Question

The length of an elastic string is a metre, when the longitudinal tension is 4 N and the length is $b$ m, when the tension is 5 N. When the longitudinal tension is 9 N, the length of the string (in metre) is

• A. $2b-\frac{a}{2}$
• B. $5b-4a$
• C. $4a-3b$
• D. $a-b$

Answer 1

Answer: (B)

If $L$ is the initial length, then the increase in length by a tension $F$ is given by $l=\frac{FL}{\pi r^2\gamma }$
Hence, $a=L+l=L+\frac{4L}{\pi r^2\gamma }=L+4C$ ...(i)
and $b=L+\frac{5L}{\pi r^2\gamma }+L+5C$ ...(ii)
$\left(\text{where},C=\frac{L}{\pi r^2\gamma }\right)$
Thus, on solving eqn (i) and (ii), we get
$L=5a-4b$ and $C=b-a$
Hence, for $F=9N$, we get
$x=L+\frac{9L}{\pi r^2\gamma }=L+9C$
$=(5a-4b)+9(b-a)=5b-4a$