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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    The following observations were taken for determining surface tensiton T of water by capillary method : Diameter of capilary, \[D=1.25\times {{10}^{-2}}m\] rise of water, \[h=1.45\times {{10}^{-2}}m\]. Using \[g=9.80\,m/{{s}^{2}}\]and the simplified relation\[\text{T=}\frac{\text{rhg}}{\text{2}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\,\text{N/m,}\] the possible error in surface tension is closest to:    JEE Main Solved Paper-2017

    A)  2.4%                    

    B)  10%

    C)  0.15%                  

    D)  1.5%

    Correct Answer: D

    Solution :

     \[\text{T=}\frac{rhg}{2}\times {{10}^{3}}\]                 \[\frac{\Delta T}{T}=\frac{\Delta r}{r}+\frac{\Delta h}{h}+0\]                 \[100\times \frac{\Delta T}{T}=\left( \frac{{{10}^{-2}}\times .01}{1.25\times {{10}^{-2}}}+\frac{{{10}^{-2}}\times .01}{1.45\times {{10}^{-2}}} \right)100\] \[=(0.8+0.689)\] \[=(1.489)\] \[100\times \frac{\Delta T}{T}=1.489%\] \[\simeq 1.5%\]


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