The poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area
by 4%. The percentage increase in the length is :
(A) 1 % (B) 2% (C) 2.5% (D) 4%

Question

Arun · Answer

Poisson's ratio = 0 . 5 Since, density is constant therefore change in volume is zero, we have V = A × l = c o n s t a n t lo g V = lo g A + lo g l or A d A ​ + l d l ​ = 0 l d l ​ = − A d A ​ ∴ Percentage increase in length = 4 %

Vikas TU · Answer

Given Poisson's ratio (⊂) = 0.5.

Here, density is constant and hence change in volume is 0.

Given that there is a decrease in cross-sectional area by 4%

So, (dA/A) = 4.

We know that dA/A = 2 * (dr/r).

= > 4 = 2 * (dr/r)

= > 2 = (dr/r).

Hence,

= > dl/l

= > (dr/r)/⊂

= > 2/0.5

= > 4.

Therefore, the % increase in length = 4% (or) 0.04.

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The poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in the length is : (A) 1 % (B) 2% (C) 2.5% (D) 4%

The poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area
by 4%. The percentage increase in the length is :
(A) 1 % (B) 2% (C) 2.5% (D) 4%

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The poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in the length is : (A) 1 % (B) 2% (C) 2.5% (D) 4%

The poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in the length is : (A) 1 % (B) 2% (C) 2.5% (D) 4%

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The poisson’s ratio of a material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area
by 4%. The percentage increase in the length is :
(A) 1 % (B) 2% (C) 2.5% (D) 4%