Today's question deals with geometry!
The diameter of a hollow cone is equal to the diameter of a spherical ball. If the ball is placed at the base of a cone, what portion of ball will be outside the cone?
Now in case of cone what will happen is very clear from figure itself. Tangent touches the ball at the point that is in the lower halve. So ball is less than 50 % inside. So more than 50 % of ball will remain outside.
So option 3.
The diameter of a hollow cone is equal to the diameter of a spherical ball. If the ball is placed at the base of a cone, what portion of ball will be outside the cone?
- <50%
- 50%
- >50%
- 100%
Key concept : Tangent and radius are perpendicular to each other at point of contact.
Consider the following image in which 50% of ball is inside. This is possible only in case of cylinder. Because then only radius and tangent will be perpendicular. Consider the radius ( or diameter) passing through center and completely horizontal. This divides the ball in two equal halves. It can be easily understood that only in case of cylinder 50 % of ball will be in side.
Now in case of cone what will happen is very clear from figure itself. Tangent touches the ball at the point that is in the lower halve. So ball is less than 50 % inside. So more than 50 % of ball will remain outside.
So option 3.
