NEET Sample Paper NEET Sample Test Paper-90

  • question_answer
    A proton and an \[\alpha \]-particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes \[25\,\,\mu \] second to make 5 revolutions, then the time period for the \[\alpha \]-particle would be

    A) \[50\,\,\mu \]sec 

    B)        \[25\,\,\mu \]sec

    C) \[10\,\,\mu \]sec 

    D)         \[5\,\,\mu \]sec

    Correct Answer: C

    Solution :

    [c] Time taken by proton to make one revolution \[=\frac{25}{5}=5\mu \,\,\sec \] As \[T=\frac{2\pi m}{qB};\]so \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\times \frac{{{q}_{1}}}{{{q}_{2}}}\] or \[{{T}_{2}}={{T}_{1}}\frac{{{m}_{2}}\,{{q}_{1}}}{{{m}_{1}}\,{{q}_{2}}}=\frac{5\times 4\,{{m}_{1}}}{{{m}_{1}}}\times \frac{q}{2q}=10\mu \sec .\]


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