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Physics · PYQ NEET UG 2019(2)
हिन्दीA body of mass \( m \) is kept on a rough horizontal surface (coefficient of friction = \( \mu \)). A horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given by \( F \), where \( F \) is,
Correct answer · Option D
\( \vec{F} \le mg\sqrt{1+\mu^2} \)
Explanation
The resultant force \( F = \sqrt{N^2 + f^2} \). Here \( N = mg \) and \( f \le \mu mg \). Thus \( F = \sqrt{(mg)^2 + f^2} \le \sqrt{(mg)^2 + (\mu mg)^2} = mg\sqrt{1+\mu^2} \).
