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 . . × 10-5. (Use Young's modulus 1011 N / m 2 and g =10 m / s 2 )

Question

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 . . × 10-5. (Use Young's modulus 1011 N / m 2 and g =10 m / s 2 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d0fbab5a43562625950439456a0f3153-.png" /> =( mg +( mv 2/ R )/ AY ) =(20+(2(5)2/0.5)/3 × 10-6 × 1011)=30 × 10-5

=AYmg+Rmv2​​ =3×10−6×101120+0.52(5)2​​=30×10−5

$= \frac{m g + \frac{m v ^{2}}{R}}{A Y}$
$= \frac{20 + \frac{2 ( 5 ) ^{2}}{0.5}}{3 \times 1 0 ^{- 6} \times 1 0 ^{11}} = 30 \times 1 0^{- 5}$