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JEE Main202118 Mar 2021Evening ShiftPhysicsMathematics in PhysicsActual

The radius of a sphere is measured to be ( 7 . 50 ± 0 . 85 ) cm . Suppose the percentage error in its volume is x . The value of x , to the nearest x , is ___ .

Correct answer

0

Step-by-step solution

∵ v = 4 3 π r 3 taking log   & then differentiate d V V = 3 d r r = 3 × 0 . 85 7 . 5 × 100 % = 34 %

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