Question

A 5.0 cm cube of substance ; has its upper face displaced by 0.65 cm, by a tangential force of 0.25 N . Calculate the modulus of rigidity of the substance.

• A. 769.2 $Nm^{-2}$
• B. 754.3 $Nm^{-2}$
• C. 735.2 $Nm^{-2}$
• D. 714.3 $Nm^{-2}$

Answer 1

Answer: (A)

$\eta = \frac{FL}{Al} , A = L^2 , \eta = \frac{FL}{L^2l} = \frac{F}{Ll}$
Here, $L = 5.0 \times 10^{-2} m $
$l = 0.65 \times 10^{-2} \, m , F = 0.25 \, N$
$\eta = \frac{0.25}{5.0 \times 10^{-2} \times 0.65 \times 10^{-2}} = \frac{0.25 \times 10^4}{3.25}$
= 769.2 $Nm^{-2}$