Solved papers for JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

done JEE Main Paper (Held On 9 April 2016) Total Questions - 3

• question_answer1) To know the resistance G of a galvanometer by half deflection method, a battery of emf \[{{V}_{E}}\] and resistance R is used to deflect the galvanometer by angle\[\theta .\]If a shunt of resistance S is needed to gat half deflection then G, R and S are related by the equation :   JEE Main Online Paper (Held On 09 April 2016)
  A)
  2S = G                                  
  done clear
  B)
  2G = S                
  done clear
  C)
  S(R+G) = RG                      
  done clear
  D)
  2S(R+G) = RG
  done clear
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• question_answer2) In the circuit shown, the resistance r is a variable resistance. If for r = fR, the heat generation in r is maximum then the value of f is :   JEE Main Online Paper (Held On 09 April 2016)
  A)
  1                                             
  done clear
  B)
  \[\frac{3}{4}\]
  done clear
  C)
  \[\frac{1}{4}\]                                   
  done clear
  D)
  \[\frac{1}{2}\]
  done clear
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• question_answer3) \[A50\Omega \]resistance is connected to a battery of 5V. A galvanometer of resistance \[100\Omega \]is to be used as ammeter to measure current through the resistance, for this a resistance \[{{r}_{s}}\] is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the current without the ammeter in the circuit ?   JEE Main Online Paper (Held On 09 April 2016)
  A)
  \[{{r}_{s}}=0.5\Omega n\]parallel with the galvanometer
  done clear
  B)
  \[{{r}_{s}}=0.5\Omega \]in series with the galvanometer
  done clear
  C)
  \[{{r}_{s}}=1\Omega \] in series with galvanometer
  done clear
  D)
  \[{{r}_{s}}=1\Omega \] in parallel with galvanometer
  done clear
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