Solved papers for JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

done JEE Main Paper (Held On 11 April 2015) Total Questions - 2

• question_answer1) Two long straight parallel wires, carrying (adjustable) currents \[\,{{I}_{1}}\]and \[\,{{I}_{2}},\] are kept at a distance d apart. If the force 'F' between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other, the graph showing the dependence of 'F', on the product \[\,{{I}_{1}}\,{{I}_{2}},\]would be : [JEE Main Online Paper (Held On 11 April 2015)]  
  A)
   
  done clear
  B)
   
  done clear
  C)
   
  done clear
  D)
   
  done clear
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• question_answer2) A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle \[2{{\theta }_{0}}\] at the centre of the circle (of which it forms an arch) then the tension in the wire is : [JEE Main Online Paper (Held On 11 April 2015)]  
  A)
   \[IBR\]
  done clear
  B)
  \[\frac{IBR}{\sin {{\theta }_{0}}}\]
  done clear
  C)
  \[\frac{IBR}{2\sin {{\theta }_{0}}}\]
  done clear
  D)
  \[\frac{IBR{{\theta }_{0}}}{\sin {{\theta }_{0}}}\]
  done clear
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