JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 1)
For a solid rod, the Young's modulus of elasticity is $3.2 \times 10^{11} \mathrm{Nm}^{-2}$ and density is $8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$. The velocity of longitudinal wave in the rod will be.
$3.65 \times 10^3 \mathrm{~ms}^{-1}$
$6.32 \times 10^3 \mathrm{~ms}^{-1}$
$18.96 \times 10^3 \mathrm{~ms}^{-1}$
$145.75 \times 10^3 \mathrm{~ms}^{-1}$
Explanation
$v=\sqrt{\frac{Y}{\rho}}=\sqrt{\frac{3.2 \times 10^{11}}{8 \times 10^{3}}}$
$$ \begin{aligned} & =\sqrt {0.4 \times {{10}^8}} \\\\ & = \sqrt {40 \times {{10}^6}} \\\\ & =6.32 \times 10^{3} \mathrm{~m} / \mathrm{s} \end{aligned} $$
$$ \begin{aligned} & =\sqrt {0.4 \times {{10}^8}} \\\\ & = \sqrt {40 \times {{10}^6}} \\\\ & =6.32 \times 10^{3} \mathrm{~m} / \mathrm{s} \end{aligned} $$
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