Mechanical Properties of Solids

JEE Main level, numerical answer, medium

34 of 118

Question 34 of 118

Two wires each of radius 0.2cm0.2cm 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 ______10–6m{10}^{–6}m. [Young's modulus for steel =2×1011Nm–2=2\times {10}^{11}N{m}^{–2} and g=10ms–2g=10m{s}^{–2}]

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Answer: 20

The force acting on steel wire FF is (2+1.14)g=3.14g\left(2+1.14\right)g=3.14g
From the relation of Young's modulus, it can be written
FA=Y(ΔLL)\dfrac{F}{A}=Y\left(\dfrac{\Delta L}{L}\right)
⇒ΔL=(FLAY)=3.14×g×1.6π×(0.2)2×2×1011×10−4\Rightarrow \Delta L=\left(\dfrac{FL}{AY}\right)=\dfrac{3.14\times g\times 1.6}{\pi \times {\left(0.2\right)}^{2}\times 2\times {10}^{11}\times {10}^{-4}}
⇒∆L=2×10−5\Rightarrow ∆L=2\times {10}^{-5}
⇒∆L=20×10−6m\Rightarrow ∆L=20\times {10}^{-6}m

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