Practice portal › Mechanical Properties of Solids › Breaking Stress and Dynamical Systems
Asked in JEE Main 31st Jan 2nd Shift 2024 · Wires in accelerating and rotating systems
Given: m₁=2 kg, m₂=4 kg, r=4.0×10⁻⁵ m, Y=2.0×10¹¹ N m⁻².
The wire over a smooth pulley carries one tension throughout, T=(2m₁m₂g)/(m₁+m₂)=(2×2×4×10)/6=(80)/3 N.
A=π r²=π(4.0×10⁻⁵)²=1.6×10⁻⁹π m², so AY=1.6×10⁻⁹π×2.0×10¹¹=320π N.
Strain =T/(AY)=(80)/(3×320π)=1/(12π).
Comparing with 1/(απ) gives α=12.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer