MHT CET · Physics · Mechanical Properties of Solids
A transverse wave is travelling with velocity ' \(\mathrm{V}^{\prime}\) through a metal wire of length ' \(\mathrm{L}\) '
and density ' \(\rho\) '. The tensile stress in the wire is
- A \(\mathrm{V} \rho^{2}\)
- B \(\frac{\mathrm{v}^{2}}{\rho}\)
- C \(\frac{\rho}{\mathrm{V}^{2}}\)
- D \(\mathrm{~V}^{2} \rho\)
Answer & Solution
Correct Answer
(D) \(\mathrm{~V}^{2} \rho\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=\sqrt{\frac{T}{m}}\) where \(\mathrm{T}\) is the tension and \(\mathrm{m}\) is mass per unit length.
\(\mathrm{m}=\mathrm{A} . p\) where \(\mathrm{A}\) is an area of cross-section.
\(\therefore V=\sqrt{\frac{T}{A \rho}}\)
\(\therefore V^{2}=\frac{T}{A \rho}\)
\(\therefore \frac{T}{A}=V^{2} \rho\)
\(\mathrm{m}=\mathrm{A} . p\) where \(\mathrm{A}\) is an area of cross-section.
\(\therefore V=\sqrt{\frac{T}{A \rho}}\)
\(\therefore V^{2}=\frac{T}{A \rho}\)
\(\therefore \frac{T}{A}=V^{2} \rho\)
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