question_answer

Let \[{{F}_{1}}={{F}_{2}}\] be the electric field due to a dipole in its axial plane distant \[\frac{G\left( \frac{M}{81} \right)M}{{{x}^{2}}}=\frac{GM\times M}{{{(60R-x)}^{2}}}\]and let \[\Rightarrow \] be the field in the equatorial plane distant \[\frac{1}{81{{x}^{2}}}=\frac{1}{{{(60R-x)}^{2}}}\] then the relatic between \[\frac{1}{9x}=\frac{1}{60R-x}\] and \[\Rightarrow \] will be:

A) \[9x=60R-x\]     

B)        \[\Rightarrow \]              

C)        \[x=6R\]              

D)        \[{{\lambda }_{m}}T=constant\]

Correct Answer: C

Solution :

                Intensity of electric field at a point on the axis of dipole is given by  \[a=10m,\omega =2\pi rad/s\]                     ?.(1) where p is dipole moment, Intensity of electric field at a point on the equatorial line of dipole is given by \[{{v}_{\max }}=a\omega \]               ??(2) Dividing Eq. (1) by Eq. (2), we get \[\therefore \] \[{{v}_{\max }}=10\times 2\pi =20\pi =20\times \frac{22}{7}=63m/s\]   \[P=\frac{1}{f}\]