A rubber cord of length 40 cm and area of cross-section 4 × 10⁻⁶ m² gains 30 J of energy when it is elongated through 10 cm. The Young's modulus of rubber is N × 10⁸ N m⁻². Find N.
Physics · Mixed Chapter
JEE Main Practice

Question

A rubber cord of length 40 cm and area of cross-section 4 × 10⁻⁶ m² gains 30 J of energy when it is elongated through 10 cm. The Young's modulus of rubber is N × 10⁸ N m⁻². Find N.

G1 Physics · Accepted Answer

Step 1: L = 0.40 m, A = 4 × 10⁻⁶ m², ΔL = 0.10 m, U = 30 J. Step 2: For linear elastic extension, U = ½FΔL and F = YAΔL/L. Step 3: Hence U = YA(ΔL)²/(2L). Step 4: Y = 2UL/[A(ΔL)²] = (2 × 30 × 0.40)/[(4 × 10⁻⁶)(0.10)²] = 6 × 10⁸ N m⁻². Final answer: N = 6.