JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A uniform sphere of weight W and radius 5cm is being held by a string as shown in the figure. The tension in the string will be:                                              [JEE ONLINE 09-04-2013]

    A)             \[12\frac{W}{5}\]

    B) \[5\frac{W}{12}\]

    C)             \[13\frac{W}{12}\]

    D) \[13\frac{W}{12}\]

    Correct Answer: D

    Solution :

    \[T\cos \theta =w\]
    \[T=\frac{w}{\cos \theta }\]
    \[\cos \theta =\frac{AB}{AC}=\frac{AB}{AD+BC}=\frac{AB}{8+5}=\frac{AB}{13}\]
    From right angle triangle \[AB=12\]
    So                    \[T=\frac{w\times 13}{12}\]

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