8.Mechanical Properties of Solids
easy

To double the length of a iron wire having $0.5\,c{m^2}$ area of cross-section, the required force will be $(Y = {10^{12}}\,dyne/c{m^2})$

A

$1.0 \times {10^{ - 7}}N$

B

$1.0 \times {10^7}N$

C

$0.5 \times {10^{ - 7}}N$

D

$0.5 \times {10^{12}}$dyne

Solution

(d) If length of wire doubled then strain $= 1$

$Y = {\rm{stress}}$ ==> $F = Y \times A$$ = {10^{12}} \times 0.5$$ = 0.5 \times {10^{12}}dyne$

Standard 11
Physics

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