The slant height of the right circular cone is $10$ cm and its height in $8$ cm It is cut by plane parallel to its base passing through the mid-point of the height The ratio of the volume cone to that of the frustum of the cone cut is

2 : 1

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3 : 2

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4 : 3

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8 : 7

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Solution

The correct option is D $8 : 7$
It is given that
$O M = 8, O P = 4, a n d O B = 10$
$∴ B M^{2} = O B^{2} = O M^{2}$
$= 100 - 64 = 36$
$\Rightarrow P M = 6$ so that $P C = 3$
Now volume of cone OAB
$= \frac{1}{3} . π {.6}^{2} .8 = 96 π$
Volume of cone OCD $= \frac{1}{3} . π {.3}^{2} .4 = 12 π$
$∴$ Volume of frustum $A B C D = 96 π - 12 π = 84 π$
Reqd. ratio $= \frac{96 π}{84 π} = \frac{8}{7} \Rightarrow 8 : 7$

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The slant height of the right circular cone is 10cm and its height in 8cm It is cut by plane parallel to its base passing through the mid-point of the height The ratio of the volume cone to that of the frustum of the cone cut is

The correct option is D 8:7It is given thatOM=8,OP=4,andOB=10∴BM2=OB2=OM2=100−64=36⇒PM=6 so that PC=3Now volume of cone OAB=13.π.62.8=96πVolume of cone OCD =13.π.32.4=12π∴ Volume of frustum ABCD=96π−12π=84πReqd. ratio =96π84π=87⇒8:7

The slant height of the right circular cone is $10$ cm and its height in $8$ cm It is cut by plane parallel to its base passing through the mid-point of the height The ratio of the volume cone to that of the frustum of the cone cut is