BVP Medical BVP Medical Solved Paper-2002

  • question_answer If the coefficient of friction of a plane inclined at 45°, is 0.5, then acceleration of a body sliding freely on it is:

    A) \[\frac{9.8}{2\sqrt{2}}m/{{s}^{2}}\]

    B) \[\frac{9.8}{\sqrt{2}}m/{{s}^{2}}\]

    C) \[9.8m/{{s}^{2}}\]                           

    D) \[4.9m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

                    Acceleration of the body on inclined plane. \[a=g(\sin \theta -\mu \cos \theta )\] \[=g(\sin {{45}^{o}}-\mu \cos {{45}^{o}})\] \[=g\left( \frac{1}{\sqrt{2}}-0.5\frac{1}{\sqrt{2}} \right)\] \[=\frac{g}{\sqrt{2}}(1-0.5)=\frac{g}{2\sqrt{2}}\] \[=\frac{9.8}{2\sqrt{2}}m/{{s}^{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner