Mechanical Properties of Solids

JEE Main level, single correct, medium

17 of 118

Question 17 of 118

Two wires AA and BB made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections 3.0×10−5 m23.0 \times 10^{-5} \mathrm{~m}^{2} and 4.5×10−5 m24.5 \times 10^{-5} \mathrm{~m}^{2}, respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of AA to that of BB is x:3x: 3. The value of xx is ____\_\_\_\_.

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Answer: (D) 5

Young's modulus: Y=σϵ=F/AΔL/LY = \frac{\sigma}{\epsilon} = \frac{F/A}{\Delta L/L}.
When the same force is applied and both wires stretch by the same extension ΔL\Delta L, the ratio of Young's moduli is:
YAYB=F/AAΔL/LA×ΔL/LBF/AB=ABAA×LALB\frac{Y_A}{Y_B} = \frac{F/A_A}{\Delta L/L_A} \times \frac{\Delta L/L_B}{F/A_B} = \frac{A_B}{A_A} \times \frac{L_A}{L_B}
=4.5×10−53.0×10−5×6.05.4=1.5×109=53=\frac{4.5 \times 10^{-5}}{3.0 \times 10^{-5}} \times \frac{6.0}{5.4} = 1.5 \times \frac{10}{9} = \frac{5}{3}.
Therefore x:3=5:3x : 3 = 5 : 3, giving x=5x = 5.

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