Solved papers for JEE Main & Advanced AIEEE Solved Paper-2010

done AIEEE Solved Paper-2010 Total Questions - 2

• question_answer1)   In the circuit shown below, the key K is closed at\[t=0.\]the current through the battery is ?       AIEEE  Solved  Paper-2010
  A)
  \[\frac{V({{R}_{1}}+{{R}_{2}})}{{{R}_{1}}{{R}_{2}}}\]at \[t=0\]and \[\frac{V}{{{R}_{2}}}\]at\[t=\infty \]
  done clear
  B)
  \[\frac{V{{R}_{1}}{{R}_{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}\]at \[t=0\]and \[\frac{V}{{{R}_{2}}}\]at \[t=\infty \]
  done clear
  C)
  \[\frac{V}{{{R}_{2}}}\]at\[t=0\]and \[\frac{V({{R}_{1}}+{{R}_{2}})}{{{R}_{1}}{{R}_{2}}}\]at\[t=\infty \]
  done clear
  D)
  \[\frac{V}{{{R}_{2}}}\]at \[t=0\]and \[\frac{V{{R}_{1}}{{R}_{2}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}\]at\[t=\infty \]
  done clear
  View Answer play_arrow

• question_answer2)   A rectangular loop has a sliding connector PQ of length\[l\]and resistance\[R\,\Omega \]and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents\[{{I}_{1}},{{I}_{2}}\]and I are ?       AIEEE  Solved  Paper-2010
  A)
  \[{{I}_{1}}={{I}_{2}}=\frac{B/v}{6R},I=\frac{B/v}{3R}\]
  done clear
  B)
  \[{{I}_{1}}=-{{I}_{2}}=\frac{B/v}{R},I=\frac{2B/v}{R}\]
  done clear
  C)
  \[{{I}_{1}}={{I}_{2}}=\frac{B/v}{3R},I=\frac{2B/v}{3R}\]
  done clear
  D)
  \[{{I}_{1}}={{I}_{2}}=I=\frac{B/v}{R}\]
  done clear
  View Answer play_arrow