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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
  • 6402 Views
  • 97 Answers
  Answer
97 Answers

  • B

    0

  • 2000N

    1

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

    -1

  • 👶💜✌

    1

  • 500

    -1

  • Good

    0

  • 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.

    0

  • 1000N

    -1

  • 1000 N

    0

  • 1000N

    0

  • 1000

    0

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

    0

  • 2000N

    -1

  • 500

    -1

  • D) 250

    -1

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

    -1

  • 2000 N bcoz 500÷0.5^2=f÷1^2

    -1

  • A

    -1

  • 1000 because diameter double tension also double

    -2

  • provide more information

    -1
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