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JEE • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Mixed difficulty

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Practice format and chapter quiz

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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181 Medium
181 of 265
Application +4 | -1
Two steel wires have equal volumes, their diameters are in the ratio 2:12:1. When same force is applied on them, the elongation produced will be in the ratio of
View answer and explanation
Correct option: C
1:161:16
Option C is correct because the elongation is inversely proportional to the fourth power of the radius when the volume and force are constant.
182 Medium
182 of 265
Understanding +4 | -1
Breaking stress of a steel wire of radius 1 mm1 \mathrm{~mm} and length 1 m1 \mathrm{~m} is 1012 N/m210^{12} \mathrm{~N} / \mathrm{m}^2. Breaking stress of another steel wire of radius 2 mm2 \mathrm{~mm} and length 0.5 m0.5 \mathrm{~m} will be
View answer and explanation
Correct option: B
1012 N/m210^{12} \mathrm{~N} / \mathrm{m}^2
1012 N/m210^{12} \mathrm{~N} / \mathrm{m}^2 is the correct answer because breaking stress is a material property and is independent of the dimensions of the wire.
183 Medium
183 of 265
Application +4 | -1
The ratio of two specific heats of gas Cp/Cv{C_p}/{C_v} for argon is 1.6\mathrm{1}\mathrm{.6} and for hydrogen is 1.4\mathrm{1}\mathrm{.4}. Adiabatic elasticity of argon at pressure PP is EE. Adiabatic elasticity of hydrogen will also be equal to EE at the pressure
View answer and explanation
Correct option: B
87P\frac{8}{7}P
Option B (87P\frac{8}{7}P) is correct because it equates the adiabatic elasticity of argon and hydrogen, using the given ratio of specific heats to find the pressure at which their elasticities are equal.
184 Medium
184 of 265
Understanding +4 | -1
A: When a beam is bent only tensile strain is produced.
R: The depression produced in a rectangular beam is directly proportional to its width.
View answer and explanation
Correct option: D
If both A and R are false
Option D (If both A and R are false) is correct because bending a beam produces both tensile and compressive strain, and the depression is inversely proportional to the width.
185 Medium
185 of 265
Remembering +4 | -1
Under elastic limit the stress is
View answer and explanation
Correct option: B
Directly proportional to strain
Option B (Directly proportional to strain) is correct because Hooke's Law states that stress is directly proportional to strain within the elastic limit.
186 Medium
186 of 265
Application +4 | -1
A copper bar of length LL and area of cross-section AA is placed in a chamber at atmospheric pressure. If the chamber is evacuated, the percentage change in its volume will be (compressibility of copper is 8  × 1012 m 2N−1and  1  atm=105  Nm2)8\,\, \times \,{10^{12}}\,\mathrm{m}{\,^2}{\mathrm{N}^{ - 1}}\mathrm{and}\,\,1\,\,\mathrm{atm} = {10^5}\,\,\mathrm{N}{\mathrm{m}^2})
View answer and explanation
Correct option: B
8  ×  10−58\,\, \times \,\,{10^{ - 5}}
Option B (8×10−58 \times 10^{-5}) is correct because it represents the percentage change in volume calculated using the compressibility of copper and the change in pressure.
187 Medium
187 of 265
Understanding +4 | -1
What is the nature of intermolecular forces?
View answer and explanation
Correct option: C
Attractive and repulsive both depend on the intermolecular separation
The correct answer is C (Attractive and repulsive both depend on the intermolecular separation) because intermolecular forces can be attractive at larger distances and repulsive at very short distances.
188 Medium
188 of 265
Application +4 | -1
KK is the force constant of a spring. The work done in increasing its extension from l1{l_1} to l2{l_2} will be
View answer and explanation
Correct option: D
K2(l22−l12)\frac{K}{2}\left( {l_2^2 - l_1^2} \right)
Option D (K2(l22−l12)\frac{K}{2}\left( {l_2^2 - l_1^2} \right)) is correct because it represents the difference in potential energy stored in the spring at the two extensions.
189 Medium
189 of 265
Remembering +4 | -1
The value of Poisson’s ratio lies between
View answer and explanation
Correct option: A
−1  to  12- 1\,\,\mathrm{to}\,\,\frac{1}{2}
Option A (−1,,to 12- 1\\,\\,\mathrm{to}\\\,\frac{1}{2}) is correct because Poisson's ratio is theoretically limited to this range due to thermodynamic considerations.
190 Medium
190 of 265
Understanding +4 | -1
Assertion: The stretching of a coil is determined by its Shear modulus.
Reason: Shear modulus changes only shape of a body keeping its dimensions unchanged.
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 the stretching of a coil primarily involves changing its shape without significantly altering the length of the wire, which is related to the shear modulus.
191 Medium
191 of 265
Application +4 | -1
Equal torsional torques act on two rods xandyx\mathrm{ }and\mathrm{ }y having equal length. The diameter of rod yy is twice the diameter of rod x.   If  θx   and   θyx.{\mkern 1mu} \,{\mkern 1mu} \mathrm{If}\,\,{\theta _x}{\mkern 1mu} {\mkern 1mu} \,and{\mkern 1mu} \,{\mkern 1mu} {\theta _y} are the angles of twist, then θxθy=\frac{{{\mathrm{\theta }_x}}}{{{\mathrm{\theta }_y}}} =
View answer and explanation
Correct option: D
16
Option D (16) is correct because the angle of twist is inversely proportional to the fourth power of the diameter, so doubling the diameter reduces the twist by a factor of 16.
192 Medium
192 of 265
Application +4 | -1
The upper end of a wire of diameter 12 mm12 \mathrm{~mm} and length 1 m1 \mathrm{~m} is clamped and its other end is twisted through an angle of 30∘30^{\circ}. The angle of shear is
View answer and explanation
Correct option: B
0.18∘0.18^{\circ}
Option B (0.18∘0.18^{\circ}) is correct because it is calculated using the formula relating the angle of twist, radius, length, and angle of shear.
193 Medium
193 of 265
Application +4 | -1
The upper end of a wire of radius 4  mm\mathrm{4}\,\,\mathrm{mm} and length 100  cm\mathrm{100}\,\,\mathrm{cm} is clamped and its other end is twisted through and angle of 300{30^0}. Then angle of shear is
View answer and explanation
Correct option: B
0.12∘0.12^\circ
Option B (0.12∘0.12^\circ) is correct because the angle of shear is calculated using the formula ϕ=rlθ\phi = \frac{r}{l}\theta, where r is the radius, l is the length, and θ\theta is the angle of twist.
194 Medium
194 of 265
Application +4 | -1
The bulk modulus of a spherical object is ′B′.'B'.If it is subjected to uniform pressure ′p′,'p',the fractional decreases in radius is
View answer and explanation
Correct option: A
p3B\frac{p}{{3B}}
Option A (p3B\frac{p}{{3B}}) is correct because it is derived from the definition of bulk modulus and the relationship between volume and radius for a sphere.
195 Medium
195 of 265
Application +4 | -1
If a bar is made of copper whose coefficient of linear expansion is one and a half times that of iron, the ratio of the force developed in the copper bar to the iron bar of identical lengths and cross-sections, when heated through the same temperature range (Young’s modulus for copper may be taken equal to that of iron) is
View answer and explanation
Correct option: A
32\frac{3}{2}
Option A (32\frac{3}{2}) is correct because the force developed is proportional to the coefficient of linear expansion, and the coefficient for copper is 1.5 times that of iron.
196 Medium
196 of 265
Application +4 | -1
A wire of length L\mathrm{L}, area of cross section A\mathrm{A} is hanging from a fixed support. The length of the wire changes to L1{\mathrm{L}_\mathrm{1}} when mass M\mathrm{M} is suspended from its free end. The expression for Young’s modulus is:
View answer and explanation
Correct option: D
MgLA (L1−L)\frac{{\mathrm{MgL}}}{{\mathrm{A}\,\mathrm{(}{\mathrm{L}_\mathrm{1}} - \mathrm{L)}}}
Option D (MgLA (L1−L)\frac{{\mathrm{MgL}}}{{\mathrm{A}\,\mathrm{(}{\mathrm{L}_\mathrm{1}} - \mathrm{L)}}}) is correct because it is the correct formula for Young's modulus, which is stress divided by strain.
197 Medium
197 of 265
Understanding +4 | -1
The general form of potential energy curve for atoms or molecules can be represented by the following equation U(R)=ARn−BRmU\left( R \right) = \frac{A}{{{R^n}}} - \frac{B}{{{R^m}}} . Here, R is the interatomic or molecular distance, A and B are coefficients, nn and mm are the exponents. In the above equation
View answer and explanation
Correct option: B
Second term represents the attractive part of the potential
Option B (Second term represents the attractive part of the potential) is correct because the negative sign in front of the second term indicates an attractive potential, as the potential energy decreases with increasing distance.
198 Medium
198 of 265
Analyzing +4 | -1
Which of the following relations is true
View answer and explanation
Correct option: D
σ=0.5Y−ηη\sigma = \frac{{0.5Y - \eta }}{\eta }
Option D is correct because the correct relation between Young's modulus (Y), shear modulus (η\eta), bulk modulus (K), and Poisson's ratio (σ\sigma) is Y=2η(1+σ)Y = 2\eta(1 + \sigma) which can be rearranged to σ=0.5Y−ηη\sigma = \frac{{0.5Y - \eta}}{\eta}.
199 Medium
199 of 265
Application +4 | -1
Find the density of the metal under a pressure of 20,000 N cm−220,000 \mathrm{~N} \mathrm{~cm}^{-2}. Given density of the metal =11 g cm−3=11 \mathrm{~g} \mathrm{~cm}^{-3}, bulk modulus of the metal =8×109 Nm−2=8 \times 10^9 \mathrm{~Nm}^{-2}.
View answer and explanation
Correct option: A
11.28 g cm−311.28 \mathrm{~g} \mathrm{~cm}^{-3}
Option A (11.28 g cm−311.28 \mathrm{~g} \mathrm{~cm}^{-3}) is correct because it is calculated by considering the bulk modulus and the pressure applied to the metal.
200 Medium
200 of 265
Understanding +4 | -1
If the tension on a wire is removed at once, then:
View answer and explanation
Correct option: D
Its temperature increases
Option D (Its temperature increases) is correct because the potential energy stored in the wire due to tension is converted into heat when the tension is released.
201 Medium
201 of 265
Application +4 | -1
A wire of length 4 m4 \mathrm{~m} is hanged vertically from ceiling if it is stretched due to its own weight, then find out the energy stored per unit volume. [Y=2×107 N/m2\left[Y=2 \times 10^7 \mathrm{~N} / \mathrm{m}^2\right. and r=103r=10^3 kg/m3\mathrm{kg} / \mathrm{m}^3 ]
View answer and explanation
Correct option: D
10 J/m310 \mathrm{~J} / \mathrm{m}^3
Option D (10 J/m³) is correct because it is calculated using the formula for potential energy density in terms of Young's modulus, density, gravity, and length.
202 Medium
202 of 265
Application +4 | -1
When change in volume of a cube under a pressure at atm of 10 is 1% ; compressibility of material of black in m2/N{m^2}/N is (Take 1atm =105Pa= {10^5}Pa)
View answer and explanation
Correct option: B
10−8{10^{ - 8}}
Option B (10−8{10^{ - 8}}) is correct because compressibility is the reciprocal of the bulk modulus, which relates pressure to the fractional change in volume.
203 Medium
203 of 265
Application +4 | -1
Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is 0.5  cm0.5\,\,cm, the elongation (l)\left( l \right) of each wire is Ys(steel)=2.0×1011N/m2{Y_s}\left( {\mathrm{steel}} \right) = 2.0 \times {10^{11}}N/{m^2}, Yc(copper)=1.2×1011N/m2{Y_c}\left( {\mathrm{copper}} \right) = 1.2 \times {10^{11}}N/{m^2}
View answer and explanation
Correct option: A
ls=0.75  cm,lc=1.25  cm{l_s} = 0.75\,\,cm,{l_c} = 1.25\,\,cm
Option A (ls=0.75  cm,lc=1.25  cm{l_s} = 0.75\,\,cm,{l_c} = 1.25\,\,cm) is correct because it satisfies both the elongation ratio based on Young's modulus and the given difference in elongations.
204 Easy
204 of 265
Understanding +4 | -1
A long piece of rubber is wider than it is thick. When is is stretched in length by some amount:
View answer and explanation
Correct option: D
Both its thickness and width decrease
Option D is correct because when a rubber piece is stretched, both its thickness and width decrease due to the Poisson effect.
205 Medium
205 of 265
Analyzing +4 | -1
Assertion: If length of a rod is doubled, the breaking load remains the same.
Reason: Breaking load is equal to the elastic limit.
View answer and explanation
Correct option: C
If assertion is true but reason is false
The correct answer is C (If assertion is true but reason is false) because breaking load depends on the cross-sectional area, not the length, and it's related to breaking stress, not the elastic limit.
206 Medium
206 of 265
Understanding +4 | -1
Young’s modulus of perfectly rigid body material is
View answer and explanation
Correct option: A
Infinite
Option A (Infinite) is correct because a perfectly rigid body experiences no deformation under stress, resulting in an infinite Young's modulus.
207 Medium
207 of 265
Application +4 | -1
A wire of Young’s modulus 1.5×1012  Nm−21.5 \times {10^{12}}\;\mathrm{N}{\mathrm{m}^{ - 2}} is stretched by a force so as to produce a strain of 2×104.2 \times {10^4}. The energy stored per unit volume is
View answer and explanation
Correct option: D
3  ×104  Jm−33\mathrm{\;} \times {10^4}\mathrm{\;J}{\mathrm{m}^{ - 3}}
Option D (3  ×104  Jm−33\mathrm{\;} \times {10^4}\mathrm{\;J}{\mathrm{m}^{ - 3}}) is correct because it is the energy stored per unit volume calculated using the given Young's modulus and strain.
208 Medium
208 of 265
Remembering +4 | -1
Which is the most elastic
View answer and explanation
Correct option: C
Quartz
Option C (Quartz) is correct because quartz exhibits a higher resistance to deformation under stress compared to the other options, making it more elastic.
209 Medium
209 of 265
Application +4 | -1
If longitudinal strain for a wire is 0.03\mathrm{0}\mathrm{.03} and its Poisson’s ratio is 0.5\mathrm{0}\mathrm{.5}, then its lateral strain is
View answer and explanation
Correct option: C
0.015
Option C (0.015) is correct because lateral strain is calculated by multiplying Poisson's ratio by longitudinal strain.
210 Medium
210 of 265
Application +4 | -1
If a spring extends by xx on loading, then the energy stored by the spring is (if TT is tension in the spring and kk is spring constant)
View answer and explanation
Correct option: B
T22k\frac{{{T^2}}}{{2k}}
Option B (T22k\frac{{{T^2}}}{{2k}}) is correct because the energy stored in a spring is related to the tension and spring constant by this formula.
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