Figure shows a composite rod of cross-sectional area `10^(-4) m^(2)` made by joining three rods AB, BC and CD of different materials end to end. The composite rod is suspended vertically and an object of 10 kg is hung by it. `L _(AB) =0.1 m, _(BC) =0.2 m and L _(CD) =0.15 m.`
Calculate displacement of `B,C and D Y_(AB) =2.5 xx 10 ^(10) Pa, Y_(BC) = 4 xx 10 ^(10) Pa and Y_(CD) = 1 xx 10 ^(10) Pa.`

Explanation

`M = 10 kg` Young.s Modulus `Y = ( Mg //A)/(Delta l //l)`
`L _(AB) = 0.1 m therefore Delta l ( Mg l)/( AY) " m "...(1)`
`L _(BC) = 0.2m`
`L _(CD) = 0.15 m`
`A = 10 ^(-4) m ^(2)`
(1) Let increase in length of rod `AB= Delta L _(AB)`
`therefore Delta L _(AB) = (MgL _(AB))/(AY _(AB))= (10 xx 9.8 xx 0.1)/( 10^(-4) xx 2.5 xx 10^(10))`
`Delta L _(AB) = 3.92 xx 10 ^(-6) m`
`therefore ` Displacemtn of end `B = 3.92 xx 10 ^(-6) m`
(2) For rod BC
`Delta L _(BC) = (Mg L _(BC))/( AY _(BC))`
` = (10 xx 9.8 xx 0.2)/( 10 ^(-4) xx 4 xx 10 ^(10)) `
`Delta L _(BC) = 4.9 xx 10 ^(-6) m`
`therefore` Displacement of and C
`=Delta L _(AB)+ Delta L _(BC)`
`= 3.92 xx 10 ^(-6) + 4.9 xx 10 ^(-6) = 8.82 xx 10 ^(-5) m`
(3) For rod CD,
`Delta L _(CD) = (Mg L_(CD))/( AY _(CD)) = (10 xx 9.8 xx 0.15)/(10 ^(-4) xx 1 xx 10 ^(10))= 14.7 xx 10 ^(-6) m`
`therefore` Displacement of point D
`= Delta L_(AB) +Delta L _(BC) +Delta L _(CD)`
`= 8.82 xx 10 ^(-6) + 14.7 xx 10 ^(-6)`
`= 23.52 xx 10 ^(-6) m`