Question

The possible error involved in measuring each dimension of a rectangular box is ±0.02 inches. The dimensions of the box are 8 inches by 5 inches by 12 inches. Approximate the propagated error and the relative error in the calculated volume of the box.

Answer 1

Answer:

Step-by-step explanation:

Given

Possible dimension error is $\pm 0.02 in.$

length of box $L=8 in.$

Width $W=5 in.$

height $h=12 in.$

Volume V is given by

$V=LWh$

$dV=LWdh+LhdW+WhdL$

and it is given

$dL=dW=dh=\pm 0.02\ in.$

since error always add therefore

$dV=(8\times 5+5\times 12+8\times 12)\cdot 0.02$

$dV=3.92 in.^3$

Propagated error is $3.92 in.^3$

relative error$=\frac{dV}{V}$

$=\frac{3.92}{480}$

$=0.00816$

$=0.816 \%$