Question

A steel wire of length 4m and diameter 5mm is stretched by 5kg weight. Find the increase in its length , if the young's modulus is 2.4×10^12 dyne cm-2

Answer 1

Solution:-

Given

l = 4cm

2r = 5mm => r= 2.5mm = 0.25cm

f=5kg-wt

f=  = 5×1000 g-wt  

f= 5000g-wt ×960D

y= 2.4×10^12 dyne cm∧2

Δl =?

Y= $\frac{F}{\pi R^{2} Y}$×$\frac{l}{Δl}$

Δl= Fl/$\pi$r∧2y

=>5000×980×400/22/7×(0.25)∧2×2.4×10∧12

=>Δl= 0.0041cm

=============

@GauravSaxena01

Answer 2

Answer:

The increase in length will be 0.0041 cm

Explanation:

Given,

length = 4 m = 400 cm

diameter = 5 mm = 0.5 cm

=> radius, r = 0.5/2 = 0.25 cm

=> Area = πr² = 3.14 x 0.25 x 0.25 ≈ 0.2 cm²

Force = 5 kg wt = 5000 g x 980 cm/s² = 49 x 10⁵

young's modulus = 2.4 x 10¹² dyne cm²

We know that young's modulus is given by the relation,

Y = (F/A)/(ΔL/L) = FL/AΔL

=> ΔL = FL/YA

= (49 x 10⁵ x 400)/(2.4 x 10¹² x 0.2)

= 0.0041 cm

Hence the increase in length will be 0.0041 cm