2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    A neutral water molecule\[\text{(}{{\text{H}}_{\text{2}}}\text{O)}\]in its vapour state has an electric dipole moment of magnitude\[6.4\times {{10}^{-30}}C-m\]. How far apart are the molecules centres of positive and negative charges?

    A)  4m             

    B)  4 mm

    C)  \[4\mu \]m

    D)  4 pm

    Correct Answer: D

    Solution :

    There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is \[p=q(2I)=10e\,(2I)\] Hence length of the dipole ie, distance between centres of positive and negative charges is \[2l=\frac{p}{10e}=\frac{6.4\times {{10}^{-30}}}{10\times 1.6\times {{10}^{-19}}}\] \[=4\times {{10}^{-12}}m\] \[=4pm\]  


You need to login to perform this action.
You will be redirected in 3 sec spinner