Calculus - 14

+2 votes

The radius of sphere is measured as 10cm with an error 0.06cm. Find approximate error in the volume.

(A) 24\pi

(B) 72\pi

(C) 36\pi

(D) 12\pi

EDIT: Marks for all for now

asked May 30 in Basic Maths by gbmentor (83,420 points) 1 flag
edited May 31 by getgatebook

5 Answers

0 votes
Sir, here u didn't mention...whether we have to find approximation error for area or value or some other....

Please check this question...

For is 24 pie
answered May 31 by harish29pro20 (840 points)
I dont know about the question but can you explain it to me if the question is asked for the volume as you have mentioned some answer 24pi.
Volume = 4/3 pi R^3
R =10
r = 0.06
Delta v = dv/dr * r
Solve it
4/3 * pi * 3*10*10*0.06
24 pi

But bro here they didn't mention whether area or volume or some other...But method is this
updated the question.
Thankyou brother for your answer.
This is some generalized method to solve such type of question or this is your approach?
U can use it for almost every problem....
Okay. Got it. Thanks
Marks will be given to all in this question or not?
0 votes
Here not considered for volume or for area so I think question should be wrong
answered May 31 by tsnikhilsharmagate2018 (30,120 points)
Yes is wrong.
0 votes
it was not mentioned volume
answered May 31 by tssoumambanerjee-nit (1,100 points)
0 votes
this question is incomplete ,I think it should not be taken for evaluation .How some one can do a question which is not complete great imagination
answered May 31 by tsankitjha910 (3,690 points)
For this marks will be given for all.
+2 votes

\\ V=4/3*\pi*R^3\\ \\ \frac{\partial V}{\partial R}=4/3*\pi *3*R^2\\ \\ \partial V=4/3*\pi *3*10*10*\partial R\\ \\ \partial V=4/3*\pi*3*10*10*0.06=24\pi

Therefore Answer is A

answered Jun 3 by tskushagra-guptacse (15,660 points)