Free Physics of Solids MCQs with Answers

221. Two wires of some nature have L2=3L1 and D1 = 3D2 for some extension F1/F2

• A. 1 : 9
• B. 9 : 1
• C. 1 : 27
• D. 27 : 1

Explanation: The force ratio F1/F2 can be determined by considering the equation for the force required to cause a certain extension in a wire, which is proportional to its cross-sectional area and inversely proportional to its length. Given that L2 = 3L1 and D1 = 3D2, the cross-sectional area ratio A1/A2 is (D1/D2)2 = 9:1. Thus, the force ratio F1/F2 becomes (3L1L2) / (1/9) = 27:1. Therefore, option D is correct, as it accurately reflects these calculations. The other options do not correctly factorise the relationships between length, diameter, and force.

222. Breaking strength of a wire is 4 x 10^5 N if its diameter is halved, what is the strength?

• A. 4 x 10^5N
• B. 2 x 10^5N
• C. 1 x 10^5N
• D. None

Explanation: The breaking strength of a wire is directly proportional to its cross-sectional area. When the diameter of the wire is halved, the cross-sectional area is reduced by a factor of four (since the area is proportional to the square of the diameter). Therefore, the breaking strength also reduces by a factor of four, resulting in a new breaking strength of 1x105N. Option A is incorrect because it assumes no change in strength, and Option B is incorrect because it assumes the strength is halved, not quartered. Option D is incorrect as it suggests no correct option exists, which is false.

223. A spring is cut into 1:2, the spring constant of larger part will be.

• A. 2K
• B. 3K
• C. 3/2 K
• D. 1/2 K

Explanation: When a spring is cut into parts, the spring constant for each part changes based on its length. The spring constant (k) is inversely proportional to the length of the spring (L), meaning k ∝ 1/L. If a spring with spring constant K is cut into two parts in the ratio of 1:2, the larger part is twice the length of the smaller part. Therefore, the spring constant of the larger part becomes 3/2 K. Option A (2K) and Option D (1/2 K) are incorrect as they suggest the spring constant is directly proportional to length, which it is not. Option B (3K) is incorrect due to a miscalculation.

224. The ratio of the lengths of two wires A and B of the same material is 1 : 2 and the ratio of their diameter is 2 : 1. They are stretched by the same force, then the ratio of the increase in length will be

• A. 2 : 1
• B. 1 : 4
• C. 1 : 8
• D. 8 : 1

Explanation: To determine the ratio of increase in length for wires A and B, we use the formula for elongation: ΔL = (F*L)/(A*Y), where F is the force, L is the initial length, A is the cross-sectional area, and Y is the Young's modulus. Given that both wires are made of the same material and are subjected to the same force, their Young's modulus is constant and can be ignored when comparing elongations.For wire A, let its length be L, and its diameter be d. Thus, its cross-sectional area is A = π(d/2)2.For wire B, the length is 2L, and the diameter is d/2. Its cross-sectional area is A = π(d/4)2 = πd2/16.The ratio of increase in length for wires A and B is given by the ratio of their elongations: ΔLA/ΔLB = (F*L/AA)/(F*2L/AB) = (L/AA)/(2L/AB).Substituting for the areas, we get (1/π(d/2)2)/(2/π(d/4)2), which simplifies to (1/(d2/4))/(2/(d2/16)) = 16/8 = 1/8. Hence, the correct answer is 1 : 8.

225. The Young's modulus of a wire of length L and radius r is Y N/m2. If the length and radius are reduced to L/2 and r/2, then its Young's modulus will be

• A. Y/2
• B. Y
• C. 2Y
• D. 4Y

Explanation: The Young's modulus (Y) is an intrinsic property of a material that describes its stiffness, and it is independent of the dimensions, such as length and radius of the wire. Therefore, even if the length and radius of the wire are reduced to L/2 and r/2, respectively, the Young's modulus remains unchanged. It only depends on the material itself and not its size or shape. Option A is incorrect because reducing dimensions does not alter Young's modulus. Option C and Option D incorrectly suggest that Young's modulus increases with reduced dimensions, which is not the case. The correct answer is Option B, as Young's modulus remains Y.

226. Hook's law defines

• A. Stress
• B. Strain
• C. Modulus of elasticity
• D. Elastic limit

Explanation: Hooke's law states that the strain in a solid is proportional to the applied stress within the elastic limit of that solid. This relationship is characterised by the modulus of elasticity, which quantifies how much a material will deform under a certain amount of stress. Therefore, Hooke's law defines the modulus of elasticity. Stress and strain are related concepts within Hooke's law, but they are not what the law defines. The elastic limit is the boundary beyond which Hooke's law no longer applies.

227. If the temperature increases, the modulus of elasticity

• A. Decreases
• B. Increases
• C. Remains constant
• D. Becomes zero

Explanation: The modulus of elasticity, also known as Young's modulus, measures a material's resistance to deformation. As the temperature increases, atoms in the solid vibrate more intensely, weakening the intermolecular forces. This increased atomic movement reduces the stiffness of the material, hence decreasing the modulus of elasticity. This explains why Option A: Decreases is correct. Option B is incorrect because increased atomic vibrations typically reduce, not increase, the modulus. Option C is incorrect since the modulus changes with temperature. Option D is incorrect because the modulus doesn't become zero unless the material becomes a liquid.

228. A wire is loaded by 6 kg at its one end, the increase in length is 12mm. If the radius of the wire is doubled and all other magnitudes are unchanged, then increase in length will be

• A. 6 mm
• B. 3 mm
• C. 24 mm
• D. 48 mm

Explanation: The increase in length of a wire under a load is given by the formula ΔL = (F*L) / (A*Y), where F is the force, L is the original length, A is the cross-sectional area, and Y is Young's modulus. Doubling the radius of the wire increases the cross-sectional area by a factor of four (since A = πr²). This means the increase in length, ΔL, will be reduced to a quarter of its original value, from 12 mm to 3 mm. Thus, the correct answer is 3 mm. The other options are incorrect because they do not account for the change in cross-sectional area correctly.The increase in length (ΔL) of a wire under load is given by:ΔL = (FL) / (AY)where F is the force (load), L is the original length, A is the cross-sectional area, and Y is Young's modulus.Since the radius is doubled, the cross-sectional area (A = πr2) becomes 4 times the original area.Given:Original ΔL = 12 mmNew area (A') = 4ASince ΔL ∝ 1/A, the new increase in length (ΔL') will be:ΔL' = ΔL / 4= 12 mm / 4= 3 mmSo, the increase in length will be 3 mm.

229. In a wire of length L, the increase in its length is l. If the length is reduced to half, the increase in its length will be

• A. l
• B. 2l
• C. l/2
• D. None of the above

Explanation: When the length of the wire is halved, the increase in its length (strain) under the same force will also be halved. Strain is proportional to the original length, so reducing the length by half results in a proportional reduction in the extension. Thus, the increase in length becomes l/2. Option A is incorrect as it does not account for the change in original length. Option B is incorrect because it incorrectly suggests the increase would double rather than halve. Option D is incorrect because Option C accurately describes the situation.ΔL = (FL) / (AY)Given:Original ΔL = l, original length = LIf the length is reduced to half (L' = L/2), and other factors remain constant:ΔL' = (F(L/2)) / (AY)= (1/2) × (FL) / (AY)= (1/2) × lSo, the increase in length will be l/2.

230. If the length of a wire is reduced to half, then it can hold the

• A. Half load
• B. Same load
• C. Double load
• D. One fourth

Explanation: The load a wire can support is directly related to its cross-sectional area. When the length of a wire is reduced by half, and assuming constant volume, its cross-sectional area doubles. This is because the volume of the wire (which remains constant) is equal to the cross-sectional area multiplied by its length. Since the cross-sectional area is inversely proportional to the length when volume is constant, halving the length doubles the cross-sectional area, allowing the wire to hold double the load. Option A is incorrect because the wire's increased cross-sectional area allows it to hold more than half the load. Option B is incorrect because the wire can actually hold more than the same load due to the increased cross-sectional area. Option D is incorrect as the load capacity increases rather than decreases to one-fourth.