Question

4. In the se
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 found to be 0.050cm. Then
the Young's modulus of the wire is​

Answer 1

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

Increases length = 0.050 cm

Weight = 1.5 kg

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

Using formula of Young's modulus

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$

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

Where, F = force

A = area of cross section

l = length

$\Delta l$ = increases length

Put the value into the formula

$Y=\dfrac{1.5\times9.8\times156\times10^{-2}}{\pi\times(0.027\times10^{-2})^2\times0.050\times10^{-2}}$

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

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

# Learn more

Topic : Young's modulus of the wire

https://brainly.in/question/4865733

https://brainly.in/question/6497069