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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 211-240 of 265 questions
211 Medium
211 of 265
Application +4 | -1
A 0.10.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m1\mathrm{ }m and its cross-sectional area is 4.9×10−7m2.4.9 \times {10^{ - 7}}{\mathrm{m}^{2.}} If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140  rads−1140\,\,\mathrm{rad}{\mathrm{s}^{ - 1}}. If the Young’s modulus of the material of the wire is n×109 Nm−2n \times {10^9}\,\mathrm{N}{\mathrm{m}^{ - 2}}, the value of nn is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the Young's modulus is calculated using the formula for the angular frequency of SHM in a wire, and solving for n.
212 Medium
212 of 265
Understanding +4 | -1
What happens when the applied load increases and upto breaking stress in the experiment to determine the Young's modulus of elasticity?
View answer and explanation
Correct option: A
The area of wire goes on decreasing and wire extends and breaks.
Option A is correct because as the load increases, the wire stretches (extends), and its cross-sectional area decreases until it reaches the breaking stress and fractures.
213 Medium
213 of 265
Understanding +4 | -1
A: Strain is a dimensionless quantity.
R: Strain is internal force per unit area of a body
View answer and explanation
Correct option: C
If A is true but R is false
Option C is correct because strain is dimensionless (change in dimension/original dimension), and the reason is false as strain is not internal force per unit area.
214 Medium
214 of 265
Understanding +4 | -1
In Searle's apparatus, used for study of extension of wires verses load, we have two wires. During the experiment, we study the extension in one wire. The use of second wire is
View answer and explanation
Correct option: B
To compensate for the change in length caused by changes in temperature of atmosphere during experimentation.
Option B is correct. The second wire in Searle's apparatus compensates for length changes due to temperature variations during the experiment.
215 Medium
215 of 265
Understanding +4 | -1
If G={G}= shear modulus and Y=Y= Young's modulus, then in general:
View answer and explanation
Correct option: A
Y>G{Y} > {G}
Option A (Y>G{Y} > {G}) is correct because Young's modulus (Y) typically represents resistance to tensile stress, while shear modulus (G) represents resistance to shear stress, and Y is generally larger than G for most materials.
216 Medium
216 of 265
Understanding +4 | -1
Statement I: With increase of temperature, elastic property of a substance decreases.
Statement II: Elasticity is due to intermolecular forces which decreases with the increase of inter molecular distance.
View answer and explanation
Correct option: A
Statement I is true and statement II is true
Option A (Statement I is true and statement II is true) is correct because increased temperature weakens intermolecular forces, thus decreasing elasticity.
217 Medium
217 of 265
Application +4 | -1
A 5 m long aluminium wire Y=7×1010  Nm−2  Y = 7 \times {10^{10}}\;\mathrm{N}{\mathrm{m}^{ - 2\;}} of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in the copper wire (Y=12×102Nm  −2)  \left( {Y = 12 \times {{10}^2}\mathrm{Nm}{\mathrm{\;}^{ - 2}}} \right)\; of the same length under the same weight, the diameter should now be (in mm)
View answer and explanation
Correct option: C
2.3
Option C (2.3) is correct because the diameter is inversely proportional to the square root of Young's modulus for the same elongation, length, and weight. Calculating the new diameter using the given Young's moduli results in approximately 2.3 mm.
218 Medium
218 of 265
Understanding +4 | -1
An elastic metal rod will change its length when it
View answer and explanation
Correct option: D
Both (2) and (3)
Option D (Both (2) and (3)) is correct because stress is developed in both cases when the metal rod is pulled or stretched.
219 Medium
219 of 265
Analyzing +4 | -1
A: It is the breaking stress and not the breaking strength which depends on the material
R: Breaking strength =Breaking stressArea=\frac{\text {Breaking stress}}{\text {Area}}
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 both the assertion and the reason are incorrect statements about breaking stress and breaking strength.
220 Medium
220 of 265
Understanding +4 | -1
A: Adiabatic elasticity of a gas is greater than isothermal elasticity
R: EadiabaticEisothermal =γ\frac{E_{\text {adiabatic}}}{E_{\text {isothermal }}}=\gamma
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 is correct. Adiabatic elasticity is greater than isothermal elasticity because the ratio is \γ\gamma, which is always greater than 1, and the reason correctly explains the assertion.
221 Medium
221 of 265
Application +4 | -1
The breaking stress of a wire is 6×108 N/m26 \times 10^8 \mathrm{~N} / \mathrm{m}^2 and density is 3×103Kg/m33 \times 10^3 \mathrm{Kg} / \mathrm{m}^3. What should be the maximum length of the wire so that it may not break under its own weight?
View answer and explanation
Correct option: A
20 km20 \mathrm{~km}
Option A (20 km) is correct because it is the maximum length calculated using the breaking stress, density, and gravity.
222 Medium
222 of 265
Understanding +4 | -1
A rod elongated by ll when a body of mass MM is suspended from it. The work done is
View answer and explanation
Correct option: B
12Mgl\frac{1}{2}M\mathrm{g}l
Option B (12Mgl\frac{1}{2}M\mathrm{g}l) is correct because the work done is equal to the average force multiplied by the elongation.
223 Medium
223 of 265
Application +4 | -1
A copper wire of negligible mass, 1m\mathrm{1 m} length and cross-sectional area 10−6{10^{ - 6}} is kept on a smooth horizontal table with one end fixed. A ball of mass 1\mathrm{1} kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rad s−1{s^{ - 1}}. If the elongation in the wire is 10−3m{10^{ - 3}}\mathrm{m}, then the Young’s modulus is
View answer and explanation
Correct option: A
4  ×1011 Nm−24\,\, \times {10^{11}}\,\mathrm{N}{\mathrm{m}^{ - 2}}
Option A (4×1011 Nm−24 \times {10^{11}}\,\mathrm{N}{\mathrm{m}^{ - 2}}) is correct because it is derived from the formula relating Young's modulus to the force, length, area, and elongation.
224 Medium
224 of 265
Understanding +4 | -1
The pressure on an object of bulk modulus B undergoing hydraulic compression due to a stress exerted by surrounding fluid having volume strain (ΔVV)2\left(\frac{\Delta V}{V}\right)^2 is
View answer and explanation
Correct option: D
B(ΔVV)\mathrm{B}\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)
Option D (B(ΔVV)\mathrm{B}(\frac{\Delta \mathrm{V}}{\mathrm{V}})) is correct because the pressure is directly proportional to the bulk modulus and the volume strain.
225 Medium
225 of 265
Application +4 | -1
The maximum elongation of a steel wire of 1 m1 \mathrm{~m} length if the elastic limit of steel and its Young's modulus, respectively, are 8×108 N m−28 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2} and 2×1011 N m−2,2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}, is:
View answer and explanation
Correct option: D
4 mm4 \mathrm{~mm}
Option D (4 mm4 \mathrm{~mm}) is correct because it is calculated using the relationship between stress, Young's modulus, and strain.
226 Medium
226 of 265
Understanding +4 | -1
Assertion: The force exerted by a spring is directly proportional to its displacement from its equilibrium position.

Reason: The spring force follows Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.
View answer and explanation
Correct option: A
Both Assertion and Reason are true, and Reason is the correct explanation for Assertion
Option A (Both Assertion and Reason are true, and Reason is the correct explanation for Assertion) is correct because Hooke's Law directly explains why the spring force is proportional to displacement.
227 Medium
227 of 265
Analyzing +4 | -1
Two rods of copper and steel having same area of cross - section but different lengths l1l_{1} and l2l_{2} respectively are connected to each other. The combined length of two rods fits between two walls without any stress on walls. If α1\alpha_{1} and α2\alpha_{2} are their coefficient of linear expansion, Y1Y_{1} and Y2Y_{2} are their Young's modulus respectively, then (There is no buckling of rods while heating) With increase in temperature:
View answer and explanation
Correct option: D
combined length remains l1+l2l_{1}+l_{2} while length of one rod increases and that of other decreases
Option D is correct because the combined length remains l1+l2l_1 + l_2 due to the rigid walls, but the individual lengths change, with one increasing and the other decreasing to maintain the overall length.
228 Medium
228 of 265
Analyzing +4 | -1
Statement-I: If the volume of a body remains unchanged, When subjected to tensile strain, the value of Poisson's ratio is 1/21 / 2
Statement-II: Phosphor bronze has low Young's modulus and high rigidity modulus
View answer and explanation
Correct option: B
Statement I is true and statement II is false
Option B (Statement I is true and statement II is false) is correct because a constant volume under tensile strain implies a Poisson's ratio of 1/2, while phosphor bronze has a high Young's modulus and low rigidity modulus.
229 Medium
229 of 265
Application +4 | -1
The approximate depth of an ocean is 3000 m3000 \mathrm{~m}. The compressibility of water is 5×10−10 Pa−15 \times 10^{-10} \mathrm{~Pa}^{-1} and density of water is 103 kg/m310^3 \mathrm{~kg} / \mathrm{m}^3. What fractional compression of water will be obtained at the bottom of the ocean.
View answer and explanation
Correct option: B
1.5×10−21.5 \times 10^{-2}
Option B (1.5×10−21.5 \times 10^{-2}) is correct because it is the fractional compression calculated using the formula involving compressibility, pressure, and density.
230 Medium
230 of 265
Application +4 | -1
A steel rod y=2×1011 N/m2y = 2 \times {10^{11}}\,N/{m^2} and α=10−5/  ∘C\alpha = {10^{ - 5}}/\,{\,^ \circ }C of length 1m and area of cross section 1 cm21\,c{m^2} is heated from 0∘C{0^ \circ }C to 200∘C{200^ \circ }C with out being allowed to extend or bend. What is the tension produced in the rod?
View answer and explanation
Correct option: D
4×104 N4 \times {10^4}\,N
Option D (4×104 N4 \times {10^4}\,N) is correct because the tension produced in the rod is calculated using the formula involving Young's modulus, area, coefficient of linear expansion, and change in temperature.
231 Medium
231 of 265
Remembering +4 | -1
Energy stored in stretching a string per unit volume is
View answer and explanation
Correct option: A
12  ×  stress  ×strain\frac{1}{2}\,\, \times \,\,\mathrm{stress}\,\, \times \mathrm{strain}
Option A (12×stress×strain\frac{1}{2}\\ \times \\ \text{{stress}}\\ \times \\ \text{{strain}}) is correct because the energy stored per unit volume in a stretched string is given by half the product of stress and strain.
232 Medium
232 of 265
Remembering +4 | -1
Stress to strain ratio is equivalent to
View answer and explanation
Correct option: A
modulus of elasticity
Option A (modulus of elasticity) is correct because the modulus of elasticity is defined as the ratio of stress to strain.
233 Medium
233 of 265
Understanding +4 | -1
When a spiral spring is stretched by force, strain produced is
View answer and explanation
Correct option: D
Both (1) and (3)
Option D is correct because stretching a spiral spring produces both longitudinal and shear strain.
234 Medium
234 of 265
Application +4 | -1
Two wires of same material and length but radii in the ratio 1 : 4 are stretched by two forces to produce equal elongation. The ratio of two forces is
View answer and explanation
Correct option: C
0.052777778
Option C (01:16) is correct because the force required to produce equal elongation is proportional to the square of the radius.
235 Medium
235 of 265
Understanding +4 | -1
Statement I: Glassy solids have sharp melting point.
Statement II: The bonds between the atoms of glassy solids broken at the same temperature.
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 glassy solids do not have a sharp melting point, and their bonds break over a range of temperatures.
236 Medium
236 of 265
Application +4 | -1
The volume of a solid at 1 atmosphere pressure is 104 cm310^4 \mathrm{~cm}^3. If the pressure is increased to 51 atmosphere then percentage change in its volume will be (K=1012\left(\mathrm{K}=10^{12}\right. dyne/ cm2)\left.\mathrm{cm}^2\right)-
View answer and explanation
Correct option: C
0.005%0.005 \%
Option C (0.005%0.005 \%) is correct because it is calculated using the bulk modulus formula and the given pressure change.
237 Medium
237 of 265
Application +4 | -1
In above question, the work done in the two wires is
View answer and explanation
Correct option: A
0.5 J, 0.03 J
Option A (0.5 J, 0.03 J) is correct because it calculates the work done based on the change in length and the force applied to each wire.
238 Medium
238 of 265
Understanding +4 | -1
Statement I: Stress is the internal force per unit area of the body.
Statement II: Rubber is more elastic than steel.
View answer and explanation
Correct option: B
Statement I is true and statement II is false
Option B is correct because stress is defined as the internal force per unit area, and steel is more elastic than rubber because it requires more force to deform it.
239 Medium
239 of 265
Understanding +4 | -1
Assertion: A material which stretches more is more elastic.
Reason: Rubber is more elastic than steel.
View answer and explanation
Correct option: D
Both Assertion &\& Reason are False
Option D is correct because a material that stretches more is less elastic, and steel is more elastic than rubber.
240 Medium
240 of 265
Understanding +4 | -1
The specific heat at constant pressure and at constant volume for an ideal gas are Cp{C_p} and Cv{C_v} and its adiabatic and isothermal elasticities are Eϕ{E_\phi } and Eθ{E_\theta } respectively. The ratio of Eϕ{E_\phi } to Eθ{E_\theta } is
View answer and explanation
Correct option: B
Cp/Cv{C_p}/{C_v}
Option B (Cp/Cv{C_p}/{C_v}) is correct because the ratio of adiabatic to isothermal elasticity is equal to the ratio of specific heats, which is γ=Cp/Cv\gamma = {C_p}/{C_v}.
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