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If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.
NOTE: IT SEEMS THAT r + delta r = 7 but answer comes only by taking initial value r = 7
V=(4pi/3)r3, dr=0.02m,,(r+dr)3=(r3+3r2dr+3r(dr)2 +(dr)3)≈r3(1+3(dr/r)) dV=(4pi/3)(3dr/r)=(4pi/3)*3*0.02/7
as r= 7m dr = 0.02m
V = (4pi/3)r3
differentiating both sides
dV = (4pi)/3*3r2dr
=4*22/7*49*0.02
= 15.32m3
Hope it is usefull...
then plz don't forget to approve by clicking 'Yes' below....
Dear manoj jain,
Given that,
r = 7m dr = 0.02m
V = (4π/3)r3
Now, differentiating both sides
dV = (4π)/3*3r2dr
=4 * 3.14 * 49 * 0.02
Hope this helped you immensely...
All the Very Best & Good Luck to you dear..
Regards,
AskIITians Expert,
Godfrey Classic Prince
IIT-Madras
Please approve my answer if you liked it by clicking on "Yes" given below...!!
according to question initial r is 6.98 or 7.02
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