A right-angled cone (with base radius 5 cm and height 12 cm), as shown in the figure below, is rolled on the ground keeping the point P fixed until the point Q (at the base of the cone, as shown) touches the ground again.
By what angle (in radians) about P does the cone travel?

Question

A right-angled cone (with base radius 5 cm and height 12 cm), as shown in the figure below, is rolled on the ground keeping the point P fixed until the point Q (at the base of the cone, as shown) touches the ground again.

By what angle (in radians) about P does the cone travel?

Prepp Team · Accepted Answer

To find the angle about point P by which the cone travels, we need to analyze the rolling motion of the cone.

Step-by-Step Solution:

First, observe that the cone is rolled on the ground, keeping point P fixed, until point Q touches the ground again. This means the cone makes a complete rotation about point P.
The slant height of the cone forms the radius of the sector when unfolded. To find the slant height, use the Pythagorean theorem:

$l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \text{cm}$

The circumference of the base of the cone is given by:

$C = 2\pi r = 2\pi \times 5 = 10\pi \, \text{cm}$

When the cone is rolled, the arc length covered by the base of cone equals the circumference of the base.
The angle of rotation about point P is given by the ratio of the arc length (circumference) to the circumference of the circle with radius $l$ (when developed).

$\text{Arc length} = 10\pi$

$\text{Circumference of circle with radius } l = 2\pi \times l = 2\pi \times 13 = 26\pi$

The angle in radians is:

$\theta = \frac{\text{Arc length}}{\text{circumference with radius} \, l} = \frac{10\pi}{26\pi} = \frac{10}{26} = \frac{5}{13}$

Therefore, the angle by which the cone travels about point P is $\frac{10\pi}{13}$ radians.

Thus, the correct answer is $\frac{10\pi}{13}$.