Physics Q46 – JEE-MAIN 2026

Home Physics Mechanical Properties of SolidsQuestion #286

A cube has side length

5 cm

and modulus of rigidity

10^{5} {N/m}^{2}

. The displacement produced by a force of

10 N

in the upper face of cube is _______ mm.

Enter Calculated Numerical Value:

Clue 1 of 3: Understanding Shear Deformation

When a tangential force is applied to the top surface of a body, while the bottom surface is fixed, the body undergoes a shear deformation. This deformation is characterized by a displacement of the top layer relative to the bottom layer.

🧠 Full Solution Path

Step 1: Convert Units and Calculate Area✦ Active

First, convert the given side length of the cube from centimeters to meters and calculate the area of the upper face where the force is applied.

L = 5 cm = 0.05 m

Area of upper face, A = L^{2} = (0.05 m)^{2} = 0.0025 m^{2}

Step 2: Calculate Shear Stress and Shear Strain○ Expand

Next, calculate the shear stress ( $τ$ ) using the applied force and the area. Then, use the modulus of rigidity ( $G$ ) to find the shear strain ( $γ$ ). The modulus of rigidity is given by $G = \frac{τ}{γ}$ .

τ = \frac{F}{A} = \frac{10 N}{0.0025 m^{2}} = 4000 {N/m}^{2}

γ = \frac{τ}{G} = \frac{4000 {N/m}^{2}}{10^{5} {N/m}^{2}} = 0.04

Step 3: Calculate Displacement and Convert to mm○ Expand

Finally, use the shear strain and the side length to calculate the displacement ( $Δ x$ ), using the formula $γ = \frac{Δ x}{L}$ . Convert the result from meters to millimeters.

Δ x = γ \cdot L = 0.04 \cdot 0.05 m = 0.002 m

Δ x = 0.002 m \times 1000 mm/m = 2 mm

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before performing calculations to avoid errors.

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Physics Question 46 – JEE-MAIN 2026

Clue 1 of 3: Understanding Shear Deformation

🧠 Full Solution Path