A metal wire of uniform cross-sectional area A and length L, has mass m. It is suspended vertically from a ceiling. If its Young's modulus is Y, then the elongation Deltal of wire due to its own weight will be .........

A metal wire of uniform cross-sectional area A and length L, has mass

1.	A metal wire of uniform cross-sectional area A and length L, has mass m. It is suspended vertically from a ceiling. If its Young's modulus is Y, then the elongation Deltal of wire due to its own weight will be .........
Answer» zero `(mgL)/(2AY)` `(mgL)/(AY)` `(2mgL)/(AY)` Solution :Let us consider an element length dx. Now consider small element from a support (celling) at x distance. The length of wire below this element `= L - x` Hence the tension at this point = WEIGHT of wire below this point `thereforeT = ((L -x )W)/(L)` The elongation in this element `= ("original length" XX "stress")/(Y)` `= (Tdx)/(AY) = ((L -x ) Edx)/(LAY)` `THEREFORE ` Total elongation in length `= int _(0) ^(L) ((L -x) W dx )/( LAY) = (W)/( LAY) int _(0) ^(L) (L-x) dx` `= (W)/( LAY) [ Lx - (x ^(2))/(2) ] _(0) ^(L) = (WL )/( 2 AY) = (mgL)/(2 AY)`