JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    The adjacent figure is the part of a horizontally stretched net. section AB is stretched with a force of 10 N. The tensions in the sections BC and BF are                                   [KCET 2005]

    A)             10 N, 11 N

    B)               10 N, 6 N

    C)           10 N, 10 N      

    D)             Can't calculate due to insufficient data

    Correct Answer: C

    Solution :

                    By drawing the free body diagram of point B             Let the tension in the section BC and BF are \[{{T}_{1}}\] and \[{{T}_{2}}\]respectively.             From Lami's theorem             \[\frac{{{T}_{1}}}{\sin 120{}^\circ }=\frac{{{T}_{2}}}{\sin 120{}^\circ }=\frac{T}{\sin 120{}^\circ }\]               Þ \[T={{T}_{1}}={{T}_{2}}=10\ N.\]


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