J & K CET Engineering J and K - CET Engineering Solved Paper-2004

  • question_answer
    If the radius of the earth contracts to half of its present day value without change in mass, then the length of the day will be

    A)  \[24\text{ }h\]         

    B)  \[48\text{ }h\]

    C)  \[\text{6 }h\]           

    D)  \[\text{12 }h\]

    Correct Answer: C

    Solution :

    From law of conservation of angular momentum, \[J=I\omega =\text{constant}\] \[\therefore \] \[{{I}_{1}}{{\omega }_{1}}={{I}_{2}}{{\omega }_{2}}\] where I is moment of inertia and co the angular velocity. Moment of inertia of earth assuming it be a sphere of radius \[R=\frac{2}{5}M{{R}^{2}}\] Also,         \[\omega =\frac{2\pi }{T}\] \[\therefore \] \[\left( \frac{2}{5}MR_{1}^{2} \right)\left( \frac{2\pi }{{{T}_{1}}} \right)=\left( \frac{2}{5}MR_{2}^{2} \right)\left( \frac{2\pi }{{{T}_{2}}} \right)\] \[\Rightarrow \] \[\frac{R_{1}^{2}}{{{T}_{1}}}=\frac{R_{2}^{2}}{{{T}_{2}}}\] \[\Rightarrow \] \[\frac{R_{1}^{2}}{24}={{\left( \frac{{{R}_{1}}}{2} \right)}^{2}}\times \frac{1}{{{T}_{2}}}\] \[\Rightarrow \] \[{{T}_{2}}=6h\]


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