QJEEM20Ph5bc0: JEE Main 2020 Physics - Class 11 JEEM20Ph5bc0 Solution: A wire of density 9 × 10 –3 kg cm –3 is stretched between two…

Question

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A wire of density 9

\times

10^–3 kg cm^–3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9

\times

10^–4. The lowest frequency of the transverse vibrations in the wire is : (Young’s modulus of wire Y = 9

\times

10¹⁰ Nm^–2), (to the nearest integer), _________

Accepted Answer

ρ

_wire = 9

\times

10^–3 kg cm^–3

=

\frac{9 \times 10^{- 3}}{10^{- 6}}

kg/m³ = 9000 kg/m²

f =

\frac{1}{2 l} \sqrt{\frac{T}{μ}} =

\frac{1}{2 l} \sqrt{\frac{T}{ρ_{w i r e} A}}

=

\frac{1}{2 l} \sqrt{\frac{Y Δ l}{ρ_{w i r e} A}}

= \frac{1}{2 \times 1} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{- 4}}{9000 \times 1}}

= 35 Hz

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