Question

A horizontal metal rod of diameter 4 cm projects 6 cm from a wall. The shear modulus of the metal is \(3 \times 10^{10}\, N\,m^{-2}\). An object of mass 1100 kg is suspended from the free end of the rod. The free end of the rod moves down by a distance. (Take \(g = 10\,m\,s^{-2}\), neglect mass of rod)

• A. \(2.5 \times 10^{-6}\, m\)
• B. \(6.75 \times 10^{-5}\, m\)
• C. \(2.25 \times 10^{-6}\, m\)
• D. \(1.75 \times 10^{-5}\, m\)

Hint:

For shear problems, always use \( \Delta x = \frac{FL}{AG} \) and carefully track powers of 10 in area.

Answer 1

The Correct Option is D

Step 1: Specimen dimensions and properties.
Diameter = 4 cm → radius = 0.02 m; length L = 6 cm = 0.06 m; G = 3×10¹⁰ N m⁻²; load F = 1100×10 = 11000 N.

Step 2: Lateral displacement formula.
Δx = F L / (A G).

Step 3: Cross-sectional area.
A = π r² = π (0.02)² ≈ 1.256×10⁻³ m².

Step 4: Substituting values.
Δx = (11000 × 0.06) / (1.256×10⁻³ × 3×10¹⁰) = 660 / (3.768×10⁷).

Step 5: Result.
Δx ≈ 1.75×10⁻⁵ m.