As shown in the figure, a light uniform rod PQ of length 150 cm is suspended from the ceiling horizontally using two metal wires A and B tied to the ends of the rod. The ratios of the radii and the Young’s modulus of the materials of the two wires A and B are respectively 2 : 3 and 3 : 2. The position at which a weight should be suspended from the rod such that the elongations of the two wires become equal is
Show Hint
In problems involving elongations, relate force distribution with material properties and geometric ratios, and apply equilibrium conditions.
Step 1: Understand the problem setup
A rod of length 150 cm is suspended horizontally by two wires A and B attached at ends P and Q respectively. A weight is hung at some point along PQ so that the elongations of both wires are equal. Step 2: Given ratios
- Radius ratio: \( r_A : r_B = 2 : 3 \)
- Young’s modulus ratio: \( E_A : E_B = 3 : 2 \) Step 3: Relationship between elongation, force, and material properties
Elongation \(\Delta L\) in a wire is given by:
\[
\Delta L = \frac{F L}{A E}
\]
where \(F\) is force, \(L\) length, \(A\) cross-sectional area (\( \propto r^2 \)), and \(E\) Young’s modulus. Step 4: Express elongations for both wires
Let the weight be suspended at distance \(x\) from P.
The tension in wire A = \(W \times \frac{150 - x}{150}\), tension in wire B = \(W \times \frac{x}{150}\) due to lever principle.
Area of wire A: \(A_A \propto r_A^2 = 2^2 = 4\)
Area of wire B: \(A_B \propto r_B^2 = 3^2 = 9\) Step 5: Equal elongation condition
\[
\Delta L_A = \Delta L_B \implies \frac{F_A L}{A_A E_A} = \frac{F_B L}{A_B E_B}
\]
\(L\) cancels out. Substitute \(F_A\) and \(F_B\):
\[
\frac{(150 - x)}{4 \times 3} = \frac{x}{9 \times 2}
\]
Simplify denominators:
\[
\frac{150 - x}{12} = \frac{x}{18}
\]
Cross multiply:
\[
18(150 - x) = 12x \implies 2700 - 18x = 12x \implies 2700 = 30x \implies x = 90 \text{ cm}.
\]
Step 6: Conclusion
The weight should be suspended 90 cm from point P for equal elongations.
