A cube of aluminium of side 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

Question

A cube of aluminium of side 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be (A) 0.02 (B) 0.1 (C) 0.005 (D) 0.002

Tardigrade Admin · Accepted Answer

Given, length of cube, L=0.1 m, Shearing force, F=100 N and displaced distance,x=0.02 cm =0.02 × 10-2 m Shearing strain, φ=(x/L) =(0.02 × 10-2/0.1 m )=0.002

Q. A cube of aluminium of side $0.1 m$ is subjected to a shearing force of $100 N$ . The top face of the cube is displaced through $0.02 c m$ with respect to the bottom face. The shearing strain would be

$0.02$

$0.1$

$0.005$

$0.002$

Given, length of cube, $L = 0.1 m$ ,
Shearing force, $F = 100 N$
and displaced distance, $x = 0.02 c m$
$= 0.02 \times 1 0^{- 2} m$
Shearing strain, $ϕ = \frac{x}{L}$
$= \frac{0.02 \times 1 0 ^{- 2}}{0.1 m} = 0.002$

Q. A cube of aluminium of side 0.1m is subjected to a shearing force of 100N. The top face of the cube is displaced through 0.02cm with respect to the bottom face. The shearing strain would be

0.02

0.1

0.005

0.002