% error..

If the percentage error in measuring the surface area of a sphere is x%,then find the error in its volume.

6 Answers

1
Rocky Crazy ·

Nyone..help..

3
iitimcomin ·

if x is really really small ,,,, and ur talkin abt surface area

then its 3x/2

1
Rocky Crazy ·

Sorry yaar yup its area..

No x is not very small..its not given..

1
Rocky Crazy ·

Ny1 wid a try\\\

106
Asish Mahapatra ·

percentage error in measuring surface area is x%

CASE I: x is small

Then % error in measuring length is x/2%
So % error in volume is 3x/2%

CASE II: x is not small

Let the % error in measuring length be a

Then if the length is measured as l, the actual length is l(1±a/100)

Let actual length be l(1+a/100) then surface area = l2(1+a/50+a2/10000)

So %error = (a/50+a2/10000)*100 = 2a+a2/100 = x ...(i)

volume = lact3 = l3(1+3a/100 + 3a2/10000 + a3/106)

So % error = (3a/100 + 3a2/10000 + a3/106)*100 = 3a + 3a2/100 + a3/10000

u can use the value of a obtained in (i)

1
Rocky Crazy ·

Great work Asish

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