Question

QUE-6-> find the estimation of younges modules

γ=(4mgL)/(d²l)

for the specimen following observation were recorded L=2.89± 0.001 m=3.00± 1 d=0.082 g=9.81 l=0.087 calculate the maximum %age rror for the calculation of γ

Answer 1

we know that

Y = (4mgL) / (Ï€d²l)

here

m = 3 ± 1 kg

g = 9.8 m/s²

L = 2.89 ± 0.001 m

d = 0.082 m

l = 0.087 m

thus,

Y = (4x3x9.81x2.89) / (3.14x0.082²x0.087)

or

Y = 340.210 / 1.836x10^-3

finally,

Y = 1.852x10⁵ N/m²

now,

the corresponding error equation will be

(ΔY / Y) = ± [(Δm/m) + (ΔL/L)]

so, the maximum percentage error will be

(ΔY / Y) x 100 = ± [(Δm/m) + (ΔL/L)] x 100

% error = ± [1/3 + 0.001/2.89] x 100

or

% error = ± 33.36 %

which seems a bit too large, you might want to have a look the error values again....