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

Answer 1

Answer: (B)

$5 b - 4a$

Answer 2

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 $b = L + \frac{5L}{\pi r^2 \gamma} + L + 5 C$ ...(ii)
$\left({where} , C = \frac{L}{\pi r^2 \gamma} \right)$
Thus, on solving eqn (i) and (ii), we get
$L = 5a - 4b$ and $C = b - a$
Hence, for $F = 9 \, N$, we get
$x = L + \frac{9L}{\pi r^2 \gamma} = L + 9C$
$= (5a - 4b) + 9(b -a) = 5b - 4a$