JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    A uniform solid sphere of mass m is lying at rest between a vertical wall and a fixed inclined plane as shown. There is no friction between sphere and the vertical wall but coefficient of friction between the sphere and the fixed inclined plane is \[\text{ }\!\!\mu\!\!\text{ =1/2}\]. Then the magnitude of frictional force exerted by fixed inclined plane on sphere is (g is acceleration due to gravity)

    A)  \[\frac{mg}{2}\]                                              

    B)  \[\frac{\sqrt{3}}{4}mg\]

    C)  mg                                        

    D)  0

    Correct Answer: D

    Solution :

     If student use, \[f=\mu \,mg\cos \theta \]\[=\frac{1}{2}\times mg\cos {{30}^{o}}=\frac{\sqrt{3}}{4}mg\] Hence option [B] but correct answer is [D]. All forces on sphere pass through its centre except the force of friction exerted by inclined plane. Since net torque on sphere in equilibrium about its centre is zero, the torque on sphere due to frictional force about its centre must be zero. Hence frictional force on cylinder is zero.


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