A copper cube 0.30m on a side is subjected to a shearing force of F = 6.0 × 106 N. Assuming that the shear modulus for copper is 4.2 × 1010 N/m2. The angle through which the cube shear is (approximately)

l = b = h = 0.3m ; A = l x b = 0.3 x 0.3 A = 9 × 10–2m2 ; F = 6 × 106 Nη = 4.2 x 1010 N/m2η=FAθ⇒θ=6 x 1069 x 10-2 x 4.2 x 1010= 0.1587 x 10-2 rad ⇒θ = 0.09o

A copper cube 0.30m on a side is subjected to a shearing force of F = 6.0 × 106 N. Assuming that the shear modulus for copper is 4.2 × 1010 N/m2. The angle through which the cube shear is (approximately)

l = b = h = 0.3m ; A = l x b = 0.3 x 0.3 A = 9 × 10–2m2 ; F = 6 × 106 Nη = 4.2 x 1010 N/m2η=FAθ⇒θ=6 x 1069 x 10-2 x 4.2 x 1010= 0.1587 x 10-2 rad ⇒θ = 0.09o

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A copper cube 0.30m on a side is subjected to a shearing force of F = 6.0 × 10⁶ N. Assuming that the shear modulus for copper is 4.2 × 10¹⁰N/m². The angle through which the cube shear is (approximately)

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0.32^o

0.09^o

0.15^o

0.21^o

answer is A.

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Detailed Solution

l = b = h = 0.3m ; A = l x b = 0.3 x 0.3
A = 9 × 10^–2m² ; F = 6 × 10⁶ N
$η$ = 4.2 x 10¹⁰ N/m²
$η = \frac{F}{Aθ} \Rightarrow θ = \frac{6 x 10^{6}}{9 x 10^{- 2} x 4.2 x 10^{10}}$
= 0.1587 x 10^-2 rad