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QJEEM23Phf474: JEE Main 2023 Physics - Class 11 JEEM23Phf474 Solution: For a solid rod, the Young's modulus of elasticity is and…

For a solid rod, the Young's modulus of elasticity is 3.2×1011Nm−23.2 \times 10^{11} \mathrm{Nm}^{-2} and density is 8×103 kg m−38 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} . The velocity of longitudinal wave in the rod will be.

Multiple Choice Options

  • A 3.65×103 ms−13.65 \times 10^3 \mathrm{~ms}^{-1}
  • B 6.32×103 ms−16.32 \times 10^3 \mathrm{~ms}^{-1}
  • C 18.96×103 ms−118.96 \times 10^3 \mathrm{~ms}^{-1}
  • D 145.75×103 ms−1145.75 \times 10^3 \mathrm{~ms}^{-1}

Step-by-Step Verified Solution

v=Yρ=3.2×10118×103v=\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{3.2 \times 10^{11}}{8 \times 10^{3}}}

=0.4×108=40×106=6.32×103 m/s

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