**Answer the following.**

A wire gets stretched by 4 mm due to a certain load. If the same load is applied to a wire of same material with half the length and double the diameter of the first wire, what will be the change in its length?

#### Solution

**Given:** l_{1} = 4 mm = 4 × 10^{-3} m

L_{2} = `"L"_1/2,` D_{2} = 2D, r_{2} = 2r_{1}

**To find:** Change in length (l_{2})

**Formula: **Y = `"FL"/("A"l) = "FL"/(pi"r"^2l)`

**Calculation:**

From formula,

`"Y"_1 = ("F"_1"L"_1)/(pi"r"_1^2l_1)` .....(i)

`"Y"_2 = ("F"_2"L"_2)/(pi"r"_2^2l_2)` .....(ii)

Dividing equation (ii) by equation (i),

`"Y"_2/"Y"_1 = (("F"_2"L"_2)/(pi"r"_2^2l_2))/(("F"_1"L"_1)/(pi"r"_1^2l_1))` .....(iii)

Since same load is applied on same wire, Y_{1} = Y_{2} and F_{1} = F_{2 }

∴ `"L"_1/("r"_1^2 l_1) = "L"_2/("r"_2^2l_2)` .....[From (iii)]

`l_2 = ("L"_2 xx "r"_1^2 xx l_1)/("r"_2^2 xx "L"_1)`

`= ("L"_2 xx "r"_1^2 xx l_1)/(4"r"_1^2 xx 2 xx "L"_2) .....(because "L"_1 = 2"L"_2, "r"_2 = 2"r"_1)`

`= l_1/8`

`= (4 xx 10^-3)/8`

= 0.5 × 10^{-3} m

= **0.5 mm**

The new change in length of the wire is 0.5 mm.