MHT CET · Physics · Mechanical Properties of Solids
The length of an elastic string is \(a\) metre when the longitudinal tension is \(4 \mathrm{~N}\) and \(b\) metre when the longitudinal tension is \(5 \mathrm{~N}\). The length of the string in metre when longitudinal tension is \(9 \mathrm{~N}\) is
- A \(a-b\)
- B \(5 b-4 a\)
- C \(2 b-\frac{1}{4} a\)
- D \(4 a-3 b\)
Answer & Solution
Correct Answer
(B) \(5 b-4 a\)
Step-by-step Solution
Detailed explanation
Let \(L\) is the original length of the wire and \(k\) is force constant of wire. Final length = initial length \(+\) elongation
\(L^{\prime}=L+\frac{F}{k}\)
For first condition \(a=L+\frac{4}{k}\) For second condition \(b=L+\frac{5}{k}\)
By solving Eqs. (i) and (ii), we get
\(L=5 a-4 b \text { and } k=\frac{1}{b-a}\)
Now, when the longitudinal tension is \(9 \mathrm{~N}\), length of the string
\(\begin{array}{l}=L+\frac{9}{k}=5 a-4 b+9(b-a) \\=5 b-4 a\end{array}\)
\(L^{\prime}=L+\frac{F}{k}\)
For first condition \(a=L+\frac{4}{k}\) For second condition \(b=L+\frac{5}{k}\)
By solving Eqs. (i) and (ii), we get
\(L=5 a-4 b \text { and } k=\frac{1}{b-a}\)
Now, when the longitudinal tension is \(9 \mathrm{~N}\), length of the string
\(\begin{array}{l}=L+\frac{9}{k}=5 a-4 b+9(b-a) \\=5 b-4 a\end{array}\)
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