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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2003

  • question_answer
    If the surface tension of liquid is\[5\times {{10}^{-2}}N/m\] and weight of liquid column is\[6.28\times {{10}^{-4}}N,\]then radius of capillary is

    A)  \[2\times {{10}^{-3}}m\]       

    B)  \[2.5\times {{10}^{-3}}m\]

    C)  \[2\times {{10}^{-4}}\]      

    D)  \[4\times {{10}^{-3}}m\]

    Correct Answer: A

    Solution :

     \[F=2\pi rT\] \[\Rightarrow \] \[r=\frac{F}{2\pi T}\] \[=\frac{6.28\times {{10}^{-4}}}{2\times 3.14\times 5\times {{10}^{-2}}}=2\times {{10}^{-3}}m\]


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