Question

A steel wire is loaded by 2 kg weight. If the radius of the wire is doubled, then its extension will become​

Answer 1

Answer:

remain unchanged .

Explanation:

I hope it helps you.

Answer 2

Answer: Extension will become one fourth.

Given:

• Steel wire is loaded by 2 kg weight.
• Radius of the wire is doubled.

To Find: Extension will become.

Explanation:

Step 1: Young’s modulus, numerical constant, named for the 18th-century English physician and physicist Thomas Young, that describes the elastic properties of a solid undergoing tension or compression in only one direction, as in the case of a metal rod that after being stretched or compressed lengthwise returns to its original length. Young's modulus is a measurement of a material's capacity to endure changes in length when subjected to compression or tension along its length. Sometimes referred to as the modulus of elasticity, Young’s modulus is equal to the longitudinal stress divided by the strain. For a metal bar under tension, stress and strain may be explained as follows.

Step 2: When a force F is applied to both ends of a metal bar with a cross-sectional area of A, the bar extends from its initial length $L_0$ to a new length $L_n$. (At the same time, the cross section shrinks.) The stress is calculated as the cross-sectional area divided by the tensile force, or F/A. The length change, $L_n - L_0$, divided by the initial length, $L_0$, yields the strain or relative deformation, or ($L_n- L_0$)/$L_0$. (Strength has no dimensions.) Consequently, Young's modulus may be mathematically represented as

Young’s modulus = stress/strain =$(FL_0)/A(L_n - L_0)$.

Step 3: A steel wire is loaded by 2 kg weight. As we know that the change in length of material is depend upon the radius of metal bar. So, if we double the radius of the wire then the extension will become decrease.

$Y=\frac{FL_0}{A\triangle L} \\\\\triangle L = \frac{FL_0}{AY}$

For a wire $A=\pi r^2$. If radius is doubled i.e.

$A_{new}=\pi (2r)^2\\\\A_{new}=4\pi r^2$

Thus, extension will become

$\triangle L_{new} = \frac{FL_0}{A_{new}Y}\\\\\triangle L_{new} = \frac{FL_0}{4AY}\\\\\triangle L_{new} = \frac{\triangle L}{4}$

Which means new extension will become one fourth of the previous one.

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