Question

2 wires of equal cross sectional area k e made up of aluminium and other made up of brass are joint end to end. When combination of wire is keot ynder tension the elongation of wire is found to be equal. Find the ration of length of 2 wires

Answer 1

Firat we have to recall the expression for Young's Modulus of a wire.

$Y=\dfrac {FL}{A\ \Delta L}\quad\longrightarrow\quad (1)$

where,

• F = force or tension acting on the wire linearly.
• L = length of the wire.
• A = cross sectional area of the wire.
• ∆L = elongation or extension produced in the wire.

In this case about the rods made by aluminium and brass,

• cross sectional area (A) of the wires are equal.
• the tensions (F) acting on the wires are same.
• elongation (∆L) produced in the wires are equal.

Then, from (1), we have the relation,

$Y\propto L$

thereby forming the equation,

$\large\boxed {\dfrac {L_a}{L_b}=\dfrac {Y_a}{Y_b}}$

Thus, the ratio of lengths of the wires is the same as that of the Young's Moduli of the wires.