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QJEEM23Phe4a0: JEE Main 2023 Physics - Class 11 JEEM23Phe4a0 Solution: Two wires each of radius 0.2 cm and negligible mass, one made…

Two wires each of radius 0.2 cm and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is __________ ×\times 10 −6^{-6} m. [Young's modulus for steel = 2 ×\times 10 11^{11} Nm −2^{-2} and g = 10 ms −2^{-2} ]

JEE Main 2023 (Online) 10th April Morning Shift Physics - Properties of Matter Question 67 English

Step-by-Step Verified Solution

JEE Main 2023 (Online) 10th April Morning Shift Physics - Properties of Matter Question 67 English Explanation

amp;r1=r2=0.2 cm=2×10−3 mamp;y1 (steel) =2×1011 N m−2amp;l1=1.6 m\begin{aligned} & r_1 =r_2=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} \\\\ & y_1 \text { (steel) } =2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} \\\\ & l_1 =1.6 \mathrm{~m}\end{aligned}

At equilibrium,

amp;T2=1.14 g=11.4 Namp;T2+2g=T1amp;∴T1=11.4+20=31.4 N\begin{aligned}&T_2 =1.14 \mathrm{~g}=11.4 \mathrm{~N} \\\\&T_2+2 g =T_1 \\\\&\therefore T_1 =11.4+20=31.4 \mathrm{~N}\end{aligned}

amp;Δl1=T1l1Y1A1=amp;31.4×1.62×1011×π(2×10−3)21.1=amp;168×10−5=amp;2×10−5 m=20×10−6 m\begin{aligned}& \Delta l_1=\frac{T_1 l_1}{Y_1 A_1} \\\\= & \frac{31.4 \times 1.6}{2 \times 10^{11} \times \pi\left(2 \times 10^{-3}\right)^2} 1.1 \\\\= & \frac{16}{8} \times 10^{-5} \\\\= & 2 \times 10^{-5} \mathrm{~m}=20 \times 10^{-6} \mathrm{~m}\end{aligned}

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