Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is 24/π × 10² Nm-2. What is the minimum radius of the wire? (Take g = 10 ms-2) (a) 125 cm (b) 1250 cm (c) 12.5 cm (d) 1.25 cm

Answer: (c). Tension in an Atwood machine: T = (2m₁ m₂ g)/(m₁ + m₂) = (2 × 3 × 5 × 10)/(8) = 37.5 N. Minimum area: π r² = T/S = (37.5)/((24/π) × 10²) r² = (37.5)/(2400) = 0.015625 ⇒ r = 0.125 m So the answer is (c) 12.5 cm.

Two blocks of masses 3 kg and 5 kg are connected by a metal wire… | JEE Main 2021 (26 Aug, Shift 2)

Q10JEE Main 2021 · 26 AUG · SHIFT 2Medium

Thermal Stress & Breaking Stress

Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{\pi} \times 10^{2}$ Nm-2. What is the minimum radius of the wire? (Take g = 10 ms-2)

Show answer and solution

Answer: (c)

Tension in an Atwood machine: $T = \dfrac{2m_1 m_2 g}{m_1 + m_2} = \dfrac{2 \times 3 \times 5 \times 10}{8} = 37.5\text{ N}$ .
Minimum area: $\pi r^2 = \dfrac{T}{S} = \dfrac{37.5}{(24/\pi) \times 10^2}$
$r^2 = \dfrac{37.5}{2400} = 0.015625 \Rightarrow r = 0.125\text{ m}$
So the answer is (c) 12.5 cm.

Report a mistake

Thermal Stress & Breaking Stress question from JEE Main 2021 (26 Aug, Shift 2)

More questions on Thermal Stress & Breaking Stress