Question 7 of 13advanced🔧 ApplyNumerical4 marks

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.

Correct Answer

The ratio of diameters is d_copper : d_iron = Yiron/Ycopper\sqrt{Y_iron/Y_copper} : 1 ≈\approx 1.31 : 1

Exercise: EXERCISES | Q: 8.10
For More Understanding

Explanation

This problem involves the concept of Young's modulus and elastic extension. Since the rigid bar is supported symmetrically and each wire must have the same tension, all three wires must elongate by the same amount (otherwise the bar would tilt). Using the relation ΔL = TL/(A·Y) and equating elongations for copper and iron wires, we derive that the cross-sectional areas must be inversely proportional to their Young's moduli. The Young's modulus values from NCERT are: Y_copper = 1.1×101.1 \times 10¹¹ N/m² and Y_iron = 1.9×101.9 \times 10¹¹ N/m². The copper wire needs a larger diameter because copper has lower Young's modulus (less stiff), so it requires larger cross-section to produce same elongation under same tension.

Solution Steps

  1. Step 1: Since the bar is rigid and horizontal, all three wires must have the same elongation ΔL for equal tension.

  2. Step 2: For a wire under tension: ΔL = TL/(A·Y), where T = tension, L = length, A = cross-sectional area, Y = Young's modulus.

  3. Step 3: For same elongation: ΔL_copper = ΔL_iron, which gives: TL/(A_c·Y_c) = TL/(A_i·Y_i)

  4. Step 4: Simplifying: A_c·Y_c = A_i·Y_i, or A_c/A_i = Y_i/Y_c

  5. Step 5: Using A = πd²/4, we get: d_c²/d_i² = Y_i/Y_c, therefore d_c/d_i = Yi/Yc\sqrt{Y_i/Y_c}

  6. Step 6: Using Y_copper = 1.1×101.1 \times 10¹¹ N/m² and Y_iron = 1.9×101.9 \times 10¹¹ N/m²: d_copper/d_iron = 1.9/1.1\sqrt{1.9/1.1} = 1\sqrt{1}.727 ≈\approx 1.31

  7. Step 7: Therefore, the ratio of diameters is d_copper : d_iron ≈\approx 1.31 : 1

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