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Young's Modulus of Elasticity

A wire that o...

Question

A wire that obeys Hooke's law is of length $l_{1}$ when it is in equilibrium under a tension $F_{1}$ . Its length becomes $l_{2}$ when the tension is increased of $F_{2}$ . The energy stored in the wire during this process is

(F_{2} - F_{1}) (l_{2} - l_{1})

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\frac{1}{4} (F_{2} + F_{1}) (l_{2} - l_{1})

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\frac{1}{4} (F_{2} - F_{1}) (l_{2} - l_{1})

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\frac{1}{2} (F_{2} - F_{1}) (l_{2} - l_{1})

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Solution

The correct option is D $\frac{1}{2} (F_{2} - F_{1}) (l_{2} - l_{1})$
$\frac{\frac{F_{1}}{A} = \frac{Δ l_{1}}{l}}{\frac{F_{2}}{A} = \frac{Δ l_{2}}{l}}$
$\frac{F_{1}}{F_{2}} = \frac{l_{1} - l}{l_{2} - l}$
$F_{1} l_{2} - F_{1} l = F_{2} l_{1} - F_{2} l$
$(F_{2} - F_{1}) l = F_{2} l_{1} - F_{1} l_{2}$
$l = \frac{F_{2} l_{1} - F_{1} l_{2}}{F_{2} - F_{1}}$
Energy $= \frac{1}{2} s t r e s s \times s t r a i n$
$= \frac{1}{2} \times \frac{(F_{2} - F_{1})}{A} \times (l_{2} - l_{1})$

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A wire that obeys Hooke's law is of length l1 when it is in equilibrium under a tension F1. Its length becomes l2 when the tension is increased of F2. The energy stored in the wire during this process is

The correct option is D 12(F2−F1)(l2−l1)F1A=Δl1lF2A=Δl2lF1F2=l1−ll2−lF1l2−F1l=F2l1−F2l(F2−F1)l=F2l1−F1l2l=F2l1−F1l2F2−F1Energy =12stress×strain=12×(F2−F1)A×(l2−l1)

A wire that obeys Hooke's law is of length $l_{1}$ when it is in equilibrium under a tension $F_{1}$ . Its length becomes $l_{2}$ when the tension is increased of $F_{2}$ . The energy stored in the wire during this process is