The length of an elastic string is a metre when the tension is 4 N and b metre when the tension is 5 N. The length in metre, when the tension is 9 N is
A
(a+b)
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B
(4b−5a)
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C
(5b−4a)
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D
(9b−9a)
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Solution
The correct option is C(5b−4a) Let l be the natural length and K(=YAI) be the force constant of wire.
As (Δl=FK), a=l+4K and b=l+5K
on simplification, 1K=(b−a) and l=(5a−4b)