Verified Solution • 2025–26 Session moderate

QNEET-2013-Ph-1607: NEET 2013 Physics - Class 11 NEET-2013-Ph-1607 Solution: If the ratio of diameters lengths and Young's modulus of steel…

If the ratio of diameters lengths and Young's modulus of steel and copper wires shown in the figure are p q and s respectively then the corresponding ratio of increase in their lengths would be (5q7sp2)(\frac{5q}{7sp^2}) (7q5sp2)(\frac{7q}{5sp^2}) (2q5sp)(\frac{2q}{5sp}) (7q5sp)(\frac{7q}{5sp})

Multiple Choice Options

  • A (5q7sp2)\left( {{{5q} \over {7s{p^2}}}} \right) ( 7 s p 2 5 q _ )
  • B (7q5sp2)\left( {{{7q} \over {5s{p^2}}}} \right) ( 5 s p 2 7 q _ )
  • C (2q5sp)\left( {{{2q} \over {5sp}}} \right) ( 5 s p 2 q _ )
  • D (7q5sp)\left( {{{7q} \over {5sp}}} \right) ( 5 s p 7 q _ )

Step-by-Step Verified Solution

From formula, Increase in length _ L = F L A Y = 4 F L � D 2 Y _ L S _ L C = F S F C ( D C D S ) 2 Y C Y S L S L C = 7 5 _ ( 1 p ) 2 ( 1 s ) q = 7 q ( 5 s p 2 )

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