StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A cylindrical wire of radius 1 mm and length 1 m (Young's modulus = 2 x 10¹¹ N/m², Poisson's ratio mu = pi/10) is stretched by a force of 100 N. What does its radius become?

  1. 0.99998 mm
  2. 0.99999 mm
  3. 0.99997 mm
  4. 0.99995 mm

Correct answer: 0.99995 mm

Solution

Longitudinal strain = 100/(pi*1e-6*2e11) = 1.59e-4; Delta r/r = -(pi/10)(1.59e-4) = -5e-5, so r = 0.99995 mm.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →