JEE Advanced · Physics · 10. Properties of Solids
A \(0.1 \mathrm{~kg}\) mass is suspended from a wire of negligible mass. The length of the wire is \(1 \mathrm{~m}\) and its cross-sectional area is \(4.9 \times 10^{-7} \mathrm{~m}^2\). If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency \(140 \mathrm{rad} \mathrm{s}^{-1}\). If the Young's modulus of the material of the wire is \(n \times 10^9 \mathrm{Nm}^{-2}\), the value of \(n\) is
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
\(\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}\)
Putting, \(\omega=140 \operatorname{rad} \mathrm{s}^{-1}\) in above equation we get,
\(n=4\)
\(\therefore\) Answer is 4 .
Putting, \(\omega=140 \operatorname{rad} \mathrm{s}^{-1}\) in above equation we get,
\(n=4\)
\(\therefore\) Answer is 4 .
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