AP EAMCET · PHYSICS · Mechanical Properties of Solids
A one metre steel wire of negligible mass and area of cross-section \(0.01 \mathrm{~cm}^2\) is kept on a smooth horizontal table with one end fixed. A ball of mass \(1 \mathrm{~kg}\) is attached to the other end. The ball and the wire are rotating with an angular velocity of \(\omega\). If the elongation of the wire is \(2 \mathrm{~mm}\), then \(\omega\) is (Young's modulus of steel \(=2 \times 10^{11} \mathrm{Nm}^{-2}\) )
- A \(5 \mathrm{rad} \mathrm{s}^{-1}\)
- B \(10 \mathrm{rad} \mathrm{s}^{-1}\)
- C \(15 \mathrm{rad} \mathrm{s}^{-1}\)
- D \(20 \mathrm{rad} \mathrm{s}^{-1}\)
Answer & Solution
Correct Answer
(D) \(20 \mathrm{rad} \mathrm{s}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, elongation of the wire, \(\Delta l=2 \mathrm{~mm}\) \[ =2 \times 10^{-3} \mathrm{~m} \] Mass of the ball, \(m=1 \mathrm{~kg}\) Length of wire, \(l=1 \mathrm{~m}\) Area of cross-sectional of wire, \[ A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m} \] Young's modulus…
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