Two metallic wires \( P \) and \( Q \) have the same volume and are made of the same material. If their areas of cross-section are in the ratio \( 4 : 1 \) and a force \( F_1 \) is applied to \( P \), producing an extension of \( \Delta l \), the force required to produce the same extension in \( Q \) is \( F_2 \).The value of \( \frac{F_1}{F_2} \) is _______.
Correct Answer: 16
Solution and Explanation
Given that the volumes of wires \( P \) and \( Q \) are the same and they are made of the same material, we can express volume as \( V = A \times l \), where \( A \) is the area of cross-section and \( l \) is the length.
Let \( A_P \) and \( A_Q \) be the areas of cross-sections of \( P \) and \( Q \), and \( l_P \) and \( l_Q \) be the respective lengths. Given the ratio \( \frac{A_P}{A_Q} = 4 \), we have:
\( A_P = 4A_Q \)
Since the volumes are equal:
\( A_P \cdot l_P = A_Q \cdot l_Q \)
Substituting \( A_P = 4A_Q \):
\( 4A_Q \cdot l_P = A_Q \cdot l_Q \)
Thus, \( l_Q = 4l_P \).
Now, consider the Young's modulus formula:
\( \text{Stress} = \frac{F}{A} \) and \( \text{Strain} = \frac{\Delta l}{l} \)
Young's modulus \( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/l}{A/\Delta l} \), hence \( F = \frac{YA\Delta l}{l} \).
For wire \( P \): \( F_1 = \frac{YA_P \Delta l}{l_P} \)
For wire \( Q \) to have the same extension \( \Delta l \): \( F_2 = \frac{YA_Q \Delta l}{l_Q} \)
Substituting \( A_Q = \frac{A_P}{4} \) and \( l_Q = 4l_P \):
\( F_2 = \frac{Y(A_P/4)\Delta l}{4l_P} = \frac{YA_P \Delta l}{16l_P} \)
Thus, the ratio \( \frac{F_1}{F_2} = \frac{\frac{YA_P \Delta l}{l_P}}{\frac{YA_P \Delta l}{16l_P}} = 16 \).
The computed value is \( 16 \).
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