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The length of an elastic string is $a$ metre when the longitudinal tension is $4\,N$ and $b$ metre when the longitudinal tension is $5\,N$. The length of the string in metre when longitudinal tension is $9\,N$ is:
A. $a - b$
B. $5b - 4a$
C. $2b - \dfrac{1}{4}a$
D. $4a - 3b$

Last updated date: 13th Jul 2024
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Hint:In order to answer this question, you must be aware about the concept of Tension and force constant. Force constant is a proportionality constant. The greater the constant of force, the greater the restored force.

Complete step by step answer:
Let $L$ be the original length of the wire and $k$ be the force constant of the wire.
Final length = Original length + Elongation
${L^1} = L + \dfrac{F}{k}$
In the first case where tension is $4\,N$, $a = L + \dfrac{4}{k}$ ……. (1)
In the second case where tension is $5\,N$, $b = L + \dfrac{5}{k}$ ……. (2)
By solving eq. (1) and (2), we get
$L = 5a - 4b$ and $k = \dfrac{1}{{b - a}}$
Now, when the longitudinal tension is 9N, length of the string will be
${L^1} = L + \dfrac{9}{k}$
$\Rightarrow {L^1} = 5a - 4b + 9(b - a)$
$\therefore {L^1} = 5b - 4a$

Hence, option B is correct.

Note:Tension is described as the pulling force transmitted axially by the means of a string. It can also be defined as the action – reaction pair of forces acting at each end of the said elements.While considering a rope, the tension force is felt by every section of the rope in both the directions, apart from the endpoints.