Answer
% error in external diameter is 5% and % change in internal diameter is 0.5%
Work Step by Step
We are given that the % change in internal volume is 1%
% change in external surface area is 1%
let the diameter of the cylinder be x(internal) and y(external)
then radius $r=\frac{x}{2}$
Volume of cylinder $V=\pi r^2 h=\pi (\frac{x}{2})^2h=\pi (\frac{x^2}{4})h$
on differentiating the above equation:
$dV=\frac{\pi}{4}(2xhdx+x^2(dh))$
height is not changing so $dh=0$
$dV=\frac{\pi}{4}(2xh(dx))=\frac{\pi}{2}(xh(dx))$
so
%change in volume =change in volume/original volume
$1=\frac{dV}{V}$
$\frac{\frac{\pi}{2}(xh(dx))}{\pi (\frac{x^2}{4})h}=1 $
${\frac{\pi}{2}(xh(dx))}={\pi (\frac{x^2}{4})h}$
$\frac{dx}{x}=\frac{1}{2}$
so the internal diameter of the cylinder is changing by 0.5%
for exterior diameter change
external surface area of cylinder
$S=2\pi r h=2\pi (\frac{y}{2})h=\pi ({y})h$
on differentiating the above euation
$dS={\pi}(hdy+y(dh))$
height is not changing so $dh=0$
$dS={\pi}(h(dy))$
so
%change in surface area =change in surface area/original surface area
$5=\frac{dS}{S}$
$\frac{{\pi}(h(dy))}{\pi ({y})h}=5 $
$\frac{dy}{y}=5$
so % error in external diameter is 5% and % change in internal diameter is of 0.5%
