Two wires of same material & length are stored by the same force. Their masses are in the ratio 3:2. Find ratio of their elongation

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Solution

$Δ L = \frac{F L}{A Y} = \frac{F L^{2}}{V Y}$ where $V = v o l u m e$

$⟹ Δ L = \frac{F L ρ}{M Y}$ where $ρ = d e n s i t y$

$⟹ Δ L \propto \frac{1}{M}$

$⟹ Δ L_{1} : Δ L_{2} = M_{2} : M_{1} = 2 : 3$

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Two wires of same material & length are stored by the same force. Their masses are in the ratio 3:2. Find ratio of their elongation

ΔL=FLAY=FL2VY where V=volume⟹ΔL=FLρMY where ρ=density⟹ΔL∝1M⟹ΔL1:ΔL2=M2:M1=2:3

$Δ L = \frac{F L}{A Y} = \frac{F L^{2}}{V Y}$ where $V = v o l u m e$

$⟹ Δ L = \frac{F L ρ}{M Y}$ where $ρ = d e n s i t y$

$⟹ Δ L \propto \frac{1}{M}$

$⟹ Δ L_{1} : Δ L_{2} = M_{2} : M_{1} = 2 : 3$