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4-1.Newton's Laws of Motion
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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 $\mathrm{x}_{1}$ along the plane. But when the inclination is decreased to $30^{\circ}$ and the same object the shot with the same velocity, it can travel $x_{2}$ distance. Then $x_{1}: x_{2}$ will be
A$1: \sqrt{2}$
B$\sqrt{2}: 1$
C$1: \sqrt{3}$
D$1: 2 \sqrt{3}$
(NEET-2019)
Solution
$v^{2}=u^{2}-2 a s$
$\Rightarrow \mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}}=\frac{\mathrm{u}^{2}}{2 \mathrm{gsin} \theta}$
$\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}$
$\Rightarrow \frac{x_{1}}{x_{2}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}}=\frac{\mathrm{u}^{2}}{2 \mathrm{gsin} \theta}$
$\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}}=\frac{\sin \theta_{2}}{\sin \theta_{1}}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}$
$\Rightarrow \frac{x_{1}}{x_{2}}=\frac{1}{\sqrt{3}}$
Standard 11
Physics
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