Manipal Medical Manipal Medical Solved Paper-2001

  • question_answer
    A bar magent has a magnetic moment equal to\[5\times {{10}^{-5}}\Omega m\]. It is suspended in a magnetic field, which has a magnetic induction\[B=8\pi \times {{10}^{-4}}\]tesla. Then the magnet .vibrates with period of vibration equal to 15 sec. Then the moment of inertia of magnet will be:

    A)  \[7.16\times {{10}^{-7}}kg\text{ }{{m}^{2}}\]

    B)  \[14.32\times {{10}^{7}}kg\text{ }{{m}^{2}}\]

    C)  \[3.58\times {{10}^{-7}}kg\text{ }{{m}^{2}}\]

    D)  none of these

    Correct Answer: A

    Solution :

     Here: Magnetic moment of bar magnet \[M=5\times {{10}^{-50}}\Omega -m\] Magnetic induction\[B=8\pi \times {{10}^{-4}}\]testa Period of oscillation of the magnet is given by, \[T=2\pi \sqrt{\frac{I}{MB}}\] \[{{T}^{2}}=4{{\pi }^{2}}\frac{I}{MB}\] Or \[I=\frac{{{T}^{2}}}{4{{\pi }^{2}}}.MB\] \[=\frac{{{15}^{2}}}{4\times {{(3.14)}^{2}}}\times 5\times {{10}^{-5}}\times 8\pi \times {{10}^{-4}}\] \[=7.16\times {{10}^{-7}}kg-{{m}^{2}}\]


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