Modulus of Rigidity

Answer: A. The modulus of rigidity (shear modulus) G is defined as shearing stress divided by shearing strain, as stated in NCERT Class 11 Physics, Chapter 8, page 173.

Why B is wrong: B describes the bulk modulus K, not the modulus of rigidity. Bulk modulus deals with volumetric compression, not tangential deformation. (trap: confusing K with G)

Why C is wrong: C describes Young's modulus, not the modulus of rigidity. The deformation types are different: longitudinal vs shear. (trap: confusing Y with G)

Why D is wrong: D mixes tensile stress with volumetric strain — no standard elastic modulus is defined this way. Each modulus pairs matched stress and strain types. (trap: confusing Y with G)

Answer: D. Fluids (liquids and gases) cannot sustain static shear stress, so their modulus of rigidity is zero. NCERT Class 11 Physics, Chapter 8, page 173 notes this distinction between solids and fluids.

Why A is wrong: Steel is a solid with a well-defined shear modulus. Only fluids have G = 0. (trap: confusing Y with G by thinking all moduli apply universally)

Why B is wrong: Copper is a solid with a definite shear modulus. Ductility does not mean zero G. (trap: confusing Y with G)

Why C is wrong: Rubber is a solid (though highly elastic) and does resist shear — it has a finite, if small, shear modulus. (trap: confusing high elasticity with zero rigidity)

Answer: A. Shear deformation changes the shape (angles between faces change) but preserves volume. This is the defining characteristic that separates shear from compression.

Why B is wrong: Shape changes — that is exactly what shear deformation does. The opposite face displaces laterally, tilting the solid. (trap: confusing Y with G — longitudinal stretch also changes shape but additionally changes length)

Why C is wrong: The face along the force direction does not maintain its original effective length during shear; the lateral displacement alters geometry. (trap: confusing shear geometry with simple extension)

Why D is wrong: The angle between adjacent faces changes by the shear angle φ — this IS the shear strain. (trap: confusing Y with G — in tension, the right angles in cross-section are preserved)

Answer: D. D is correct. Area of the face A = (0.20 m)² = 4.0 × 10⁻² m². Shearing stress = F/A = (5.0 × 10⁵)/(4.0 × 10⁻²) = 1.25 × 10⁷ Pa. Shearing strain = Δx/L = (1.0 × 10⁻⁵)/0.20 = 5.0 × 10⁻⁵. G = stress/strain = (1.25 × 10⁷)/(5.0 × 10⁻⁵) = 2.5 × 10¹¹ Pa (NCERT Class 11 Physics, Chapter 8, page 173).

Why A is wrong: A is wrong because 2.5 × 10¹⁰ Pa comes from a decimal slip in the area, taking (0.20)² as 0.4 m² instead of 0.04 m². That makes the stress, and so the modulus, ten times too small.

Why B is wrong: B is wrong because 5.0 × 10¹⁰ Pa comes from not squaring the side: using A = 0.20 m² instead of (0.20 m)² = 0.04 m². The face of a cube is an area, side squared. (trap: confusing Y with G formula structure)

Why C is wrong: C is wrong because 1.25 × 10⁷ Pa is the shearing STRESS, F/A, reported as the modulus. Both are in pascals, which is exactly why this slips through; the modulus is stress divided by strain, and the strain here is 5.0 × 10⁻⁵.

Answer: C. Shearing stress = G × shearing strain = 8.0 × 10¹⁰ × 2.0 × 10⁻⁴ = 1.6 × 10⁷ Pa. Direct application of the definition G = stress/strain, rearranged.

Why A is wrong: 4.0 × 10¹⁴ results from multiplying 8.0 × 10¹⁰ by 2.0 × 10⁻⁴ with a power-of-ten error (adding exponents as 10 + 4 = 14 instead of 10 + (−4) = 6, then mishandling the coefficient). (trap: arithmetic slip in scientific notation)

Why B is wrong: 4.0 × 10⁶ comes from dividing G by strain instead of multiplying, and an additional factor error. G = stress/strain means stress = G × strain, not G/strain. (trap: confusing Y with G by inverting the formula)

Why D is wrong: 1.6 × 10⁶ results from a power-of-ten slip — writing 10⁶ instead of 10⁷. Always recheck exponent arithmetic: 10¹⁰ × 10⁻⁴ = 10⁶, then 8 × 2 = 16 = 1.6 × 10¹, giving 1.6 × 10⁷. (trap: exponent counting error)

Answer: B. Uniform compression from all sides is volumetric deformation → bulk modulus K = −V(dP/dV). The description explicitly mentions pressure and volume change, which are the defining variables for K. NCERT Class 11 Physics, Chapter 9 distinguishes the three moduli by deformation type.

Why A is wrong: Young's modulus applies to longitudinal stretch/compression along one axis — not uniform compression from all sides. The problem specifies all-sided compression, ruling out Y. (trap: Y vs G vs K confusion — applying the most familiar modulus by default)

Why C is wrong: Modulus of rigidity applies to tangential force causing angular deformation. There is no mention of shear or tangential force here — the compression is uniform and volumetric. (trap: Y vs G vs K confusion)

Why D is wrong: Poisson's ratio σ is a dimensionless ratio of lateral to longitudinal strain, not a modulus. It cannot be used to compute stress or strain magnitudes. (trap: listing Poisson's ratio among moduli)

Answer: C. C is correct. The shear strain is φ = (F/A)/G, and the displacement of the end face is Δx = φL = FL/(AG). Same material means the same G, and the forces are equal. Area goes as the square of the diameter, so A_A : A_B = 2² : 1² = 4 : 1. Then Δx_A/Δx_B = (L_A/L_B) × (A_B/A_A) = (1/2) × (1/4) = 1/8. Note the distinction: the STRAINS alone would be in the ratio 1 : 4, because strain does not depend on length; it is the displacement that picks up the factor of L.

Why A is wrong: A is wrong because 1 : 4 is the ratio of shear STRAINS, not of displacements. Shear strain φ = F/(AG) does not involve length at all, so it goes only as 1/A: (1/4) : 1. The question asks how far the faces move, and that is φ × L, which brings the length ratio in.

Why B is wrong: B is wrong because 1 : 2 uses only the length ratio and ignores the areas. Wire A is twice the diameter, so four times the area, which cuts its displacement by a further factor of 4. (trap: incomplete substitution)

Why D is wrong: D is wrong because 2 : 1 inverts the length ratio and leaves out the area. Wire A is both shorter and thicker, and both of those make it deform LESS than B, so its displacement must be the smaller of the two.

Answer: B. Problem (iii) describes a tangential force on one face with the opposite face fixed — classic shear deformation → use G. Problem (i) is longitudinal stretch → Y. Problem (ii) is uniform compression → K. Identifying deformation type is the first step; G applies only to shear.

Why A is wrong: Problem (i) is longitudinal stretching (pull along length), which requires Young's modulus Y, not G. (trap: confusing Y with G — both deal with a force along a surface, but the deformation geometry differs)

Why C is wrong: Problem (ii) describes uniform fluid compression — volumetric deformation requiring bulk modulus K, not G. No tangential force is involved. (trap: confusing K with G)

Why D is wrong: Problem (i) is tension → Y, not G. The modulus of rigidity applies only to problem (iii) where the force is tangential to the surface. (trap: confusing Y with G — grouping any 'force on a solid' as rigidity)