KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    A body slides down an inclined plane of inclination 6. The coefficient of friction down the plane varies in direct proportion to the distance moved down the plane \[(\mu =k\,x)\]. The body will move down the plane with a

    A) constant acceleration\[~=g\text{ }sin\text{ }\theta \]

    B) constant acceleration\[=(g\,\sin \,\theta -\mu \,g\,\cos \theta )\]

    C) constant retardation \[=(\mu \,g\,\,\cos \,\theta -g\sin \theta )\]

    D) variable acceleration that first decreases from \[g\text{ }sin\text{ }\theta \] to zero and then becomes negative.

    Correct Answer: D

    Solution :

    Force of friction is \[F=\mu R=\mu \,mg\cos \theta \]
    When the body slides down, the downward force along the plane = component \[mg\sin \theta \]of the weight mg. since the force of friction acts upwards along the plane, the effective downward force = \[mg\sin \theta -\mu mg\cos \theta =mg(sin\theta -\mu cos\theta )\]
    \[\therefore \]Acceleration = force/ mass = \[g(sin\theta -\mu cos\theta )\]
    \[=g(\sin \theta -kx\cos \theta )\]
    Hence, the acceleration varies with x and decreases as x increases. Thus, the correct choice is [D].

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