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There is an error of $\pm 0.04\;cm$ in the measurement of the diameter of a sphere. When the radius is $10\;cm$, the percentage error in the volume of the sphere is :

\[\begin {array} {1 1} (1)\;\pm 1.2 & \quad (2)\;\pm 1.0 \\ (3)\;\pm 0.8 & \quad (4)\;\pm 0.6 \end {array}\]

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$ (4)\;\pm 0.6 $
answered Nov 7, 2013 by pady_1
 
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