HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# A Steel Wire Having Cross-sectional Area 2 Mm2 is Stretched by Ion. Find the Lateral Strain Produced in the Wire. (Given : Y for Steel = 2 X 1011 N / M2, Poisson'S Ratio δ = 0.29) - HSC Science (Electronics) 12th Board Exam - Physics

#### Question

A steel wire having cross-sectional area 2 mm2 is stretched by ION. Find the lateral strain produced in the wire. (Given : Y for steel = 2 x 1011 N / m2, Poisson's ratio δ = 0.29)

#### Solution

Given: A = 2 mm2 = 2 x  10–6 m2

F = 10 N

Ysteel = 2 x 1011 N/m2

δ = 0.29

To find: Lateral strain

Formula:

1) Y = longitudinalstress/longitudinalstrain

2) δ = lateralstrain/longitudinalstrain

Calculation: Longitudinal stress = F/A

= 10/(2xx10^(-6))

= 5 x  106 N/m2

Using formula i.,

longitudinal strain = longitudinalstress/Y

= (5 xx 10^6)/(2xx10^11)

= 2.5 x  10–5

Using formula ii,

lateral strain = δ x longitudinal strain

= 0.29 x  2.5 x  10–5

lateral strain = 7.25 x 10–6

Lateral strain produced in the wire is 7.25 x 10–6.

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#### APPEARS IN

2015-2016 (July) (with solutions)
Question 3.3 | 3.00 marks

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Solution A Steel Wire Having Cross-sectional Area 2 Mm2 is Stretched by Ion. Find the Lateral Strain Produced in the Wire. (Given : Y for Steel = 2 X 1011 N / M2, Poisson'S Ratio δ = 0.29) Concept: Definition of Stress and Strain.
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