AP EAMCET · PHYSICS · Mechanical Properties of Solids
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 moduli 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
- A 90 cm from P
- B 100 cm from P
- C 40 cm from Q
- D 45 cm from Q
Answer & Solution
Correct Answer
(A) 90 cm from P
Step-by-step Solution
Detailed explanation
Given: \(\frac{r_A}{r_B} = \frac{2}{3}\), \(\frac{Y_A}{Y_B} = \frac{3}{2}\) For equal elongation: \(\Delta L_A = \Delta L_B \Rightarrow \frac{F_A L}{A_A Y_A} = \frac{F_B L}{A_B Y_B}\) \(\frac{F_A}{\pi r_A^2 Y_A} = \frac{F_B}{\pi r_B^2 Y_B}\)…
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