A right circular cone is sliced into a smaller cone and a frustum of a cone by a plane perpendicular...........

Quantitative Aptitude (CDS) Mensuration 3D

A right circular cone is sliced into a smaller cone and a frustum of a cone by a plane perpendicular to its axis. The volume of the smaller cone and the frustum of the cone are in the ratio 64 : 61. Then their curved surface areas are in the ratio

4 :1

16 : 9

64 : 61

81: 64

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Explanation:

For small cone, radius = r, height = h, slant height = l For big cone, radius = R, height = H, slant height = L

The triangles formed in the smaller and bigger cones are similar, hence, r/R = h/H = l/L --- eq (1)

Now, (Volume of small cone/ Volume of Frustum) = 64/61 = k(constant)

Thus, volume of big cone = 64k + 61k = 125k

Volume of small come; V1/ Volume of big cone, V2 = (1/3 πr²h)/ (1/3 πR²H)

Also, V1/ V2 = 64m/125m = 64/125 So, r²h/ R²H = 64/125

From eq (1), r³/ R³ = 64/125; r/R = 4/5

Now, Ratio of curved surface area of small cone/ Ratio of curved surface area of big cone = πrl/ πRL = (4/5)*(4/5) (from eq 1) = 16/25 = k constant

So, Curved surface area of frustum=25k – 16k=9k Thus, Ratio of curved surface area of small cone/ Ratio of curved surface area of frustum

=16k/9k=16:9