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Physics Helpline
L K Satapathy
Direct Current - 15
Resistance of Parts of a wire
Physics Helpline
L K Satapathy
Direct Current - 15
Cells in Parallel
Question : A metallic wire of length 100cm is cut into two unequal parts A and B.
The part A is stretched uniformly to double its length such that its new resistance is
equal to that of the part B. Then the length of the part B is
(a) 20cm (b) 40cm (c) 80cm (d) 90cm
Let the resistance / centimeter of the wire = k
4 (100 ) 400 4A BR R k x kx x x       
Answer :
Correct option = (c)
Let the length of part B = x cm BR kx 
 The length of part A = (100 – x) cm (100 )AR k x   
When A is stretched to double its length , its area of cross section is halved
 The resistance of A becomes 4 4 (100 )A AR R k x    
5 4 [0 ]00 8 Anx sx cm   
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L K Satapathy
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Direct Current QA 15

  • 1. Physics Helpline L K Satapathy Direct Current - 15 Resistance of Parts of a wire
  • 2. Physics Helpline L K Satapathy Direct Current - 15 Cells in Parallel Question : A metallic wire of length 100cm is cut into two unequal parts A and B. The part A is stretched uniformly to double its length such that its new resistance is equal to that of the part B. Then the length of the part B is (a) 20cm (b) 40cm (c) 80cm (d) 90cm Let the resistance / centimeter of the wire = k 4 (100 ) 400 4A BR R k x kx x x        Answer : Correct option = (c) Let the length of part B = x cm BR kx   The length of part A = (100 – x) cm (100 )AR k x    When A is stretched to double its length , its area of cross section is halved  The resistance of A becomes 4 4 (100 )A AR R k x     5 4 [0 ]00 8 Anx sx cm   
  • 3. Physics Helpline L K Satapathy For More details: www.physics-helpline.com Subscribe our channel: youtube.com/physics-helpline Follow us on Facebook and Twitter: facebook.com/physics-helpline twitter.com/physics-helpline