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pdf bow guru antra goal get a chance to learn from india s best teachers crack jee advanced 2021 vedantu s eklavya 2021 is a specialized program for jee advanced ...

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                                   BOW
                                   GURU      ANTRA
                                   GOAL
       Get a Chance to Learn from India’s Best Teachers & Crack JEE ADVANCED 2021
       Vedantu’s Eklavya 2021 is a Specialized 
       Program for JEE ADVANCED, designed 
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       & many more.
                   Anand Prakash
                     Co-Founder, Vedantu
                          IIT- Roorkee
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                   & Their Achievements
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         Chirag  Jain                       731             41
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                 JEE-Main-26-08-2021-Shift-1 (Memory Based) 
                                                           
                                                  PHYSICS 
                                                           
              Question: In an L R circuit at steady state energy stored in inductor is 64 J and power 
              consumed by circuit is 640 J. Find time constant for this LR circuit. 
                                     
              Options: 
              (a) 0.2 
              (b) 0.1 
              (c) 0.8 
              (d) 0.25 
              Answer: (a) 
              Solution: 
              In steady state, energy stored in inductor 
               1 LI2 = 64J... 1  
               2            ( )
              Power consumed in steady state, 
                2
               IR=640J...(2)  
              Dividing (1) by (2), we get 
                L = 1  
               2R 10
                  L   2
               ⇒==0.2
                  R 10          
               
               
              Question: If 'E' represents energy, 'G' represents gravitational constant, 'M' represents mass 
              & 
                                                      −−2 2 33
                                                   
              'L' represents angular momentum; find  GMEL   
                                                   
              Options: 
                     4−67
                  
              (a)  MLT   
                  
                     222
                  
              (b)  M LT   
                  
                     −−33−4
                  
              (c)  M LT   
                  
                     −−422
                  
              (d)  M LT  
                  
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