9 =4 9^ 2 =324 array Let V be the error in V due to error r in r. Then, array l V = dV dr r \\ V =324 0.03=9.72 cm ^ 3 array
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