2. Solved Papers
  3. JCECE Medical

JCECE Medical JCECE Medical Solved Paper-2015

  • question_answer
    A uniform rod of mass m and length L is rotated about an axis passing through the point P as shown in figure. The magnitude of angular momentum of the rod about the rotational axis yy passing through the point P is

    A)  \[\frac{M{{L}^{2}}}{18}\]

    B)  \[\frac{M{{(x+y)}^{2}}}{9}\]

    C)  \[\frac{M({{L}^{2}}+{{x}^{2}})}{9}\]

    D)  \[\frac{M({{L}^{2}}+2{{y}^{2}})}{18}\]

    Correct Answer: B

    Solution :

     We have \[\frac{x}{y}=\frac{2}{1}\]or \[\frac{x+y}{y}=\frac{3}{1}\] or \[y=\frac{x+y}{3}=\frac{L}{3}\] \[\Rightarrow \] \[x=2y=\frac{2L}{3}\] Now,    \[{{I}_{p}}={{I}_{com}}+M{{(d)}^{2}}\] \[=\frac{M{{(L)}^{2}}}{12}+M{{\left( \frac{L}{2}-\frac{L}{3} \right)}^{2}}\] \[=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{36}\] \[=\frac{4M{{L}^{2}}}{36}=\frac{M{{L}^{2}}}{9}=\frac{M{{(x+y)}^{2}}}{9}\]


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