A copper cube 0.30m on a side is subjected to a shearing force of $F = 6.0 \times 10^{6} N$ . Assuming that the shear modulus for copper is $4.0 \times 10^{10} N /$ . The angle through which the cube shear is (approximately)

0.09

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0.15

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0.21

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0.32

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Solution

The correct option is A $0.09$
solution shearing $= \frac{F}{A} = \frac{F}{a^{2}}$
shearing strain $= t a n θ$
shear modulus = $\frac{s h e a r i n g s t r e s s}{s h e a r i n g s t r a i n}$
$\Rightarrow 4 \times 10^{10} = \frac{6 \times 10^{6}}{\frac{(0.3)^{2}}{t a n θ}}$
$\Rightarrow t a n θ = \frac{6 \times 10^{6}}{4 \times 10^{10} \times 9 \times 10^{- 2}}$
$\Rightarrow θ = t a n^{- 1} (\frac{1 \times 10^{- 2}}{6}) = 0.095$
Option - A is correct

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A copper cube 0.30m on a side is subjected to a shearing force of F=6.0×106N. Assuming that the shear modulus for copper is 4.0×1010N/. The angle through which the cube shear is (approximately)

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A copper cube 0.30m on a side is subjected to a shearing force of $F = 6.0 \times 10^{6} N$ . Assuming that the shear modulus for copper is $4.0 \times 10^{10} N /$ . The angle through which the cube shear is (approximately)