Two wires of same material and radius have their lengths in ratio $1 : 2$ . If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio

2 : 1

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1 : 1

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1 : 2

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1 : 4

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Solution

The correct option is D $1 : 1$
$\begin{matrix} Given, two wires of same material and same radius have lengths in the ratio 1 : 2 . According to Hooke's formula, \begin{matrix} F & = Y A (\frac{Δ L}{L}) Strain & = \frac{Δ L}{L} = \frac{F}{Y A} \end{matrix} \end{matrix}$

$\frac{(S t r a i n) A}{(Strain)_{B}} = \frac{F_{A}}{Y_{A} A_{A}} \times \frac{Y_{B} \cdot A_{B}}{F_{B}} = \frac{1}{1}$

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Two wires of same material and radius have their lengths in ratio 1:2. If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio

The correct option is D 1:1 Given, two wires of same material and same radius have lengths in the ratio 1:2 . According to Hooke's formula, F=YA(ΔLL) Strain =ΔLL=FYA(Strain)A(Strain)B=FAYAAA×YB⋅ABFB=11

Two wires of same material and radius have their lengths in ratio $1 : 2$ . If these wires are stretched by the same force, the strain produced in the two wires will be in the ratio