CBSE Class 10 Answered

a right circular is divided by a plane parallel to its base into small cone of volume V1 at the top and a frustum of volume V2 as second part at the bottom . if V1:V2=1:3, find the ratio of the height of the altitude of small cone and that of the frustum

Asked by Rishikesh Jadhav | 09 Feb, 2014, 04:30: PM

Expert Answer

Volume of smaller cone =

$V subscript 1 equals fraction numerator pi r subscript 1 squared h subscript 1 over denominator 3 end fraction space w h e r e space r subscript 1 space i s space t h e space b a s e space r a d i u s space a n d space h subscript 1 space i s space t h e space h e i g h t space o f space o f space t h e space s m a l l e r space c o n e space r e s p e c t i v e l y.$

Volume of the frustum =

$V subscript 2 equals fraction numerator pi h subscript 2 over denominator 3 end fraction open parentheses r subscript 2 squared plus r subscript 1 r subscript 2 plus r subscript 1 squared close parentheses comma space w h e r e space r subscript 1 space i s space t h e space r a d i u s space o f space t h e space u p p e r space b a s e comma space r subscript 2 space i s space t h e r a d i u s space o f space t h e space l o w e r space b a s e space a n d space h subscript 2 space i s space t h e space h e i g h t space o f space t h e space f u r s t u m space o f space t h e space c o n e.$

Given that, V subscript 1 over V subscript 2 equals 1 third

That is,

$V subscript 1 over V subscript 2 equals fraction numerator begin display style fraction numerator pi r subscript 1 squared h subscript 1 over denominator 3 end fraction end style over denominator begin display style fraction numerator pi h subscript 2 open parentheses r subscript 2 squared plus r subscript 1 r subscript 2 plus r subscript 1 squared close parentheses over denominator 3 end fraction end style end fraction rightwards double arrow 1 third equals fraction numerator r subscript 1 squared h subscript 1 over denominator h subscript 2 open parentheses r subscript 2 squared plus r subscript 1 r subscript 2 plus r subscript 1 squared close parentheses end fraction rightwards double arrow h subscript 1 over h subscript 2 equals fraction numerator open parentheses r subscript 2 squared plus r subscript 1 r subscript 2 plus r subscript 1 squared close parentheses over denominator 3 r subscript 1 squared end fraction$

Now consider the following diagram

Here, triangles ABC and ADE are similar triangles.

Thus, we have

$fraction numerator A B over denominator A D end fraction equals fraction numerator B C over denominator D E end fraction rightwards double arrow fraction numerator h subscript 1 over denominator h subscript 1 plus h subscript 2 end fraction equals r subscript 1 over r subscript 2 rightwards double arrow fraction numerator h subscript 1 plus h subscript 2 over denominator h subscript 1 end fraction equals r subscript 2 over r subscript 1 rightwards double arrow 1 plus h subscript 2 over h subscript 1 equals r subscript 2 over r subscript 1 rightwards double arrow h subscript 2 over h subscript 1 equals r subscript 2 over r subscript 1 minus 1 rightwards double arrow h subscript 2 over h subscript 1 equals fraction numerator r subscript 2 minus r subscript 1 over denominator r subscript 1 end fraction rightwards double arrow h subscript 1 over h subscript 2 equals fraction numerator r subscript 1 over denominator r subscript 2 minus r subscript 1 end fraction$

Therefore, we have,

$fraction numerator r subscript 1 over denominator r subscript 2 minus r subscript 1 end fraction equals fraction numerator r subscript 2 squared plus r subscript 1 r subscript 2 plus r subscript 1 squared over denominator 3 r subscript 1 squared end fraction rightwards double arrow 3 r subscript 1 cubed equals r subscript 2 cubed plus r subscript 1 r subscript 2 squared plus r subscript 1 squared r subscript 2 minus r subscript 1 r subscript 2 squared minus r subscript 1 squared r subscript 2 minus r subscript 1 cubed rightwards double arrow 4 r subscript 1 cubed equals r subscript 2 cubed rightwards double arrow 3 root of 4 r subscript 1 equals r subscript 2$

Substituting this value in the ratio of altitudes, we have

$h subscript 1 over h subscript 2 equals fraction numerator r subscript 1 over denominator r subscript 2 minus r subscript 1 end fraction equals fraction numerator r subscript 1 over denominator 3 root of 4 r subscript 1 minus r subscript 1 end fraction equals fraction numerator 1 over denominator 3 root of 4 minus 1 end fraction$