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CMC Medical CMC-Medical Ludhiana Solved Paper-2012

  • question_answer
    A mass of 10 kg connected at the end of a rod of negligible mass is rotating in a circle of radius 30 cm with an angular velocity of 10 s\[\text{rad}{{\text{s}}^{-1}}\]. If the mass is brought to rest in 10 s by a brake, what is the magnitude of the torque applied?

    A)  0.5 N-m              

    B)  0.9 N-m

    C)  1.2N-m                               

    D)  2.3 N-m

    Correct Answer: B

    Solution :

                    Given, \[{{\omega }_{1}}=10\,\,\text{rad}{{\text{s}}^{-1}},\] \[{{\omega }_{2}}=0,\] \[t=10\,s\] \[\therefore \]  \[\alpha =\frac{{{\omega }_{2}}-{{\omega }_{1}}}{t}=\frac{0-10}{10}\]     \[=-1\,\,rad\,{{s}^{-2}}\] (Negative sign shows retardation) Now, moment of inertia \[I=m{{r}^{2}},\] \[=10\times {{(0.3)}^{2}}=0.9\,kg\text{-}{{m}^{2}}\] \[\therefore \]Torque, \[\tau =I\,\alpha =0.9\times 1=0.9\,N\text{-}m\]


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