Grade 10Mechanics

The density of the material of cylindrical rod was determined by the formula d = m/pi r²L . The percentage error in m , r ,and l are 2% , 1.5% and 0.8% respectively . Calculate the maximum possible percentage error in determination of density.

Aiden Mathew Denny

8 Years agoGrade 10

3 Answers

Khimraj

Approved Tutor Answer8 Years ago

density is given by

d = m/ $\pi$ r²L

error maximum error is calculated as

$\Delta$ d/d = $\Delta$ m/m + 2 $\Delta$ r/r + $\Delta$ L/L

error in density = 2 + 3 + 0.8 = 5.8%

Hope it clears. If you like answer then please approve it.

ajay kumar jha

6 Years ago

given : d=m/pir^2l and % error in m=2%,r=1.5% and l=0.8%

maximum error: deltad/d =delta m/m+2deltar/r+deltal/l= 2+1.5*2+0.8=5.8%

hope it will be helpful for you

Vikas TU

6 Years ago

The density of the material of a cylindrical rod is determined by formula , D = m/πr²l

To calculate fractional error in D using formula,

∆D/D = ∆m/m + 2∆r/r + ∆l/l

and hence, % error in D = % error in m + 2 × % error in r + % error in l

given,

percentage error in m = 2%
percentage error in r = 1.5 %
percentage error in l = 0.8%
so, % error in D = 2 % + 2 × 1.5 % + 0.8%

= 2% + 3% + 0.8%

= 5.8%

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The density of the material of cylindrical rod was determined by the formula d = m/pi r2L . The percentage error in m , r ,and l are 2% , 1.5% and 0.8% respectively . Calculate the maximum possible percentage error in determination of density.

Aiden Mathew Denny

3 Answers

Khimraj

ajay kumar jha

Vikas TU

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