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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 14

  • question_answer
    A glass capillary tube of inner diameter 0.28 mm is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that water level in the tube is same as that in the vessel in \[N{{m}^{-2}}\]is (Surface tension of water = 0.07 \[\text{N}\,{{\text{m}}^{-1}}\]and atmospheric pressure \[\text{=1}{{\text{0}}^{5}}\,\text{N}\,{{\text{m}}^{-2}}\])

    A) \[\text{1}{{\text{0}}^{3}}\]                 

    B) \[99\times {{10}^{3}}\]

    C) \[100\times {{10}^{3}}\]          

    D) \[101\times {{10}^{3}}\]

    Correct Answer: D

    Solution :

    [d]: Given : \[d=0.28\text{ }mm=0.28\times {{10}^{-3}}m\] \[r=0.14\times {{10}^{-3}}m\] Surface tension, \[S=0.07N{{m}^{-1}}\] Atmospheric pressure \[={{10}^{5}}N\,{{m}^{-2}}\] \[S=\frac{rh\rho g}{2}=0.07\] \[0.07=\frac{P\times 0.14\times {{10}^{-3}}}{2}\]\[(\because P=h\rho g)\] \[P=1\times {{10}^{3}}N{{m}^{-2}}\] Total pressure = P + Atmospheric pressure \[={{10}^{3}}N{{m}^{-2}}+{{10}^{5}}N{{m}^{-2}}\] \[={{10}^{3}}(1+100)N{{m}^{-2}}\] \[=101\times {{10}^{3}}N{{m}^{-2}}\]


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