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PhysicsInteger+4 / -1Question 23 of 90

JEE Main 2022 · 28 Jul · Shift 2, Physics question 23

A string of area of cross-section 4 mm2^2 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\times10^{-5}.

(use young's modulus 101110^{11} N/m2^2 and g = 10 m/s2^2)

Answer and worked solution

Official answer: 30

Tension at the bottom mg+mv2r=20+100=120mg+\frac{mv^2}r=20+100=120 N.

Strain 1204×10−6×1011=3×10−4=30×10−5\frac{120}{4\times10^{-6}\times10^{11}}=3\times10^{-4}=30\times10^{-5}.

Answer: 30

Topics:Stress and strainMotion in a vertical circle
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