A cube of aluminium of sides 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

0.02

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.005

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.002

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
0.002

Shearing strain $ϕ = \frac{x}{L} = \frac{0.02 c m}{10 c m} ∴ ϕ = 0.002$

Suggest Corrections

Similar questions

Q. A cube of aluminium of sides 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 cube of aluminium of sides 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

Q. A metal cube of side

10 c m

is subjected to a shearing stress of

10^{4} N m^{- 2}

. The modulus of rigidity if the top of the cube is displaced by

0.05 c m

with respect to its bottom is

Q. A cube of side

40 m m

has its upper face displaced by

0.1 m m

by a tangential force of

8 k N

. The shearing modulus of cube is

Q. A 0.05 m cube has its upper face displaced by 0.2 cm by a tangential force of 8 N, Calculate the shearing strain, shearing stress and modulus of rigidity of the material of the cube.

Join BYJU'S Learning Program

A cube of aluminium of sides 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

The correct option is D 0.002 Shearing strain ϕ=xL=0.02 cm10 cm ∴ϕ=0.002

The correct option is D
0.002

Shearing strain $ϕ = \frac{x}{L} = \frac{0.02 c m}{10 c m} ∴ ϕ = 0.002$