JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2023
- Medium
- Numerical
The length of wire becomes \(l_1\) and \(l_2\) when \(100\,N\) and \(120\,N\) tensions are applied respectively. If \(10l_2=11l_1\), the natural length of wire will be \(\frac{1}{x} l_1\). Here the value of \(x\) is ........
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
Let the original length be ' \(\ell_0\) '
When \(T _1=100\,N\), Extension \(=\ell_1-\ell_0\)
When \(T _2=120\,N\), Extension \(=\ell_2-\ell_0\)
Then \(100= K \left(\ell_1-\ell_0\right)\)
And \(120= K \left(\ell_2-\ell_0\right)\)
\(\frac{1}{2} \Rightarrow \frac{5}{6}=\frac{\ell_1-\ell_0}{\ell_2-\ell_0}\)
\(5 \ell_2-5 \ell_0=6 \ell_1-6 \ell_0\)
\(\ell_0=6 \ell_1-5 \ell_2\)
\(\ell_0=6 \ell_1-5\left(\frac{11 \ell_1}{10}\right)\)
\(\ell_0=6 \ell_1-\frac{11 \ell_1}{2}\)
\(\ell_0=\frac{\ell_1}{2}\)
\(\therefore x=2\)
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