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

• A. (a+b)
• B. (4b-5a)
• C. (5b-4a)
• D. (9b-9a)

Answer 1

The correct option is C (5b-4a)

Let l be the natural length and $K(=\frac{YA}{l})$ be the force constant of wire. Then
$a=l+\frac{4}{K}(F=K\Delta lor\Delta l=\frac{F}{K})$
and $b=l+\frac{5}{K}$
or $\frac{1}{K}=(b-a)$
and $l(5a-4b)$
Now when T=9N
$l^{\prime }=l+\frac{9}{K}=(5a-4b)+(b-a)$
$=(5b-4a)$