Free Physics of Solids MCQs with Answers

51. The ultimate strength of a sample is the stress at which the sample:

• A. Remains underwater
• B. Returns to its original shape when the stress is removed
• C. Bends at 180 degrees
• D. Breaks

Explanation: The ultimate strength of a sample is the maximum stress that the sample can withstand before it breaks or fractures.

52. Very weak magnetic fields are detected by

• A. Squids
• B. MRI
• C. Magnetometer
• D. Oscilloscope

Explanation: Superconducting Quantum Interference Devices (SQUIDs) are highly sensitive detectors used to measure very weak magnetic fields, often at the level of a few femto teslas

53. In a simple cube, one atom or molecule lies at its

• A. Four corners
• B. Nine corners
• C. Eight corners
• D. Six corners

Explanation: a simple cube structure, each corner is shared by eight adjacent cubes. Therefore, one-eighth of each corner atom belongs to the unit cell, resulting in eight corner atoms for a unit cell.

54. The bulk modulus is a proportionality constant that relates the pressure acting on an object to

• A. The fractional change in volume
• B. The shear
• C. The fractional change in length
• D. The spring constant

Explanation: The bulk modulus (K) is a measure of the resistance of a material to uniform compression. It relates the pressure acting on an object to the fractional change in volume of the object.

55. The Young's modulus for a perfectly rigid body is

• A. zero
• B. infinite
• C. 1
• D. none of these

Explanation: Young's modulus (Y) is a measure of the stiffness of a material. For a perfectly rigid body, which does not deform under any stress, the Young's modulus is considered to be infinite because it experiences no strain regardless of the applied stress

56. The force required to stretch a steel wire 1 sq. cm in cross section to double its length is (given y=2×10^11N/m2)

• A. 10^7 N
• B. 10^11N
• C. 2 × 10^7 N
• D. 2 x 10^11 N

Explanation: The force (F) required to stretch a wire can be calculated using Hooke's Law: F=Y⋅A⋅ΔL/L , where:Y is the Young's modulus,A is the cross-sectional area,ΔL is the change in length,L is the original length.Given that the cross-sectional area remains constant, we have A=1cm^2.The change in length (ΔL) is double the original length, so ΔL=2L.Substituting the values into Hooke's Law, we get: F =2×10^7N

57. If the work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of the same material but double the radius and half the length by 1 mm is

• A. 16 J
• B. 4 J
• C. 8 J
• D. 1/4 J

Explanation: Work Done in Stretching a Wire:The work done in stretching a wire depends on two factors:Force (F): The force applied to stretch the wire.Distance (d): The amount by which the wire is stretched.We can express the work done (W) as the product of force and distance:W = F * dRelationship Between Force and Cross-Sectional Area:For a wire of a given material, the force required to stretch it is proportional to the cross-sectional area of the wire. This is because a wider wire can distribute the stretching force over a larger area, making it more resistant to stretching.Impact of Doubling the Radius:If the radius of the wire is doubled, the cross-sectional area increases by a factor of 4 (since area is proportional to the square of the radius). This means the force required to stretch the wire also increases by a factor of 4.Impact of Halving the Length:If the length of the wire is halved, the distance it needs to be stretched to achieve the same 1 mm elongation is also halved.Combining the Effects:When we combine the effects of doubling the radius and halving the length:The force required to stretch the wire increases by 4.The distance the wire needs to be stretched is halved.Overall Work Done:Since the force increases by 4 and the distance is halved, the overall work done (which is the product of force and distance) remains the same:New work = (4 * original force) * (0.5 * original distance) = original workTherefore, the work done in stretching the second wire (with double the radius and half the length) by 1 mm is also 2 J, which is the same as the work done in stretching the first wire.

58. A copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied which stretches their combined length by 1 cm. Then the two wires have

• A. the same stress and strain
• B. the same strain but different stresses
• C. the same stress but different strain
• D. different stresses and strains

Explanation: Since the wires are connected end to end and are subjected to the same force, they experience the same change in length. Therefore, they have the same strain. However, since the two wires are made of different materials (copper and steel), they have different Young's moduli. Thus, they experience different stresses for the same strain

59. The maximum stress which a body can bear is called its

• A. ultimate tensile strength
• B. permanent stress
• C. elastic stress
• D. all type of stress

Explanation: The ultimate tensile strength (UTS) is the maximum stress that a material can withstand while being stretched or pulled before necking, which leads to fracture. It is an important property of materials, especially in engineering and material sciences, as it indicates the maximum load that a material can bear under tension

60. Such substances which break soon after they cross the elastic limit is called

• A. weak substances
• B. brittle substances
• C. ductile substances
• D. polymeric substances

Explanation: Brittle substances are those materials that break or fracture soon after they exceed their elastic limit. They lack plastic deformation and undergo very little or no elongation before fracturing. Examples of brittle substances include ceramics, certain types of glass, and some metals under certain conditions