AP EAMCET2014MathematicsApplication of Derivatives
If there is an error of 0.04 ~cm in the measurement of the diameter of a sphere, then the approximate percentage error in its volume, when the radius is 10 ~cm , is
Options
- A1.2
- B0.06
- C0.006
- D0.6
Correct answer
D. 0.6
Step-by-step solution
Given, r= 0.04 2 =0.02 Volume of sphere V= 4 3 r^3 On differentiating w.r.t. r , we get aligned & d U d r = 4 3 3 r^2=4 r^2 & V= d U d r r=4 r^2 r & aligned Relative per cent error aligned & V V 100= 4 r^2 r 4 3 r^3 100 & = 3 r r 100 & = 3 ( 0.02) 10 100 & = 6 10 = 0.6 aligned