Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle \(60 \degree\)with horizontal, it can travel a distance \(x_1\) along the plane. But when the inclination is decreased to \(30 \degree\) and the same object is shot with the same velocity, it can travel \(x_2\) distance. Then \(x_1:x_2\) will be:

• A. \(1:\sqrt2\)
• B. \(\sqrt2:1\)
• C. \(1:\sqrt3\)
• D. \(1:2\sqrt3\)

Answer 1

The Correct Option is C

To determine the ratio \(x_1:x_2\) of the distances traveled by an object on an inclined plane at two different angles, we must understand how the inclination angle affects the range along the plane.

The range of a projectile on an inclined plane is affected by the inclination of the plane and the initial velocity of projection. The range on an inclined plane is given by:

R = \frac{v^2 \sin(2\theta)}{g \cos\alpha}

where:

• v is the initial velocity of projection.
• \theta is the angle of projection with respect to the inclined plane.
• \alpha is the angle of the inclined plane with the horizontal.
• g is the acceleration due to gravity.

In this question, the incline angles are \(60^\circ\) and \(30^\circ\). We need to find the ratio of distances traveled, \(x_1\) and \(x_2\), when the same object is projected with the same initial velocity on planes inclined at these angles.

For \alpha = 60^\circ:

{x_1} = \frac{u^2 \sin(2\theta)}{g \cos{60^\circ}}

For \alpha = 30^\circ:

{x_2} = \frac{u^2 \sin(2\theta)}{g \cos{30^\circ}}

Now, divide the two equations to get the ratio:

\frac{x_1}{x_2} = \frac{\cos{30^\circ}}{\cos{60^\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}

Thus, the ratio \(x_1:x_2\) is 1:\sqrt{3}.

Therefore, the correct answer is:

\(1:\sqrt3\)