A 100 N block is suspended by copper and steel wires of equal cross-sectional area 0.5 cm². Their lengths are √3 m and 1 m, and they make angles 30° and 60° with the ceiling. Let Δl₁ and Δl₂ be the elongations of the copper and steel wires. Take Y₁=1×10¹¹ N m⁻² for copper and Y₂=2×10¹¹ N m⁻² for steel. Find Δl₁/Δl₂.
Physics · Properties of solid and fluid
JEE Advanced 2019

Question

A 100 N block is suspended by copper and steel wires of equal cross-sectional area 0.5 cm². Their lengths are √3 m and 1 m, and they make angles 30° and 60° with the ceiling. Let Δl₁ and Δl₂ be the elongations of the copper and steel wires. Take Y₁=1×10¹¹ N m⁻² for copper and Y₂=2×10¹¹ N m⁻² for steel. Find Δl₁/Δl₂.

Physics · Properties of solid and fluid

JEE Advanced 2019

G1 Physics · Accepted Answer

Let tension T₁ act in the copper wire and T₂ in the steel wire. Horizontal equilibrium gives T₁cos30°=T₂cos60°, so T₁/T₂=1/√3. Using Δl=TL/(AY), Δl₁/Δl₂=(T₁/T₂)(√3/1)(Y₂/Y₁)=2.