The length of a wire becomes l₁ and l₂ when 100 N and 120 N tensions are applied respectively. If 10 l₂ = 11 l₁, the natural length of wire will be 1/x l₁. Here the value of x is _____________. Numerical answer type. Work it out, then open the answer.

Answer: 2. L = (l₁T₂ - l₂T₁)/(T₂ - T₁) = (120 l₁ - 100 l₂)/(20) = 6l₁ - 5l₂ With l₂ = 1.1 l₁: L = 6l₁ - 5.5l₁ = 1/2l₁ So x = 2.

The length of a wire becomes l₁ and l₂ when 100 N and 120 N tensions… | JEE Main 2023 (11 Apr, Shift 1)

Q11JEE Main 2023 · 11 APR · SHIFT 1Medium

Wire Combinations & Special Cases

The length of a wire becomes $l_1$ and $l_2$ when $100\text{ N}$ and $120\text{ N}$ tensions are applied respectively. If $10\,l_2 = 11\,l_1$ , the natural length of wire will be $\frac{1}{x}\,l_1$ . Here the value of $x$ is _____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

$L = \frac{l_1T_2 - l_2T_1}{T_2 - T_1} = \frac{120\,l_1 - 100\,l_2}{20} = 6l_1 - 5l_2$
With $l_2 = 1.1\,l_1$ : $L = 6l_1 - 5.5l_1 = \frac{1}{2}l_1$
So $x = 2$ .

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