As shown in the figure, forces of 105 N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be :

Properties of Matter PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q53 with Solution

View written solutionFree

Correct answer: A

For a cube under tangential force,

\text{Shear stress} = \frac{F}{A}, \qquad \text{Shear strain} = \frac{x}{L}

and for the same material,

\text{modulus of rigidity } \eta = \frac{\text{shear stress}}{\text{shear strain}}

So,

\eta = \frac{F/A}{x/L} = \frac{FL}{Ax}

Since the material is same and conditions are similar, $\eta$ remains constant.

Side of cube:

L_1 = 10\text{ cm}

Area of face:

A_1 = (10)^2 = 100\text{ cm}^2

Force:

F_1 = 10^5\text{ N}

Displacement:

x_1 = 0.5\text{ cm}

Side:

L_2 = 20\text{ cm}

Area:

A_2 = (20)^2 = 400\text{ cm}^2

Force is applied under similar conditions, so take the same tangential force:

F_2 = 10^5\text{ N}

Let displacement be $x_2$ .

\frac{F_1L_1}{A_1x_1} = \frac{F_2L_2}{A_2x_2}

Since $F_1 = F_2$ ,

\frac{L_1}{A_1x_1} = \frac{L_2}{A_2x_2}

So,

x_2 = x_1\cdot \frac{L_2}{L_1}\cdot \frac{A_1}{A_2}

Substitute values:

x_2 = 0.5 \times \frac{20}{10} \times \frac{100}{400}

x_2 = 0.5 \times 2 \times \frac{1}{4}

x_2 = 0.25\text{ cm}

The displacement is

0.25\text{ cm}

So the correct option is A.