A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that $\mathrm{g}=3.1 \pi \mathrm{~ms}^{-2}$, what will be the tensile stress that would be developed in the wire?

(D) $3.1 \times 10^{6} \mathrm{Nm}^{-2}$

(A) $6.2 \times 10^{6} \mathrm{Nm}^{-2}$

(B) $4.8 \times 10^{6} \mathrm{Nm}^{-2}$

(C) $5.2 \times 10^{6} \mathrm{Nm}^{-2}$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Elasticity & SHM

A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that

g = 3.1 π ms^{- 2}

, what will be the tensile stress that would be developed in the wire?

Answer: (D)

Tensile stress is the load per unit cross-sectional area,

σ = \frac{m g}{π R ^{2}}

. With

m = 4

kg,

g = 3.1 π m/s^{2}

and

R = 2.0

= 2 \times 1 0^{- 3}

m, the chosen value of

g

lets

π

cancel:

σ = \frac{4 \times 3.1 π}{π ( 2 \times 1 0 ^{- 3} ) ^{2}} = \frac{4 \times 3.1}{4 \times 1 0 ^{- 6}} = 3.1 \times 1 0^{6} N/m^{2}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option