Solved papers for JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

done JEE Main Paper (Held On 12 April 2014) Total Questions - 2

• question_answer1) A spherically symmetric charge distribution is characterized by a charge density having the following variations: \[\rho (r)={{\rho }_{o}}\left( 1-\frac{r}{R} \right)\]for r < R\[\rho (r)=0\]for \[r\ge R\] Where r is the distance from the centre of the charge distribution \[{{\rho }_{o}}\]is a constant. The electric field at an internal point (r < R) is:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )
  A)
  \[\frac{{{\rho }_{o}}}{4{{\varepsilon }_{0}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]               
  done clear
  B)
  \[\frac{{{\rho }_{o}}}{{{\varepsilon }_{o}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]
  done clear
  C)
  \[\frac{{{\rho }_{o}}}{3{{\varepsilon }_{o}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]               
  done clear
  D)
  \[\frac{{{\rho }_{o}}}{12{{\varepsilon }_{o}}}\left( \frac{r}{3}-\frac{{{r}^{2}}}{4R} \right)\]
  done clear
  View Answer play_arrow

• question_answer2) The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: \[K(x)={{K}_{o}}+\lambda x\] (\[\lambda =\] a constant) The capacitance C, of the capacitor, would be related to its vacuum capacitance Co for the relation :   [JEE Main Online Paper ( Held On 12 Apirl  2014 )
  A)
  \[C=\frac{\lambda d}{\ln (1+{{K}_{o}}\lambda d)}{{C}_{o}}\]      
  done clear
  B)
  \[C=\frac{\lambda }{d.ln(1+{{K}_{o}}\lambda d)}{{C}_{o}}\]
  done clear
  C)
  \[C=\frac{\lambda d}{ln(1+\lambda d/{{K}_{o}})}{{C}_{o}}\]
  done clear
  D)
  \[C=\frac{\lambda }{d.ln(1+{{K}_{o}}/\lambda d)}{{C}_{o}}\]
  done clear
  View Answer play_arrow