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A metal cube of side \( 10 \mathrm{cm} \) is subjected to a shearing stress of \( 10^{4} N m^{-2} \). The modulus of rigidity if the top of the cube is displaced by \( 0.05 \mathrm{cm} \) with respect to its bottom is

1. \( 2 \times 10^{6} \mathrm{Nm}^{-2} \)
2. \( 10^{5} N m^{-2} \)
3. \( 1 \times 10^{7} N m^{-2} \)
4. \( 4 \times 10^{5} \mathrm{Nm}^{-2} \)

Correct answer: \( 2 \times 10^{6} \mathrm{Nm}^{-2} \)

Solution

Shear strain is the relative displacement divided by the height of the cube: \(0.05\text{ cm}/10\text{ cm} = 0.005\). The modulus of rigidity is \(\tau/\gamma = 10^4/0.005 = 2\times10^6\,\text{N m}^{-2}\).

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