Verified Solution • 2025–26 Session moderate

Q8.10: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF SOLIDS 8.10 Solution: A rigid bar of mass 15 kg is supported symmetrically by three…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume that because the three wires carry the same tension they must have the same cross‑section, overlooking that the rigid bar forces all three wires to stretch the same amount. The correct condition is equal elongation ($\Delta L = TL/(AY)$), which leads to $A_{Cu}Y_{Cu}=A_{Fe}Y_{Fe}$ and hence the diameter ratio depends on the Young's moduli, not just the load.

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Step-by-Step Verified Solution

**Answer:** dCu:dFe=1.31:1d_{Cu} : d_{Fe} = 1.31 : 1 **Step-by-Step Explanation:** For equal tension and equal elongation:
TLACuYCu=TLAFeYFe\frac{T L}{A_{Cu} Y_{Cu}} = \frac{T L}{A_{Fe} Y_{Fe}}
So ACuYCu=AFeYFeA_{Cu} Y_{Cu} = A_{Fe} Y_{Fe}. Since A=πd2/4A = \pi d^2/4:
dCu2dFe2=YFeYCu\frac{d_{Cu}^2}{d_{Fe}^2} = \frac{Y_{Fe}}{Y_{Cu}}
Using YFe=1.9×1011Y_{Fe} = 1.9 \times 10^{11} Pa and YCu=1.1×1011Y_{Cu} = 1.1 \times 10^{11} Pa:
dCudFe=1.91.1=1.727=1.31\frac{d_{Cu}}{d_{Fe}} = \sqrt{\frac{1.9}{1.1}} = \sqrt{1.727} = 1.31
Hence dCu:dFe=1.31:1d_{Cu} : d_{Fe} = 1.31 : 1 [citation:16][citation:24].

All Questions in this Chapter

Q8.1 Q8.2 Q8.3 Q8.4 Q8.5 Q8.6 Q8.7 Q8.8 Q8.9 Q8.10 Q8.11 Q8.12 Q8.13 Q8.14 Q8.15 Q8.16

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