# When a 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$x_2 \: \: distance .\: \: Then\: \: x_1 : x_ 2 \: \: will\: \: be :$Option 1)$1 : \sqrt 2$Option 2)$\sqrt 2 : 1$Option 3)$1 : \sqrt 3$Option 4)$1 : 2 \sqrt 3$

$\Delta KE = - \Delta PE \\\\ \frac{1}{2} mV^2 = mgh \\\\ h_1 = h_2 \\\\ h_ 1 = x \sin 60 \\\\ h_ 2= x_ 2 \sin 30 \\\\ \frac{x_1 }{x_2 } = \frac{\sin 30 }{\sin 60}$

$= \frac{1}{2} \times \frac{1 }{\sqrt 3 /2 } = 1 / \sqrt 3$

Option 1)

$1 : \sqrt 2$

Option 2)

$\sqrt 2 : 1$

Option 3)

$1 : \sqrt 3$

Option 4)

$1 : 2 \sqrt 3$

Exams
Articles
Questions