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

Let, R be the outer radius

R be the inner radius

Here, h = 14 cm

2πRh - 2πrh = 88

⟹ 2πh(R – r) = 88

⟹ 2 * 22/7 * 14(R – r) = 88

⟹ (R – r) = 1cm .....1

Volume of tube $=\pi * R^{2} * h-\pi * r^{2} \star h$

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

$176=22 / 7^{*} 14\left(R^{2}-r^{2}\right)$

$\Rightarrow\left(R^{2}-r^{2}\right)=4$

⟹ (R + r)(R – r) = 4

Here, (R - r) = 1

⟹ (R + r) (1) = 4

⟹ (R + r) = 4 cm

⟹ R = 4 – r  ....2

Here, R – r = 1

⟹ R = 1 + r

Substitute R value in eq 2

⟹ 1 + r = 4 – r

⟹ 2r = 3

⟹ r = 3/2

= 1.5 cm

Substitute ‘r’ value in eq 1

⟹ R – 1.5 = 1

⟹ R = 1 + 1.5

⟹ R = 2.5 cm

Hence, the value of inner radii is 1.5 cm and radius of outer radii is 2.5 cm