If the percentage error in measuring the radius of a sphere is 2%, what will be the percentage error in the calculation of its volume? MCQ with Answer and Explanation
If the percentage error in measuring the radius of a sphere is 2%, what will be the percentage error in the calculation of its volume?
A. 4%
B. 6%
C. 2%
D. 8%
Answer: Option B
Solution (By JKSSB Mock Tests)
The volume of a sphere is given by V = (4/3)πr³. The formula for relative error in volume is (ΔV/V) = 3 × (Δr/r). Since the percentage error in radius (Δr/r × 100) is 2%, the percentage error in volume will be 3 × 2% = 6%. Constants like 4/3 and π do not contribute to the error.
Explanation:
Sound travels to cliff and back. Total distance = speed × time = 340 × 2 = 680 m. So distance to cliff = 680/2 = 340 m. Use formula d = (v × t)/2. Time for round trip. Minimum distance for echo 17 m (approx) for 0.1 s persistence.
Explanation:
The relationship between phase difference (phi) and path difference (delta x) is given by the formula phi = (2pi / lambda) * delta x. Given a path difference of lambda / 2, substitute this into the formula: phi = (2pi / lambda) * (lambda / 2) = pi radians. This indicates strictly destructive interference.
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