The difference between inside and outside surfaces of a cylindrical tube 14 cm long, is 88 cm2.

Let the inner and outer radii of the tube be r cm and R cm, respectively.

Length of the cylindrical tube, h = 14 cm

Outer curved surface of the cylinder − Inner curved surface of the cylinder = 88 cm2        (Given)

$\therefore 2 \pi R h-2 \pi r h=88$

$\Rightarrow 2 \times \frac{22}{7} \times 14 \times(R-r)=88$

$\Rightarrow R-r=\frac{88 \times 7}{2 \times 22 \times 14}$

$\Rightarrow R-r=1 \mathrm{~cm} \quad \ldots \ldots(1)$

Volume of the tube = 176 cm3                (Given)

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

$\Rightarrow \frac{22}{7} \times(R-r) \times(R+r) \times 14=176$

$\Rightarrow R+r=\frac{176 \times 7}{22 \times 1 \times 14} \quad\left[\begin{array}{ll}\text { Using } & \text { (1) }\end{array}\right.$

$\Rightarrow R+r=4 \mathrm{~cm} \quad \ldots \ldots(2)$

Adding (1) and (2), we get

2R = 5

⇒ R = 2.5 cm

Putting R = 2.5 cm in (2), we get

2.5 + r = 4

⇒ r = 4 − 2.5 = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm, respectively.