JEE Mains · Physics · STD 11- 8. mechanical properties of solids
A string of area of cross-section \(4\,mm ^{2}\) and length \(0.5\) 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 \(\ldots . . \times 10^{-5}\). (Use Young's modulus \(10^{11}\,N / m ^{2}\) and \(g =10\,m / s ^{2}\) )
- A \(29\)
- B \(300\)
- C \(30\)
- D \(303\)
Answer & Solution
Correct Answer
(C) \(30\)
Step-by-step Solution
Detailed explanation
Strain \(=\) F/AY \(=\frac{ mg +\frac{ mv ^{2}}{ R }}{ AY }\) \(=\frac{20+\frac{2(5)^{2}}{0.5}}{3 \times 10^{-6} \times 10^{11}}=30 \times 10^{-5}\)
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