xii) If the length of a cylinder is 1 =(4.00+0.001) cm, radius r = (0.

1.	xii) If the length of a cylinder is 1 =(4.00+0.001) cm, radius r = (0.0250+0.001) cm and mass m = (6.25+0.01)gm. Calculate the percentage error in thedetermination of density.
Answer» wow this question answersing give me few minutes ACCEPT AS BEST 🙂🙂🙂🙂🙂🙂🙂 length of cylinder , l = (4.00 ± 0.001) cm radius of base of cylinder , r = (0.025 ± 0.001) cm mass of cylinder, m = (6.25 ± 0.01) gm we know, density , d = m/v and volume of cylinder , v = πr²l so, density , d = m/(πr²l) then fractional error can be written as, ∆d/d = ∆m/m + 2 × ∆r/r + ∆l/l given, ∆m = 0.01g , m = 6.25 g , ∆r = 0.001, r = 0.025 , ∆l = 0.001 and l = 4.00 so, ∆d/d = (0.01)/(6.25) + 2 × (0.001)/(0.025) + (0.001)/(4.00) = 0.0016 + 0.08 + 0.00025 = 0.08185 percentage error of density = 100 × fractional error of density = 100 × 0.08185 = 8.185 % 8.185% is the correct answer of the given question

xii) If the length of a cylinder is 1 =(4.00+0.001) cm, radius r = (0.0250+0.001) cm and mass m = (6.25+0.01)gm. Calculate the percentage error in thedetermination of density.

Answer»

wow this question answersing give me few minutes

ACCEPT AS BEST 🙂🙂🙂🙂🙂🙂🙂

length of cylinder , l = (4.00 ± 0.001) cm

radius of base of cylinder , r = (0.025 ± 0.001) cm

mass of cylinder, m = (6.25 ± 0.01) gm

we know, density , d = m/v

and volume of cylinder , v = πr²l

so, density , d = m/(πr²l)

then fractional error can be written as,

∆d/d = ∆m/m + 2 × ∆r/r + ∆l/l

given, ∆m = 0.01g , m = 6.25 g , ∆r = 0.001, r = 0.025 , ∆l = 0.001 and l = 4.00

so, ∆d/d = (0.01)/(6.25) + 2 × (0.001)/(0.025) + (0.001)/(4.00)

= 0.0016 + 0.08 + 0.00025

= 0.08185

percentage error of density = 100 × fractional error of density

= 100 × 0.08185

= 8.185 %

8.185% is the correct answer of the given question

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