JEE Main 2021 · 20 July · evening · Single correct · +4 / −1
The length of a metal wire is l₁, when the tension in it is T₁ and is l₂ when the tension is T₂. The natural length of the wire is :
- (a) √l₁l₂
- (b) (l₁T₂ - l₂T₁)/(T₂ - T₁)
- (c) (l₁T₂ + l₂T₁)/(T₂ + T₁)
- (d) (l₁ + l₂)/2
Answer and solution
Answer (b) (l₁T₂ - l₂T₁)/(T₂ - T₁)
Solution
Physics sir · 5 steps
Write Hooke's law at both tensions, then divide so the unknown stiffness of the wire cancels.
- Hooke's law. Let the natural length be L. Within the elastic limit, tension is proportional to extension, so T = k(l - L), where k = YA/L is the same constant for both cases.
- Divide the two. Write T₁ = k(l₁ - L) and T₂ = k(l₂ - L). Dividing one by the other removes k, so you do not need Y or A.
T₁/T₂ = (l₁ - L)/(l₂ - L)
- Solve for L. Cross-multiply to get T₁ l₂ - T₁ L = T₂ l₁ - T₂ L. Then L(T₂ - T₁) = l₁ T₂ - l₂ T₁, so L = (l₁ T₂ - l₂ T₁)/(T₂ - T₁).
- Sanity check. A stretched wire is always longer than its natural length, so L must be shorter than both l₁ and l₂. Options (a), (c) and (d) are all averages that lie between l₁ and l₂, so none of them can be L.
- So. The natural length is (l₁ T₂ - l₂ T₁)/(T₂ - T₁), which is option (b).
The idea to keep. Two loads and two lengths? Write T = k(l - L) for each and take the ratio. Then check that your natural length comes out shorter than every stretched length.
Answer from the NTA final answer key. Solution by PrepWiser.