Question

The density of cylindrical rod was measured by the formula d = 4 2 the percentage in m, d, and l are 1%, 1.5% and 0.5%. Calculate the % error in the calculate value of density.

Answer 1

Answer:

4.2%

Explanation:

Given The density of cylindrical rod was measured by the formula ρ = 4m/πD^2l . The percentage errors in m, d, and l are 1%, 1.5% and 0.5%. Calculate the % error in the calculated value of density.

We know that

    Density ρ = 4m/πD^2l

Now we know that

     Δρ/ρ max = Δm / m + 2ΔD/D + Δl/l

 But given  Δm/m = 1% ΔD/D = 1.5% and Δl / l = 0.5%

Therefore maximum percentage error in the calculated value of density will be

   Δρ/ρ max = 1% + 2 x 1.5% + 0.5%

                    = 1 + 3 + 0.5

                     = 4.5%

Answer 2

Answer:

The value of density is 4.5%.

Explanation:

Given that,

The percentage values

$m =1\%$

$d=1.5\%$

$l=0.5\%$

Suppose the density is

$\rho=\dfrac{4m}{\pi D^2 l}$

Error in the density is given by

$\dfrac{\Delta \rho}{\rho}=\dfrac{\Delta m}{m}+2\dfrac{\Delta D}{D}\times100+\dfrac{\Delta l}{l}\times100$

Put the value in to the formula

$\dfrac{\Delta \rho}{\rho}\times100=1\%+2(1.5\%)+0.5\%$

$\dfrac{\Delta \rho}{\rho}\times100=4.5\%$

Hence, The value of density is 4.5%.