Verified Solution • 2025–26 Session moderate

QJEEM24Ph23e7: JEE Main 2024 Physics - Class 11 JEEM24Ph23e7 Solution: A wire of cross sectional area A, modulus of elasticity 2 × 10…

A wire of cross sectional area A, modulus of elasticity 2×1011 Nm−2 and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ=1100 radian on the points of support. The value of A is ________ ×10−4 m2 (consider x<<L ).

(given : g=10 m/s2)

JEE Main 2024 (Online) 6th April Evening Shift Physics - Properties of Matter Question 19 English

Step-by-Step Verified Solution

ΔIspring =(L2+x2)−L=x22LT=kΔI=kn22L

JEE Main 2024 (Online) 6th April Evening Shift Physics - Properties of Matter Question 19 English Explanation

⇒2Tθ=mg2(YAl)x22lθ=mgA=mgθ3Y=2010−6×2×1011=10×10−5=1×10−4

Related Learning Resources & Exam Preparation