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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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Showing 241-265 of 265 questions
241 Medium
241 of 265
Analyzing +4 | -1
Assertion: Two wires AA and BB have the same cross-sectional area and are made of the same material but the length of wire AA is twice that of BB. For a given load, the strain in wire AA is twice that in BB.
Reason: For a given load, the extension in a wire is not proportional to its length.
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. Strain is directly proportional to length when other factors are constant.
242 Medium
242 of 265
Remembering +4 | -1
In magnitude hydraulic stress is equal to:
View answer and explanation
Correct option: B
Hydraulic pressure
Option B (Hydraulic pressure) is correct because hydraulic stress is defined as the hydraulic pressure applied to the object.
243 Easy
243 of 265
Understanding +4 | -1
Assertion(A): Bulk modulus of an incompressible fluid is infinite.

Reason(R): Density of incompressible fluid remains constant.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the constant density doesn't directly cause the infinite bulk modulus; it's the zero volume change under pressure.
244 Medium
244 of 265
Application +4 | -1
The Marina trench is located in the Pacific Ocean and at one place it is nearly eleven km\mathrm{km} beneath the surface of water. The water pressure at the bottom of the trench is about 1.1×108 Pa1.1 \times 10^8 \mathrm{~Pa}. A steel ball of initial volume 0.32 m30.32 \mathrm{~m}^3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
View answer and explanation
Correct option: A
2.02×10−4 m32.02 \times 10^{-4 } \mathrm{~m}^3
Option A (2.02×10−4 m32.02 \times 10^{-4 } \mathrm{~m}^3) is correct because it is calculated using the bulk modulus formula with the given pressure, volume, and bulk modulus of steel.
245 Medium
245 of 265
Understanding +4 | -1
Assertion(A): Upto the elastic limit, strain ∝\propto stress.

Reason(R): Upto elastic limit, material returns to its original shape and size, when external force is removed.
View answer and explanation
Correct option: D
If Assertion is false but Reason is true.
Option D (If Assertion is false but Reason is true) is correct because the assertion is true only up to the proportional limit, not the elastic limit.
246 Medium
246 of 265
Analyzing +4 | -1
Assertion:- Elastic restoring forces may be conservative
Reason:- The value of strain for same stress are different while increasing the load and while decreasing the load.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because both the assertion and the reason are true statements, but the reason does not correctly explain the assertion. Elastic restoring forces can be conservative, and hysteresis can cause different strain values for the same stress.
247 Medium
247 of 265
Application +4 | -1
The area of cross section of a steel wire (Y=2.0  ×1011 N/m2)\left( {Y = 2.0\,\, \times {{10}^{11}}\,N/{m^2}} \right) is 0.1  cm20.1\,\,c{m^2}. The force required to double its length will be
View answer and explanation
Correct option: D
2×106N2 \times {10^6}N
Option D (2×106N2 \times {10^6}N) is correct because it correctly applies Young's modulus formula to calculate the force required to double the wire's length.
248 Medium
248 of 265
Understanding +4 | -1
Consider the two statements.

I. Mach number is the ratio of strain to stress.

II. Sound waves in air are longitudinal while light waves in air are transverse.

Choose the correct option among the following.
View answer and explanation
Correct option: B
Statement I is false while statement II is true
Option B (Statement I is false while statement II is true) is correct because Mach number is a speed ratio, not a stress/strain ratio, and sound waves are longitudinal while light waves are transverse.
249 Medium
249 of 265
Application +4 | -1
When a spring is stretched by a distance xx , it exerts a force given by F=[−5x−16x3]NF = \left[ { - 5x - 16{x^3}} \right]N.
The work done, when the spring is stretched from 0.1  m  to  0.2  m0.1\,\,m\,\,\mathrm{to}\,\,0.2\,\,m is
View answer and explanation
Correct option: A
8.7×10−2J8.7 \times {10^{ - 2}}J
Option A (8.7×10−2J8.7 \times {10^{ - 2}}J) is correct because it is the work done calculated by integrating the force function over the given displacement.
250 Medium
250 of 265
Understanding +4 | -1
Statement I: Strain is a unitless quantity.
Statement II: Strain is equivalent to force.
View answer and explanation
Correct option: D
Statement I is false and statement II is false
Option D is correct because strain is a dimensionless quantity representing deformation, and it is not equivalent to force.
251 Medium
251 of 265
Understanding +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: B
Solid are least compressible
Option B (Solid are least compressible) is correct because solids have the strongest intermolecular forces, making them the least compressible.
252 Medium
252 of 265
Understanding +4 | -1
If the area of cross-section is halved and length of wire having young's modulus YY is doubled, then its young's modulus will become
View answer and explanation
Correct option: A
YY
Young's modulus is a material property and is independent of the dimensions of the wire. Therefore, it remains Y.
253 Medium
253 of 265
Application +4 | -1
An elastic material of Young’s modulus YY is subjected to a stress SS. The elastic energy stored per unit volume of the material is
View answer and explanation
Correct option: B
S22Y\frac{{{S^2}}}{{2Y}}
Option B (S22Y\frac{{{S^2}}}{{2Y}}) is correct because the elastic energy stored per unit volume is half the square of the stress divided by Young's modulus.
254 Medium
254 of 265
Understanding +4 | -1
What is the Young's modulus of elasticity for a perfectly rigid body?
View answer and explanation
Correct option: A
Infinity
Option A (Infinity) is correct because a perfectly rigid body has zero strain for any amount of stress, leading to an infinite Young's modulus.
255 Medium
255 of 265
Application +4 | -1
The pressure of a medium is changed from 1.01×1051.01 \times {10^5} Pa to 1.165  ×1051.165\,\, \times {10^5} Pa and change in volume is 10  %10\,\,\% keeping temperature constant. The Bulk modulus of the medium is
View answer and explanation
Correct option: D
1.55  ×  1051.55\,\, \times \,\,{10^5} Pa
Option D (1.55 × 1051.55\, \times \,{10^5} Pa) is correct because it's the bulk modulus calculated using the given pressure and volume changes.
256 Medium
256 of 265
Understanding +4 | -1
Statement I: A hollow shaft is found to be stronger than a solid shaft made of same material.
Statement II: The torque required to produce a given twist in hollow cylinder is greater than that required to twist a solid cylinder of same size and material.
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 a hollow shaft is stronger and requires more torque than a solid shaft of the same size and material.
257 Medium
257 of 265
Understanding +4 | -1
A: Diamond is more elastic than rubber.
R: When same deforming force is applied diamond deforms less than rubber.
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 because diamond is more elastic than rubber, and this is because diamond deforms less than rubber when the same deforming force is applied.
258 Medium
258 of 265
Application +4 | -1
A mass of 100 gram100 \mathrm{~gram} is attached to the end of a rubber string 49 cm49 \mathrm{~cm} long and having an area of cross-section 20 mm220 \mathrm{~mm}^2. The string is whirled round, horizontally at a constant speed of 40 rps40 \mathrm{~rps} in a circle of radius 51 cm51 \mathrm{~cm}. Find Young's modulus of rubber.
View answer and explanation
Correct option: A
3.95×109 Nm−23.95 \times 10^9 \mathrm{~Nm}^{-2}
Option A (3.95×109 Nm−23.95 \times 10^9 \mathrm{~Nm}^{-2}) is correct because it is calculated using the formula for Young's modulus, considering the centripetal force.
259 Medium
259 of 265
Application +4 | -1
A wire of length L and cross-section A is made of material of Young’s modulus Y. It is stretched by an amount x, the work done is
View answer and explanation
Correct option: C
Yx2A2L\frac{{Y{x^2}A}}{{2L}}
Option C (Yx2A2L\frac{{Y{x^2}A}}{{2L}}) is correct because it represents the work done in stretching the wire, derived from the formula relating Young's modulus, stress, and strain.
260 Medium
260 of 265
Application +4 | -1
A metal rod of Young’s modulus 2×1010 Nm−22 \times {10^{10}}\,N{m^{ - 2}} undergoes an elastic strain of 0.2%0.2\%. The energy per unit volume stored in the rod in joule/m3joule/{m^3} is
View answer and explanation
Correct option: A
400
Option A (400) is correct because it is the result of substituting the given values into the strain energy per unit volume formula.
261 Medium
261 of 265
Application +4 | -1
Cross sectional area of a wire (Y=5×1011\left(\mathrm{Y}=5 \times 10^{11}\right. N/m2\mathrm{N} / \mathrm{m}^2 ) is 0.1 cm20.1 \mathrm{~cm}^2. Find the force required to make its length double.
View answer and explanation
Correct option: B
5×106 N5 \times 10^6 \mathrm{~N}
Option B (5×106 N5 \times 10^6 \mathrm{~N}) is correct because it is the force calculated using Young's modulus formula to double the length of the wire.
262 Easy
262 of 265
Understanding +4 | -1
With rise in temperature, the Young’s modulus of elasticity:
View answer and explanation
Correct option: A
Decreases
Option A is correct because Young's modulus generally decreases with increasing temperature due to increased atomic vibrations.
263 Medium
263 of 265
Application +4 | -1
A 14.5 kg14.5 \mathrm{~kg} mass, fastened to the end of a steel wire of unstretched length 1 m1 \mathrm{~m}, is whirled in a vertical circle with an angular velocity of 2 rev/s2 \mathrm{~rev} / \mathrm{s} at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm20.065 \mathrm{~cm}^2. The elongation of the wire when the mass is at the lowest point of its path is;

[ Ysteel =2×1011 N/m2 ] \text {[ } Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { ] }
View answer and explanation
Correct option: C
1.87 mm1.87 \mathrm{~mm}
Option C (1.87 mm1.87 \mathrm{~mm}) is correct because it is the elongation calculated using Young's modulus, the force due to gravity and centripetal acceleration, and the given dimensions of the wire.
264 Easy
264 of 265
Understanding +4 | -1
When a force is applied at one end of an elastic wire, it produces a strain EE in the wire. If YY is the Young's modulus of the material of the wire, the amount of energy stored per unit volume of the wire is given by
View answer and explanation
Correct option: D
12(Y×E2)\frac{1}{2}\left(Y \times E^2\right)
Option D is correct because the energy stored per unit volume in a wire is given by 12×Y×E2\frac{1}{2} \times Y \times E^2, where Y is Young's modulus and E is strain.
265 Medium
265 of 265
Application +4 | -1
A wire increases by 10−310^{-3} of its length when a stress of 1×108 Nm−21 \times 10^8 \mathrm{~Nm}^{-2} is applied to it. What is the Young's modulus of the material of the wire?
View answer and explanation
Correct option: A
1011 Nm−210^{11} \mathrm{~Nm}^{-2}
Option A (1011 Nm−210^{11} \mathrm{~Nm}^{-2}) is correct because Young's modulus is calculated by dividing stress by strain.
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