Question

The length of an elastic string is $a$ metre when the tension is $4\text{N}$ and $b$ metre when the tension is $5\text{N}$. The length in metre, when the tension is $9\text{N}$ 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 $K(=\frac{YA}{I})$ be the force constant of wire.
As $(\Delta l=\frac{F}{K}),$
$a=l+\frac{4}{K}$ and $b=l+\frac{5}{K}$
on simplification,
$\frac{1}{K}=(b-a)$ and $l=(5a-4b)$

Now when $T=9\text{N},$ length of the wire is

$l^{\prime }=l+\frac{9}{K}=(5a-4b)+9(b-a)$
$l^{\prime }=5b-4a$