A wire fixed at the upper end stretches by length $l$ by applying a force $F$ . The work done in stretching is:

\frac{F}{2 l}

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F l

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\frac{2 F}{l}

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\frac{F l}{2}

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Solution

The correct option is D $\frac{F l}{2}$
Let the area be $A$ and length $L$ of the string then the work done by the force is :
$W = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e = \frac{1}{2} \times \frac{F}{A} \times \frac{l}{L} \times A \times L = \frac{F l}{2}$

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A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is:

The correct option is D Fl2Let the area be A and length L of the string then the work done by the force is :W=12×stress×strain×volume=12×FA×lL×A×L=Fl2

A wire fixed at the upper end stretches by length $l$ by applying a force $F$ . The work done in stretching is:

The correct option is D $\frac{F l}{2}$
Let the area be $A$ and length $L$ of the string then the work done by the force is :
$W = \frac{1}{2} \times s t r e s s \times s t r a i n \times v o l u m e = \frac{1}{2} \times \frac{F}{A} \times \frac{l}{L} \times A \times L = \frac{F l}{2}$