# The Edge of an Aluminium Cube is 10 Cm Long. One Face of the Cube is Firmly Fixed to a Vertical Wall. a Mass of 100 Kg is Then Attached to the Opposite Face of the Cube. the Shear Modulus of Aluminium is 25 Gpa. What is the Vertical Deflection of this Face? - Physics

The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

#### Solution 1

Edge of the aluminium cube, L = 10 cm = 0.1 m

The mass attached to the cube, m = 100 kg

Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa

Shear modulus, η = "Shear stress"/"Shear strain" = (F/A)/(L/(triangleL))

Where

F = Applied force = mg = 100 × 9.8 = 980 N

A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2

ΔL = Vertical deflection of the cube

:.triangleL = FL/(Aη)

= (980xx0.1)/(10^(-2)xx(25xx10^9))

= 3.92 × 10–7 m

The vertical deflection of this face of the cube is 3.92 ×10–7 m.

#### Solution 2

Here, side of cube, L = 10 cm =10/100= 0.1 m

∴ Area of each face, A = (0.1)2 = 0.01 m2

Tangential force acting on the face,

F = 100 kg  = 100 x 9.8 = 980 N

Shear Modulus, η = 25 Gpa = 25 xx 10^9 Nm^(-2)

Since shear modulus is given as

eta  = "Tangential stress"/"Shearing strain"

:."Shearing strain" = "Tangential stress"/"Shear modulus"

= F/(Aeta) = 980/(0.01xx25xx10^9) = 3.92 xx 10^(-6)

Now "Lateral Strain"/"Side of cube" =  "Shearing strain"

:.Lateral Strain =  Shearing stain x Side of the cube

= 3.92 xx 10^(-6) xx 0.1

=3.92 xx 10^(-7) m = 4 xx 10^(-7) m

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 9 Mechanical Properties of Solids
Q 6 | Page 244