Two wires of same material and length but cross-sections in the ratio $1 : 2$ are used to suspend the same loads. The extensions in them will be in the ratio

1 : 2

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2 : 1

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4 : 1

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1 : 4

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Solution

The correct option is C $2 : 1$
Let the cross section of wire be A , 2A
$Δ L_{1} = \frac{F l}{A Y}$
$Δ L_{2} = \frac{F l}{2 A Y}$
so, $\frac{Δ L_{1}}{Δ_{2}} = 2 / 1$

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Two wires of same material and length but cross-sections in the ratio 1:2 are used to suspend the same loads. The extensions in them will be in the ratio

The correct option is C 2:1Let the cross section of wire be A , 2AΔL1=FlAYΔL2=Fl2AYso, ΔL1Δ2=2/1

Two wires of same material and length but cross-sections in the ratio $1 : 2$ are used to suspend the same loads. The extensions in them will be in the ratio

The correct option is C $2 : 1$
Let the cross section of wire be A , 2A
$Δ L_{1} = \frac{F l}{A Y}$
$Δ L_{2} = \frac{F l}{2 A Y}$
so, $\frac{Δ L_{1}}{Δ_{2}} = 2 / 1$