JEE Mains · Physics · STD 11 - 14. waves and sound
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......\(HZ\) (Young's modulus of wire \(Y =9 \times 10^{10}\, Nm ^{-2}\) ), (to the nearest integer),
- A \(35\)
- B \(55\)
- C \(20\)
- D \(40\)
Answer & Solution
Correct Answer
(A) \(35\)
Step-by-step Solution
Detailed explanation
\(\rho_{\text {wire }}=9 \times 10^{-3} \frac{ kg }{ cm ^{3}}=\frac{9 \times 10^{-3}}{10^{-6}} kg / m ^{3}\) \(=9000 kg / m ^{2}\) \(( A = CSA\) of wire \()\) \(\left( Y =9 \times 10^{10} Nm ^{2}\right)\) \(\left(\right.\) Strain \(\left.=4.9 \times 10^{-4}\right)\)…
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