Radhika Batra
Last Activity: 11 Years ago
9
Maximum percentage error in Y is given by
Y = W/((?D^2)/4)×L/X
(?Y/Y)_max=2(?D/D)+?X/X+?L/L
= 2(0.001/0.05)+(0.001/0.125)+(0.1/110) = 0.0489
So, maximum percentage error = 4.89%
It is given that
W = 50 N; D = 0.05 cm; = 0.05 × 10–2 m;
X = 0.125 cm = 0.125 × 10–2m;
L = 110 cm = 110 × 10–2 m
Y = (50×4×110×?10?^(-2))/(3.14(0.05×?10?^(-2) )×(0.125×?10?^(-2) ) )
= 2.24 × 1011 N/m2
? ?Y = 0.0489 × 2.24 × 1011 = 1.09 × 1010 N/m2
= 11 × 109 N/m2 = (9 + 2) × 109 N/m2
? x = 9