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If error in measuring the radius of a sp...

If error in measuring the radius of a sphere is 1% then the error in calculating the volume of sphere would be

A

0.03

B

0.05

C

0.07

D

0.01

Text Solution

Verified by Experts

The correct Answer is:
A
Given, Error in radius = 1%
We know that, Volume of sphere, V =`4/3 pi r^3`
Take log on both sides,
`log V = 3log r + log (4/3 pi)`
On differentiating,
`(Delta V)/(V) = 3 xx (Delta r)/(r )`
`(Delta V)/(V) = 3 xx (Delta r)/(r)`
`(Delta V)/(V) xx 100 = 3 xx ((Delta r)/(r ) xx 100)`
` = 3 xx 1 = 3%`
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