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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A steel wire can withstand a load up to 2940 N2940 \text{ N}. A load of 150 kg150 \text{ kg} is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is (2008 E)

A

3030^\circ

B

6060^\circ

C

8080^\circ

D

8585^\circ

Step-by-Step Solution

Let the length of the wire be ll and the maximum angle of displacement be θ0\theta_0. When the load is released from angle θ0\theta_0, it falls through a vertical height h=llcosθ0=l(1cosθ0)h = l - l\cos\theta_0 = l(1 - \cos\theta_0). By conservation of mechanical energy, the kinetic energy at the lowest (equilibrium) position is equal to the loss in potential energy: 12mv2=mgh=mgl(1cosθ0)\frac{1}{2}mv^2 = mgh = mgl(1 - \cos\theta_0) v2=2gl(1cosθ0)v^2 = 2gl(1 - \cos\theta_0) At the equilibrium position, the tension TT is maximum and provides the required centripetal force: Tmg=mv2lT - mg = \frac{mv^2}{l} T=mg+ml[2gl(1cosθ0)]T = mg + \frac{m}{l}[2gl(1 - \cos\theta_0)] T=mg+2mg(1cosθ0)=mg(32cosθ0)T = mg + 2mg(1 - \cos\theta_0) = mg(3 - 2\cos\theta_0) Given maximum tension Tmax=2940 NT_{\text{max}} = 2940 \text{ N}, mass m=150 kgm = 150 \text{ kg}, and taking acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2: 2940=150×9.8×(32cosθ0)2940 = 150 \times 9.8 \times (3 - 2\cos\theta_0) 2940=1470(32cosθ0)2940 = 1470(3 - 2\cos\theta_0) 2=32cosθ02 = 3 - 2\cos\theta_0 2cosθ0=1    cosθ0=122\cos\theta_0 = 1 \implies \cos\theta_0 = \frac{1}{2} θ0=60\theta_0 = 60^\circ

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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