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To Determine Young's Modules Of Elasticity Of Material Of Given Wire - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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A metallic wire of length 2.5 meters and diameter 0.8 mm is subjected to a tensile force of 800 N, causing it to elongate by 2.2 mm. The density of the material of the wire is 7.8 g/cm3. Calculate the Young’s modulus of elasticity of the material of the wire.

A copper wire of length 2.0 m and diameter 1.2 mm is suspended vertically from a ceiling. A mass of 4.5 kg is attached to the lower end of the wire. Given that the density of copper is 8.92 g/cm3 , calculate the elongation of the wire due to the weight of the attached mass.

 

A metallic wire of length 2.5 meters and diameter 0.8 mm is subjected to a tensile force of 800 N, causing it to elongate by 2.2 mm. The density of the material of the wire is 7.8 g/cm3 . Calculate the Young’s modulus of elasticity of the material of the wire.

A wire of length L=2.0 \mathrm{~m} and diameter d=0.5 \mathrm{~mm} is suspended vertically from a ceiling. A weight W=20 \mathrm{~N} is attached to the lower end of the wire, causing it to stretch. If the extension of the wire is e=2.5 \mathrm{~mm}, calculate the Young's modulus (Y) of the material.

An experiment is conducted to determine Young’s modulus (Y ) of a material using the concept of elasticity. A wire of length L = 2 m and diameter d = 0.5 mm is subjected to a load F = 500 N, resulting in an extension e = 1.5 mm. Calculate Young’s modulus of the material.

The 18th division of the Vernier scale exactly coincides with one of the main scale divisions During Searle's experiment, zero of the Vernier scale lies between 2.20 \times 10^{-2} \mathrm{~m} and 2.25 \times 10^{-2} \mathrm{~m} of the main scale. When an additional load of 4 \mathrm{~kg} is applied to the wire, the zero of the Vernier scale still lies between 3.20 \times 10^{-2} \mathrm{~m} and 3.25 \times 10^{-2} \mathrm{~m} of the main scale, but now the 55 \mathrm{th} division of the Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 1.5 \mathrm{~m} and its cross-sectional area is 6 \times 10^{-7} \mathrm{~m}^{2}. The least count of the Vernier scale is 1.0 \times 10^{-5} \mathrm{~m}. The maximum percentage error in Young's modulus of the wire is:

In a laboratory experiment, a metallic wire of length L=2.5 \mathrm{~m} and diameter d=0.5 \mathrm{~mm} is subjected to a tensile force F=250 \mathrm{~N}. The extension x produced in the wire is measured as 1.5 \mathrm{~mm}. The cross-sectional area of the wire is given by A=\frac{\pi d^{2}}{4}.

The Young's modulus Y of the material is:

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Concepts Covered - 1

To determine Young's modules of elasticity of material of given wire

Aim-
To determine Young's modulus of elasticity of the material of a given wire.

Apparatus-
Searle's apparatus, two long steel wires of same length and diameter, a meter scale, a screw gauge, eight
 0.5 kg slotted weights and a 1 kg hanger.

Theory-

If a wire of length L and radius r be loaded by a weight Mg and if l be the increase in length.

Then, Normal stress $=\frac{M g}{\pi r^2}$
and Longitudinal strain $=\frac{l}{L}$
Hence, Young's modulus $=\frac{\text { Normal stress }}{\text { Longitudinal strain }}$
or $\quad Y=\frac{M g / \pi r^2}{l / L}$
or $\quad Y=\frac{M g L}{\pi r^2 l}$
where
$Y=$ Young's modulus
$M=$ Mass
L= length of wire
r- radius of the wire
$l=$ Extention
Knowing $L$ and $r$, and finding $l$ for known $M g, Y$ ean be calculated.

Diagram-

 Procedure-

1. Take two steel wires of the same length and diameter and tight their ends in torsion screws A, B and C, D
as shown in diagram. Wire AB becomes an experimental wire and wire CD becomes auxiliary wire.
2. Suspend a 1 kg dead load from the hook of frame F2.

3. Suspend a 1 kg hanger and eight 0.5 kg slotted weights from the hook of frame F1, The experimental
wire becomes taut.
4. Remove kinks from the experimental wire by pressing the wire between nails of the right-hand thumb and first
finger (through a handkerchief) and moving them along the length of the wire.

5. Remove all slotted weights from the hanger. Now each wire is equally loaded with 1 kg weight.
6. Measure the length of the experimental wire from tip A to tip B using a meter scale.
7. Find the pitch and the least count of the screw gauge.
8. Measure the diameter of the experimental wire at five different places, equally spaced
along the length (near two ends, two-quarter distance from ends and middle). At each place, measure
diameter along with two mutually perpendicular directions. Record the observations in the table.

9. Note the breaking stress for steel fable of constants. Multiply that by the cross-section area of
the wire to find breaking load of the wire. The maximum load is not to exceed one-third of the breaking
load.
10. Find the pitch and the least count of the spherometer screw attached to frame F1.
11. Adjust the spherometer screw such that the bubble in the spirit level is exactly in the center. Note the
reading of the spherometer disc. This reading is recorded against zero load.

12. Gently slip a 0.5 kg slotted weight in the hanger and wait for two minutes to allow the wire to extend
fully. The bubble moves up from the center.
13. Rotate the spherometer screw to bring the bubble back to the centre. Note the reading of the
spherometer disc. This reading is recorded against 1 kg load in the load increasing column.
14. Repeat steps 12 and 13 till all the eight 0.5 kg slotted weights have been used (now total load on
the experimental wire is 5 kg which must be one-third of the breaking load).

15. Now remove one slotted weight (load decreasing), wait for two minutes to allow the wire to contract
fully. The bubble moves down from the center.
16. Repeat step 13, The reading is recorded against load in the hanger in load decreasing column.

17. Repeat steps 15 and 16 till all the eight slotted weights are removed (now load on the experimental wire is 1 kg the initial load).
(Observations for the same load in load increasing and in load decreasing column must not differ much.
Their mean is taken to eliminate the error.)
18. Record your observations.

 

Result-
1. The Young's modulus for steel as determined by Searle's apparatus $=\ldots . . \mathrm{Nm}^{-2}$
2. Straight-line graph between load and extension shows that stress $\propto$ strain. This verifies Hooke's Law.
Percentage Error
Actual value of $Y$ for steel (from tables)

$$
=\ldots \ldots \mathrm{Nm}^{-2}
$$


Difference in values $\quad=\ldots \ldots \mathrm{Nm}^{-2}$
Percentage error $=\frac{\text { Difference in values }}{\text { Actual value }}=\ldots \ldots \%$
It is within limits of experimental error.

 

 

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