Waves

JEE Main level, single correct, medium

126 of 142

Question 126 of 142

The speed of a longitudinal wave in a metallic bar is 400 m/s400 \mathrm{~m} / \mathrm{s}. If the density and Young's modulus of the bar material are increased by 0.5%0.5 \% and 1%1 \%, respectively then the speed of the wave is changed approximately to ____\_\_\_\_ m/s\mathrm{m} / \mathrm{s}.

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Answer: (B) 401

The speed of a longitudinal wave in a solid is given by v=Yρv = \sqrt{\frac{Y}{\rho}} where YY is Young's modulus and ρ\rho is density.
Initial condition: v0=Yρ=400v_0 = \sqrt{\frac{Y}{\rho}} = 400 m/s
After changes: ρ′=1.005ρ\rho' = 1.005\rho (0.5% increase) and Y′=1.01YY' = 1.01Y (1% increase)
New speed: v′=Y′ρ′=1.01Y1.005ρ=1.011.005⋅Yρv' = \sqrt{\frac{Y'}{\rho'}} = \sqrt{\frac{1.01Y}{1.005\rho}} = \sqrt{\frac{1.01}{1.005}} \cdot \sqrt{\frac{Y}{\rho}}
v′v0=1.011.005=1.00498...≈1.00249\frac{v'}{v_0} = \sqrt{\frac{1.01}{1.005}} = \sqrt{1.00498...} \approx 1.00249
v′=400×1.00249≈401v' = 400 \times 1.00249 \approx 401 m/s

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