Correct option is (a) 4.5 cm
Let, R be the external radius and r be the inner radius of the metallic pipe. h = 14 cm
Outer surface area−inner surface area = 44 cm2
∴ 2πRh − 2πrh = 44 cm2
The volume of metal used = 99 cm3
∴ External volume − internal volume = 99 cm3
∴ πR2h − πr2h = 99 cm3
∴ πh(R2 − r2) = 99 cm3
∴ \(\frac{22}7\) × 14 (R + r)(R − r) = 99 cm3 [∵ a2 − b2 = (a + b)(a − b)] and [∵ R − r = 1/2]
∴ External radius = 2.5 cm
R + r = 29
r + 2.5 = 4.5
r = 4.5 − 2.5 = 2 cm
∴ Internal radius = 2 cm
Hence,
The sum of the inner and outer radii of the pipe = 2.5 + 2 = 4.5 cm