The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cros

Question

The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross-section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of:

Accepted Answer

1. **Formula:** Young's Modulus ($Y$) is defined as $Y = \frac{F L}{A \Delta L}$, where $F$ is the force (weight added), $L$ is the original length, $A$ is the area of cross-section, and $\Delta L$ is the elongation .
2. **Rearrange for Force:** We need to find the ratio of weights ($F$). Rearranging the formula: $F = \frac{Y A \Delta L}{L}$.
3. **Given Conditions:**
   - Lengths are equal ($L_s = L_b = L$).
   - Areas are equal ($A_s = A_b = A$).
   - Lower ends are at the same level, meaning elongation is equal ($\Delta L_s = \Delta L_b = \Delta L$).
   - Young's modulus of steel is twice that of brass ($Y_s = 2Y_b$).
4. **Proportionality:** Since $A, L,$ and $\Delta L$ are constant for both wires, the force is directly proportional to Young's modulus ($F \propto Y$).
5. **Calculation:**
   $$ \frac{F_{	ext{steel}}}{F_{	ext{brass}}} = \frac{Y_{	ext{steel}}}{Y_{	ext{brass}}} = \frac{2Y_b}{Y_b} = \frac{2}{1} $$

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