The diameter of one of the bases of a truncated cone is 100 mm. If the diameter of this base is increased by `21%` such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by `21%`. The radius the other base (in mm) is

Options

A. 65
B. 55
C. 45
D. 35

Answer

B

Explanation

Let initially 2 bases have radii 5 cm and r cm.
Finally base have radii `(1.21 xx 5)` and r
Ratios of volumes = `(V_(2))/(V_(1))=1.21`
`V_(2)=(pih)/3[(6.05)^(2)+(6.05)r+r^(2)]`
`V_(1)=(pih)/3[5^(2)+5r+r^(2)]`
`V_(2)/V_(1)=1.21 rArr((6.05)^(2)+(6.05)r+r^(2))/(5^(2)+5r+r^(2))`
`rArr r^(2)=(6.3525)/21 `
`rArr r=11/2 cm = 55 mm`