A) \[YAL\Delta t\]
B) \[YA\alpha \Delta t\]
C) \[\frac{YA\alpha \Delta t}{A}\]
D) \[Y\alpha A\Delta t\]
Correct Answer: B
Solution :
Young?s elasticity coefficient \[Y=\frac{Fl}{A\Delta l}\] According to definition of linear expansion coefficient \[\Delta l=\alpha l\Delta t\] where, \[l=\] initial length \[Y=\frac{Fl}{A\,\alpha \,\,l\Delta t}\] or \[Y=\frac{F}{A\alpha \Delta t}\] \[F=YA\alpha \Delta t\] where, F = force applied on wire A = area of cross-section of wire \[\Delta t\] = increase in temperature of wire.
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