Question

Answer the following.

A composite wire is prepared by joining a tungsten wire and steel wire end to end. Both the wires are of the same length and the same area of cross-section. If this composite wire is suspended to a rigid support and a force is applied to its free end, it gets extended by 3.25 mm. Calculate the increase in the length of tungsten wire and steel wire separately.
`("Y"_"Tungsten"= 4.1 × 10^11 "Pa, Y"_"Steel"= 2 × 10^11 "Pa")`

Answer 1

Given: l_S + l_T = 3.25 mm, Y_T = 4.11 × 10¹¹ Pa, Y_s = 2 × 10¹¹ Pa

To find: Extension in tungsten wire (l_T)
Extension in steel wire (l_s)

Formula: Y = `"FL"/("A"l)`

Calculation: From formula,

`"Y" prop 1/l`    .....(∵ F, L and A are same for both the wires)

∴ `("Y"_"S")/("Y"_"T") = l_"T"/l_"S"`

∴ `(2 xx 10^11)/(4.11 xx 10^11) = l_"T"/l_"S"`

∴ `l_"T"/l_"S" = 0.487`

But l_S + l_T = 3.25

l_S + 0.487 l_S = 3.25

l_S (1 + 0.487) = 3.25

l_S = 2.186 mm

∴ l_T = 3.25 - 2.186 = 1.064 mm

The extension in tungsten wire is 1.064 mm and the extension in steel wire is 2.186 mm.