Question

In the Searle's method to determine the
Young's modulus of a wire, a steel wire of
length 156cm
and diameter 0.054 cm is taken as
experimental wire. The average increase in
length for 1.5 kg wt is found to be 0.050cm. Then
the Young's modulus of the wire is​

Answer 1

Answer:

5.52 x 10³ N/mm²

Explanation:

stress s = 15 N / π(27)²×10∧-8 mm² = (5/9π)x10∧8 N/mm²

strain ∈ = 5x10∧-4mm ÷1560mm = 5/156 x 10∧-5

the Young's modulus of the wire is​

Y = s/∈ = (5/9π)x10∧8 N/mm²÷5/156 x 10∧-5 = (156/9π)x 10³ N/mm²

Y = 5.52 x 10³ N/mm²

Answer 2

Answer:

The young's modulus of the wire is $2.002\times10^{11}\ N/m^2$

Explanation:

Given that,

Length = 156 cm

Diameter = 0.054 cm

Radius $r=\dfrac{d}{2}=\dfrac{0.054}{2}=2.7\times10^{-4} m$

Increase length = 0.050 cm kg

Weight = 1.5 kg

We need to calculate the young's modulus of the wire

Using formula of young's modulus

$y=\dfrac{Fl}{A\Delta l}$

Where, l = length

$\Delta l$ = increases length

A = area of cross section

F = force

Put the value into the formula

$y=\dfrac{1.5\times9.8\times156\times10^{-2}}{\pi\times(2.7\times10^{-4})^2\times5\times10^{-4}}$

$y=2.002\times10^{11}\ N/m^2$

Hence, The young's modulus of the wire is $2.002\times10^{11}\ N/m^2$