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Question

The length of an elastic string is ‘a’ metre when the tension is 4N and ‘b’ metre when the tension is 5N. The length, in metre, when the tension is 9 N, is
  1. (5b-4a)
  2. (9b-9a)
  3. (a+b)
  4. (4b-5a)


Solution

The correct option is A (5b-4a)
Let l be the natural length and K(=YAl) be the force constant of wire. Then
a=l+4K(F=KΔl or Δl=FK)
and b=l+5K
or 1K=(ba)
and l(5a4b)
Now when T=9N
l=l+9K=(5a4b)+(ba)
=(5b4a)

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