Two wires A and B of equal cross-sectional area but made of different materials are joined together in series.
The ratio of Young's modulus of wire A to wire B is:
YA : YB = 20 : 11
When the combined wire is subjected to a certain tension, the elongations produced in both wires are equal.
If the length of wire A is 2.2 m, find the length of wire B.
Options:

(1) 1.1 m
(2) 1.21 m
(3) 4.44 m
(4) 2.22 m

Question

Two wires A and B of equal cross-sectional area but made of different materials are joined together in series.
The ratio of Young's modulus of wire A to wire B is:
YA : YB = 20 : 11
When the combined wire is subjected to a certain tension, the elongations produced in both wires are equal.
If the length of wire A is 2.2 m, find the length of wire B.
Options:

(1) 1.1 m
(2) 1.21 m
(3) 4.44 m
(4) 2.22 m

Shiksha Nation · Accepted Answer

For a wire under tension, elongation is given by:
&Delta;L = FL / AY
where:

F = Applied force
L = Length of wire
A = Cross-sectional area
Y = Young's modulus

Since the two wires are connected in series:

The same force acts on both wires.
Cross-sectional areas are equal.
Elongations are equal.

Therefore:
F LA / A YA = F LB / A YB
Cancelling common terms:
LA / YA = LB / YB
Substituting:
2.2 / 20 = LB / 11
LB = (2.2 &times; 11) / 20
= 1.21 m
Final Answer
Option (2) 1.21 m

Verified Answer