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JEE • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 265 questions

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31 Medium
31 of 265
Application +4 | -1
Forces of 100 N each are applied in opposite directions on the upper and lower faces of a cube of side 20 cm20\,cm. The upper face is shifted parallel to itself by 0.25 cm0.25\,cm. If the side of the cube were 10 cm, then the displacement would be
View answer and explanation
Correct option: B
0.5 cm0.5\,cm
Option B (0.5 cm0.5\,cm) is correct because the displacement is inversely proportional to the side of the cube, so halving the side doubles the displacement.
32 Medium
32 of 265
Remembering +4 | -1
Thermal stress, If developed in a solid on heating is given by
View answer and explanation
Correct option: A
yΔlly\frac{{\Delta l}}{l}
Option A (yΔlly\frac{{\Delta l}}{l}) is correct because thermal stress is related to Young's modulus and strain.
33 Medium
33 of 265
Remembering +4 | -1
Dimensions of stress are
View answer and explanation
Correct option: D
[ M L−1 T−2 ]\mathrm{[}\,\mathrm{M}\,{\mathrm{L}^{ - 1}}\,{\mathrm{T}^{ - 2}}\,]
Option D ([ M L−1 T−2 ]\mathrm{[}\,\mathrm{M}\,{\mathrm{L}^{ - 1}}\,{\mathrm{T}^{ - 2}}\,\mathrm{]}) is correct because stress is force per unit area, and its dimensions are derived accordingly.
34 Medium
34 of 265
Analyzing +4 | -1
Statement I: A steel wire of natural length 'ℓ\ell' breaks, when clongated by length 'xx'. If original length is halved and elongated by x/2x / 2 it breaks.
Statement II: Breaking force is directly proportional to the length of the wire.
View answer and explanation
Correct option: B
Statement I is true and statement II is false
Option D (Statement I is true and statement II is false) is correct because breaking force depends on the area of cross-section, not the length of the wire.
35 Medium
35 of 265
Application +4 | -1
A uniform wire, fixed at its upper end, hangs vertically and supports a weight at its lower end. If its radius is rr, its length LL and the Young’s modulus for the material of the wire is EE, the extension is
1. directly proportional to EE

2. inversely proportional to rr

3. directly proportional to LL
View answer and explanation
Correct option: A
If only 33 is correct
Option A (If only 33 is correct) is correct because the extension is directly proportional to the length of the wire, as shown by the formula.
36 Medium
36 of 265
Understanding +4 | -1
Assertion: Steel is more elastic than rubber.
Reason: Under given deforming force, steel is deformed less than rubber.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion
Option A (If both assertion and reason are true and reason is the correct explanation of the assertion) is correct because steel's higher elasticity means it deforms less under the same force, making both the assertion and reason true and linked.
37 Medium
37 of 265
Understanding +4 | -1
Assertion: Rise in temperature of a material causes decrease in the elasticity of the material.
Reason: The presence of impurities always increases the elasticity of the material.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D is correct because both the assertion and the reason are false. Temperature increase usually decreases elasticity, but not always, and impurities can increase or decrease elasticity.
38 Easy
38 of 265
Understanding +4 | -1
The elastic potential energy of a stretched wire is given by
View answer and explanation
Correct option: B
U=AY2Ll2U=\frac{A Y}{2 L} l^2
Option B is correct because the elastic potential energy stored in a stretched wire is given by the formula U = (AY/2L)l2, where A is the cross-sectional area, Y is Young's modulus, L is the original length, and l is the extension.
39 Easy
39 of 265
Application +4 | -1
A wire extends by 1mm\mathrm{1 mm} when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be
View answer and explanation
Correct option: A
8\mathrm{8}
Option A (8) is correct because the elongation is inversely proportional to the area (or square of the radius) and directly proportional to the force. Doubling the force and halving the radius results in 8 times the elongation.
40 Medium
40 of 265
Application +4 | -1
A stress of 3.18  ×108Nm−23.18\,\, \times {10^8}N{m^{ - 2}} is applied to a steel rod of length 1  m1\,\,m along its length. Its Young’s modulus is 2  ×  1011Nm−22\,\, \times \,\,{10^{11}}N{m^{ - 2}}. Then the elongation produced in the rod in mmmm is
View answer and explanation
Correct option: D
1.59
Option D (1.59) is correct because it is the elongation calculated using Young's modulus formula with the given stress and length.
41 Easy
41 of 265
Understanding +4 | -1
According to Hooke’s law of elasticity, if stress is increased, them the ratio of stress to strain
View answer and explanation
Correct option: B
Remains constant
Option B (Remains constant) is correct because Hooke's law states that the ratio of stress to strain is constant and equal to the modulus of elasticity.
42 Medium
42 of 265
Application +4 | -1
An elevator cable is to have a maximum stress of 7×107Nm−27 \times 10^7 \mathrm{Nm}^{-2} to allow for appropriate safety factors. Its maximum upward acceleration is 1.5 ms−21.5 \mathrm{~ms}^{-2}. If the cable has to support the total weight of 2000 kg2000 \mathrm{~kg} of a loaded elevator, the area of cross-section of the cable should be
View answer and explanation
Correct option: A
3.23 cm23.23 \mathrm{~cm}^2
Option A (3.23 cm²) is correct because it is the minimum area required to support the load with the given maximum stress and acceleration.
43 Medium
43 of 265
Understanding +4 | -1
If the load is increased beyond the _____ point, the strain increases rapidly for even a small change in the stress.
View answer and explanation
Correct option: B
Yield point
Option B (Yield point) is correct because beyond the yield point, a small change in stress leads to a large increase in strain.
44 Medium
44 of 265
Understanding +4 | -1
A: Lateral strain is directly proportional to the longitudinal strain within the elastic limit.
R: Poisson's ratio for a given material at a constant temperature is constant.
View answer and explanation
Correct option: A
If both A &\& R are true and the R is the correct explanation of the A
Option A (If both A &R\& R are true and the R is the correct explanation of the A) is correct because Poisson's ratio defines the relationship between lateral and longitudinal strain, making the reason a correct explanation for the assertion.
45 Medium
45 of 265
Understanding +4 | -1
In a wire stretched by hanging a weight from its end, the elastic potential energy per unit volume in terms of longitudinal strain σ\sigma and modulus of elasticity YY is
View answer and explanation
Correct option: A
Yσ22\frac{{Y{\sigma ^2}}}{2}
Option A (Yσ22\frac{{Y{\sigma ^2}}}{2}) is correct because the elastic potential energy per unit volume is given by half the product of stress and strain, which simplifies to the given expression.
46 Medium
46 of 265
Remembering +4 | -1
Which of the following relation is true?
View answer and explanation
Correct option: D
(1+σ)2η=Y\left( {1 + \sigma } \right)2\eta = Y
Option D ((1 + σ)2η = Y) is correct because it represents the correct relationship between Young's modulus (Y), shear modulus (η), and Poisson's ratio (σ).
47 Medium
47 of 265
Application +4 | -1
A solid sphere of radius 10 cm10 \mathrm{~cm} is subjected to a uniform pressure =5×108 Nm−2=5 \times 10^8 \mathrm{~Nm}^{-2}. Determine the consequent change in volume. Bulk modulus of the material of the sphere is equal to 3.14×1011 Nm−23.14 \times 10^{11} \mathrm{~Nm}^{-2}.
View answer and explanation
Correct option: A
6.67×10−6 m36.67 \times 10^{-6} \mathrm{~m}^3
Option A (6.67×10−6 m36.67 \times 10^{-6} \mathrm{~m}^3) is correct because it is the change in volume calculated using the bulk modulus formula.
48 Medium
48 of 265
Remembering +4 | -1
In which year Robert Hooke presented his law of elasticity?
View answer and explanation
Correct option: C
1676
Option C (1676) is correct because Robert Hooke presented his law of elasticity in 1676.
49 Medium
49 of 265
Understanding +4 | -1
Shape is related to:
View answer and explanation
Correct option: A
Shear stress
Option A is correct because shear stress is the force acting tangentially on a surface, causing deformation in shape.
50 Medium
50 of 265
Understanding +4 | -1
Assertion: Rigid body can be elastic.
Reason: If a force is applied on the rigid body, its dimension may change.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because a rigid body, by definition, cannot be elastic, and its dimensions do not change under the application of force.
51 Easy
51 of 265
Application +4 | -1
If to break a wire of 1 m1 \mathrm{~m} length, minimum 40 kg40 \mathrm{~kg}-wt is required, then to break a wire of same material 6 m6 \mathrm{~m} in length and double in radius, the breaking weight required will be
View answer and explanation
Correct option: C
160 kg160 \mathrm{~kg}-wt
Option C (160 kg160 \mathrm{~kg}-wt) is correct because the breaking weight is proportional to the cross-sectional area of the wire, which depends on the square of the radius.
52 Medium
52 of 265
Application +4 | -1
The elastic limit of brass is 379MPa379MPa. What should be the minimum diameter of a brass rod, if it is to support a 400N400N load without exceeding its elastic limit
View answer and explanation
Correct option: C
1.16  mm1.16\,\,mm
Option C (1.16 mm) is correct because it calculates the minimum diameter required to support the load without exceeding the elastic limit.
53 Medium
53 of 265
Understanding +4 | -1
Statement-I: Young's modulus, bulk modulus and shear modulus have the units of pressure.
Statement-II: The value of Young's modulus depends area of body.
View answer and explanation
Correct option: B
Statement I is true and statement II is false
Option B is correct because Young's modulus, bulk modulus, and shear modulus all have units of pressure, but Young's modulus is a material property and does not depend on the area of the body.
54 Medium
54 of 265
Remembering +4 | -1
Within proportional limit, the slope of the stress stain curve yields,
View answer and explanation
Correct option: B
Modulus of elasticity
Option B (Modulus of elasticity) is correct because the slope of the stress-strain curve within the proportional limit defines the modulus of elasticity.
55 Medium
55 of 265
Understanding +4 | -1
The Young’s modulus of a wire of length LL and radius rr is YN/m2Y\mathrm{ }N/{m^2}. If the length and radius are reduced to L/2L/2 and r/2r/2, then its Young’s modulus will be
View answer and explanation
Correct option: B
YY
Option B (YY) is correct because Young's modulus is a material property and is independent of the dimensions of the wire.
56 Medium
56 of 265
Remembering +4 | -1
Which of the following substances has the highest elasticity?
View answer and explanation
Correct option: B
Steel
Option B (Steel) is correct because steel has the highest Young's modulus among the given materials, indicating greater resistance to deformation and thus higher elasticity.
57 Medium
57 of 265
Analyzing +4 | -1
Statement I: Two identical solid balls, one of ivory and the other of wet-clay are dropped from the same height on the floor. Both the balls will rise to same height after bouncing.
Statement II: Ivory and wet-clay have same elasticity.
View answer and explanation
Correct option: D
Statement I is false and statement II is false
Option D (Statement I is false and statement II is false) is correct because wet-clay is nearly inelastic and will not rebound to the same height, and ivory and wet-clay do not have the same elasticity.
58 Medium
58 of 265
Understanding +4 | -1
A stretched rubber has
View answer and explanation
Correct option: B
Increased potential energy
Option B (Increased potential energy) is correct because stretching a rubber band stores potential energy within it due to the elastic deformation.
59 Medium
59 of 265
Understanding +4 | -1
Statement I: Silver is a ductile material,
Statement II: For a ductile material yield point and breaking point are separated by larger distance than for brittle materials on the stress-strain curve.
View answer and explanation
Correct option: A
Statement I is true and statement II is true
Option A is correct because silver is indeed ductile, and ductile materials have a larger separation between their yield and breaking points on the stress-strain curve compared to brittle materials.
60 Medium
60 of 265
Understanding +4 | -1
Bulk modulus of a material, whose Poison's ratio equal to 0.5 is
View answer and explanation
Correct option: B
Infinity
Option B (Infinity) is correct because a Poisson's ratio of 0.5 indicates a perfectly incompressible material, which has an infinite bulk modulus.
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