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Physics · MECHANICAL PROPERTIES OF SOLIDS
हिन्दीA square lead slab of side \( 50 \text{ cm} \) and thickness \( 10 \text{ cm} \) has its lower face fixed to the floor. A shearing force of \( 9.0 \times 10^4 \text{ N} \) is applied parallel to its upper narrow face of area \( 50 \text{ cm} \times 10 \text{ cm} \). If the shear modulus of lead is \( 5.6 \times 10^9 \text{ N m}^{-2} \), the lateral displacement of the upper edge relative to the lower fixed face is:
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Correct answer · Option D
\( 0.16 \text{ mm} \)
Explanation
Area \( A = 0.5 \times 0.1 = 0.05 \text{ m}^2 \). Shearing stress \( = \frac{9.0 \times 10^4}{0.05} = 1.8 \times 10^6 \text{ N m}^{-2} \). Displacement \( \Delta x = \frac{\text{Stress} \times L}{G} = \frac{1.8 \times 10^6 \times 0.5}{5.6 \times 10^9} = 1.6 \times 10^{-4} \text{ m} = 0.16 \text{ mm} \).
