The solution to the given question is that the radius of the wire will be 0.378 millimetre approximately
At equilibrium,
Lets assume Tension T in the wire
mass of the block is given as 29 Kg
T sin 12° + T sin 12° = mg
T sin 12° + T sin 12° = 29 × 9.8
2T sin 12° = 29 × 9.8
T = ( 29 × 9.8 )/2 sin 12°
T = 141.1 / sin 12°
T = 141.1 / 0.2
T = 705.5 Newton
Stress = K × Strain
where K is the elastic modulus of the metal
elastic modulus is given as 7.0 × 10¹⁰ N/m²
T/A = 7.0 × 10¹⁰ × (Δl/l)
A = T × (l/Δl) × (1/ (7.0 × 10¹⁰) )
A = π r² = T / ( (7.0 × 10¹⁰) × ( (1/cos 12°) - 1) )
π r² = 705.5 / ( (7.0 × 10¹⁰) × 0.02234 )
r = √( 705.5 / ( π × (7.0 × 10¹⁰) × 0.02234 ) ) metre
r = √( 705.5 / ( 4910.33 × 10⁶ ) ) metre
r = √( 0.1436 × 10⁻⁶ ) metre
r = ( 0.378 × 10⁻³ ) metre
r = 0.378 millimetre
Thus we have find the radius of the wire by applying stress strain relation which came out to be 0.378 millimetre.
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