Let the outer and inner radii be R cm and r cm respectively. Also, it is given that h = 14 cm. Then,
outside surface area `= (2pi Rh)` sq units
`= (2 xx (22)/(7) xx R xx 14) cm^(2)`
`= (88R) cm^(2)`
Inside surface area `= (2pi rh)` sq units
`= (2 xx (22)/(7) xx r xx 14) cm^(2) = (88r) cm^(2)`
`:. (88R - 88r) = 44 rArr 88 (R - r) = 44 rArr (R - r) = (1)/(2)`..(i)
External volume `= (pi R^(2)h)`
`= ((22)/(7) xx R^(2) xx 14) cm^(3) = (44R^(2)) cm^(3)`
Internal voluem `= (pi r^(2) h)` cubic units
`= ((22)/(7) xx r^(2) xx 14) cm^(3) = (44r^(2)) = cm^(3)`
Volume of metal = (external volume) - (internal volume)
`= (44R^(2) - 44 r^(2)) cm^(3) = 44 (R^(2) - r^(2)) cm^(3)`
`:. 44(R^(2) - r^(2)) = 99 rArr (R^(2) - r^(2)) = (99)/(44) rArr (R^(2) - r^(2)) = (9)/(4)`...(ii)
On dividing (ii) by (i), we get `(R + r) = (9)/(2)`..(iii)
On solving (i) and (iii), we get `R = (5)/(2) and r = 2`
Hence, the outer radius = 2.5 cm and the inner radius = 2 cm