WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    Two blocks of equal masses m are released from the top of a smooth fixed wedge as shown in the figure. The acceleration of the centre of mass of the two blocks is :

    A)  g                                            

    B)  \[\frac{g}{2}\]

    C)  \[\frac{3g}{4}\]                               

    D)  \[\frac{g}{\sqrt{2}}\]

    Correct Answer: B

    Solution :

                     The acceleration of the centre of mass of theblock. \[=\frac{g}{2}\sqrt{{{(\sin \angle ACB)}^{2}}+{{(\sin \angle ACB)}^{2}}}\] \[=\frac{g}{2}\sqrt{{{\sin }^{2}}{{30}^{o}}+{{\sin }^{2}}{{60}^{o}}}\] \[=\frac{g}{2}\sqrt{{{(0.5)}^{2}}+{{(0.866)}^{2}}}=\frac{g}{2}\]


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