A rope 1 cm in diameter breaks if tension in it exceeds 500 N. The maximum tension that may be given to a similar rope of diameter 2 cm is

A) 2000 N B) 1000 N C) 500 N D) 250 N

• Deepshikha
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• B

  • Deepika
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• 2000N

  • Arpan
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• Tension =2000N Becz D=2 tension wl b 4 times

  • Megha
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• 👶💜✌

  • Ajaib
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  1

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• 500

  • Priyanka
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• Good

  • Yagpal
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• as you know that young's modulus = (tension*original length)/(cross sectional area*change in length) since in this question all the quantities except tension and area are constant we can say that tension∝area of cross section. thus tension∝D² (as area = πD²/4 ; D is diameter) so if for D =1 tension = 500N then for D = 2 tension will be four times i.e tension = 2000N.

  • Deepak
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• 1000N

  • Pathan
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• 1000 N

  • Rishikesh
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  0

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• 1000N

  • Binawade
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  0

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• 1000

  • Sahil
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• as b.s=plg. t1/t2=l1/l2. 500/t2=1/2. t2=1000.

  • Suraj
  • 07 मई
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• 2000N

  • Devansh
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• 500

  • Will
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• D) 250

  • Kavita
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• 2000 breaking force(Tmax)=pressure x area hence,Tmax is proportional to r^2 which is square of cross sectional radius

  • Amit
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• 2000 N bcoz 500÷0.5^2=f÷1^2

  • Omeshwari
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• A

  • Siddharth
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• 1000 because diameter double tension also double

  • Vikas
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• provide more information

  • Milind
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