A cube of aluminium of side 6cm is subjected to a tangential force...

Lateral displacement, $x = 0 . 012cm$

$∴$ Shearing strain,

$\theta = \dfrac{x}{L} = \dfrac{0 . 012}{6} = 0 . 002$

Shearing stress, $\dfrac{F}{a} = η\theta$

$= 2 \times {10}^{11} \times 0 . 002$

$= 4 \times {10}^{8}{dynecm}^{- 2}$

Shearing force, $F =$ shearing stress $\times$ area

$= 4 \times {10}^{8} \times ( L{)}^{2}$

$= 4 \times {10}^{8} \times ( 6{)}^{2}$

$= 1 . 44 \times {10}^{10}dyne$

$⇒ K = 1 . 44$