In the question number 46, if the mass M is hung the free end of the wire, then the extension produc in the wire is

In the question number 46, if the mass M is hung the free end of the w

1.	In the question number 46, if the mass M is hung the free end of the wire, then the extension produc in the wire is
Answer» `(mugL^(2) + MgL)/(2YA)` `(2mugL^(2) + MgL)/(YA)` `(mugL^(2) + 2MgL)/(2YA)` `(mugL^(2) + MgL)/(YA)` Solution :Consider a SMALL element oflengthdx at a distance x fromthe load as shown in thewire. TENSIONIN the wire at a distance x from THELOWER end is`T(x) = mugx + Mg` Let dl be increase in lengthof the elemenet Then `Y= (T(x)//A)/(dl//dx)` `dl =(T(x)dx)/(YA) = ((mugx+Mg)/(YA))dx ""("USING (i)")` `therefore` Totalextensionproduced in thewire is `l = underset(0)overset(L)int ((mugx+Mg)/(YA))dx = (1)/(Y) [(mugx)/(2) +Mgx]_(0)^(L) = (1)/(YA) [(mugL^(2))/(L)+Mgl] =(mugx^(2)+2MgL)/(2YA)`