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Two wires of same material and same exte...

Two wires of same material and same extension `Deltal` have lengths l and 3l and diameters 3d and d. What will be the ratio of the forces applied on the two?

A

`1//27`

B

27

C

9

D

3

Text Solution

AI Generated Solution

The correct Answer is:
To find the ratio of the forces applied on two wires of the same material and same extension, we can use the relationship defined by Young's modulus. Let's break down the solution step by step. ### Step 1: Understand the relationship of Young's Modulus Young's modulus (Y) is defined as the ratio of stress to strain: \[ Y = \frac{\text{Stress}}{\text{Strain}} \] Where: - Stress = \(\frac{F}{A}\) (Force per unit area) - Strain = \(\frac{\Delta L}{L_0}\) (Change in length/original length) ### Step 2: Express the force in terms of Young's modulus From the definition of Young's modulus, we can express the force (F) as: \[ F = Y \cdot A \cdot \frac{\Delta L}{L_0} \] Where: - \(A\) is the cross-sectional area of the wire. ### Step 3: Calculate the cross-sectional area The cross-sectional area \(A\) of a wire with diameter \(d\) is given by: \[ A = \frac{\pi d^2}{4} \] ### Step 4: Set up the equations for both wires Let: - Wire 1: Length \(l\), Diameter \(3d\) - Wire 2: Length \(3l\), Diameter \(d\) For Wire 1: \[ A_1 = \frac{\pi (3d)^2}{4} = \frac{9\pi d^2}{4} \] For Wire 2: \[ A_2 = \frac{\pi d^2}{4} \] ### Step 5: Substitute into the force equation For Wire 1: \[ F_1 = Y \cdot A_1 \cdot \frac{\Delta L}{l} = Y \cdot \frac{9\pi d^2}{4} \cdot \frac{\Delta L}{l} \] For Wire 2: \[ F_2 = Y \cdot A_2 \cdot \frac{\Delta L}{3l} = Y \cdot \frac{\pi d^2}{4} \cdot \frac{\Delta L}{3l} \] ### Step 6: Find the ratio of the forces Now, we can find the ratio \( \frac{F_1}{F_2} \): \[ \frac{F_1}{F_2} = \frac{Y \cdot \frac{9\pi d^2}{4} \cdot \frac{\Delta L}{l}}{Y \cdot \frac{\pi d^2}{4} \cdot \frac{\Delta L}{3l}} \] The \(Y\), \(\frac{\pi}{4}\), and \(\Delta L\) cancel out: \[ \frac{F_1}{F_2} = \frac{9d^2}{\frac{d^2}{3}} = 9 \cdot 3 = 27 \] ### Conclusion Thus, the ratio of the forces applied on the two wires is: \[ \frac{F_1}{F_2} = 27:1 \]
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