Thermal Properties of Matter

JEE Main level, single correct, medium

35 of 126

Question 35 of 126

When the temperature of a metal wire is increased from 0ºCto10ºC0ºCto10ºC, its length increases by 0.020.02%.The percentage change in its mass density will be closed to:

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Answer: (A) 0.060.06

Δℓ=ℓαΔt\Delta \ell =\ell \alpha \Delta t
α=ΔℓℓΔT=0.02100×10;α=2×10−5\alpha =\dfrac{\Delta \ell }{\ell \Delta T}=\dfrac{0.02}{100\times 10};\alpha =2\times {10}^{-5}
γ=3α=6×10−5\gamma =3\alpha =6\times {10}^{-5}
ΔVV=γ×ΔT;ΔVV×100=(6×10−5×10×100)=6×10−2\dfrac{\Delta V}{V}=\gamma \times \Delta T;\dfrac{\Delta V}{V}\times 100=\left(6\times {10}^{-5}\times 10\times 100\right)=6\times {10}^{-2}
Volume increase by 0.060.06% therefore density decrease by 0.060.06%

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