If the error in measurement of radius of a sphere is 2%, the percentage error in the calculated volume of the sphere will be: MCQ with Answer and Explanation
If the error in measurement of radius of a sphere is 2%, the percentage error in the calculated volume of the sphere will be:
A. 4%
B. 2%
C. 6%
D. 8%
Answer: Option C
Solution (By JKSSB Mock Tests)
Volume of sphere V = (4/3)πr³. For error propagation, percentage error in V = 3 × (percentage error in r). Given error in r is 2%, so error in V = 3 × 2% = 6%. This follows the power rule: if Q = xⁿ, then % error in Q = |n| × % error in x. This principle is crucial for experimental error analysis. Exam tip: Remember that volume depends on cube of radius, so errors amplify threefold. This type of question tests understanding of error propagation in derived quantities.
Explanation:
Apparent frequency change depends on relative velocity between source and observer with respect to medium. Both motions contribute: f' = f(v±v_o)/(v∓v_s). Memory tip: 'Approaching ⇒ higher pitch; receding ⇒ lower pitch'. Conceptual question testing Doppler effect fundamentals, frequently examined in competitive exams.
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