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CBSE Class 10 Science 2016 Term 1 Paper

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Question : 35 of 36
Marks: +1, -0
Find the least count of a milliammeter in which there are 2020 divisions between 400400   mA\;\mathrm{mA} and 500  mA500\;\mathrm{mA} marks.
Solution:  
   Least count    =     Range      No. of division   \;\text{ Least count}\;\;=\;\frac{\;\text{ Range }\;}{\;\text{ No. of division }\;}
  =  500−40020=  10020=5  mA\;=\;\frac{500-400}{20}=\;\frac{100}{20}=5\;\mathrm{mA}
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