The breaking stress of the material of a wire is 6 × 106 Nm-2 Then density ρ of the material is 3 × 103 kg m3 If the wire is to break under its own weight, the length (in metre) of the wire made of that material should be (take 54g = 10 ms-2)

Question

The breaking stress of the material of a wire is 6 × 106 Nm-2 Then density ρ of the material is 3 × 103 kg m3 If the wire is to break under its own weight, the length (in metre) of the wire made of that material should be (take 54g = 10 ms-2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Breaking stress =( textForce/ textarea) The breaking force will be its own weight F=mg =V ρ g = area × ℓρ g Breaking stress = 6× 106 =(area ×ℓ×ρ g/area) or ℓ=(6×106/3×103×10)=200 m

Q. The breaking stress of the material of a wire is $6 \times 1 0^{6} N m^{- 2}$ Then density $ρ$ of the material is $3 \times 1 0^{3} k g m^{3}$ If the wire is to break under its own weight, the length (in metre) of the wire made of that material should be (take $54 g = 10 m s^{- 2})$

Breaking stress $= \frac{Force}{area}$
The breaking force will be its own weight
$F = m g = V ρ g$ = area $\times ℓ ρ g$
Breaking stress $= 6 \times 1 0^{6} = \frac{a re a \times ℓ \times ρ g}{a re a}$
or $ℓ = \frac{6 \times 1 0 ^{6}}{3 \times 1 0 ^{3} \times 10} = 200 m$

Q. The breaking stress of the material of a wire is 6×106Nm−2 Then density ρ of the material is 3×103kgm3 If the wire is to break under its own weight, the length (in metre) of the wire made of that material should be (take 54g=10ms−2)

Breaking stress =areaForce​ The breaking force will be its own weight F=mg=Vρg = area ×ℓρg Breaking stress =6×106=areaarea×ℓ×ρg​ or ℓ=3×103×106×106​=200m

Q. The breaking stress of the material of a wire is $6 \times 1 0^{6} N m^{- 2}$ Then density $ρ$ of the material is $3 \times 1 0^{3} k g m^{3}$ If the wire is to break under its own weight, the length (in metre) of the wire made of that material should be (take $54 g = 10 m s^{- 2})$

Breaking stress $= \frac{Force}{area}$
The breaking force will be its own weight
$F = m g = V ρ g$ = area $\times ℓ ρ g$
Breaking stress $= 6 \times 1 0^{6} = \frac{a re a \times ℓ \times ρ g}{a re a}$
or $ℓ = \frac{6 \times 1 0 ^{6}}{3 \times 1 0 ^{3} \times 10} = 200 m$