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If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be :

$\begin {array} {1 1} (1)\;2 \% & \quad (2)\;4 \% \\ (3)\;6 \% & \quad (4)\;8 \% \end {array}$

 

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A)
(3) is correct. Solution: Given error 2% For a sphere, V = 4\3 × π r^3 Taking log both sides, Log V = 3 log r + log (4\3 π) Diff. Both sides w.r.to V d\dV log V = 3 d\dV r + d\dV log (4\3 π) 1\V = 3\r dr\dV + 0 dV\V = 3×2% = 6%. ANSWER (given dr\r = 2%).
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