Free Physics of Solids MCQs with Answers

241. The only elastic modulus that applies to fluids is

• A. Young's modulus
• B. Shear modulus
• C. Modulus of rigidity
• D. Bulk modulus

Explanation: The correct answer is the Bulk modulus. This modulus is crucial for understanding how fluids respond to changes in pressure, which involves volume changes. Unlike solids, fluids cannot sustain shear stress, so moduli such as Young's modulus, shear modulus, and modulus of rigidity, which relate to shape change under stress, do not apply to fluids.

242. The compressibility of a material is

• A. Product of volume and its pressure
• B. The change in pressure per unit change in volume strain
• C. The fractional change in volume per unit change in pressure
• D. None of the above

Explanation: Compressibility is a measure of how much the volume of a material decreases under pressure. It is defined as the fractional change in volume per unit change in pressure, which is why Option C is correct. Option A confuses compressibility with a thermodynamic property, while Option B describes the bulk modulus, which is the reciprocal of compressibility. Option D is incorrect because a correct definition is provided.

243. The ratio of lengths of two rods A and B of same material is 1 : 2 and the ratio of their radii is 2 : 1, then the ratio of modulus of rigidity of A and B will be

• A. 4 : 1
• B. 16 : 1
• C. 8 : 1
• D. 1 : 1

Explanation: The modulus of rigidity, also known as shear modulus, is a property that only depends on the material composition, not on its dimensions, such as length or radius. Since rods A and B are made of the same material, their modulus of rigidity will be the same; hence, the ratio is 1:1. Other options suggest a dependence on dimensions, which is incorrect for modulus of rigidity.

244. Modulus of rigidity of a liquid

• A. Non zero constant
• B. Infinite
• C. Zero
• D. Cannot be predicted

Explanation: The modulus of rigidity, also known as shear modulus, is a property that measures a material's ability to withstand shear stress. It is applicable to solids, where deformation under shear stress can be measured. However, liquids cannot sustain shear stress; they flow instead of deforming elastically. Therefore, the concept of modulus of rigidity does not apply to liquids, and it cannot be predicted. Options A, B, and C are incorrect because they attempt to apply a concept to a state of matter where it does not hold.

245. Shearing stress causes change in

• A. Length
• B. Breadth
• C. Shape
• D. Volume

Explanation: Shearing stress involves forces that act parallel to the surface of a material, causing it to deform by sliding layers over one another, which changes the shape. It does not change the volume or individual dimensions like length or breadth. Instead, it results in a distortion that is characteristic of a change in shape.

246. The work done in stretching an elastic wire per unit volume is or strain energy in a stretched string is

• A. Stress × Strain
• B. 2 × Stress × Strain
• C. 1/2 × Strain × Stress
• D. Stress/Strain

Explanation: The correct answer is 1/2 × Strain × Stress. This formula is derived from the work done on an elastic wire or material. When a material is deformed elastically, the work done is stored as strain energy, and the correct calculation involves multiplying stress by strain and then taking half of that product. The factor of 1/2 comes from the average value of stress over the deformation process.

247. Two wires of same diameter of the same material having the length L and 2L. If the force F is applied on each, the ratio of the work done in the two wires will be

• A. 1 : 2
• B. 1 : 4
• C. 2 : 1
• D. 1 : 1

Explanation: The work done on a wire is given by the expression W = (1/2) × F × ΔL, where ΔL is the change in length. For wires of the same material and cross-sectional area, ΔL is proportional to the original length when the same force is applied. Therefore, for the first wire of length L, ΔL1 = L and for the second wire of length 2L, ΔL2 = 2L. Consequently, the work done on the first wire, W1, is proportional to L, and the work done on the second wire, W2, is proportional to 2L. This gives a ratio of W1:W2 = 1:2. The other options are incorrect as they either assume an incorrect proportionality or do not account for the difference in length.

248. A weight of 10 kg is hung is fixed to the ceiling and is 1 meter above the floor. The wire was elongated by 1 mm. The energy stored in the wire due to stretching is

• A. Zero
• B. 0.05 joule
• C. 100 joule
• D. 500 joule

Explanation: The energy stored in a stretched wire is given by the formula U = (1/2) × F × e, where F is the force applied, and e is the elongation. Here, the force F is the weight of the object, which is 10 kg × 9.8 m/s2 = 98 N. The elongation e is 1 mm, which is 0.001 m. Substituting these values, the energy U = 0.5 × 98 N × 0.001 m = 0.049 J, which rounds to 0.05 J. Therefore, option B is correct. Other options are incorrect as they either suggest no energy storage or provide unrealistic values of energy.

249. The ratio of Young's modulus of the material of two wires is 2 : 3. If the same stress is applied on both, then the ratio of elastic energy per unit volume will be

• A. 3 : 2
• B. 2 : 3
• C. 3 : 4
• D. 4 : 3

Explanation: The elastic energy per unit volume (u) is given by the formula u = (1/2) * stress2 / Young's modulus. Since the same stress is applied to both wires, the energy depends inversely on Young's modulus. Thus, if the ratio of Young's modulus is 2:3, the ratio of elastic energy per unit volume will be the inverse, which is 3:2. Therefore, Option A is correct. Options B, C, and D are incorrect because they either assume a direct relationship or provide a ratio that doesn't align with the inverse relationship.

250. When strain is produced in a body within elastic limit, its internal energy

• A. Remains constant
• B. Decreases
• C. Increases
• D. None of the above

Explanation: When strain is applied to a body within the elastic limit, work is done to deform the body. According to the principles of elasticity, this work is stored as potential energy, thus increasing the internal energy of the system. Therefore, the internal energy increases. Option A is incorrect because internal energy does not remain constant; it changes due to the work done during deformation. Option B is incorrect because the energy does not decrease; instead, it increases as potential energy is accumulated. Option D is incorrect because the correct behaviour of internal energy is described in one of the given options.