A) \[\Delta l\] versus\[\frac{1}{l}\]
B) \[\Delta l\]versus \[{{l}^{2}}\]
C) \[\Delta l\]versus\[\frac{1}{{{l}^{2}}}\]
D) \[\Delta l\]versus \[l\]
Correct Answer: B
Solution :
Young's modulus \[Y=\frac{Stress}{Strain}\] or \[Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}\] \[\Rightarrow \]\[\Delta l=\frac{Fl}{AY}\] (i) The volume of wire V= Al or \[A=\frac{V}{l}\] (ii) Substituting the value of A from (ii) in equation (i) We get \[\Delta l=\frac{F{{l}^{2}}}{VY}\Rightarrow \Delta l\propto {{l}^{2}}\] Hence the graph of \[\Delta l\]versus\[{{l}^{2}}\]will give a straight line.You need to login to perform this action.
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