Question:
The diameter of a hollow cone is equal to the diameter of a spherical ball. If the ball is placed at the base of the cone, what portion of the ball will be outside the cone?
The diameter of a hollow cone is equal to the diameter of a spherical ball. If the ball is placed at the base of the cone, what portion of the ball will be outside the cone?
Show Hint
The cone narrows away from its base, so the ball cannot sink so that its centre is level with the base circle. Find where its centre ends up.
Updated On: Nov 5, 2026
- 50%
- less than 50%
- more than 50%
- 100%
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Set up the picture:
Let the ball have radius r. The cone has base radius r (same diameter as the ball) and some height H. The ball is placed in the opening of the cone, with the cone's pointed end away from the ball.
Step 2: Could the ball's centre be exactly in the plane of the base?
If it were, the ball's widest circle (radius r) would coincide with the base rim, and just inside the cone the ball would be wider than the cone, because the cone's cross-section shrinks steadily while the ball's cross-section shrinks only slowly at first. The ball would cut through the wall. So it cannot sink that far.
Step 3: Where does the ball rest?
The ball rests touching the inside wall of the cone. Looking at a cross-section, the wall is a straight line from the rim to the apex. The centre of the ball is at distance r from this line. Working it out, the centre stays at a height \( \sqrt{H^2 + r^2} - H \) above the plane of the base, which is greater than 0.
Step 4: Conclusion:
Only the part of the ball below the plane of the base is inside the cone. Since the centre is above that plane, less than half the ball lies below it. So more than half of the ball is outside the cone.
Final Answer:
The correct option is (3): more than 50%.
Let the ball have radius r. The cone has base radius r (same diameter as the ball) and some height H. The ball is placed in the opening of the cone, with the cone's pointed end away from the ball.
Step 2: Could the ball's centre be exactly in the plane of the base?
If it were, the ball's widest circle (radius r) would coincide with the base rim, and just inside the cone the ball would be wider than the cone, because the cone's cross-section shrinks steadily while the ball's cross-section shrinks only slowly at first. The ball would cut through the wall. So it cannot sink that far.
Step 3: Where does the ball rest?
The ball rests touching the inside wall of the cone. Looking at a cross-section, the wall is a straight line from the rim to the apex. The centre of the ball is at distance r from this line. Working it out, the centre stays at a height \( \sqrt{H^2 + r^2} - H \) above the plane of the base, which is greater than 0.
Step 4: Conclusion:
Only the part of the ball below the plane of the base is inside the cone. Since the centre is above that plane, less than half the ball lies below it. So more than half of the ball is outside the cone.
Final Answer:
The correct option is (3): more than 50%.
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