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Physics Question 32 – JEE-MAIN 2025

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A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of 105 N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement θ of the rod axis from its original position would be : (shear moduli, G=1010 N/m2)

When a shear force is applied, the object undergoes a change in shape, characterized by an angular displacement or shear strain.

Step 1: Calculate the Cross-sectional Area✦ Active

The radius of the cylindrical rod is r=4 cm =0.04 m. The area A over which the shear force is applied is the cross-sectional area of the top surface.

A=πr2=π(0.04 m)2=π(16×10−4) m2
Step 2: Calculate Shear Stress○ Expand

The shear force F=105 N is applied at the top. Shear stress τ is the force per unit area.

τ=FA=105 Nπ(16×10−4) m2=105×10416π N/m2=10916π N/m2
Step 3: Calculate Angular Displacement (Shear Strain)○ Expand

The shear modulus G=1010 N/m2 is given. The angular displacement θ is the shear strain, which is related to shear stress and shear modulus by G=τθ.

θ=τG=10916π1010=10916π×1010=116π×10=1160π radians

Comparing this result with the given options, option 4 is the correct answer.

💡 Teacher's Secret Hint

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

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