Question

determine Young's modulus of material of a wire experimentally




using the above figure

Answer 1

here's your answer $<b><I>$

consists of two long strips wires of same length and equal radius suspended side by side from of fixed rigid support the wire ( called reference wire )
carries a millimetre scaleM and a pant to place a weight the wire be (called the experimental wire) of uniform area of cross section also carries a pan in Which known weights are placed a vernier scaleV is attached to a pointer at the bottom of the experimental wire B and the main scale M is fixed to the reference wire A
the weight placed in the pan exerts a downward force and stretch the experimental wire under a tensile stress the elevation of the wire is measured by the vernier arrangement
both the reference and the experimental wires are given an inertial small note to keep the wire straight and Vernier reading is noted
now the experimental wire is gradually loaded with the more weights to bring it under a tensile stress and the vernier reading is noted again
the difference between the two vernier readings gives the elongation produced in the wire
let R and L be the inertial radius and the length of the experimental wire respectively the area of cross section of the wire would be
${\pi}{{r}^{2} }$
let M be the mass that produced on elongation
∆L.
in the wire
thus the applied force is equal to mass ( mg)
then youngs modulus of the material of the experiment write us given below
$\frac{Mg}{ {\pi}^{2} }$.$\frac{L}{∆L}$= $\frac{mg \times l}{\pi {r}^{2} }$

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Answer 2

heya !!.


here is your answer⬇️⬇️


There are two wires which have same length.

The right wire carries millimetre scale M and the wire will be in uniform motion which carries pans in which vernier scale is attached to point B and and M is fixed to A , In the point A the force will be downward which is exerted by pans . It have a tensile stress. These is to keep wire straight and vernier is loaded more weight in this experiment.

so,

Let R and L be intial radius , length of wire will be pier^2.

Mg /pie^2 L/ aL = mg×l/ pier^2.


hope my answer helps you