The difference between the outer and inner

The height of the hollow cylinder is 14 cm. Let the inner and outer radii of the hollow cylinder are r cm and R cm respectively. The difference between the outer and inner surface area of the hollow cylinder is

$=2 \pi R \times 14-2 \pi r \times 14$

$=28 \pi(R-r) \mathrm{cm}^{2}$

By the given condition, this difference is 88 square cm. Hence, we have

$28 \pi(R-r)=88$

$\Rightarrow R-r=\frac{44 \times 7}{14 \times 22}$

$\Rightarrow R-r=\frac{4 \times 7}{14 \times 2}$

$\Rightarrow R-r=1$

The volume of the metal used in making the cylinder is

$V_{1}=\pi\left\{(R)^{2}-(r)^{2}\right\} \times 14 \mathrm{~cm}^{3}$

By the given condition, the volume of the metal is 176 cubic cm. Hence, we have

$\pi\left\{(R)^{2}-(r)^{2}\right\} \times 14=176$

$\Rightarrow R^{2}-r^{2}=\frac{176 \times 7}{14 \times 22}$

$\Rightarrow R^{2}-r^{2}=4$

$\Rightarrow(R-r)(R+r)=4$

$\Rightarrow 1 \times(R+r)=4$

$\Rightarrow R+r=4$

Hence, we have two equations with unknowns R and r

$R-r=1$

$R+r=4$

Adding the two equations, we have

$(R-r)+(R+r)=1+4$

$\Rightarrow 2 R=5$

$\Rightarrow R=2.5$

Then from the second equation, we have

$r=4-2.5=1.5$

Therefore, the outer and inner diameters of the hollow cylinder are 5cm and 3cm respectively.