Question

A load of 4.0kg is suspended from a ceiling through a steel wire of length 20m and radious 2.0mm it is found that the length of the wire increase by 0.031mm as equilibrium is achieved find young's modulus of steel take g=3.1πm/s²

Answer 1

Hii friend,

# Answer- Y = 2 × 10^12 N/m2

# Given-

m=4kg

l=20m

dl=0.031mm=3.1×10-5m

r=2mm=2×10^-3m

# Formula-

Youngs modulus is given by

Y = stress/strain

Y = (F/A)/(dl/l)

Y = (mg/πr^2)/(dl/l)

Y = (4×3.1π/π×4×10^-6)/(3.1×10-5/20)

Y = 2 × 10^12 N/m2

Youngs modulus = 2 × 10^12 N/m2.

Hope that was helpful...

Answer 2

Explanation:

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The answer is in the above picture.

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