Question

Let $y=l^2-\frac{l^3}{z}$ where $l=2.0\pm 0.1,z=1.0\pm 0.1$ then the value of y is given by

• A. $+2\pm 0.8$
• B. $-4\pm 1.6$
• C. $-4\pm 0.8$
• D. $-2\pm 1.6$

Answer 1

The correct option is B $-4\pm 1.6$

$y=l^2-\frac{l^3}{z}dy=2ldl-\left(\frac{z.3l^{2}dl-l^{3}dz}{z^2}\right)=(2l-\frac{3l^2}{z})dl+\frac{l^3}{z^2}dz=(2\times 2-\frac{3\times 2^2}{1})(\pm 0.1)+\frac{8}{1}(\pm 0.1)y=l^2-\frac{l^3}{z}=2^2-\frac{2^3}{1}=4-8=-4\therefore y=-4\pm 1.6$