Question

A uniform metal wire has length ' \(L\) ', mass ' \(M\) ' and area of cross-section ' A '. It is under tension ' \(T\) ' and ' \(V\) ' is the speed of transverse wave along the wire. The density of the wire is

• A. \(\mathrm{TV}^{2}\text{ }\mathrm{A}\)
• B. \(\frac{\mathrm{T}}{\mathrm{V}^{2}\text{ }\mathrm{A}}\)
• C. \(\frac{\mathrm{A}}{\mathrm{TV}^{2}}\)
• D. \(\frac{\mathrm{V}^{2}}{\mathrm{AT}}\)

Hint:

Wave speed on a string is the square root of tension over linear density, and linear density equals density times area.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Speed of a transverse wave on a string is \(V = \sqrt{\dfrac{T}{\mu}}\), where \(\mu = \dfrac{M}{L}\) is mass per unit length.

Step 2: Link to Density
\(\mu = \rho A\), because mass per length equals density times cross-sectional area.

Step 3: Solve
\(V^2 = \dfrac{T}{\rho A}\), so \(\rho = \dfrac{T}{V^2 A}\).

Step 4: Compare the options
This is option B.

Final Answer:
The density is T divided by V squared times A. \[ \boxed{\text{(B) }\dfrac{\mathrm{T}}{\mathrm{V}^{2}\,\mathrm{A}}} \]