BHU PMT BHU PMT (Mains) Solved Paper-2010

  • question_answer

    Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

    A simple pendulum with a bob of mass m is suspended from the roof of a car moving with a horizontal acceleration a. Then (1) the string makes an angle of\[{{\tan }^{-1}}\left( \frac{a}{g} \right)\] with the vertical (2) the string makes an angle of\[{{\tan }^{-1}}\left( 1-\frac{a}{g} \right)\] with the vertical     (3) the tension in the string is\[\sqrt{{{a}^{2}}+{{g}^{2}}}\] (4) The tension in the string is\[\sqrt{{{g}^{2}}-{{a}^{2}}}\]

    A)  1, 2 and 3 are correct

    B)  1 and 2 are correct

    C)   2 and 4 are correct

    D)  1 and 3 are correct

    Correct Answer: D

    Solution :

                     From the figure \[T\cos \theta =mg\] \[T\sin \theta =ma\] \[\tan \theta =\frac{a}{g}\] \[\theta ={{\tan }^{-1}}\left( \frac{a}{g} \right)\] The tension of the string \[=\sqrt{T{{\cos }^{2}}\theta +T{{\sin }^{2}}\theta }\] \[=\sqrt{{{(mg)}^{2}}+{{(ma)}^{2}}}\] \[=m\sqrt{{{g}^{2}}+{{a}^{2}}}\] Hence, option  is correct.


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