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) Numerical answer type. Work it out, then open the answer.

Answer: 30. At the bottom: T - mg = mv²/r T = 2 × 10 + (2 × 5²)/(0.5) = 20 + 100 = 120 N Strain = T/AY = 1204 × 10⁻⁶ × 10¹¹ = 3 × 10⁻⁴ = 30 × 10⁻⁵ Answer: 30

A string of area of cross-section 4 mm² and length 0.5 m is connected… | JEE Main 2022 (28 Jul, Shift 2)

Q15JEE Main 2022 · 28 JUL · SHIFT 2Medium

Elastic Behaviour, Stress & Strain

A string of area of cross-section $4\text{ mm}^2$ and length $0.5\text{ m}$ is connected with a rigid body of mass $2\text{ kg}$ . The body is rotated in a vertical circular path of radius $0.5\text{ m}$ . The body acquires a speed of $5\text{ m/s}$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _________ $\times 10^{-5}$ .

(use Young's modulus $10^{11}\text{ N/m}^2$ and $g = 10\text{ m/s}^2$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 30

At the bottom: $T - mg = \frac{mv^2}{r}$
$T = 2 \times 10 + \frac{2 \times 5^2}{0.5} = 20 + 100 = 120\ \text{N}$
Strain $= \frac{T}{AY} = \frac{120}{4 \times 10^{-6} \times 10^{11}} = 3 \times 10^{-4} = 30 \times 10^{-5}$
Answer: $30$

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