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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
    A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \[25.5{{\operatorname{KVm}}^{-1}}\]. The density of liquid is\[1.26\times {{10}^{3}}\]. The radius of the drop is (neglect buoyancy)     JEE Main Online Paper ( Held On 23  April 2013 )

    A) \[4.3\times {{10}^{-7}}\operatorname{m}\]       

    B)        \[7.8\times {{10}^{-7}}\operatorname{m}\]

    C) \[0.078\times {{10}^{-7}}\operatorname{m}\]

    D)        \[3.4\times {{10}^{-7}}\operatorname{m}\]

    Correct Answer: B

    Solution :

    \[F=qE=mg(q=6e=6\times 1.6\times {{10}^{-19}})\] Density (d) \[=\frac{\text{mass}}{\text{volume}}=\frac{m}{\frac{4}{3}\pi {{r}^{3}}}\]or\[{{r}^{3}}=\frac{m}{\frac{4}{3}\pi d}\] Putting the value of d and\[m\left( =\frac{qE}{g} \right)\]and solving we get \[r=7.8\times {{10}^{-7}}m\]


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