99 Percentile Qs Bank for JEE MainPhysicsMechanical Properties of Solids
A rubber cube of side 5 ~cm has one face fixed, while a tangential force 1800 ~N is applied on its opposite face. If modulus of rigidity of rubber is 2.4 10^6 Nm ⁻² , then the lateral displacement of the strained face is ________
Options
- A3 mm
- B5 mm
- C15 mm
- D1.5 mm
Correct answer
C. 15 mm
Step-by-step solution
Shear stress = Tangential force Area = 1800 ~N (0.05)^2 Shear strain = aligned & = x h = Lateral displacement Height of cube = x 5 ~cm & = x ~m 0.05 ~m aligned Modulus of rigidity, = Shear stress Shear strain 2.4 10^6 N m ^2 = 1800 25 10⁻⁴ 0.05 x ~N / m ^2 or aligned x & = 1800 0.05 25 10⁻⁴ 2.4 10^6 & =15 10⁻³ ~m & =15 ~mm aligned