Question

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

• A. $1:\sqrt{3}$
• B. $1: 2\sqrt{3}$
• C. $1:\sqrt{2}$
• D. $\sqrt{2}:1$

Answer 1

The Correct Option is A

To solve the problem of finding the ratio of distances $x_1$ to $x_2$ traveled by the object on inclined planes at angles $60^\circ$ and $30^\circ$, we need to analyze the motion on inclined planes.

When an object is projected up a smooth inclined plane, the distance it travels depends on the initial velocity and the inclination angle. Considering the conservation of energy, the distance $x$ traveled on an inclined plane can be expressed as:

x = \frac{u^2}{2g \sin \theta}

Where:

• $u$ is the initial velocity of the object.
• $g$ is the acceleration due to gravity.
• $\theta$ is the angle of inclination.

From the formula, it is evident that the distance traveled is inversely proportional to $\sin \theta$.

First, calculate $x_1$ for $\theta = 60^\circ$:

x_1 = \frac{u^2}{2g \sin 60^\circ} = \frac{u^2}{2g \cdot \frac{\sqrt{3}}{2}} = \frac{u^2}{g \sqrt{3}}

Next, calculate $x_2$ for $\theta = 30^\circ$:

x_2 = \frac{u^2}{2g \sin 30^\circ} = \frac{u^2}{2g \cdot \frac{1}{2}} = \frac{u^2}{g}

Now, find the ratio \frac{x_1}{x_2}:

\frac{x_1}{x_2} = \frac{\frac{u^2}{g \sqrt{3}}}{\frac{u^2}{g}} = \frac{1}{\sqrt{3}}

The ratio $x_1 : x_2$ is thus $1 : \sqrt{3}$.

Therefore, the correct answer is:

$1:\sqrt{3}$