A steel wire of length 4m and diameters 5mm is stretched by 5kg weight find the change in it's diameter if $y = 0.4 \times 10^{12} d y n e / c m^{2}$ and $σ = 0.3$

Open in App

Solution

$Y = 0.4 \times 10^{12} d y n e / c m^{2} = 4 \times 10^{10} n / m^{2}$

$s t r a i n = \frac{F}{A Y} = \frac{5 \times 9.8}{π \frac{(5 \times 10^{- 3})^{2}}{4} \times 4 \times 10^{10}}$

$⟹ s t r a i n = 6.24 \times 10^{- 5}$

$σ = \frac{\frac{- 2 Δ D}{D}}{s t r a i n}$

$⟹ Δ D = - 4.68 \times 10^{- 5} m = - 0.047 m m$
(-ve sign shows that diameter decreases)

Suggest Corrections

Similar questions

Q. A uniform cylindrical wire of length

4 m

and diameter

0.6 m m

is stretched by a certain force such that its length is increased by

4 m m

. If the Poisson's ratio of the material is

0.3

then, calculate the change in diameter of the wire.

Q. What is the percentage increase in length of a wire having diameter

1.5 mm

, stretched by a force of

5 kg-wt

?
(Young's modulus of elasticity of wire

(Y) = 15 \times 10^{11} {dyne/cm}^{2} and g = 10 {m/s}^{2}

)

Q. A vertical steel post of diameter

25 c m

and length

2.5 m

supports a weight of

8000 k g

. Find the change in length produced.

(Given

Y = 2 \times 10^{11} P a

)

Q. When a wire is stretched, its length increases by 0.3% and the diameter decreases by 0.1%. Poisson's ratio of the material of the wire is about

A wire 4 m long and 0.3 mm diameter is stretched by a force of 100N. If the extension in the wire is 0.3 mm, calculate the potential energy stored in the wire.

Join BYJU'S Learning Program

A steel wire of length 4m and diameters 5mm is stretched by 5kg weight find the change in it's diameter if y=0.4×1012dyne/cm2 and σ=0.3

Y=0.4×1012dyne/cm2=4×1010n/m2strain=FAY=5×9.8π(5×10−3)24×4×1010⟹strain=6.24×10−5σ=−2ΔDDstrain⟹ΔD=−4.68×10−5m=−0.047mm(-ve sign shows that diameter decreases)

A steel wire of length 4m and diameters 5mm is stretched by 5kg weight find the change in it's diameter if $y = 0.4 \times 10^{12} d y n e / c m^{2}$ and $σ = 0.3$