If error in measurement of radius of a sphere is 2% . What will be the error in volume of the sphere.
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100- 2= 98
initial= 100³= 1000000
final= 98³=941192
% error=( 1000000-941192)/1000000×100=58808/100000× 100=5.88%
initial= 100³= 1000000
final= 98³=941192
% error=( 1000000-941192)/1000000×100=58808/100000× 100=5.88%
smitchandi:
approx can be 5.9 & not directly 6
ok bhai , thik me mera galat he , ok ,sry
it's OK
in which standard you are???
9th
dude it's question of standard 11th
here approax don't work
ok
gn
nice of you as atleast you were making efforts to solve it, *Thank You* .
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