There are two wires of same material. their radii and lengths are both in the ratio 1:2. if the extensions produced are equal, what is the ratio of the loads?

1:2

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2:1

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1:4

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4:1

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Solution

The correct option is A 1:2
Given,

$l_{1} : l_{2} = 1 : 2$

$r_{1} : r_{2} = 1 : 2$

$Δ l_{1} = Δ l_{2}$

Area, $A = π r^{2}$

$A \propto r^{2}$

$A_{1} : A_{2} = 1 : 4$

The material are same, so the modulus is also equal,

$Y_{1} = Y_{2}$

$\frac{F_{1} / A_{1}}{Δ l / l_{1}} = \frac{F_{2} / A_{1}}{Δ l / l_{2}}$

$\frac{F_{1}}{F_{2}} = \frac{l_{2}}{l_{1}} . \frac{A_{1}}{A_{2}}$

$\frac{F_{1}}{F_{2}} = \frac{2}{1} \times \frac{1}{4} = \frac{1}{2}$

$F_{1} : F_{2} = 1 : 2$

The correct option is A.

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There are two wires of same material. their radii and lengths are both in the ratio 1:2. if the extensions produced are equal, what is the ratio of the loads?

The correct option is A 1:2Given,l1:l2=1:2r1:r2=1:2Δl1=Δl2Area, A=πr2A∝r2A1:A2=1:4The material are same, so the modulus is also equal,Y1=Y2F1/A1Δl/l1=F2/A1Δl/l2F1F2=l2l1.A1A2F1F2=21×14=12F1:F2=1:2The correct option is A.