A string of area of cross-section 4 mm² and length 0.5 m is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m / s at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _________ × 10⁻⁵.
(use Young's modulus 10¹¹ N / m² and g=10 m / s²)

Solution

Physics sir · 4 steps

At the bottom the string holds up the weight and also supplies the centripetal force, so add both.

1. Tension at bottom. At the lowest point the tension pulls up and gravity pulls down. Their difference is the net force towards the centre, so T - mg = mv²/r.

   T = mg + mv²/r = 2 × 10 + (2 × 5²)/0.5 = 20 + 100 = 120 N

2. Find the stress. The area is 4 mm² = 4 × 10⁻⁶ m². Stress = T/A = 120/(4 × 10⁻⁶) = 3 × 10⁷ N m⁻².
3. Find the strain. Strain = stress/Y = (3 × 10⁷)/10¹¹ = 3 × 10⁻⁴.
4. So. 3 × 10⁻⁴ = 30 × 10⁻⁵, so the blank is 30.

The common slip: 25. Leaving out the weight and taking T = mv²/r = 100 N gives a strain of 25 × 10⁻⁵, so 25 instead of 30.

The idea to keep. In a vertical circle, the tension at the bottom is mg + mv²/r, its largest value. Find the force first, then stress = F/A and strain = stress/Y.

Answer from the NTA final answer key. Solution by PrepWiser.