Advertisement Remove all ads

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?

Advertisement Remove all ads

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?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 9 Mechanical Properties of Solids
Q 6 | Page 244
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×