Free Physics of Solids MCQs with Answers

171. In a wire stretched by hanging a weight from its end, the elastic potential energy per unit volume in terms of longitudinal strain σ and modulus of elasticity Y is:

• A. Yσ²/2
• B. Yσ/2
• C. 2Yσ²/2
• D. Y²σ/2

Explanation: The elastic potential energy per unit volume in a stretched wire can be expressed in terms of the longitudinal strain σ and the modulus of elasticity, or Young's Modulus, Y. The formula is derived from the basic relationship of potential energy in elastic deformation. This accounts for the fact that energy stored is proportional to the square of the strain. The correct option is Option A, Yσ²/2. Options B, C, and D are incorrect as they either omit necessary components or misconfigure the relationship between these variables.

172. A ball falling in a lake of 200 m shows a decrease of 0.1% in its volume. The bulk modulus of elasticity of the material of the ball is: (take g =10 m/s2)

• A. 109 N/m2
• B. 2 x 10^9 N/m2
• C. 3 x 10^9 N/m2
• D. 4 x 10^9 N/m2

Explanation: The bulk modulus of elasticity is defined as the ratio of the change in pressure to the fractional change in volume. Here, the pressure change can be calculated as ΔP = ρgh, where ρ is the density of water (1000 kg/m3), g is the acceleration due to gravity (10 m/s2), and h is the depth (200 m). Thus, ΔP = 1000 × 10 × 200 = 2 × 106 N/m2. The volume change is given as 0.1%, or 0.001 in decimal form. Using the formula for bulk modulus, B = ΔP / (ΔV/V), we find B = (2 × 106 N/m2) / 0.001 = 2 × 109 N/m2. Therefore, option B is correct. Options A, C, and D do not match the calculation based on the given parameters.

173. The rubber cord catapult has a cross-sectional area of 1 mm2 and a total unstretched length of 10 cm. It is stretched to 12 cm and then released to project a stone of mass 5 g. Taking Young's modulus Y of rubber as 5 x 10^8 N/m2, the velocity of projection will be:

• A. 20 cm/s
• B. 20 m/s
• C. 2 m/s
• D. None of these

Explanation: The potential energy stored in the stretched rubber cord is given by the formula U = (1/2) × Y × (A/L) × ΔL2, where Y is Young's modulus, A is the cross-sectional area, L is the original length, and ΔL is the change in length. Substituting the given values: U = (1/2) × 5 × 108 N/m2 × (1 × 10-6 m2 / 0.1 m) × (0.02 m)2. Calculating this gives us the potential energy. The kinetic energy of the stone when projected is (1/2)mv2, where m is the mass of the stone and v is its velocity. Equating the potential energy to kinetic energy and solving for v gives us v = 20 m/s. Other options are incorrect due to either incorrect calculations or unit conversion errors.

174. A U-tube in which the cross-sectional area of the limb on the left is one-quarter, and the limb on the right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water?

• A. 1.2 cm
• B. 2.35 cm
• C. 0.56 cm
• D. 0.8 cm

Explanation: To solve this problem, we use the principle of fluid statics. When water is added to the left limb, it applies pressure that causes the mercury in the right limb to rise. The pressure exerted by the water column is balanced by the pressure due to the rise in mercury in the right limb. The relationship is given by the formula:Pressure by water = Pressure by mercury.Considering the cross-sectional area difference and the densities, we find that the rise in mercury level in the right limb is 0.56 cm. The other options do not correctly account for the density and cross-sectional area differences, leading to incorrect calculations.

175. A piston of a cross-sectional area of 100 cm2 is used in a hydraulic press to exert a force of 107 dynes on the water. The cross-sectional area of the other piston which supports an object having a mass of 2000 kg is:

• A. 100 cm2
• B. 109 cm2
• C. 2 x 10^4 cm2
• D. 2 x 10^10cm2

Explanation: The problem involves a hydraulic press, which operates on Pascal's Law: the pressure applied to a confined fluid is transmitted undiminished in all directions. Given that the first piston applies a force of 107 dynes over 100 cm2, the pressure can be calculated as P = F/A = 107 dynes / 100 cm2. This pressure is transmitted to the second piston, which must support a mass of 2000 kg (corresponding to a force due to gravity of approximately 2 × 108 dynes). Solving for the area using A = F/P gives 2 × 104 cm2, making option C correct.P₁ = P₂ ⇒ F₁/A₁ = F₂/A₂⇒ 10⁷/10² = 2000 × 10³ × 10³ / A₂∴ A₂ = 2 × 10⁴ cm² (g = 980 ≈ 10³ cm/s²)Options A and B do not account for the mass and resulting force, while option D is an impractical overestimation.

176. There are _ types of solids.

• A. One
• B. Two
• C. Three
• D. Four

Explanation: Solids are primarily classified into two main categories based on the arrangement of their constituent particles: Crystalline solids: These have a regular, repeating, three-dimensional structure (a crystal lattice). Amorphous solids: These have a random, non-repeating arrangement of particles, often described as supercooled liquids or pseudo-solids.

177. The matter which has definite shape is

• A. Solid
• B. Gas
• C. Liquid
• D. Plasma

Explanation: The correct answer is Solid because solids have a definite shape, unlike liquids, gases, and plasma. Solids maintain their shape due to the strong intermolecular forces between their particles, which are closely packed in a fixed arrangement. In contrast, gases and plasma do not have a definite shape or volume, and liquids have a definite volume but take on the shape of their container.

178. Which one of the following is BCC?

• A. Fe
• B. Cr
• C. Tungsten
• D. All of the above

Explanation: The correct answer is 'All of the above' because Iron (Fe), Chromium (Cr), and Tungsten all have a Body-Centred Cubic (BCC) crystal structure. BCC is one of the most common crystal structures found in metals, where each atom is at the centre of a cube formed by eight other atoms. Iron is BCC at room temperature but transitions to FCC at higher temperatures. Chromium and Tungsten maintain their BCC structure across a wide range of temperatures, contributing to their unique properties like hardness and high melting points.

179. Which one of the following is FCC?

• A. Zn
• B. Cadmium
• C. Magnesium
• D. All of these

Explanation: The face-centred cubic (FCC) structure is a type of crystal structure where atoms are located at each of the corners and the centres of all the cube faces of the unit cell. Common metals with an FCC structure include copper, aluminium, and nickel. However, Zinc, Cadmium, and Magnesium all crystallise in a hexagonal close-packed (HCP) structure, meaning none of them is FCC. Thus, the correct answer is 'All of these', as none of the options represent an FCC structure.

180. Which one of the following are amorphous?

• A. Ruby stone
• B. Glass
• C. Both a & b
• D. Charcoal

Explanation: The correct answer is Charcoal. Charcoal is an amorphous solid because it does not have a well-defined crystalline structure. Glass is also an amorphous solid; however, the option indicating 'Both a & b' is incorrect because ruby stone is crystalline. Ruby stone is a crystalline form of aluminium oxide. Therefore, option C is incorrect. Option B is incorrect because it does not include charcoal, which is also amorphous.