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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • question_answer
    Radius of a capillary is \[2\times {{10}^{-3}}m\]. A liquid of weight \[6.28\,\times {{10}^{-4}}\,N\] may remain in the capillary, then the surface tension of liquid will be

    A)  \[5\times {{10}^{-3}}\,N/m\]    

    B)                         \[5\times {{10}^{-2}}\,N/m\]

    C)  5 N/m                 

    D)  50 N/m

    Correct Answer: B

    Solution :

    Given,\[r=2\times {{10}^{3}}m.\]                 \[F=W={{V}_{\rho g}}=6.28\times {{10}^{-4}}m\] Surface tension\[(T)=\frac{F}{2\pi r}=\frac{6.28\times {{10}^{-4}}}{2\times 3.14\times 2\times {{10}^{-3}}}\] \[=0.05N/m=5\times {{10}^{-2}}N/m\]


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