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JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let \[{{P}_{2}}\]be the pressure inside the inner bubble and\[{{P}_{0}},\]the pressure outside the outer bubble. Radius of another bubble with pressure difference \[{{P}_{2}}-{{P}_{0}}\] between its inside and outside would be [JEE Main 16-4-2018]

    A)  6 cm              

    B)  12 cm

    C)  4.8 cm           

    D)  2.4 cm

    Correct Answer: D

    Solution :

     As we enter a soap bubble the pressure increases by                 \[\frac{4T}{R}\] Here is radius of bubble, and is surface tension.                 \[{{P}_{2}}={{P}_{0}}+\frac{4T}{4}+\frac{4T}{6}\] \[{{p}_{2}}={{P}_{0}}+\frac{5}{12}\times 4T\] \[{{p}_{2}}-{{P}_{0}}=\frac{4T}{\frac{12}{5}}\] So for a single bubble to have a pressure difference of\[{{P}_{2}}-{{P}_{0}}\] \[R=\frac{12}{5}=2.4\,cm\]


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