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Elastic Moduli - Determination of Young’s Modulus of the Material of a Wire

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Notes

Determination of Young’s Modulus of the Material of a Wire:

  • Two strings were hung from a support and two pans were attached to both the strings.

  • Weights are kept on both the pans.

  • When the number of weights in the second pan was increased, the string got stretched and moved in a downward direction.

  • The change in length was measured by the metre scale which was kept on reference wire.

  • Using this experiment, the Young’s modulus value was calculated

  • `"Y"= "longitudinal stress"/ "longitudinal strain" = σ/ ε`

  • = `("F"/"A")/ ("ΔL"/"L")`

  • Where original length = L and ΔL = change in length, F=mg (acting downwards) and A (area of cross-section of wire) = `pir^2`

  • = `("mg"/ (πr^2))/ (Δ"L"/"L")`

  • `Y="mgL"/(pi"r"^2Delta"L")`

  • This is the way to calculate the Young’s modulus.

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Related QuestionsVIEW ALL [13]

The dimension ML−1T−2 can correspond to

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 1011 N m−2. Make appropriate approximations. 

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by l. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2f. Find the increase in length of this wire.

The elastic limit of steel is 8 × 108 N m−2 and its Young modulus 2 × 1011 N m−2. Find the maximum elongation of a half-metre steel wire that can be given without exceeding the elastic limit.

 

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of the strains developed. Y of steel = 2 × 1011N m−2. Y of copper = 1.3 × 10 11 N m−2. 

A steel wire of original length 1 m and cross-sectional area 4.00 mm2 is clamped at the two ends so that it lies horizontally and without tensions. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression ? Y of the steel = 2.0 × 1011 N m−2. Take g = 10 m s−2. 

In figure the upper wire is made of steel and the lower of copper. The wires have equal cross section. Find the ratio of the longitudinal strains developed in the two wires.

Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross section = 2 cm2. 

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