The length of wire becomes $\ell_{1}$ and $\ell_{2}$ when 100 N and 120 N tensions are applied respectively. If $\ell_{2}=\frac{11}{10}\ell_{1}$, the natural length of wire will be $\frac{1}{x}\ell_{1}$. Here the value of $x$ is ______.

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Elasticity & SHM

The length of wire becomes

ℓ_{1}

and

ℓ_{2}

when 100 N and 120 N tensions are applied respectively. If

ℓ_{2} = \frac{11}{10} ℓ_{1}

, the natural length of wire will be

\frac{1}{x} ℓ_{1}

. Here the value of

x

is ______.

Answer: 2

Taking the wire to obey Hooke's law,

F = K (ℓ - ℓ_{0})

for natural length

ℓ_{0}

. Then

100 = K (ℓ_{1} - ℓ_{0})

and

120 = K (ℓ_{2} - ℓ_{0})

. Dividing,

\frac{100}{120} = \frac{ℓ _{1} - ℓ _{0}}{ℓ _{2} - ℓ _{0}}

, which rearranges to

ℓ_{0} = 6 ℓ_{1} - 5 ℓ_{2}

. With

ℓ_{2} = \frac{11}{10} ℓ_{1}

ℓ_{0} = 6 ℓ_{1} - \frac{11}{2} ℓ_{1} = \frac{1}{2} ℓ_{1}

. Comparing with

ℓ_{0} = \frac{1}{x} ℓ_{1}

gives

x = 2

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer